\(\int \log (a \sin (x)) \, dx\) [161]

Optimal result

Integrand size = 5, antiderivative size = 47 \[ \int \log (a \sin (x)) \, dx=\frac {i x^2}{2}-x \log \left (1-e^{2 i x}\right )+x \log (a \sin (x))+\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i x}\right ) \]

[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))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2628, 3798, 2221, 2317, 2438} \[ \int \log (a \sin (x)) \, dx=x \log (a \sin (x))+\frac {1}{2} i \operatorname {PolyLog}\left (2,e^{2 i x}\right )+\frac {i x^2}{2}-x \log \left (1-e^{2 i x}\right ) \]

[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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 2628

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

Rule 3798

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*} \text {integral}& = 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 \text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \log (a \sin (x)) \, dx=-x \log \left (1-e^{2 i x}\right )+x \log (a \sin (x))+\frac {1}{2} i \left (x^2+\operatorname {PolyLog}\left (2,e^{2 i x}\right )\right ) \]

[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)])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (37 ) = 74\).

Time = 1.44 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.85

method	result	size
default	\(-i \left (\ln \left ({\mathrm e}^{i x}\right ) \ln \left (i a \left (1-{\mathrm e}^{2 i x}\right ) {\mathrm e}^{-i x}\right )+\frac {\ln \left ({\mathrm e}^{i x}\right )^{2}}{2}+\operatorname {dilog}\left ({\mathrm e}^{i x}\right )-\ln \left ({\mathrm e}^{i x}\right ) \ln \left ({\mathrm e}^{i x}+1\right )-\operatorname {dilog}\left ({\mathrm e}^{i x}+1\right )-\ln \left (2\right ) \ln \left ({\mathrm e}^{i x}\right )\right )\)	\(87\)
risch	\(-x \ln \left ({\mathrm e}^{i x}\right )+\frac {i \pi \operatorname {csgn}\left (a \sin \left (x \right )\right )^{3} x}{2}-\frac {i \pi \,\operatorname {csgn}\left (a \sin \left (x \right )\right ) \operatorname {csgn}\left (i a \sin \left (x \right )\right )^{2} x}{2}+\frac {i x^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (\sin \left (x \right )\right ) \operatorname {csgn}\left (a \sin \left (x \right )\right )^{2} x}{2}+\frac {i \pi \operatorname {csgn}\left (\sin \left (x \right )\right )^{3} x}{2}-\frac {i \pi x}{2}+\ln \left (a \right ) x -x \ln \left (2\right )+i \ln \left ({\mathrm e}^{i x}\right ) \ln \left ({\mathrm e}^{i x}+1\right )+\frac {i \pi \,\operatorname {csgn}\left (a \sin \left (x \right )\right ) \operatorname {csgn}\left (i a \sin \left (x \right )\right ) x}{2}+i \operatorname {dilog}\left ({\mathrm e}^{i x}+1\right )-i \operatorname {dilog}\left ({\mathrm e}^{i x}\right )+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-i x}\right ) \operatorname {csgn}\left (\sin \left (x \right )\right )^{2} x}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \operatorname {csgn}\left (\sin \left (x \right )\right )^{2} x}{2}-i \ln \left ({\mathrm e}^{i x}\right ) \ln \left ({\mathrm e}^{2 i x}-1\right )+\frac {i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 i x}-1\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-i x}\right ) \operatorname {csgn}\left (\sin \left (x \right )\right ) x}{2}+\frac {i \pi \operatorname {csgn}\left (i a \sin \left (x \right )\right )^{2} x}{2}-\frac {i \pi \operatorname {csgn}\left (i a \sin \left (x \right )\right )^{3} x}{2}+\frac {i \pi \operatorname {csgn}\left (a \sin \left (x \right )\right )^{2} \operatorname {csgn}\left (i a \right ) x}{2}-\frac {i \pi \,\operatorname {csgn}\left (\sin \left (x \right )\right ) \operatorname {csgn}\left (a \sin \left (x \right )\right ) \operatorname {csgn}\left (i a \right ) x}{2}\)	\(289\)

[In]

int(ln(a*sin(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (32) = 64\).

Time = 0.34 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.21 \[ \int \log (a \sin (x)) \, dx=x \log \left (a \sin \left (x\right )\right ) - \frac {1}{2} \, x \log \left (\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (-\cos \left (x\right ) + i \, \sin \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (-\cos \left (x\right ) - i \, \sin \left (x\right ) + 1\right ) + \frac {1}{2} i \, {\rm Li}_2\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right ) - \frac {1}{2} i \, {\rm Li}_2\left (-\cos \left (x\right ) + i \, \sin \left (x\right )\right ) + \frac {1}{2} i \, {\rm Li}_2\left (-\cos \left (x\right ) - i \, \sin \left (x\right )\right ) \]

[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))

Sympy [F]

\[ \int \log (a \sin (x)) \, dx=\int \log {\left (a \sin {\left (x \right )} \right )}\, dx \]

[In]

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

[Out]

Integral(log(a*sin(x)), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (32) = 64\).

Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.85 \[ \int \log (a \sin (x)) \, dx=\frac {1}{2} i \, x^{2} - i \, x \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) + i \, x \arctan \left (\sin \left (x\right ), -\cos \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) - \frac {1}{2} \, x \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) + x \log \left (a \sin \left (x\right )\right ) + i \, {\rm Li}_2\left (-e^{\left (i \, x\right )}\right ) + i \, {\rm Li}_2\left (e^{\left (i \, x\right )}\right ) \]

[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))

Giac [F]

\[ \int \log (a \sin (x)) \, dx=\int { \log \left (a \sin \left (x\right )\right ) \,d x } \]

[In]

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

[Out]

integrate(log(a*sin(x)), x)

Mupad [F(-1)]

Timed out. \[ \int \log (a \sin (x)) \, dx=\int \ln \left (a\,\sin \left (x\right )\right ) \,d x \]

[In]

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

[Out]

int(log(a*sin(x)), x)