\(\int \log (a \sinh (x)) \, dx\) [201]

Optimal result

Integrand size = 5, antiderivative size = 39 \[ \int \log (a \sinh (x)) \, dx=\frac {x^2}{2}-x \log \left (1-e^{2 x}\right )+x \log (a \sinh (x))-\frac {\operatorname {PolyLog}\left (2,e^{2 x}\right )}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, 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, 3797, 2221, 2317, 2438} \[ \int \log (a \sinh (x)) \, dx=x \log (a \sinh (x))-\frac {\operatorname {PolyLog}\left (2,e^{2 x}\right )}{2}+\frac {x^2}{2}-x \log \left (1-e^{2 x}\right ) \]

[In]

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

[Out]

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

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 3797

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

Rubi steps \begin{align*} \text {integral}& = x \log (a \sinh (x))-\int x \coth (x) \, dx \\ & = \frac {x^2}{2}+x \log (a \sinh (x))+2 \int \frac {e^{2 x} x}{1-e^{2 x}} \, dx \\ & = \frac {x^2}{2}-x \log \left (1-e^{2 x}\right )+x \log (a \sinh (x))+\int \log \left (1-e^{2 x}\right ) \, dx \\ & = \frac {x^2}{2}-x \log \left (1-e^{2 x}\right )+x \log (a \sinh (x))+\frac {1}{2} \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 x}\right ) \\ & = \frac {x^2}{2}-x \log \left (1-e^{2 x}\right )+x \log (a \sinh (x))-\frac {\text {Li}_2\left (e^{2 x}\right )}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \log (a \sinh (x)) \, dx=-\frac {x^2}{2}-x \log \left (1-e^{-2 x}\right )+x \log (a \sinh (x))+\frac {1}{2} \operatorname {PolyLog}\left (2,e^{-2 x}\right ) \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.06 (sec) , antiderivative size = 295, normalized size of antiderivative = 7.56

method	result	size
risch	\(-x \ln \left ({\mathrm e}^{x}\right )+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )\right )^{2} x}{2}-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )\right ) \operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i a \right ) x}{2}+\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )\right )^{2} x}{2}-\frac {i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )\right )^{3} x}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )\right ) {\operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}\right )}^{2} x}{2}-\frac {i \pi {\operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}\right )}^{3} x}{2}-x \ln \left (2\right )+\ln \left (a \right ) x +\frac {x^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (-1+{\mathrm e}^{2 x}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left (-1+{\mathrm e}^{2 x}\right )\right ) x}{2}+\frac {i \pi {\operatorname {csgn}\left (i a \left (-1+{\mathrm e}^{2 x}\right ) {\mathrm e}^{-x}\right )}^{2} \operatorname {csgn}\left (i a \right ) x}{2}+\ln \left ({\mathrm e}^{x}\right ) \ln \left (-1+{\mathrm e}^{2 x}\right )-\operatorname {dilog}\left (1+{\mathrm e}^{x}\right )-\ln \left ({\mathrm e}^{x}\right ) \ln \left (1+{\mathrm e}^{x}\right )+\operatorname {dilog}\left ({\mathrm e}^{x}\right )\)	\(295\)

[In]

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

[Out]

-x*ln(exp(x))+1/2*I*Pi*csgn(I*exp(-x))*csgn(I*exp(-x)*(-1+exp(2*x)))^2*x-1/2*I*Pi*csgn(I*exp(-x)*(-1+exp(2*x))
)*csgn(I*a*(-1+exp(2*x))*exp(-x))*csgn(I*a)*x+1/2*I*Pi*csgn(I*(-1+exp(2*x)))*csgn(I*exp(-x)*(-1+exp(2*x)))^2*x
-1/2*I*Pi*csgn(I*exp(-x)*(-1+exp(2*x)))^3*x+1/2*I*Pi*csgn(I*exp(-x)*(-1+exp(2*x)))*csgn(I*a*(-1+exp(2*x))*exp(
-x))^2*x-1/2*I*Pi*csgn(I*a*(-1+exp(2*x))*exp(-x))^3*x-x*ln(2)+ln(a)*x+1/2*x^2-1/2*I*Pi*csgn(I*(-1+exp(2*x)))*c
sgn(I*exp(-x))*csgn(I*exp(-x)*(-1+exp(2*x)))*x+1/2*I*Pi*csgn(I*a*(-1+exp(2*x))*exp(-x))^2*csgn(I*a)*x+ln(exp(x
))*ln(-1+exp(2*x))-dilog(1+exp(x))-ln(exp(x))*ln(1+exp(x))+dilog(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.46 \[ \int \log (a \sinh (x)) \, dx=\frac {1}{2} \, x^{2} + x \log \left (a \sinh \left (x\right )\right ) - x \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - x \log \left (-\cosh \left (x\right ) - \sinh \left (x\right ) + 1\right ) - {\rm Li}_2\left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - {\rm Li}_2\left (-\cosh \left (x\right ) - \sinh \left (x\right )\right ) \]

[In]

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

[Out]

1/2*x^2 + x*log(a*sinh(x)) - x*log(cosh(x) + sinh(x) + 1) - x*log(-cosh(x) - sinh(x) + 1) - dilog(cosh(x) + si
nh(x)) - dilog(-cosh(x) - sinh(x))

Sympy [F]

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

[In]

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

[Out]

Integral(log(a*sinh(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \log (a \sinh (x)) \, dx=\frac {1}{2} \, x^{2} + x \log \left (a \sinh \left (x\right )\right ) - x \log \left (e^{x} + 1\right ) - x \log \left (-e^{x} + 1\right ) - {\rm Li}_2\left (-e^{x}\right ) - {\rm Li}_2\left (e^{x}\right ) \]

[In]

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

[Out]

1/2*x^2 + x*log(a*sinh(x)) - x*log(e^x + 1) - x*log(-e^x + 1) - dilog(-e^x) - dilog(e^x)

Giac [F]

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

[In]

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

[Out]

integrate(log(a*sinh(x)), x)

Mupad [F(-1)]

Timed out. \[ \int \log (a \sinh (x)) \, dx=\int \ln \left (a\,\mathrm {sinh}\left (x\right )\right ) \,d x \]

[In]

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

[Out]

int(log(a*sinh(x)), x)