3.194 \(\int \csc ^2(x) \log (\sin (x)) \, dx\)

Optimal. Leaf size=15 \[ -x-\cot (x)-\cot (x) \log (\sin (x)) \]

[Out]

-x-cot(x)-cot(x)*ln(sin(x))

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3767, 8, 2554, 3473} \[ -x-\cot (x)-\cot (x) \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2*Log[Sin[x]],x]

[Out]

-x - Cot[x] - Cot[x]*Log[Sin[x]]

Rule 8

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

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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \csc ^2(x) \log (\sin (x)) \, dx &=-\cot (x) \log (\sin (x))+\int \cot ^2(x) \, dx\\ &=-\cot (x)-\cot (x) \log (\sin (x))-\int 1 \, dx\\ &=-x-\cot (x)-\cot (x) \log (\sin (x))\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 1.00 \[ -x-\cot (x)-\cot (x) \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2*Log[Sin[x]],x]

[Out]

-x - Cot[x] - Cot[x]*Log[Sin[x]]

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fricas [A]  time = 0.50, size = 19, normalized size = 1.27 \[ -\frac {\cos \relax (x) \log \left (\sin \relax (x)\right ) + x \sin \relax (x) + \cos \relax (x)}{\sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2*log(sin(x)),x, algorithm="fricas")

[Out]

-(cos(x)*log(sin(x)) + x*sin(x) + cos(x))/sin(x)

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giac [A]  time = 0.20, size = 19, normalized size = 1.27 \[ -x - \frac {\log \left (\sin \relax (x)\right )}{\tan \relax (x)} - \frac {1}{\tan \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2*log(sin(x)),x, algorithm="giac")

[Out]

-x - log(sin(x))/tan(x) - 1/tan(x)

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maple [C]  time = 0.72, size = 72, normalized size = 4.80 \[ -\frac {2 i {\mathrm e}^{2 i x} \ln \left (2 \sin \relax (x )\right )}{{\mathrm e}^{2 i x}-1}+i \ln \left ({\mathrm e}^{i x}+1\right )+i \ln \left ({\mathrm e}^{i x}-1\right )-\frac {2 i}{{\mathrm e}^{2 i x}-1}+\frac {2 i \ln \relax (2)}{{\mathrm e}^{2 i x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^2*ln(sin(x)),x)

[Out]

-2*I/(exp(2*I*x)-1)*ln(2*sin(x))*exp(2*I*x)-2*I/(exp(2*I*x)-1)+I*ln(exp(I*x)-1)+I*ln(exp(I*x)+1)+2*I*ln(2)/(ex
p(2*I*x)-1)

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maxima [B]  time = 0.99, size = 81, normalized size = 5.40 \[ -\frac {1}{2} \, {\left (\frac {\cos \relax (x) + 1}{\sin \relax (x)} - \frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )} \log \left (\frac {2 \, \sin \relax (x)}{{\left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )} {\left (\cos \relax (x) + 1\right )}}\right ) - \frac {\cos \relax (x) + 1}{2 \, \sin \relax (x)} + \frac {\sin \relax (x)}{2 \, {\left (\cos \relax (x) + 1\right )}} - 2 \, \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^2*log(sin(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.58, size = 57, normalized size = 3.80 \[ -2\,x-\ln \left ({\mathrm {e}}^{x\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}-\frac {\ln \left (\frac {{\mathrm {e}}^{-x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )\,2{}\mathrm {i}}{{\mathrm {e}}^{x\,2{}\mathrm {i}}-1}-\frac {2{}\mathrm {i}}{{\mathrm {e}}^{x\,2{}\mathrm {i}}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(sin(x))/sin(x)^2,x)

[Out]

- 2*x - log(exp(x*2i) - 1)*1i - (log((exp(-x*1i)*1i)/2 - (exp(x*1i)*1i)/2)*2i)/(exp(x*2i) - 1) - 2i/(exp(x*2i)
- 1)

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sympy [A]  time = 16.68, size = 17, normalized size = 1.13 \[ - x - \log {\left (\sin {\relax (x )} \right )} \cot {\relax (x )} - \frac {\cos {\relax (x )}}{\sin {\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**2*ln(sin(x)),x)

[Out]

-x - log(sin(x))*cot(x) - cos(x)/sin(x)

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