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

Optimal. Leaf size=159 \[ -\frac{i f \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{i f \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}+\frac{i (e+f x) \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x) \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac{(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac{(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0973901, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {5183, 4180, 2531, 2282, 6589} \[ -\frac{i f \text{PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac{i f \text{PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}+\frac{i (e+f x) \text{PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac{i (e+f x) \text{PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac{(e+f x)^2 \tan ^{-1}\left (e^{2 a+2 b x}\right )}{2 f}+\frac{(e+f x)^2 \tan ^{-1}(\tanh (a+b x))}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 1.70609, size = 278, normalized size = 1.75 \[ -\frac{i f \left (-2 b x \text{PolyLog}\left (2,-i e^{2 (a+b x)}\right )+2 b x \text{PolyLog}\left (2,i e^{2 (a+b x)}\right )+\text{PolyLog}\left (3,-i e^{2 (a+b x)}\right )-\text{PolyLog}\left (3,i e^{2 (a+b x)}\right )+2 b^2 x^2 \log \left (1-i e^{2 (a+b x)}\right )-2 b^2 x^2 \log \left (1+i e^{2 (a+b x)}\right )\right )}{8 b^2}-\frac{e \left (-2 i \left (\text{PolyLog}\left (2,-i e^{2 (a+b x)}\right )-\text{PolyLog}\left (2,i e^{2 (a+b x)}\right )\right )-(-4 i a-4 i b x+\pi ) \left (\log \left (1-i e^{2 (a+b x)}\right )-\log \left (1+i e^{2 (a+b x)}\right )\right )+(\pi -4 i a) \log \left (\cot \left (\frac{1}{4} (4 i a+4 i b x+\pi )\right )\right )\right )}{8 b}+e x \tan ^{-1}(\tanh (a+b x))+\frac{1}{2} f x^2 \tan ^{-1}(\tanh (a+b x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

e*x*ArcTan[Tanh[a + b*x]] + (f*x^2*ArcTan[Tanh[a + b*x]])/2 - (e*(-(((-4*I)*a + Pi - (4*I)*b*x)*(Log[1 - I*E^(
2*(a + b*x))] - Log[1 + I*E^(2*(a + b*x))])) + ((-4*I)*a + Pi)*Log[Cot[((4*I)*a + Pi + (4*I)*b*x)/4]] - (2*I)*
(PolyLog[2, (-I)*E^(2*(a + b*x))] - PolyLog[2, I*E^(2*(a + b*x))])))/(8*b) - ((I/8)*f*(2*b^2*x^2*Log[1 - I*E^(
2*(a + b*x))] - 2*b^2*x^2*Log[1 + I*E^(2*(a + b*x))] - 2*b*x*PolyLog[2, (-I)*E^(2*(a + b*x))] + 2*b*x*PolyLog[
2, I*E^(2*(a + b*x))] + PolyLog[3, (-I)*E^(2*(a + b*x))] - PolyLog[3, I*E^(2*(a + b*x))]))/b^2

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Maple [C]  time = 7.638, size = 2414, normalized size = 15.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*Pi*x^2*f*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^3+1/8*Pi*x^2*f*csgn((1+I)*(exp(2*b*x+2*a)+I)/(e
xp(2*b*x+2*a)+1))^3-1/4*Pi*x*e*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^3+1/4*Pi*x*e*csgn(I*(exp(2*b*x+2*
a)+I)/(exp(2*b*x+2*a)+1))^3+1/4*Pi*x*e*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^3-1/4*Pi*x*e*csgn(I*(
exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))*csgn((1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2-1/8*Pi*x^2*f*csgn((1
+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2+1/8*Pi*x^2*f*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)-I)
/(exp(2*b*x+2*a)+1))^2-1/8*Pi*x^2*f*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2
+1/8*Pi*x^2*f*csgn(I*(exp(2*b*x+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^2+1/4*Pi*x*e*csgn(I*(ex
p(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^2-1/8*Pi*x^2*f*csgn(I*(e
xp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))+1/8*Pi*x^2*f*csgn(I*(ex
p(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))*csgn((1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))-1/4*I/b^2*f*a^2*ln(exp(2
*b*x+2*a)+I)-1/8*Pi*x^2*f*csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2+1/8*Pi*x^
2*f*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^2+1/4*Pi*x
*e*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^2-1/4*Pi*x*e*csgn((1+I)*(exp(2*b*x
+2*a)+I)/(exp(2*b*x+2*a)+1))^2-1/4*I*ln(exp(2*b*x+2*a)-I)*f*x^2-1/2*I*ln(exp(2*b*x+2*a)-I)*e*x-1/8*Pi*x^2*f*cs
gn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))*csgn((1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2+1/4*I*f/b^2*a^
2*ln(-exp(2*b*x+2*a)+I)+1/2*I*(1/2*f*x^2+e*x)*ln(exp(2*b*x+2*a)+I)+1/8*I*f*polylog(3,I*exp(2*b*x+2*a))/b^2-1/2
*I*e/b*(b*x+a)*ln(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))-1/8*Pi*x^2*f*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*
x+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))+1/8*Pi*f*x^2+1/4*Pi*e*x-1/2*I/b*e*dilog(((-I)^(1/2)-e
xp(b*x+a))/(-I)^(1/2))-1/2*I/b*e*dilog(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))+1/2*I*e/b*dilog(1+exp(b*x+a)*(-1)^(
3/4))+1/2*I*e/b*dilog(1-exp(b*x+a)*(-1)^(3/4))-1/4*Pi*x*e*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)+I)
/(exp(2*b*x+2*a)+1))^2+1/4*Pi*x*e*csgn(I*(exp(2*b*x+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^2-1
/4*Pi*x*e*csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2-1/4*Pi*x*e*csgn(I*(exp(2*
b*x+2*a)-I)/(exp(2*b*x+2*a)+1))*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))+1/4*Pi*x*e*csgn(I*(exp(2*b*x
+2*a)+I)/(exp(2*b*x+2*a)+1))*csgn((1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))+1/2*I/b^2*f*a*dilog(((-I)^(1/2)
-exp(b*x+a))/(-I)^(1/2))+1/2*I/b^2*f*a*dilog(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))+1/2*I*e/b*(b*x+a)*ln(1+exp(b*
x+a)*(-1)^(3/4))+1/2*I*e/b*(b*x+a)*ln(1-exp(b*x+a)*(-1)^(3/4))+1/4*I*f/b^2*(b*x+a)^2*ln(1+I*exp(2*b*x+2*a))+1/
4*I*f/b^2*(b*x+a)*polylog(2,-I*exp(2*b*x+2*a))-1/2*I*f/b^2*a*dilog(1+exp(b*x+a)*(-1)^(3/4))-1/2*I*f/b^2*a*dilo
g(1-exp(b*x+a)*(-1)^(3/4))-1/4*I*f/b^2*(b*x+a)^2*ln(1-I*exp(2*b*x+2*a))-1/4*I*f/b^2*(b*x+a)*polylog(2,I*exp(2*
b*x+2*a))-1/2*I*e/b*(b*x+a)*ln(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2))-1/8*Pi*x^2*f*csgn((1-I)*(exp(2*b*x+2*a)-I)/
(exp(2*b*x+2*a)+1))^2-1/4*Pi*x*e*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^2+1/4*Pi*x*e*csgn((1+I)*(ex
p(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^3-1/8*Pi*x^2*f*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^3+1/8*Pi*x^2*
f*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^3+1/2*I/b*a*e*ln(exp(2*b*x+2*a)+I)-1/2*I*e/b*a*ln(-exp(2*b*x+2
*a)+I)+1/8*Pi*x^2*f*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x
+2*a)+1))-1/4*Pi*x*e*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*
x+2*a)+1))+1/4*Pi*x*e*csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b
*x+2*a)+1))-1/8*I*f*polylog(3,-I*exp(2*b*x+2*a))/b^2+1/2*I*f/b^2*a*(b*x+a)*ln(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/
2))+1/2*I*f/b^2*a*(b*x+a)*ln(((-I)^(1/2)+exp(b*x+a))/(-I)^(1/2))-1/2*I*f/b^2*a*(b*x+a)*ln(1+exp(b*x+a)*(-1)^(3
/4))-1/2*I*f/b^2*a*(b*x+a)*ln(1-exp(b*x+a)*(-1)^(3/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(f*x^2 + 2*e*x)*arctan((e^(2*b*x + 2*a) - 1)/(e^(2*b*x + 2*a) + 1)) - integrate((b*f*x^2*e^(2*a) + 2*b*e*x
*e^(2*a))*e^(2*b*x)/(e^(4*b*x + 4*a) + 1), x)

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Fricas [C]  time = 2.32183, size = 1894, normalized size = 11.91 \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)*arctan(tanh(b*x+a)),x, algorithm="fricas")

[Out]

1/4*(2*(b^2*f*x^2 + 2*b^2*e*x)*arctan(sinh(b*x + a)/cosh(b*x + a)) + (-2*I*b*f*x - 2*I*b*e)*dilog(1/2*sqrt(4*I
)*(cosh(b*x + a) + sinh(b*x + a))) + (-2*I*b*f*x - 2*I*b*e)*dilog(-1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a
))) + (2*I*b*f*x + 2*I*b*e)*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (2*I*b*f*x + 2*I*b*e)*dilo
g(-1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + (-I*b^2*f*x^2 - 2*I*b^2*e*x - 2*I*a*b*e + I*a^2*f)*log(1/
2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b^2*f*x^2 - 2*I*b^2*e*x - 2*I*a*b*e + I*a^2*f)*log(-1/2
*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^2*f*x^2 + 2*I*b^2*e*x + 2*I*a*b*e - I*a^2*f)*log(1/2*sq
rt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (I*b^2*f*x^2 + 2*I*b^2*e*x + 2*I*a*b*e - I*a^2*f)*log(-1/2*sqr
t(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (2*I*a*b*e - I*a^2*f)*log(I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sin
h(b*x + a)) + (2*I*a*b*e - I*a^2*f)*log(-I*sqrt(4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-2*I*a*b*e + I*a^
2*f)*log(I*sqrt(-4*I) + 2*cosh(b*x + a) + 2*sinh(b*x + a)) + (-2*I*a*b*e + I*a^2*f)*log(-I*sqrt(-4*I) + 2*cosh
(b*x + a) + 2*sinh(b*x + a)) + 2*I*f*polylog(3, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*I*f*polylog
(3, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 2*I*f*polylog(3, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*
x + a))) - 2*I*f*polylog(3, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))))/b^2

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

[Out]

Integral((e + f*x)*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 )} \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)*arctan(tanh(b*x+a)),x, algorithm="giac")

[Out]

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