3.643 \(\int \frac{\log (\cos (\frac{x}{2}))}{1+\cos (x)} \, dx\)

Optimal. Leaf size=28 \[ -\frac{x}{2}+\tan \left (\frac{x}{2}\right )+\frac{\sin (x) \log \left (\cos \left (\frac{x}{2}\right )\right )}{\cos (x)+1} \]

[Out]

-x/2 + (Log[Cos[x/2]]*Sin[x])/(1 + Cos[x]) + Tan[x/2]

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Rubi [A]  time = 0.0347725, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {2648, 2554, 12, 3473, 8} \[ -\frac{x}{2}+\tan \left (\frac{x}{2}\right )+\frac{\sin (x) \log \left (\cos \left (\frac{x}{2}\right )\right )}{\cos (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[Log[Cos[x/2]]/(1 + Cos[x]),x]

[Out]

-x/2 + (Log[Cos[x/2]]*Sin[x])/(1 + Cos[x]) + Tan[x/2]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\log \left (\cos \left (\frac{x}{2}\right )\right )}{1+\cos (x)} \, dx &=\frac{\log \left (\cos \left (\frac{x}{2}\right )\right ) \sin (x)}{1+\cos (x)}-\int -\frac{1}{2} \tan ^2\left (\frac{x}{2}\right ) \, dx\\ &=\frac{\log \left (\cos \left (\frac{x}{2}\right )\right ) \sin (x)}{1+\cos (x)}+\frac{1}{2} \int \tan ^2\left (\frac{x}{2}\right ) \, dx\\ &=\frac{\log \left (\cos \left (\frac{x}{2}\right )\right ) \sin (x)}{1+\cos (x)}+\tan \left (\frac{x}{2}\right )-\frac{\int 1 \, dx}{2}\\ &=-\frac{x}{2}+\frac{\log \left (\cos \left (\frac{x}{2}\right )\right ) \sin (x)}{1+\cos (x)}+\tan \left (\frac{x}{2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0845528, size = 32, normalized size = 1.14 \[ -\frac{\sin (x) \left (x \cot \left (\frac{x}{2}\right )-2 \left (\log \left (\cos \left (\frac{x}{2}\right )\right )+1\right )\right )}{2 (\cos (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[Cos[x/2]]/(1 + Cos[x]),x]

[Out]

-((x*Cot[x/2] - 2*(1 + Log[Cos[x/2]]))*Sin[x])/(2*(1 + Cos[x]))

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Maple [C]  time = 0.163, size = 164, normalized size = 5.9 \begin{align*}{\frac{-2\,i\ln \left ({{\rm e}^{{\frac{i}{2}}x}} \right ) }{{{\rm e}^{ix}}+1}}+{\frac{1}{{{\rm e}^{ix}}+1} \left ( -i\ln \left ({{\rm e}^{ix}}+1 \right ){{\rm e}^{ix}}+\pi \,{\it csgn} \left ( i \left ({{\rm e}^{ix}}+1 \right ) \right ){\it csgn} \left ( i{{\rm e}^{-{\frac{i}{2}}x}} \right ){\it csgn} \left ( i\cos \left ({\frac{x}{2}} \right ) \right ) -\pi \,{\it csgn} \left ( i \left ({{\rm e}^{ix}}+1 \right ) \right ) \left ({\it csgn} \left ( i\cos \left ({\frac{x}{2}} \right ) \right ) \right ) ^{2}-\pi \,{\it csgn} \left ( i{{\rm e}^{-{\frac{i}{2}}x}} \right ) \left ({\it csgn} \left ( i\cos \left ({\frac{x}{2}} \right ) \right ) \right ) ^{2}+\pi \, \left ({\it csgn} \left ( i\cos \left ({\frac{x}{2}} \right ) \right ) \right ) ^{3}-x{{\rm e}^{ix}}+i\ln \left ({{\rm e}^{ix}}+1 \right ) -2\,i\ln \left ( 2 \right ) +2\,i-x \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(cos(1/2*x))/(cos(x)+1),x)

[Out]

-2*I/(exp(I*x)+1)*ln(exp(1/2*I*x))+(-I*ln(exp(I*x)+1)*exp(I*x)+Pi*csgn(I*(exp(I*x)+1))*csgn(I*exp(-1/2*I*x))*c
sgn(I*cos(1/2*x))-Pi*csgn(I*(exp(I*x)+1))*csgn(I*cos(1/2*x))^2-Pi*csgn(I*exp(-1/2*I*x))*csgn(I*cos(1/2*x))^2+P
i*csgn(I*cos(1/2*x))^3-x*exp(I*x)+I*ln(exp(I*x)+1)-2*I*ln(2)+2*I-x)/(exp(I*x)+1)

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Maxima [B]  time = 0.955665, size = 76, normalized size = 2.71 \begin{align*} \frac{\log \left (\cos \left (\frac{1}{2} \, x\right )\right ) \sin \left (x\right )}{\cos \left (x\right ) + 1} - \frac{x \cos \left (x\right )^{2} + x \sin \left (x\right )^{2} + 2 \, x \cos \left (x\right ) + x - 4 \, \sin \left (x\right )}{2 \,{\left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(cos(1/2*x))/(1+cos(x)),x, algorithm="maxima")

[Out]

log(cos(1/2*x))*sin(x)/(cos(x) + 1) - 1/2*(x*cos(x)^2 + x*sin(x)^2 + 2*x*cos(x) + x - 4*sin(x))/(cos(x)^2 + si
n(x)^2 + 2*cos(x) + 1)

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Fricas [A]  time = 2.61238, size = 105, normalized size = 3.75 \begin{align*} -\frac{x \cos \left (\frac{1}{2} \, x\right ) - 2 \, \log \left (\cos \left (\frac{1}{2} \, x\right )\right ) \sin \left (\frac{1}{2} \, x\right ) - 2 \, \sin \left (\frac{1}{2} \, x\right )}{2 \, \cos \left (\frac{1}{2} \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(cos(1/2*x))/(1+cos(x)),x, algorithm="fricas")

[Out]

-1/2*(x*cos(1/2*x) - 2*log(cos(1/2*x))*sin(1/2*x) - 2*sin(1/2*x))/cos(1/2*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (\cos{\left (\frac{x}{2} \right )} \right )}}{\cos{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(cos(1/2*x))/(1+cos(x)),x)

[Out]

Integral(log(cos(x/2))/(cos(x) + 1), x)

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Giac [A]  time = 1.08722, size = 58, normalized size = 2.07 \begin{align*} -\frac{1}{2} \, x - \frac{2 \, \log \left (\cos \left (\frac{1}{2} \, x\right )\right ) \tan \left (\frac{1}{2} \, x\right )}{{\left (x^{2} + 1\right )}{\left (\frac{x^{2} - 1}{x^{2} + 1} - 1\right )}} + \tan \left (\frac{1}{2} \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(cos(1/2*x))/(1+cos(x)),x, algorithm="giac")

[Out]

-1/2*x - 2*log(cos(1/2*x))*tan(1/2*x)/((x^2 + 1)*((x^2 - 1)/(x^2 + 1) - 1)) + tan(1/2*x)