3.76 \(\int (e+f x)^3 \tan ^{-1}(\tanh (a+b x)) \, dx\)

Optimal. Leaf size=299 \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 i f^3 \text{PolyLog}\left (5,-i e^{2 a+2 b x}\right )}{16 b^4}+\frac{3 i f^3 \text{PolyLog}\left (5,i e^{2 a+2 b x}\right )}{16 b^4}+\frac{i (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f} \]

[Out]

-((e + f*x)^4*ArcTan[E^(2*a + 2*b*x)])/(4*f) + ((e + f*x)^4*ArcTan[Tanh[a + b*x]])/(4*f) + ((I/4)*(e + f*x)^3*
PolyLog[2, (-I)*E^(2*a + 2*b*x)])/b - ((I/4)*(e + f*x)^3*PolyLog[2, I*E^(2*a + 2*b*x)])/b - (((3*I)/8)*f*(e +
f*x)^2*PolyLog[3, (-I)*E^(2*a + 2*b*x)])/b^2 + (((3*I)/8)*f*(e + f*x)^2*PolyLog[3, I*E^(2*a + 2*b*x)])/b^2 + (
((3*I)/8)*f^2*(e + f*x)*PolyLog[4, (-I)*E^(2*a + 2*b*x)])/b^3 - (((3*I)/8)*f^2*(e + f*x)*PolyLog[4, I*E^(2*a +
2*b*x)])/b^3 - (((3*I)/16)*f^3*PolyLog[5, (-I)*E^(2*a + 2*b*x)])/b^4 + (((3*I)/16)*f^3*PolyLog[5, I*E^(2*a +
2*b*x)])/b^4

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Rubi [A]  time = 0.209102, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5183, 4180, 2531, 6609, 2282, 6589} \[ \frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 i f^3 \text{PolyLog}\left (5,-i e^{2 a+2 b x}\right )}{16 b^4}+\frac{3 i f^3 \text{PolyLog}\left (5,i e^{2 a+2 b x}\right )}{16 b^4}+\frac{i (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^3*ArcTan[Tanh[a + b*x]],x]

[Out]

-((e + f*x)^4*ArcTan[E^(2*a + 2*b*x)])/(4*f) + ((e + f*x)^4*ArcTan[Tanh[a + b*x]])/(4*f) + ((I/4)*(e + f*x)^3*
PolyLog[2, (-I)*E^(2*a + 2*b*x)])/b - ((I/4)*(e + f*x)^3*PolyLog[2, I*E^(2*a + 2*b*x)])/b - (((3*I)/8)*f*(e +
f*x)^2*PolyLog[3, (-I)*E^(2*a + 2*b*x)])/b^2 + (((3*I)/8)*f*(e + f*x)^2*PolyLog[3, I*E^(2*a + 2*b*x)])/b^2 + (
((3*I)/8)*f^2*(e + f*x)*PolyLog[4, (-I)*E^(2*a + 2*b*x)])/b^3 - (((3*I)/8)*f^2*(e + f*x)*PolyLog[4, I*E^(2*a +
2*b*x)])/b^3 - (((3*I)/16)*f^3*PolyLog[5, (-I)*E^(2*a + 2*b*x)])/b^4 + (((3*I)/16)*f^3*PolyLog[5, I*E^(2*a +
2*b*x)])/b^4

Rule 5183

Int[ArcTan[Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m + 1)*ArcTan[T
anh[a + b*x]])/(f*(m + 1)), x] - Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Sech[2*a + 2*b*x], x], x] /; FreeQ[
{a, b, e, f}, x] && IGtQ[m, 0]

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 (e+f x)^3 \tan ^{-1}(\tanh (a+b x)) \, dx &=\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}-\frac{b \int (e+f x)^4 \text{sech}(2 a+2 b x) \, dx}{4 f}\\ &=-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}+\frac{1}{2} i \int (e+f x)^3 \log \left (1-i e^{2 a+2 b x}\right ) \, dx-\frac{1}{2} i \int (e+f x)^3 \log \left (1+i e^{2 a+2 b x}\right ) \, dx\\ &=-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{(3 i f) \int (e+f x)^2 \text{Li}_2\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b}+\frac{(3 i f) \int (e+f x)^2 \text{Li}_2\left (i e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{\left (3 i f^2\right ) \int (e+f x) \text{Li}_3\left (-i e^{2 a+2 b x}\right ) \, dx}{4 b^2}-\frac{\left (3 i f^2\right ) \int (e+f x) \text{Li}_3\left (i e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac{\left (3 i f^3\right ) \int \text{Li}_4\left (-i e^{2 a+2 b x}\right ) \, dx}{8 b^3}+\frac{\left (3 i f^3\right ) \int \text{Li}_4\left (i e^{2 a+2 b x}\right ) \, dx}{8 b^3}\\ &=-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac{\left (3 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4}+\frac{\left (3 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(i x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4}\\ &=-\frac{(e+f x)^4 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{4 f}+\frac{(e+f x)^4 \tan ^{-1}(\tanh (a+b x))}{4 f}+\frac{i (e+f x)^3 \text{Li}_2\left (-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x)^3 \text{Li}_2\left (i e^{2 a+2 b x}\right )}{4 b}-\frac{3 i f (e+f x)^2 \text{Li}_3\left (-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f (e+f x)^2 \text{Li}_3\left (i e^{2 a+2 b x}\right )}{8 b^2}+\frac{3 i f^2 (e+f x) \text{Li}_4\left (-i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^2 (e+f x) \text{Li}_4\left (i e^{2 a+2 b x}\right )}{8 b^3}-\frac{3 i f^3 \text{Li}_5\left (-i e^{2 a+2 b x}\right )}{16 b^4}+\frac{3 i f^3 \text{Li}_5\left (i e^{2 a+2 b x}\right )}{16 b^4}\\ \end{align*}

Mathematica [B]  time = 4.93567, size = 600, normalized size = 2.01 \[ \frac{1}{4} x \left (6 e^2 f x+4 e^3+4 e f^2 x^2+f^3 x^3\right ) \tan ^{-1}(\tanh (a+b x))-\frac{i \left (6 b^2 e^2 f \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b^2 e^2 f \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )+12 b^2 e f^2 x \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-12 b^2 e f^2 x \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )-4 b^3 (e+f x)^3 \text{PolyLog}\left (2,-i e^{2 (a+b x)}\right )+4 b^3 (e+f x)^3 \text{PolyLog}\left (2,i e^{2 (a+b x)}\right )+6 b^2 f^3 x^2 \text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b^2 f^3 x^2 \text{PolyLog}\left (3,i e^{2 (a+b x)}\right )-6 b e f^2 \text{PolyLog}\left (4,-i e^{2 (a+b x)}\right )+6 b e f^2 \text{PolyLog}\left (4,i e^{2 (a+b x)}\right )-6 b f^3 x \text{PolyLog}\left (4,-i e^{2 (a+b x)}\right )+6 b f^3 x \text{PolyLog}\left (4,i e^{2 (a+b x)}\right )+3 f^3 \text{PolyLog}\left (5,-i e^{2 (a+b x)}\right )-3 f^3 \text{PolyLog}\left (5,i e^{2 (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1-i e^{2 (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1+i e^{2 (a+b x)}\right )+8 b^4 e^3 x \log \left (1-i e^{2 (a+b x)}\right )-8 b^4 e^3 x \log \left (1+i e^{2 (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1-i e^{2 (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1+i e^{2 (a+b x)}\right )\right )}{16 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)^3*ArcTan[Tanh[a + b*x]],x]

[Out]

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*ArcTan[Tanh[a + b*x]])/4 - ((I/16)*(8*b^4*e^3*x*Log[1 - I*E^(2*
(a + b*x))] + 12*b^4*e^2*f*x^2*Log[1 - I*E^(2*(a + b*x))] + 8*b^4*e*f^2*x^3*Log[1 - I*E^(2*(a + b*x))] + 2*b^4
*f^3*x^4*Log[1 - I*E^(2*(a + b*x))] - 8*b^4*e^3*x*Log[1 + I*E^(2*(a + b*x))] - 12*b^4*e^2*f*x^2*Log[1 + I*E^(2
*(a + b*x))] - 8*b^4*e*f^2*x^3*Log[1 + I*E^(2*(a + b*x))] - 2*b^4*f^3*x^4*Log[1 + I*E^(2*(a + b*x))] - 4*b^3*(
e + f*x)^3*PolyLog[2, (-I)*E^(2*(a + b*x))] + 4*b^3*(e + f*x)^3*PolyLog[2, I*E^(2*(a + b*x))] + 6*b^2*e^2*f*Po
lyLog[3, (-I)*E^(2*(a + b*x))] + 12*b^2*e*f^2*x*PolyLog[3, (-I)*E^(2*(a + b*x))] + 6*b^2*f^3*x^2*PolyLog[3, (-
I)*E^(2*(a + b*x))] - 6*b^2*e^2*f*PolyLog[3, I*E^(2*(a + b*x))] - 12*b^2*e*f^2*x*PolyLog[3, I*E^(2*(a + b*x))]
- 6*b^2*f^3*x^2*PolyLog[3, I*E^(2*(a + b*x))] - 6*b*e*f^2*PolyLog[4, (-I)*E^(2*(a + b*x))] - 6*b*f^3*x*PolyLo
g[4, (-I)*E^(2*(a + b*x))] + 6*b*e*f^2*PolyLog[4, I*E^(2*(a + b*x))] + 6*b*f^3*x*PolyLog[4, I*E^(2*(a + b*x))]
+ 3*f^3*PolyLog[5, (-I)*E^(2*(a + b*x))] - 3*f^3*PolyLog[5, I*E^(2*(a + b*x))]))/b^4

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Maple [C]  time = 6.493, size = 7275, normalized size = 24.3 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*arctan(tanh(b*x+a)),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \,{\left (f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2} + 4 \, e^{3} x\right )} \arctan \left (\frac{e^{\left (2 \, b x + 2 \, a\right )} - 1}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \int \frac{{\left (b f^{3} x^{4} e^{\left (2 \, a\right )} + 4 \, b e f^{2} x^{3} e^{\left (2 \, a\right )} + 6 \, b e^{2} f x^{2} e^{\left (2 \, a\right )} + 4 \, b e^{3} x e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{2 \,{\left (e^{\left (4 \, b x + 4 \, a\right )} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arctan(tanh(b*x+a)),x, algorithm="maxima")

[Out]

1/4*(f^3*x^4 + 4*e*f^2*x^3 + 6*e^2*f*x^2 + 4*e^3*x)*arctan((e^(2*b*x + 2*a) - 1)/(e^(2*b*x + 2*a) + 1)) - inte
grate(1/2*(b*f^3*x^4*e^(2*a) + 4*b*e*f^2*x^3*e^(2*a) + 6*b*e^2*f*x^2*e^(2*a) + 4*b*e^3*x*e^(2*a))*e^(2*b*x)/(e
^(4*b*x + 4*a) + 1), x)

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Fricas [C]  time = 3.09964, size = 4070, normalized size = 13.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arctan(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/8*(24*I*f^3*polylog(5, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 24*I*f^3*polylog(5, -1/2*sqrt(4*I)*(
cosh(b*x + a) + sinh(b*x + a))) - 24*I*f^3*polylog(5, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 24*I*f
^3*polylog(5, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*(b^4*f^3*x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*
f*x^2 + 4*b^4*e^3*x)*arctan(sinh(b*x + a)/cosh(b*x + a)) + (-4*I*b^3*f^3*x^3 - 12*I*b^3*e*f^2*x^2 - 12*I*b^3*e
^2*f*x - 4*I*b^3*e^3)*dilog(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (-4*I*b^3*f^3*x^3 - 12*I*b^3*e*f^
2*x^2 - 12*I*b^3*e^2*f*x - 4*I*b^3*e^3)*dilog(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (4*I*b^3*f^3*x
^3 + 12*I*b^3*e*f^2*x^2 + 12*I*b^3*e^2*f*x + 4*I*b^3*e^3)*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))
) + (4*I*b^3*f^3*x^3 + 12*I*b^3*e*f^2*x^2 + 12*I*b^3*e^2*f*x + 4*I*b^3*e^3)*dilog(-1/2*sqrt(-4*I)*(cosh(b*x +
a) + sinh(b*x + a))) + (-I*b^4*f^3*x^4 - 4*I*b^4*e*f^2*x^3 - 6*I*b^4*e^2*f*x^2 - 4*I*b^4*e^3*x - 4*I*a*b^3*e^3
+ 6*I*a^2*b^2*e^2*f - 4*I*a^3*b*e*f^2 + I*a^4*f^3)*log(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (
-I*b^4*f^3*x^4 - 4*I*b^4*e*f^2*x^3 - 6*I*b^4*e^2*f*x^2 - 4*I*b^4*e^3*x - 4*I*a*b^3*e^3 + 6*I*a^2*b^2*e^2*f - 4
*I*a^3*b*e*f^2 + I*a^4*f^3)*log(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^4*f^3*x^4 + 4*I*b^4
*e*f^2*x^3 + 6*I*b^4*e^2*f*x^2 + 4*I*b^4*e^3*x + 4*I*a*b^3*e^3 - 6*I*a^2*b^2*e^2*f + 4*I*a^3*b*e*f^2 - I*a^4*f
^3)*log(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^4*f^3*x^4 + 4*I*b^4*e*f^2*x^3 + 6*I*b^4*e^2
*f*x^2 + 4*I*b^4*e^3*x + 4*I*a*b^3*e^3 - 6*I*a^2*b^2*e^2*f + 4*I*a^3*b*e*f^2 - I*a^4*f^3)*log(-1/2*sqrt(-4*I)*
(cosh(b*x + a) + sinh(b*x + a)) + 1) + (4*I*a*b^3*e^3 - 6*I*a^2*b^2*e^2*f + 4*I*a^3*b*e*f^2 - I*a^4*f^3)*log(I
*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (4*I*a*b^3*e^3 - 6*I*a^2*b^2*e^2*f + 4*I*a^3*b*e*f^2 - I*a^4
*f^3)*log(-I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-4*I*a*b^3*e^3 + 6*I*a^2*b^2*e^2*f - 4*I*a^3*b*
e*f^2 + I*a^4*f^3)*log(I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-4*I*a*b^3*e^3 + 6*I*a^2*b^2*e^2*f
- 4*I*a^3*b*e*f^2 + I*a^4*f^3)*log(-I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-24*I*b*f^3*x - 24*I
*b*e*f^2)*polylog(4, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (-24*I*b*f^3*x - 24*I*b*e*f^2)*polylog(4
, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (24*I*b*f^3*x + 24*I*b*e*f^2)*polylog(4, 1/2*sqrt(-4*I)*(c
osh(b*x + a) + sinh(b*x + a))) + (24*I*b*f^3*x + 24*I*b*e*f^2)*polylog(4, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sin
h(b*x + a))) + (12*I*b^2*f^3*x^2 + 24*I*b^2*e*f^2*x + 12*I*b^2*e^2*f)*polylog(3, 1/2*sqrt(4*I)*(cosh(b*x + a)
+ sinh(b*x + a))) + (12*I*b^2*f^3*x^2 + 24*I*b^2*e*f^2*x + 12*I*b^2*e^2*f)*polylog(3, -1/2*sqrt(4*I)*(cosh(b*x
+ a) + sinh(b*x + a))) + (-12*I*b^2*f^3*x^2 - 24*I*b^2*e*f^2*x - 12*I*b^2*e^2*f)*polylog(3, 1/2*sqrt(-4*I)*(c
osh(b*x + a) + sinh(b*x + a))) + (-12*I*b^2*f^3*x^2 - 24*I*b^2*e*f^2*x - 12*I*b^2*e^2*f)*polylog(3, -1/2*sqrt(
-4*I)*(cosh(b*x + a) + sinh(b*x + a))))/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e + f x\right )^{3} \operatorname{atan}{\left (\tanh{\left (a + b x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*atan(tanh(b*x+a)),x)

[Out]

Integral((e + f*x)**3*atan(tanh(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (f x + e\right )}^{3} \arctan \left (\tanh \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*arctan(tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*arctan(tanh(b*x + a)), x)