### 3.201 $$\int \log (a \sinh (x)) \, dx$$

Optimal. Leaf size=39 $-\frac{1}{2} \text{PolyLog}\left (2,e^{2 x}\right )+x \log (a \sinh (x))+\frac{x^2}{2}-x \log \left (1-e^{2 x}\right )$

[Out]

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

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Rubi [A]  time = 0.0594839, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 5, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 1., Rules used = {2548, 3716, 2190, 2279, 2391} $-\frac{1}{2} \text{PolyLog}\left (2,e^{2 x}\right )+x \log (a \sinh (x))+\frac{x^2}{2}-x \log \left (1-e^{2 x}\right )$

Antiderivative was successfully veriﬁed.

[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 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 3716

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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

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]

Rubi steps

\begin{align*} \int \log (a \sinh (x)) \, dx &=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} \operatorname{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]  time = 0.0203845, size = 36, normalized size = 0.92 $\frac{1}{2} \left (\text{PolyLog}\left (2,e^{-2 x}\right )-x \left (-2 \log (a \sinh (x))+x+2 \log \left (1-e^{-2 x}\right )\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [C]  time = 0.126, size = 295, normalized size = 7.6 \begin{align*} -x\ln \left ({{\rm e}^{x}} \right ) +{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ({{\rm e}^{2\,x}}-1 \right ) \right ) \left ({\it csgn} \left ( i{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}-1 \right ) \right ) \right ) ^{2}x-{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ia \left ({{\rm e}^{2\,x}}-1 \right ){{\rm e}^{-x}} \right ) \right ) ^{3}x-{\frac{i}{2}}\pi \,{\it csgn} \left ( i \left ({{\rm e}^{2\,x}}-1 \right ) \right ){\it csgn} \left ( i{{\rm e}^{-x}} \right ){\it csgn} \left ( i{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}-1 \right ) \right ) x+{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \right ) \left ({\it csgn} \left ( i{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}-1 \right ) \right ) \right ) ^{2}x+{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( ia \left ({{\rm e}^{2\,x}}-1 \right ){{\rm e}^{-x}} \right ) \right ) ^{2}{\it csgn} \left ( ia \right ) x-{\frac{i}{2}}\pi \, \left ({\it csgn} \left ( i{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}-1 \right ) \right ) \right ) ^{3}x+\ln \left ( a \right ) x-\ln \left ( 2 \right ) x+{\frac{{x}^{2}}{2}}+{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}-1 \right ) \right ) \left ({\it csgn} \left ( ia \left ({{\rm e}^{2\,x}}-1 \right ){{\rm e}^{-x}} \right ) \right ) ^{2}x-{\frac{i}{2}}\pi \,{\it csgn} \left ( i{{\rm e}^{-x}} \left ({{\rm e}^{2\,x}}-1 \right ) \right ){\it csgn} \left ( ia \left ({{\rm e}^{2\,x}}-1 \right ){{\rm e}^{-x}} \right ){\it csgn} \left ( ia \right ) x+\ln \left ({{\rm e}^{x}} \right ) \ln \left ({{\rm e}^{2\,x}}-1 \right ) +{\it dilog} \left ({{\rm e}^{x}} \right ) -{\it dilog} \left ({{\rm e}^{x}}+1 \right ) -\ln \left ({{\rm e}^{x}} \right ) \ln \left ({{\rm e}^{x}}+1 \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.2852, size = 58, normalized size = 1.49 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.02053, size = 197, normalized size = 5.05 \begin{align*} \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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log{\left (a \sinh{\left (x \right )} \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(a*sinh(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \log \left (a \sinh \left (x\right )\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log(a*sinh(x)), x)