3.75 \(\int x^2 \tan ^{-1}(\sinh (x)) \, dx\)

Optimal. Leaf size=108 \[ i x^2 \text{PolyLog}\left (2,-i e^x\right )-i x^2 \text{PolyLog}\left (2,i e^x\right )-2 i x \text{PolyLog}\left (3,-i e^x\right )+2 i x \text{PolyLog}\left (3,i e^x\right )+2 i \text{PolyLog}\left (4,-i e^x\right )-2 i \text{PolyLog}\left (4,i e^x\right )-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x)) \]

[Out]

(-2*x^3*ArcTan[E^x])/3 + (x^3*ArcTan[Sinh[x]])/3 + I*x^2*PolyLog[2, (-I)*E^x] - I*x^2*PolyLog[2, I*E^x] - (2*I
)*x*PolyLog[3, (-I)*E^x] + (2*I)*x*PolyLog[3, I*E^x] + (2*I)*PolyLog[4, (-I)*E^x] - (2*I)*PolyLog[4, I*E^x]

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Rubi [A]  time = 0.0878868, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {5205, 4180, 2531, 6609, 2282, 6589} \[ i x^2 \text{PolyLog}\left (2,-i e^x\right )-i x^2 \text{PolyLog}\left (2,i e^x\right )-2 i x \text{PolyLog}\left (3,-i e^x\right )+2 i x \text{PolyLog}\left (3,i e^x\right )+2 i \text{PolyLog}\left (4,-i e^x\right )-2 i \text{PolyLog}\left (4,i e^x\right )-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTan[Sinh[x]],x]

[Out]

(-2*x^3*ArcTan[E^x])/3 + (x^3*ArcTan[Sinh[x]])/3 + I*x^2*PolyLog[2, (-I)*E^x] - I*x^2*PolyLog[2, I*E^x] - (2*I
)*x*PolyLog[3, (-I)*E^x] + (2*I)*x*PolyLog[3, I*E^x] + (2*I)*PolyLog[4, (-I)*E^x] - (2*I)*PolyLog[4, 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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^2 \tan ^{-1}(\sinh (x)) \, dx &=\frac{1}{3} x^3 \tan ^{-1}(\sinh (x))-\frac{1}{3} \int x^3 \text{sech}(x) \, dx\\ &=-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x))+i \int x^2 \log \left (1-i e^x\right ) \, dx-i \int x^2 \log \left (1+i e^x\right ) \, dx\\ &=-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x))+i x^2 \text{Li}_2\left (-i e^x\right )-i x^2 \text{Li}_2\left (i e^x\right )-2 i \int x \text{Li}_2\left (-i e^x\right ) \, dx+2 i \int x \text{Li}_2\left (i e^x\right ) \, dx\\ &=-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x))+i x^2 \text{Li}_2\left (-i e^x\right )-i x^2 \text{Li}_2\left (i e^x\right )-2 i x \text{Li}_3\left (-i e^x\right )+2 i x \text{Li}_3\left (i e^x\right )+2 i \int \text{Li}_3\left (-i e^x\right ) \, dx-2 i \int \text{Li}_3\left (i e^x\right ) \, dx\\ &=-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x))+i x^2 \text{Li}_2\left (-i e^x\right )-i x^2 \text{Li}_2\left (i e^x\right )-2 i x \text{Li}_3\left (-i e^x\right )+2 i x \text{Li}_3\left (i e^x\right )+2 i \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^x\right )-2 i \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^x\right )\\ &=-\frac{2}{3} x^3 \tan ^{-1}\left (e^x\right )+\frac{1}{3} x^3 \tan ^{-1}(\sinh (x))+i x^2 \text{Li}_2\left (-i e^x\right )-i x^2 \text{Li}_2\left (i e^x\right )-2 i x \text{Li}_3\left (-i e^x\right )+2 i x \text{Li}_3\left (i e^x\right )+2 i \text{Li}_4\left (-i e^x\right )-2 i \text{Li}_4\left (i e^x\right )\\ \end{align*}

Mathematica [B]  time = 0.107332, size = 356, normalized size = 3.3 \[ \frac{1}{192} i \left (192 x^2 \text{PolyLog}\left (2,-i e^x\right )+192 i \pi x \text{PolyLog}\left (2,i e^x\right )+384 x \text{PolyLog}\left (3,-i e^{-x}\right )-384 x \text{PolyLog}\left (3,-i e^x\right )-48 (\pi -2 i x)^2 \text{PolyLog}\left (2,-i e^{-x}\right )-48 \pi ^2 \text{PolyLog}\left (2,i e^x\right )+192 i \pi \text{PolyLog}\left (3,-i e^{-x}\right )-192 i \pi \text{PolyLog}\left (3,i e^x\right )+384 \text{PolyLog}\left (4,-i e^{-x}\right )+384 \text{PolyLog}\left (4,-i e^x\right )-16 x^4-32 i \pi x^3+24 \pi ^2 x^2-64 x^3 \log \left (1+i e^{-x}\right )+64 x^3 \log \left (1+i e^x\right )-96 i \pi x^2 \log \left (1+i e^{-x}\right )+96 i \pi x^2 \log \left (1-i e^x\right )-64 i x^3 \tan ^{-1}(\sinh (x))+8 i \pi ^3 x+48 \pi ^2 x \log \left (1+i e^{-x}\right )-48 \pi ^2 x \log \left (1-i e^x\right )+8 i \pi ^3 \log \left (1+i e^{-x}\right )-8 i \pi ^3 \log \left (1+i e^x\right )+8 i \pi ^3 \log \left (\tan \left (\frac{1}{4} (\pi +2 i x)\right )\right )+7 \pi ^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTan[Sinh[x]],x]

[Out]

(I/192)*(7*Pi^4 + (8*I)*Pi^3*x + 24*Pi^2*x^2 - (32*I)*Pi*x^3 - 16*x^4 - (64*I)*x^3*ArcTan[Sinh[x]] + (8*I)*Pi^
3*Log[1 + I/E^x] + 48*Pi^2*x*Log[1 + I/E^x] - (96*I)*Pi*x^2*Log[1 + I/E^x] - 64*x^3*Log[1 + I/E^x] - 48*Pi^2*x
*Log[1 - I*E^x] + (96*I)*Pi*x^2*Log[1 - I*E^x] - (8*I)*Pi^3*Log[1 + I*E^x] + 64*x^3*Log[1 + I*E^x] + (8*I)*Pi^
3*Log[Tan[(Pi + (2*I)*x)/4]] - 48*(Pi - (2*I)*x)^2*PolyLog[2, (-I)/E^x] + 192*x^2*PolyLog[2, (-I)*E^x] - 48*Pi
^2*PolyLog[2, I*E^x] + (192*I)*Pi*x*PolyLog[2, I*E^x] + (192*I)*Pi*PolyLog[3, (-I)/E^x] + 384*x*PolyLog[3, (-I
)/E^x] - 384*x*PolyLog[3, (-I)*E^x] - (192*I)*Pi*PolyLog[3, I*E^x] + 384*PolyLog[4, (-I)/E^x] + 384*PolyLog[4,
(-I)*E^x])

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Maple [C]  time = 0.227, size = 758, normalized size = 7. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(sinh(x)),x)

[Out]

1/6*Pi*x^3+1/12*Pi*x^3*csgn(I*exp(-x)*(exp(x)+I)^2)*csgn(exp(-x)*(exp(x)+I)^2)+1/12*Pi*x^3*csgn(I*exp(-x)*(exp
(x)-I)^2)*csgn(exp(-x)*(exp(x)-I)^2)^2+1/12*Pi*x^3*csgn(I*(exp(x)+I))^2*csgn(I*(exp(x)+I)^2)-1/6*Pi*x^3*csgn(I
*(exp(x)+I))*csgn(I*(exp(x)+I)^2)^2-1/12*Pi*x^3*csgn(I*(exp(x)+I)^2)*csgn(I*exp(-x)*(exp(x)+I)^2)^2-1/12*Pi*x^
3*csgn(I*exp(-x)*(exp(x)+I)^2)*csgn(exp(-x)*(exp(x)+I)^2)^2-1/12*Pi*x^3*csgn(I*(exp(x)-I))^2*csgn(I*(exp(x)-I)
^2)+1/6*Pi*x^3*csgn(I*(exp(x)-I))*csgn(I*(exp(x)-I)^2)^2+1/12*Pi*x^3*csgn(I*(exp(x)-I)^2)*csgn(I*exp(-x)*(exp(
x)-I)^2)^2-1/12*Pi*x^3*csgn(I*(exp(x)-I)^2)*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)-I)^2)+1/12*Pi*x^3*csgn(I*(e
xp(x)+I)^2)*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)+I)^2)+1/12*Pi*x^3*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)-I)
^2)^2-1/12*Pi*x^3*csgn(I*exp(-x))*csgn(I*exp(-x)*(exp(x)+I)^2)^2-1/12*Pi*x^3*csgn(I*exp(-x)*(exp(x)-I)^2)*csgn
(exp(-x)*(exp(x)-I)^2)+1/12*Pi*x^3*csgn(exp(-x)*(exp(x)+I)^2)^3+1/12*Pi*x^3*csgn(exp(-x)*(exp(x)-I)^2)^3-1/12*
Pi*x^3*csgn(exp(-x)*(exp(x)+I)^2)^2+1/3*I*x^3*ln(1+I*exp(x))-1/3*I*x^3*ln(exp(x)-I)+1/3*I*x^3*ln(exp(x)+I)-1/3
*I*x^3*ln(1-I*exp(x))-I*x^2*polylog(2,I*exp(x))-2*I*x*polylog(3,-I*exp(x))-1/12*Pi*x^3*csgn(I*exp(-x)*(exp(x)-
I)^2)^3-1/12*Pi*x^3*csgn(I*(exp(x)-I)^2)^3+1/12*Pi*x^3*csgn(I*(exp(x)+I)^2)^3+1/12*Pi*x^3*csgn(I*exp(-x)*(exp(
x)+I)^2)^3-1/12*Pi*x^3*csgn(exp(-x)*(exp(x)-I)^2)^2-2*I*polylog(4,I*exp(x))+I*x^2*polylog(2,-I*exp(x))+2*I*x*p
olylog(3,I*exp(x))+2*I*polylog(4,-I*exp(x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - 2 \, \int \frac{x^{3} e^{x}}{3 \,{\left (e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(sinh(x)),x, algorithm="maxima")

[Out]

1/3*x^3*arctan(1/2*(e^(2*x) - 1)*e^(-x)) - 2*integrate(1/3*x^3*e^x/(e^(2*x) + 1), x)

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Fricas [C]  time = 1.88555, size = 477, normalized size = 4.42 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (\sinh \left (x\right )\right ) + \frac{1}{3} i \, x^{3} \log \left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right ) + 1\right ) - \frac{1}{3} i \, x^{3} \log \left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right ) + 1\right ) - i \, x^{2}{\rm Li}_2\left (i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + i \, x^{2}{\rm Li}_2\left (-i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) + 2 i \, x{\rm polylog}\left (3, i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) - 2 i \, x{\rm polylog}\left (3, -i \, \cosh \left (x\right ) - i \, \sinh \left (x\right )\right ) - 2 i \,{\rm polylog}\left (4, i \, \cosh \left (x\right ) + i \, \sinh \left (x\right )\right ) + 2 i \,{\rm polylog}\left (4, -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^2*arctan(sinh(x)),x, algorithm="fricas")

[Out]

1/3*x^3*arctan(sinh(x)) + 1/3*I*x^3*log(I*cosh(x) + I*sinh(x) + 1) - 1/3*I*x^3*log(-I*cosh(x) - I*sinh(x) + 1)
- I*x^2*dilog(I*cosh(x) + I*sinh(x)) + I*x^2*dilog(-I*cosh(x) - I*sinh(x)) + 2*I*x*polylog(3, I*cosh(x) + I*s
inh(x)) - 2*I*x*polylog(3, -I*cosh(x) - I*sinh(x)) - 2*I*polylog(4, I*cosh(x) + I*sinh(x)) + 2*I*polylog(4, -I
*cosh(x) - I*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{atan}{\left (\sinh{\left (x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atan(sinh(x)),x)

[Out]

Integral(x**2*atan(sinh(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \arctan \left (\sinh \left (x\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(sinh(x)),x, algorithm="giac")

[Out]

integrate(x^2*arctan(sinh(x)), x)