3.348 \(\int \frac {(e+f x)^3 \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=1021 \[ -\frac {6 i \text {Li}_4\left (-i e^{c+d x}\right ) f^3}{b d^4}+\frac {6 i a^2 \text {Li}_4\left (-i e^{c+d x}\right ) f^3}{b \left (a^2+b^2\right ) d^4}+\frac {6 i \text {Li}_4\left (i e^{c+d x}\right ) f^3}{b d^4}-\frac {6 i a^2 \text {Li}_4\left (i e^{c+d x}\right ) f^3}{b \left (a^2+b^2\right ) d^4}-\frac {6 a \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) f^3}{\left (a^2+b^2\right ) d^4}-\frac {6 a \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) f^3}{\left (a^2+b^2\right ) d^4}+\frac {3 a \text {Li}_4\left (-e^{2 (c+d x)}\right ) f^3}{4 \left (a^2+b^2\right ) d^4}+\frac {6 i (e+f x) \text {Li}_3\left (-i e^{c+d x}\right ) f^2}{b d^3}-\frac {6 i a^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right ) f^2}{b \left (a^2+b^2\right ) d^3}-\frac {6 i (e+f x) \text {Li}_3\left (i e^{c+d x}\right ) f^2}{b d^3}+\frac {6 i a^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right ) f^2}{b \left (a^2+b^2\right ) d^3}+\frac {6 a (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) f^2}{\left (a^2+b^2\right ) d^3}+\frac {6 a (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) f^2}{\left (a^2+b^2\right ) d^3}-\frac {3 a (e+f x) \text {Li}_3\left (-e^{2 (c+d x)}\right ) f^2}{2 \left (a^2+b^2\right ) d^3}-\frac {3 i (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right ) f}{b d^2}+\frac {3 i a^2 (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right ) f}{b \left (a^2+b^2\right ) d^2}+\frac {3 i (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right ) f}{b d^2}-\frac {3 i a^2 (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right ) f}{b \left (a^2+b^2\right ) d^2}-\frac {3 a (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) f}{\left (a^2+b^2\right ) d^2}-\frac {3 a (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) f}{\left (a^2+b^2\right ) d^2}+\frac {3 a (e+f x)^2 \text {Li}_2\left (-e^{2 (c+d x)}\right ) f}{2 \left (a^2+b^2\right ) d^2}+\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d} \]

[Out]

6*a*f^2*(f*x+e)*polylog(3,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)/d^3+6*a*f^2*(f*x+e)*polylog(3,-b*exp(d*
x+c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2)/d^3-3*a*f*(f*x+e)^2*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)
/d^2-3*a*f*(f*x+e)^2*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2)/d^2+6*I*a^2*f^3*polylog(4,-I*exp(d
*x+c))/b/(a^2+b^2)/d^4-3*I*a^2*f*(f*x+e)^2*polylog(2,I*exp(d*x+c))/b/(a^2+b^2)/d^2-6*I*a^2*f^2*(f*x+e)*polylog
(3,-I*exp(d*x+c))/b/(a^2+b^2)/d^3+3*I*f*(f*x+e)^2*polylog(2,I*exp(d*x+c))/b/d^2+6*I*f^2*(f*x+e)*polylog(3,-I*e
xp(d*x+c))/b/d^3-6*I*a^2*f^3*polylog(4,I*exp(d*x+c))/b/(a^2+b^2)/d^4+2*(f*x+e)^3*arctan(exp(d*x+c))/b/d+3/4*a*
f^3*polylog(4,-exp(2*d*x+2*c))/(a^2+b^2)/d^4-6*I*f^3*polylog(4,-I*exp(d*x+c))/b/d^4+6*I*f^3*polylog(4,I*exp(d*
x+c))/b/d^4-3*I*f*(f*x+e)^2*polylog(2,-I*exp(d*x+c))/b/d^2-6*I*f^2*(f*x+e)*polylog(3,I*exp(d*x+c))/b/d^3-2*a^2
*(f*x+e)^3*arctan(exp(d*x+c))/b/(a^2+b^2)/d+3/2*a*f*(f*x+e)^2*polylog(2,-exp(2*d*x+2*c))/(a^2+b^2)/d^2-3/2*a*f
^2*(f*x+e)*polylog(3,-exp(2*d*x+2*c))/(a^2+b^2)/d^3+3*I*a^2*f*(f*x+e)^2*polylog(2,-I*exp(d*x+c))/b/(a^2+b^2)/d
^2+6*I*a^2*f^2*(f*x+e)*polylog(3,I*exp(d*x+c))/b/(a^2+b^2)/d^3+a*(f*x+e)^3*ln(1+exp(2*d*x+2*c))/(a^2+b^2)/d-a*
(f*x+e)^3*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)/d-a*(f*x+e)^3*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2))
)/(a^2+b^2)/d-6*a*f^3*polylog(4,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/(a^2+b^2)/d^4-6*a*f^3*polylog(4,-b*exp(d*x+
c)/(a+(a^2+b^2)^(1/2)))/(a^2+b^2)/d^4

________________________________________________________________________________________

Rubi [A]  time = 1.33, antiderivative size = 1021, normalized size of antiderivative = 1.00, number of steps used = 39, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5567, 4180, 2531, 6609, 2282, 6589, 5573, 5561, 2190, 6742, 3718} \[ -\frac {6 i \text {PolyLog}\left (4,-i e^{c+d x}\right ) f^3}{b d^4}+\frac {6 i a^2 \text {PolyLog}\left (4,-i e^{c+d x}\right ) f^3}{b \left (a^2+b^2\right ) d^4}+\frac {6 i \text {PolyLog}\left (4,i e^{c+d x}\right ) f^3}{b d^4}-\frac {6 i a^2 \text {PolyLog}\left (4,i e^{c+d x}\right ) f^3}{b \left (a^2+b^2\right ) d^4}-\frac {6 a \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) f^3}{\left (a^2+b^2\right ) d^4}-\frac {6 a \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) f^3}{\left (a^2+b^2\right ) d^4}+\frac {3 a \text {PolyLog}\left (4,-e^{2 (c+d x)}\right ) f^3}{4 \left (a^2+b^2\right ) d^4}+\frac {6 i (e+f x) \text {PolyLog}\left (3,-i e^{c+d x}\right ) f^2}{b d^3}-\frac {6 i a^2 (e+f x) \text {PolyLog}\left (3,-i e^{c+d x}\right ) f^2}{b \left (a^2+b^2\right ) d^3}-\frac {6 i (e+f x) \text {PolyLog}\left (3,i e^{c+d x}\right ) f^2}{b d^3}+\frac {6 i a^2 (e+f x) \text {PolyLog}\left (3,i e^{c+d x}\right ) f^2}{b \left (a^2+b^2\right ) d^3}+\frac {6 a (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) f^2}{\left (a^2+b^2\right ) d^3}+\frac {6 a (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) f^2}{\left (a^2+b^2\right ) d^3}-\frac {3 a (e+f x) \text {PolyLog}\left (3,-e^{2 (c+d x)}\right ) f^2}{2 \left (a^2+b^2\right ) d^3}-\frac {3 i (e+f x)^2 \text {PolyLog}\left (2,-i e^{c+d x}\right ) f}{b d^2}+\frac {3 i a^2 (e+f x)^2 \text {PolyLog}\left (2,-i e^{c+d x}\right ) f}{b \left (a^2+b^2\right ) d^2}+\frac {3 i (e+f x)^2 \text {PolyLog}\left (2,i e^{c+d x}\right ) f}{b d^2}-\frac {3 i a^2 (e+f x)^2 \text {PolyLog}\left (2,i e^{c+d x}\right ) f}{b \left (a^2+b^2\right ) d^2}-\frac {3 a (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) f}{\left (a^2+b^2\right ) d^2}-\frac {3 a (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) f}{\left (a^2+b^2\right ) d^2}+\frac {3 a (e+f x)^2 \text {PolyLog}\left (2,-e^{2 (c+d x)}\right ) f}{2 \left (a^2+b^2\right ) d^2}+\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(2*(e + f*x)^3*ArcTan[E^(c + d*x)])/(b*d) - (2*a^2*(e + f*x)^3*ArcTan[E^(c + d*x)])/(b*(a^2 + b^2)*d) - (a*(e
+ f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) - (a*(e + f*x)^3*Log[1 + (b*E^(c + d*
x))/(a + Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) + (a*(e + f*x)^3*Log[1 + E^(2*(c + d*x))])/((a^2 + b^2)*d) - ((3*I
)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])/(b*d^2) + ((3*I)*a^2*f*(e + f*x)^2*PolyLog[2, (-I)*E^(c + d*x)])
/(b*(a^2 + b^2)*d^2) + ((3*I)*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)])/(b*d^2) - ((3*I)*a^2*f*(e + f*x)^2*Poly
Log[2, I*E^(c + d*x)])/(b*(a^2 + b^2)*d^2) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b
^2]))])/((a^2 + b^2)*d^2) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b
^2)*d^2) + (3*a*f*(e + f*x)^2*PolyLog[2, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^2) + ((6*I)*f^2*(e + f*x)*PolyLog
[3, (-I)*E^(c + d*x)])/(b*d^3) - ((6*I)*a^2*f^2*(e + f*x)*PolyLog[3, (-I)*E^(c + d*x)])/(b*(a^2 + b^2)*d^3) -
((6*I)*f^2*(e + f*x)*PolyLog[3, I*E^(c + d*x)])/(b*d^3) + ((6*I)*a^2*f^2*(e + f*x)*PolyLog[3, I*E^(c + d*x)])/
(b*(a^2 + b^2)*d^3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^
3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^3) - (3*a*f^2*(e
+ f*x)*PolyLog[3, -E^(2*(c + d*x))])/(2*(a^2 + b^2)*d^3) - ((6*I)*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(b*d^4) +
((6*I)*a^2*f^3*PolyLog[4, (-I)*E^(c + d*x)])/(b*(a^2 + b^2)*d^4) + ((6*I)*f^3*PolyLog[4, I*E^(c + d*x)])/(b*d^
4) - ((6*I)*a^2*f^3*PolyLog[4, I*E^(c + d*x)])/(b*(a^2 + b^2)*d^4) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a
- Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^4) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2
+ b^2)*d^4) + (3*a*f^3*PolyLog[4, -E^(2*(c + d*x))])/(4*(a^2 + b^2)*d^4)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[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 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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, 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 5561

Int[(Cosh[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 + b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 + b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 + b^2, 0]

Rule 5567

Int[(((e_.) + (f_.)*(x_))^(m_.)*Tanh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sech[c + d*x]*Tanh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sec
h[c + d*x]*Tanh[c + d*x]^(n - 1))/(a + b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0]
&& IGtQ[n, 0]

Rule 5573

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sech[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[b^2/(a^2 + b^2), Int[((e + f*x)^m*Sech[c + d*x]^(n - 2))/(a + b*Sinh[c + d*x]), x], x] + Dist[1/(
a^2 + b^2), Int[(e + f*x)^m*Sech[c + d*x]^n*(a - b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && I
GtQ[m, 0] && NeQ[a^2 + b^2, 0] && IGtQ[n, 0]

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]

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x)^3 \text {sech}(c+d x) \, dx}{b}-\frac {a \int \frac {(e+f x)^3 \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx}{b}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {a \int (e+f x)^3 \text {sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{b \left (a^2+b^2\right )}-\frac {(a b) \int \frac {(e+f x)^3 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}-\frac {(3 i f) \int (e+f x)^2 \log \left (1-i e^{c+d x}\right ) \, dx}{b d}+\frac {(3 i f) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{b d}\\ &=\frac {a (e+f x)^4}{4 \left (a^2+b^2\right ) f}+\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {a \int \left (a (e+f x)^3 \text {sech}(c+d x)-b (e+f x)^3 \tanh (c+d x)\right ) \, dx}{b \left (a^2+b^2\right )}-\frac {(a b) \int \frac {e^{c+d x} (e+f x)^3}{a-\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}-\frac {(a b) \int \frac {e^{c+d x} (e+f x)^3}{a+\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}+\frac {\left (6 i f^2\right ) \int (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) \, dx}{b d^2}-\frac {\left (6 i f^2\right ) \int (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) \, dx}{b d^2}\\ &=\frac {a (e+f x)^4}{4 \left (a^2+b^2\right ) f}+\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {a \int (e+f x)^3 \tanh (c+d x) \, dx}{a^2+b^2}-\frac {a^2 \int (e+f x)^3 \text {sech}(c+d x) \, dx}{b \left (a^2+b^2\right )}+\frac {(3 a f) \int (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac {(3 a f) \int (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}-\frac {\left (6 i f^3\right ) \int \text {Li}_3\left (-i e^{c+d x}\right ) \, dx}{b d^3}+\frac {\left (6 i f^3\right ) \int \text {Li}_3\left (i e^{c+d x}\right ) \, dx}{b d^3}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {(2 a) \int \frac {e^{2 (c+d x)} (e+f x)^3}{1+e^{2 (c+d x)}} \, dx}{a^2+b^2}+\frac {\left (3 i a^2 f\right ) \int (e+f x)^2 \log \left (1-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}-\frac {\left (3 i a^2 f\right ) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}+\frac {\left (6 a f^2\right ) \int (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^2}+\frac {\left (6 a f^2\right ) \int (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^2}-\frac {\left (6 i f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^4}+\frac {\left (6 i f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^4}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac {6 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac {6 i f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac {(3 a f) \int (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d}-\frac {\left (6 i a^2 f^2\right ) \int (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^2}+\frac {\left (6 i a^2 f^2\right ) \int (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^2}-\frac {\left (6 a f^3\right ) \int \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^3}-\frac {\left (6 a f^3\right ) \int \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^3}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 a f (e+f x)^2 \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac {6 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac {6 i f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac {\left (3 a f^2\right ) \int (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d^2}-\frac {\left (6 a f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {b x}{-a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}-\frac {\left (6 a f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac {\left (6 i a^2 f^3\right ) \int \text {Li}_3\left (-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^3}-\frac {\left (6 i a^2 f^3\right ) \int \text {Li}_3\left (i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^3}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 a f (e+f x)^2 \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac {3 a f^2 (e+f x) \text {Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}-\frac {6 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac {6 i f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}-\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}+\frac {\left (6 i a^2 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}-\frac {\left (6 i a^2 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}+\frac {\left (3 a f^3\right ) \int \text {Li}_3\left (-e^{2 (c+d x)}\right ) \, dx}{2 \left (a^2+b^2\right ) d^3}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 a f (e+f x)^2 \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac {3 a f^2 (e+f x) \text {Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}-\frac {6 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac {6 i a^2 f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}+\frac {6 i f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac {6 i a^2 f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}-\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}-\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}+\frac {\left (3 a f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{4 \left (a^2+b^2\right ) d^4}\\ &=\frac {2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac {2 a^2 (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac {a (e+f x)^3 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac {a (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac {3 i a^2 f (e+f x)^2 \text {Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 a f (e+f x)^2 \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac {3 a f^2 (e+f x) \text {Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}-\frac {6 i f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b d^4}+\frac {6 i a^2 f^3 \text {Li}_4\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}+\frac {6 i f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b d^4}-\frac {6 i a^2 f^3 \text {Li}_4\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^4}-\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}-\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^4}+\frac {3 a f^3 \text {Li}_4\left (-e^{2 (c+d x)}\right )}{4 \left (a^2+b^2\right ) d^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 23.81, size = 3088, normalized size = 3.02 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(-8*a*d^4*e^3*E^(2*c)*x - 12*a*d^4*e^2*E^(2*c)*f*x^2 - 8*a*d^4*e*E^(2*c)*f^2*x^3 - 2*a*d^4*E^(2*c)*f^3*x^4 + 8
*b*d^3*e^3*ArcTan[E^(c + d*x)] + 8*b*d^3*e^3*E^(2*c)*ArcTan[E^(c + d*x)] + (12*I)*b*d^3*e^2*f*x*Log[1 - I*E^(c
+ d*x)] + (12*I)*b*d^3*e^2*E^(2*c)*f*x*Log[1 - I*E^(c + d*x)] + (12*I)*b*d^3*e*f^2*x^2*Log[1 - I*E^(c + d*x)]
+ (12*I)*b*d^3*e*E^(2*c)*f^2*x^2*Log[1 - I*E^(c + d*x)] + (4*I)*b*d^3*f^3*x^3*Log[1 - I*E^(c + d*x)] + (4*I)*
b*d^3*E^(2*c)*f^3*x^3*Log[1 - I*E^(c + d*x)] - (12*I)*b*d^3*e^2*f*x*Log[1 + I*E^(c + d*x)] - (12*I)*b*d^3*e^2*
E^(2*c)*f*x*Log[1 + I*E^(c + d*x)] - (12*I)*b*d^3*e*f^2*x^2*Log[1 + I*E^(c + d*x)] - (12*I)*b*d^3*e*E^(2*c)*f^
2*x^2*Log[1 + I*E^(c + d*x)] - (4*I)*b*d^3*f^3*x^3*Log[1 + I*E^(c + d*x)] - (4*I)*b*d^3*E^(2*c)*f^3*x^3*Log[1
+ I*E^(c + d*x)] + 4*a*d^3*e^3*Log[1 + E^(2*(c + d*x))] + 4*a*d^3*e^3*E^(2*c)*Log[1 + E^(2*(c + d*x))] + 12*a*
d^3*e^2*f*x*Log[1 + E^(2*(c + d*x))] + 12*a*d^3*e^2*E^(2*c)*f*x*Log[1 + E^(2*(c + d*x))] + 12*a*d^3*e*f^2*x^2*
Log[1 + E^(2*(c + d*x))] + 12*a*d^3*e*E^(2*c)*f^2*x^2*Log[1 + E^(2*(c + d*x))] + 4*a*d^3*f^3*x^3*Log[1 + E^(2*
(c + d*x))] + 4*a*d^3*E^(2*c)*f^3*x^3*Log[1 + E^(2*(c + d*x))] - (12*I)*b*d^2*(1 + E^(2*c))*f*(e + f*x)^2*Poly
Log[2, (-I)*E^(c + d*x)] + (12*I)*b*d^2*(1 + E^(2*c))*f*(e + f*x)^2*PolyLog[2, I*E^(c + d*x)] + 6*a*d^2*e^2*f*
PolyLog[2, -E^(2*(c + d*x))] + 6*a*d^2*e^2*E^(2*c)*f*PolyLog[2, -E^(2*(c + d*x))] + 12*a*d^2*e*f^2*x*PolyLog[2
, -E^(2*(c + d*x))] + 12*a*d^2*e*E^(2*c)*f^2*x*PolyLog[2, -E^(2*(c + d*x))] + 6*a*d^2*f^3*x^2*PolyLog[2, -E^(2
*(c + d*x))] + 6*a*d^2*E^(2*c)*f^3*x^2*PolyLog[2, -E^(2*(c + d*x))] + (24*I)*b*d*e*f^2*PolyLog[3, (-I)*E^(c +
d*x)] + (24*I)*b*d*e*E^(2*c)*f^2*PolyLog[3, (-I)*E^(c + d*x)] + (24*I)*b*d*f^3*x*PolyLog[3, (-I)*E^(c + d*x)]
+ (24*I)*b*d*E^(2*c)*f^3*x*PolyLog[3, (-I)*E^(c + d*x)] - (24*I)*b*d*e*f^2*PolyLog[3, I*E^(c + d*x)] - (24*I)*
b*d*e*E^(2*c)*f^2*PolyLog[3, I*E^(c + d*x)] - (24*I)*b*d*f^3*x*PolyLog[3, I*E^(c + d*x)] - (24*I)*b*d*E^(2*c)*
f^3*x*PolyLog[3, I*E^(c + d*x)] - 6*a*d*e*f^2*PolyLog[3, -E^(2*(c + d*x))] - 6*a*d*e*E^(2*c)*f^2*PolyLog[3, -E
^(2*(c + d*x))] - 6*a*d*f^3*x*PolyLog[3, -E^(2*(c + d*x))] - 6*a*d*E^(2*c)*f^3*x*PolyLog[3, -E^(2*(c + d*x))]
- (24*I)*b*f^3*PolyLog[4, (-I)*E^(c + d*x)] - (24*I)*b*E^(2*c)*f^3*PolyLog[4, (-I)*E^(c + d*x)] + (24*I)*b*f^3
*PolyLog[4, I*E^(c + d*x)] + (24*I)*b*E^(2*c)*f^3*PolyLog[4, I*E^(c + d*x)] + 3*a*f^3*PolyLog[4, -E^(2*(c + d*
x))] + 3*a*E^(2*c)*f^3*PolyLog[4, -E^(2*(c + d*x))])/(4*(a^2 + b^2)*d^4*(1 + E^(2*c))) + (a*(4*e^3*E^(2*c)*x +
6*e^2*E^(2*c)*f*x^2 + 4*e*E^(2*c)*f^2*x^3 + E^(2*c)*f^3*x^4 + (4*a*Sqrt[-(a^2 + b^2)^2]*e^3*E^(2*c)*ArcTan[(a
+ b*E^(c + d*x))/Sqrt[-a^2 - b^2]])/((a^2 + b^2)^(3/2)*d) + (4*a*Sqrt[-(a^2 + b^2)^2]*e^3*E^(2*c)*ArcTanh[(a
+ b*E^(c + d*x))/Sqrt[a^2 + b^2]])/((-a^2 - b^2)^(3/2)*d) + (2*e^3*Log[b - 2*a*E^(c + d*x) - b*E^(2*(c + d*x))
])/d - (2*e^3*E^(2*c)*Log[2*a*E^(c + d*x) + b*(-1 + E^(2*(c + d*x)))])/d + (6*e^2*f*x*Log[1 + (b*E^(2*c + d*x)
)/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e^2*E^(2*c)*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 +
b^2)*E^(2*c)])])/d + (6*e*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e*E^(
2*c)*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (2*f^3*x^3*Log[1 + (b*E^(2*c
+ d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (2*E^(2*c)*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(
a^2 + b^2)*E^(2*c)])])/d + (6*e^2*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e
^2*E^(2*c)*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (6*e*f^2*x^2*Log[1 + (b*E^(
2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*e*E^(2*c)*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c +
Sqrt[(a^2 + b^2)*E^(2*c)])])/d + (2*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d
- (2*E^(2*c)*f^3*x^3*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])])/d - (6*(-1 + E^(2*c))*f*(
e + f*x)^2*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^2 - (6*(-1 + E^(2*c))*f*(e
+ f*x)^2*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^2 - (12*e*f^2*PolyLog[3, -((b
*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*e*E^(2*c)*f^2*PolyLog[3, -((b*E^(2*c + d*x))/
(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 - (12*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2
)*E^(2*c)]))])/d^3 + (12*E^(2*c)*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d
^3 - (12*e*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*e*E^(2*c)*f^2*P
olyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 - (12*f^3*x*PolyLog[3, -((b*E^(2*c +
d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*E^(2*c)*f^3*x*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + S
qrt[(a^2 + b^2)*E^(2*c)]))])/d^3 + (12*f^3*PolyLog[4, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))
])/d^4 - (12*E^(2*c)*f^3*PolyLog[4, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^4 + (12*f^3*P
olyLog[4, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^4 - (12*E^(2*c)*f^3*PolyLog[4, -((b*E^(
2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/d^4))/(2*(a^2 + b^2)*(-1 + E^(2*c))) - (a*x*(4*e^3 + 6*e^2*
f*x + 4*e*f^2*x^2 + f^3*x^3)*Csch[c/2]*Sech[c/2]*Sech[c])/(8*(a^2 + b^2))

________________________________________________________________________________________

fricas [C]  time = 0.69, size = 1723, normalized size = 1.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-(6*a*f^3*polylog(4, (a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)
/b^2))/b) + 6*a*f^3*polylog(4, (a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((
a^2 + b^2)/b^2))/b) + 3*(a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*x + a*d^2*e^2*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x +
c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + 3*(a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*
x + a*d^2*e^2*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^
2)/b^2) - b)/b + 1) - (3*a*d^2*f^3*x^2 + 3*I*b*d^2*f^3*x^2 + 6*a*d^2*e*f^2*x + 6*I*b*d^2*e*f^2*x + 3*a*d^2*e^2
*f + 3*I*b*d^2*e^2*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) - (3*a*d^2*f^3*x^2 - 3*I*b*d^2*f^3*x^2 + 6*a*d^
2*e*f^2*x - 6*I*b*d^2*e*f^2*x + 3*a*d^2*e^2*f - 3*I*b*d^2*e^2*f)*dilog(-I*cosh(d*x + c) - I*sinh(d*x + c)) + (
a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqr
t((a^2 + b^2)/b^2) + 2*a) + (a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*log(2*b*cosh(d*x + c)
+ 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + (a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x
+ 3*a*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) +
b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) + (a*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*a*c*
d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d
*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) - (a*d^3*e^3 + I*b*d^3*e^3 - 3*a*c*d^2*e^2*f - 3*I*b*c*d^2*e^2*f + 3*a*
c^2*d*e*f^2 + 3*I*b*c^2*d*e*f^2 - a*c^3*f^3 - I*b*c^3*f^3)*log(cosh(d*x + c) + sinh(d*x + c) + I) - (a*d^3*e^3
- I*b*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*I*b*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - 3*I*b*c^2*d*e*f^2 - a*c^3*f^3 + I*b*c
^3*f^3)*log(cosh(d*x + c) + sinh(d*x + c) - I) - (a*d^3*f^3*x^3 - I*b*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 - 3*I*b*
d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x - 3*I*b*d^3*e^2*f*x + 3*a*c*d^2*e^2*f - 3*I*b*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 +
3*I*b*c^2*d*e*f^2 + a*c^3*f^3 - I*b*c^3*f^3)*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1) - (a*d^3*f^3*x^3 + I*b
*d^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*I*b*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*I*b*d^3*e^2*f*x + 3*a*c*d^2*e^2*f
+ 3*I*b*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 - 3*I*b*c^2*d*e*f^2 + a*c^3*f^3 + I*b*c^3*f^3)*log(-I*cosh(d*x + c) - I
*sinh(d*x + c) + 1) - (6*a*f^3 + 6*I*b*f^3)*polylog(4, I*cosh(d*x + c) + I*sinh(d*x + c)) - (6*a*f^3 - 6*I*b*f
^3)*polylog(4, -I*cosh(d*x + c) - I*sinh(d*x + c)) - 6*(a*d*f^3*x + a*d*e*f^2)*polylog(3, (a*cosh(d*x + c) + a
*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2))/b) - 6*(a*d*f^3*x + a*d*e*f^2)*pol
ylog(3, (a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2))/b) + (
6*a*d*f^3*x + 6*I*b*d*f^3*x + 6*a*d*e*f^2 + 6*I*b*d*e*f^2)*polylog(3, I*cosh(d*x + c) + I*sinh(d*x + c)) + (6*
a*d*f^3*x - 6*I*b*d*f^3*x + 6*a*d*e*f^2 - 6*I*b*d*e*f^2)*polylog(3, -I*cosh(d*x + c) - I*sinh(d*x + c)))/((a^2
+ b^2)*d^4)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

maple [F]  time = 0.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \tanh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -e^{3} {\left (\frac {2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {a \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + \int \frac {2 \, f^{3} x^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} + \frac {6 \, e f^{2} x^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} + \frac {6 \, e^{2} f x {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-e^3*(2*b*arctan(e^(-d*x - c))/((a^2 + b^2)*d) + a*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^2 + b^2
)*d) - a*log(e^(-2*d*x - 2*c) + 1)/((a^2 + b^2)*d)) + integrate(2*f^3*x^3*(e^(d*x + c) - e^(-d*x - c))/((b*(e^
(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))) + 6*e*f^2*x^2*(e^(d*x + c) - e^(-d*x - c))/((b*
(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))) + 6*e^2*f*x*(e^(d*x + c) - e^(-d*x - c))/((b
*(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {tanh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(c + d*x)*(e + f*x)^3)/(a + b*sinh(c + d*x)),x)

[Out]

int((tanh(c + d*x)*(e + f*x)^3)/(a + b*sinh(c + d*x)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{3} \tanh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral((e + f*x)**3*tanh(c + d*x)/(a + b*sinh(c + d*x)), x)

________________________________________________________________________________________