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

Optimal. Leaf size=74 \[ i x \text{PolyLog}\left (2,-i e^x\right )-i x \text{PolyLog}\left (2,i e^x\right )-i \text{PolyLog}\left (3,-i e^x\right )+i \text{PolyLog}\left (3,i e^x\right )+x^2 \left (-\tan ^{-1}\left (e^x\right )\right )+\frac{1}{2} x^2 \tan ^{-1}(\sinh (x)) \]

[Out]

-(x^2*ArcTan[E^x]) + (x^2*ArcTan[Sinh[x]])/2 + I*x*PolyLog[2, (-I)*E^x] - I*x*PolyLog[2, I*E^x] - I*PolyLog[3,
(-I)*E^x] + I*PolyLog[3, I*E^x]

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Rubi [A]  time = 0.061271, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 5, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {5205, 4180, 2531, 2282, 6589} \[ i x \text{PolyLog}\left (2,-i e^x\right )-i x \text{PolyLog}\left (2,i e^x\right )-i \text{PolyLog}\left (3,-i e^x\right )+i \text{PolyLog}\left (3,i e^x\right )+x^2 \left (-\tan ^{-1}\left (e^x\right )\right )+\frac{1}{2} x^2 \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*ArcTan[E^x]) + (x^2*ArcTan[Sinh[x]])/2 + I*x*PolyLog[2, (-I)*E^x] - I*x*PolyLog[2, I*E^x] - I*PolyLog[3,
(-I)*E^x] + I*PolyLog[3, I*E^x]

Rule 5205

Int[((a_.) + ArcTan[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 + u^2), x], x]
, x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m +
1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int x \tan ^{-1}(\sinh (x)) \, dx &=\frac{1}{2} x^2 \tan ^{-1}(\sinh (x))-\frac{1}{2} \int x^2 \text{sech}(x) \, dx\\ &=-x^2 \tan ^{-1}\left (e^x\right )+\frac{1}{2} x^2 \tan ^{-1}(\sinh (x))+i \int x \log \left (1-i e^x\right ) \, dx-i \int x \log \left (1+i e^x\right ) \, dx\\ &=-x^2 \tan ^{-1}\left (e^x\right )+\frac{1}{2} x^2 \tan ^{-1}(\sinh (x))+i x \text{Li}_2\left (-i e^x\right )-i x \text{Li}_2\left (i e^x\right )-i \int \text{Li}_2\left (-i e^x\right ) \, dx+i \int \text{Li}_2\left (i e^x\right ) \, dx\\ &=-x^2 \tan ^{-1}\left (e^x\right )+\frac{1}{2} x^2 \tan ^{-1}(\sinh (x))+i x \text{Li}_2\left (-i e^x\right )-i x \text{Li}_2\left (i e^x\right )-i \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^x\right )+i \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^x\right )\\ &=-x^2 \tan ^{-1}\left (e^x\right )+\frac{1}{2} x^2 \tan ^{-1}(\sinh (x))+i x \text{Li}_2\left (-i e^x\right )-i x \text{Li}_2\left (i e^x\right )-i \text{Li}_3\left (-i e^x\right )+i \text{Li}_3\left (i e^x\right )\\ \end{align*}

Mathematica [A]  time = 0.0253573, size = 105, normalized size = 1.42 \[ \frac{1}{2} x^2 \tan ^{-1}(\sinh (x))-\frac{1}{2} i \left (-2 x \left (\text{PolyLog}\left (2,-i e^{-x}\right )-\text{PolyLog}\left (2,i e^{-x}\right )\right )-2 \left (\text{PolyLog}\left (3,-i e^{-x}\right )-\text{PolyLog}\left (3,i e^{-x}\right )\right )+x^2 \left (-\left (\log \left (1-i e^{-x}\right )-\log \left (1+i e^{-x}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*ArcTan[Sinh[x]])/2 - (I/2)*(-(x^2*(Log[1 - I/E^x] - Log[1 + I/E^x])) - 2*x*(PolyLog[2, (-I)/E^x] - PolyLo
g[2, I/E^x]) - 2*(PolyLog[3, (-I)/E^x] - PolyLog[3, I/E^x]))

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Maple [C]  time = 0.388, size = 732, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*Pi*x^2+1/8*Pi*x^2*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)-I)^2)^2-1/8*Pi*x^2*csgn(I*exp(-x))*csgn(I*exp(-x)
*(exp(x)+I)^2)^2-1/4*Pi*x^2*csgn(I*(exp(x)+I))*csgn(I*(exp(x)+I)^2)^2-1/8*Pi*x^2*csgn(I*(exp(x)+I)^2)*csgn(I*e
xp(-x)*(exp(x)+I)^2)^2-1/8*Pi*x^2*csgn(exp(-x)*(exp(x)+I)^2)^2-1/8*Pi*x^2*csgn(I*exp(-x)*(exp(x)+I)^2)*csgn(ex
p(-x)*(exp(x)+I)^2)^2-1/8*Pi*x^2*csgn(I*(exp(x)-I))^2*csgn(I*(exp(x)-I)^2)+1/4*Pi*x^2*csgn(I*(exp(x)-I))*csgn(
I*(exp(x)-I)^2)^2+1/8*Pi*x^2*csgn(I*exp(-x)*(exp(x)+I)^2)*csgn(exp(-x)*(exp(x)+I)^2)+1/8*Pi*x^2*csgn(I*exp(-x)
*(exp(x)-I)^2)*csgn(exp(-x)*(exp(x)-I)^2)^2+1/8*Pi*x^2*csgn(I*(exp(x)-I)^2)*csgn(I*exp(-x)*(exp(x)-I)^2)^2+1/8
*Pi*x^2*csgn(I*(exp(x)+I))^2*csgn(I*(exp(x)+I)^2)+1/8*Pi*x^2*csgn(I*exp(-x)*(exp(x)+I)^2)^3-I*x*polylog(2,I*ex
p(x))-1/8*Pi*x^2*csgn(I*(exp(x)-I)^2)^3+1/8*Pi*x^2*csgn(exp(-x)*(exp(x)+I)^2)^3-1/8*Pi*x^2*csgn(I*(exp(x)-I)^2
)*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)-I)^2)+1/8*Pi*x^2*csgn(I*(exp(x)+I)^2)*csgn(I*exp(-x))*csgn(I*exp(-x)*
(exp(x)+I)^2)+1/8*Pi*x^2*csgn(exp(-x)*(exp(x)-I)^2)^3-1/8*Pi*x^2*csgn(exp(-x)*(exp(x)-I)^2)^2-1/8*Pi*x^2*csgn(
I*exp(-x)*(exp(x)-I)^2)^3-1/2*I*x^2*ln(1-I*exp(x))-1/2*I*x^2*ln(exp(x)-I)+1/2*I*x^2*ln(exp(x)+I)+1/2*I*x^2*ln(
1+I*exp(x))-1/8*Pi*x^2*csgn(I*exp(-x)*(exp(x)-I)^2)*csgn(exp(-x)*(exp(x)-I)^2)+1/8*Pi*x^2*csgn(I*(exp(x)+I)^2)
^3+I*x*polylog(2,-I*exp(x))-I*polylog(3,-I*exp(x))+I*polylog(3,I*exp(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.97886, size = 351, normalized size = 4.74 \begin{align*} \frac{1}{2} \, x^{2} \arctan \left (\sinh \left (x\right )\right ) + \frac{1}{2} i \, x^{2} \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) - \frac{1}{2} i \, x^{2} \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) - i \, x{\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + i \, x{\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) + i \,{\rm polylog}\left (3, i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) - i \,{\rm polylog}\left (3, -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(x*arctan(sinh(x)),x, algorithm="fricas")

[Out]

1/2*x^2*arctan(sinh(x)) + 1/2*I*x^2*log(I*cosh(x) + I*sinh(x) + 1) - 1/2*I*x^2*log(-I*cosh(x) - I*sinh(x) + 1)
- I*x*dilog(I*cosh(x) + I*sinh(x)) + I*x*dilog(-I*cosh(x) - I*sinh(x)) + I*polylog(3, I*cosh(x) + I*sinh(x))
- I*polylog(3, -I*cosh(x) - I*sinh(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*atan(sinh(x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arctan(sinh(x)), x)