3.191 \(\int \log (\sin (x)) \sin ^2(x) \, dx\)

Optimal. Leaf size=74 \[ \frac {1}{4} i \text {Li}_2\left (e^{2 i x}\right )+\frac {i x^2}{4}+\frac {x}{4}-\frac {1}{2} x \log \left (1-e^{2 i x}\right )+\frac {1}{2} x \log (\sin (x))+\frac {1}{4} \sin (x) \cos (x)-\frac {1}{2} \sin (x) \cos (x) \log (\sin (x)) \]

[Out]

1/4*x+1/4*I*x^2-1/2*x*ln(1-exp(2*I*x))+1/2*x*ln(sin(x))+1/4*I*polylog(2,exp(2*I*x))+1/4*cos(x)*sin(x)-1/2*cos(
x)*ln(sin(x))*sin(x)

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Rubi [A]  time = 0.11, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {2635, 8, 2554, 12, 6742, 3717, 2190, 2279, 2391} \[ \frac {1}{4} i \text {PolyLog}\left (2,e^{2 i x}\right )+\frac {i x^2}{4}+\frac {x}{4}-\frac {1}{2} x \log \left (1-e^{2 i x}\right )+\frac {1}{2} x \log (\sin (x))+\frac {1}{4} \sin (x) \cos (x)-\frac {1}{2} \sin (x) \cos (x) \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/4 + (I/4)*x^2 - (x*Log[1 - E^((2*I)*x)])/2 + (x*Log[Sin[x]])/2 + (I/4)*PolyLog[2, E^((2*I)*x)] + (Cos[x]*Sin
[x])/4 - (Cos[x]*Log[Sin[x]]*Sin[x])/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \log (\sin (x)) \sin ^2(x) \, dx &=\frac {1}{2} x \log (\sin (x))-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)-\int \frac {1}{2} \cot (x) (x-\cos (x) \sin (x)) \, dx\\ &=\frac {1}{2} x \log (\sin (x))-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)-\frac {1}{2} \int \cot (x) (x-\cos (x) \sin (x)) \, dx\\ &=\frac {1}{2} x \log (\sin (x))-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)-\frac {1}{2} \int \left (-\cos ^2(x)+x \cot (x)\right ) \, dx\\ &=\frac {1}{2} x \log (\sin (x))-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)+\frac {1}{2} \int \cos ^2(x) \, dx-\frac {1}{2} \int x \cot (x) \, dx\\ &=\frac {i x^2}{4}+\frac {1}{2} x \log (\sin (x))+\frac {1}{4} \cos (x) \sin (x)-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)+i \int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx+\frac {\int 1 \, dx}{4}\\ &=\frac {x}{4}+\frac {i x^2}{4}-\frac {1}{2} x \log \left (1-e^{2 i x}\right )+\frac {1}{2} x \log (\sin (x))+\frac {1}{4} \cos (x) \sin (x)-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)+\frac {1}{2} \int \log \left (1-e^{2 i x}\right ) \, dx\\ &=\frac {x}{4}+\frac {i x^2}{4}-\frac {1}{2} x \log \left (1-e^{2 i x}\right )+\frac {1}{2} x \log (\sin (x))+\frac {1}{4} \cos (x) \sin (x)-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)-\frac {1}{4} i \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i x}\right )\\ &=\frac {x}{4}+\frac {i x^2}{4}-\frac {1}{2} x \log \left (1-e^{2 i x}\right )+\frac {1}{2} x \log (\sin (x))+\frac {1}{4} i \text {Li}_2\left (e^{2 i x}\right )+\frac {1}{4} \cos (x) \sin (x)-\frac {1}{2} \cos (x) \log (\sin (x)) \sin (x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 59, normalized size = 0.80 \[ \frac {1}{8} \left (2 i \text {Li}_2\left (e^{2 i x}\right )+2 x \left (i x-2 \log \left (1-e^{2 i x}\right )+2 \log (\sin (x))+1\right )+\sin (2 x) (1-2 \log (\sin (x)))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(1 + I*x - 2*Log[1 - E^((2*I)*x)] + 2*Log[Sin[x]]) + (2*I)*PolyLog[2, E^((2*I)*x)] + (1 - 2*Log[Sin[x]])*
Sin[2*x])/8

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fricas [B]  time = 0.50, size = 120, normalized size = 1.62 \[ -\frac {1}{4} \, x \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, x \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, x \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{4} \, x \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - \frac {1}{2} \, {\left (\cos \relax (x) \sin \relax (x) - x\right )} \log \left (\sin \relax (x)\right ) + \frac {1}{4} \, \cos \relax (x) \sin \relax (x) + \frac {1}{4} \, x + \frac {1}{4} i \, {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) - \frac {1}{4} i \, {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{4} i \, {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) + \frac {1}{4} i \, {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*log(cos(x) + I*sin(x) + 1) - 1/4*x*log(cos(x) - I*sin(x) + 1) - 1/4*x*log(-cos(x) + I*sin(x) + 1) - 1/4
*x*log(-cos(x) - I*sin(x) + 1) - 1/2*(cos(x)*sin(x) - x)*log(sin(x)) + 1/4*cos(x)*sin(x) + 1/4*x + 1/4*I*dilog
(cos(x) + I*sin(x)) - 1/4*I*dilog(cos(x) - I*sin(x)) - 1/4*I*dilog(-cos(x) + I*sin(x)) + 1/4*I*dilog(-cos(x) -
I*sin(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log \left (\sin \relax (x)\right ) \sin \relax (x)^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(sin(x))*sin(x)^2, x)

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maple [B]  time = 0.33, size = 146, normalized size = 1.97 \[ -\frac {i {\mathrm e}^{-2 i x} \ln \left (2 \sin \relax (x )\right )}{8}+\frac {i {\mathrm e}^{2 i x} \ln \left (2 \sin \relax (x )\right )}{8}-\frac {i \ln \left (2 \sin \relax (x )\right ) \ln \left ({\mathrm e}^{i x}\right )}{2}+\frac {i \ln \left ({\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )}{2}-\frac {i \ln \left ({\mathrm e}^{i x}\right )^{2}}{4}+\frac {i \dilog \left ({\mathrm e}^{i x}+1\right )}{2}-\frac {i \dilog \left ({\mathrm e}^{i x}\right )}{2}+\frac {i {\mathrm e}^{-2 i x}}{16}+\frac {i \ln \relax (2) {\mathrm e}^{-2 i x}}{8}-\frac {i {\mathrm e}^{2 i x}}{16}-\frac {i \ln \relax (2) {\mathrm e}^{2 i x}}{8}-\frac {i \ln \left ({\mathrm e}^{i x}\right )}{4}+\frac {i \ln \relax (2) \ln \left ({\mathrm e}^{i x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*I*ln(2*sin(x))*exp(2*I*x)-1/16*I*exp(2*I*x)-1/2*I*ln(exp(I*x))*ln(2*sin(x))-1/4*I*ln(exp(I*x))^2+1/2*I*ln(
exp(I*x))*ln(exp(I*x)+1)-1/2*I*dilog(exp(I*x))+1/2*I*dilog(exp(I*x)+1)-1/8*I*exp(-2*I*x)*ln(2*sin(x))+1/16*I*e
xp(-2*I*x)-1/4*I*ln(exp(I*x))-1/8*I*ln(2)*exp(2*I*x)+1/8*I*ln(2)*exp(-2*I*x)+1/2*I*ln(2)*ln(exp(I*x))

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maxima [B]  time = 1.94, size = 104, normalized size = 1.41 \[ \frac {1}{4} i \, x^{2} - \frac {1}{2} i \, x \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) + \frac {1}{2} i \, x \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) - \frac {1}{4} \, x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) - \frac {1}{4} \, x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) + \frac {1}{4} \, {\left (2 \, x - \sin \left (2 \, x\right )\right )} \log \left (\sin \relax (x)\right ) + \frac {1}{4} \, x + \frac {1}{2} i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + \frac {1}{2} i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) + \frac {1}{8} \, \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*x^2 - 1/2*I*x*arctan2(sin(x), cos(x) + 1) + 1/2*I*x*arctan2(sin(x), -cos(x) + 1) - 1/4*x*log(cos(x)^2 +
sin(x)^2 + 2*cos(x) + 1) - 1/4*x*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + 1/4*(2*x - sin(2*x))*log(sin(x)) +
1/4*x + 1/2*I*dilog(-e^(I*x)) + 1/2*I*dilog(e^(I*x)) + 1/8*sin(2*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \ln \left (\sin \relax (x)\right )\,{\sin \relax (x)}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(log(sin(x))*sin(x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \log {\left (\sin {\relax (x )} \right )} \sin ^{2}{\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(sin(x))*sin(x)**2, x)

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