3.73 \(\int \tan ^{-1}(\sinh (x)) \, dx\)

Optimal. Leaf size=39 \[ i \text{PolyLog}\left (2,-i e^x\right )-i \text{PolyLog}\left (2,i e^x\right )-2 x \tan ^{-1}\left (e^x\right )+x \tan ^{-1}(\sinh (x)) \]

[Out]

-2*x*ArcTan[E^x] + x*ArcTan[Sinh[x]] + I*PolyLog[2, (-I)*E^x] - I*PolyLog[2, I*E^x]

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Rubi [A]  time = 0.0317576, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 3, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.333, Rules used = {5203, 4180, 2279, 2391} \[ i \text{PolyLog}\left (2,-i e^x\right )-i \text{PolyLog}\left (2,i e^x\right )-2 x \tan ^{-1}\left (e^x\right )+x \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[Sinh[x]],x]

[Out]

-2*x*ArcTan[E^x] + x*ArcTan[Sinh[x]] + I*PolyLog[2, (-I)*E^x] - I*PolyLog[2, I*E^x]

Rule 5203

Int[ArcTan[u_], x_Symbol] :> Simp[x*ArcTan[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/(1 + u^2), x], x] /; Inv
erseFunctionFreeQ[u, x]

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 \tan ^{-1}(\sinh (x)) \, dx &=x \tan ^{-1}(\sinh (x))-\int x \text{sech}(x) \, dx\\ &=-2 x \tan ^{-1}\left (e^x\right )+x \tan ^{-1}(\sinh (x))+i \int \log \left (1-i e^x\right ) \, dx-i \int \log \left (1+i e^x\right ) \, dx\\ &=-2 x \tan ^{-1}\left (e^x\right )+x \tan ^{-1}(\sinh (x))+i \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^x\right )-i \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^x\right )\\ &=-2 x \tan ^{-1}\left (e^x\right )+x \tan ^{-1}(\sinh (x))+i \text{Li}_2\left (-i e^x\right )-i \text{Li}_2\left (i e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0308815, size = 64, normalized size = 1.64 \[ x \tan ^{-1}(\sinh (x))+i \left (\text{PolyLog}\left (2,-i e^{-x}\right )-\text{PolyLog}\left (2,i e^{-x}\right )+x \left (\log \left (1-i e^{-x}\right )-\log \left (1+i e^{-x}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[Sinh[x]],x]

[Out]

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

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Maple [B]  time = 0.051, size = 142, normalized size = 3.6 \begin{align*} x\arctan \left ( \sinh \left ( x \right ) \right ) -i{\it dilog} \left ( -i\cosh \left ( x \right ) -i\sinh \left ( x \right ) \right ) -i \left ( \ln \left ( -i\cosh \left ( x \right ) -i\sinh \left ( x \right ) \right ) -x \right ) \ln \left ( \left ( 1-i \right ) \cosh \left ({\frac{x}{2}} \right ) + \left ( 1+i \right ) \sinh \left ({\frac{x}{2}} \right ) \right ) +i{\it dilog} \left ( i\cosh \left ( x \right ) +i\sinh \left ( x \right ) \right ) +i \left ( \ln \left ( i\cosh \left ( x \right ) +i\sinh \left ( x \right ) \right ) -x \right ) \ln \left ( \left ( 1+i \right ) \cosh \left ({\frac{x}{2}} \right ) + \left ( 1-i \right ) \sinh \left ({\frac{x}{2}} \right ) \right ) +{\frac{i}{2}} \left ( -\ln \left ( -i\cosh \left ( x \right ) -i\sinh \left ( x \right ) \right ) +\ln \left ( i\cosh \left ( x \right ) +i\sinh \left ( x \right ) \right ) \right ) x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(sinh(x)),x)

[Out]

x*arctan(sinh(x))-I*dilog(-I*cosh(x)-I*sinh(x))-I*(ln(-I*cosh(x)-I*sinh(x))-x)*ln((1-I)*cosh(1/2*x)+(1+I)*sinh
(1/2*x))+I*dilog(I*cosh(x)+I*sinh(x))+I*(ln(I*cosh(x)+I*sinh(x))-x)*ln((1+I)*cosh(1/2*x)+(1-I)*sinh(1/2*x))+1/
2*I*(-ln(-I*cosh(x)-I*sinh(x))+ln(I*cosh(x)+I*sinh(x)))*x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} x \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - 2 \, \int \frac{x e^{x}}{e^{\left (2 \, x\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(sinh(x)),x, algorithm="maxima")

[Out]

x*arctan(1/2*(e^(2*x) - 1)*e^(-x)) - 2*integrate(x*e^x/(e^(2*x) + 1), x)

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Fricas [B]  time = 1.96049, size = 217, normalized size = 5.56 \begin{align*} x \arctan \left (\sinh \left (x\right )\right ) + i \, x \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) - i \, x \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) - i \,{\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + i \,{\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(sinh(x)),x, algorithm="fricas")

[Out]

x*arctan(sinh(x)) + I*x*log(I*cosh(x) + I*sinh(x) + 1) - I*x*log(-I*cosh(x) - I*sinh(x) + 1) - I*dilog(I*cosh(
x) + I*sinh(x)) + I*dilog(-I*cosh(x) - I*sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(sinh(x)),x)

[Out]

Integral(atan(sinh(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(sinh(x)),x, algorithm="giac")

[Out]

integrate(arctan(sinh(x)), x)