3.161 \(\int \log (a \sin (x)) \, dx\)

Optimal. Leaf size=47 \[ x \log (a \sin (x))+\frac {1}{2} i \text {Li}_2\left (e^{2 i x}\right )+\frac {i x^2}{2}-x \log \left (1-e^{2 i x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2548, 3717, 2190, 2279, 2391} \[ \frac {1}{2} i \text {PolyLog}\left (2,e^{2 i x}\right )+x \log (a \sin (x))+\frac {i x^2}{2}-x \log \left (1-e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[a*Sin[x]],x]

[Out]

(I/2)*x^2 - x*Log[1 - E^((2*I)*x)] + x*Log[a*Sin[x]] + (I/2)*PolyLog[2, E^((2*I)*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 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.89 \[ x \log (a \sin (x))+\frac {1}{2} i \left (x^2+\text {Li}_2\left (e^{2 i x}\right )\right )-x \log \left (1-e^{2 i x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[a*Sin[x]],x]

[Out]

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

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fricas [B]  time = 0.78, size = 104, normalized size = 2.21 \[ x \log \left (a \sin \relax (x)\right ) - \frac {1}{2} \, x \log \left (\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \relax (x) - i \, \sin \relax (x) + 1\right ) - \frac {1}{2} \, x \log \left (-\cos \relax (x) + i \, \sin \relax (x) + 1\right ) - \frac {1}{2} \, x \log \left (-\cos \relax (x) - i \, \sin \relax (x) + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\cos \relax (x) + i \, \sin \relax (x)\right ) - \frac {1}{2} i \, {\rm Li}_2\left (\cos \relax (x) - i \, \sin \relax (x)\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\cos \relax (x) + i \, \sin \relax (x)\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-\cos \relax (x) - i \, \sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*sin(x)), x)

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maple [B]  time = 0.41, size = 76, normalized size = 1.62 \[ -i \ln \left (2 a \sin \relax (x )\right ) \ln \left ({\mathrm e}^{i x}\right )+i \ln \left ({\mathrm e}^{i x}+1\right ) \ln \left ({\mathrm e}^{i x}\right )-\frac {i \ln \left ({\mathrm e}^{i x}\right )^{2}}{2}+i \dilog \left ({\mathrm e}^{i x}+1\right )-i \dilog \left ({\mathrm e}^{i x}\right )+i \ln \relax (2) \ln \left ({\mathrm e}^{i x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(a*sin(x)),x)

[Out]

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

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maxima [B]  time = 2.24, size = 87, normalized size = 1.85 \[ \frac {1}{2} i \, x^{2} - i \, x \arctan \left (\sin \relax (x), \cos \relax (x) + 1\right ) + i \, x \arctan \left (\sin \relax (x), -\cos \relax (x) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) + 1\right ) + x \log \left (a \sin \relax (x)\right ) + i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*x^2 - I*x*arctan2(sin(x), cos(x) + 1) + I*x*arctan2(sin(x), -cos(x) + 1) - 1/2*x*log(cos(x)^2 + sin(x)^2
+ 2*cos(x) + 1) - 1/2*x*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + x*log(a*sin(x)) + I*dilog(-e^(I*x)) + I*dil
og(e^(I*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \ln \left (a\,\sin \relax (x)\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(a*sin(x)),x)

[Out]

int(log(a*sin(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*sin(x)), x)

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