AI / LLM comparison

LangGraph vs Semantic Kernel

Pricing, pros, cons, and ideal use cases — side by side.

LangGraphFreemium

A low-level orchestration framework for building stateful, multi-step agent workflows with explicit control over state, branching, and human-in-the-loop steps.

Visit LangGraph

An open-source SDK from Microsoft for integrating LLMs into applications, with a focus on enterprise-grade orchestration in C#, Python, and Java.

Visit Semantic Kernel

At a glance

LangGraphSemantic Kernel
PricingFreemiumOpen-source framework, free to use. Paid tiers apply to the surrounding LangSmith platform, not the framework itself.FreeOpen-source (MIT), free. Costs come from the model APIs you call.
CategoryAI / LLMAI / LLM
Ideal for
Engineering teams building multi-step agentsEnterprise platform teamsWorkflows needing human-in-the-loop control
.NET and Java enterprise development teamsOrgs embedding AI into existing applicationsMicrosoft-ecosystem shops

Pros & cons

LangGraph

Pros
  • Explicit control over agent state and branching
  • First-class human-in-the-loop and checkpointing
  • Strong fit for durable, long-running workflows
  • Large ecosystem and active development
Cons
  • Requires real engineering investment
  • Lower-level than no-code agent builders
  • You own deployment and observability

Semantic Kernel

Pros
  • First-class C#, Python, and Java support
  • Designed to fit into existing enterprise apps
  • Backed and maintained by Microsoft
  • Clean plugin and function abstractions
Cons
  • Smaller ecosystem than Python-first frameworks
  • Roadmap is converging with Microsoft Agent Framework
  • Less suited to rapid Python prototyping

Which should you choose?

Semantic Kernel is the lighter-weight option (Free), while LangGraph sits higher on the pricing ladder (Freemium). LangGraph is built around engineering teams building multi-step agents; Semantic Kernel leans more toward .net and java enterprise development teams. Shortlist the one whose strengths line up with your biggest constraint.

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