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

Optimal. Leaf size=786 \[ \frac{6 i a f^2 (e+f x) \text{PolyLog}\left (3,-i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac{6 i a f^2 (e+f x) \text{PolyLog}\left (3,i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac{6 b f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}-\frac{6 b f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^3 \left (a^2+b^2\right )}+\frac{3 b f^2 (e+f x) \text{PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 d^3 \left (a^2+b^2\right )}-\frac{3 i a f (e+f x)^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 i a f (e+f x)^2 \text{PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}-\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}-\frac{6 i a f^3 \text{PolyLog}\left (4,-i e^{c+d x}\right )}{d^4 \left (a^2+b^2\right )}+\frac{6 i a f^3 \text{PolyLog}\left (4,i e^{c+d x}\right )}{d^4 \left (a^2+b^2\right )}+\frac{6 b f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^4 \left (a^2+b^2\right )}+\frac{6 b f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^4 \left (a^2+b^2\right )}-\frac{3 b f^3 \text{PolyLog}\left (4,-e^{2 (c+d x)}\right )}{4 d^4 \left (a^2+b^2\right )}+\frac{b (e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac{b (e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac{b (e+f x)^3 \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac{2 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.45704, antiderivative size = 786, normalized size of antiderivative = 1., number of steps used = 29, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {5573, 5561, 2190, 2531, 6609, 2282, 6589, 6742, 4180, 3718} \[ \frac{6 i a f^2 (e+f x) \text{PolyLog}\left (3,-i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac{6 i a f^2 (e+f x) \text{PolyLog}\left (3,i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac{6 b f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}-\frac{6 b f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^3 \left (a^2+b^2\right )}+\frac{3 b f^2 (e+f x) \text{PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 d^3 \left (a^2+b^2\right )}-\frac{3 i a f (e+f x)^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 i a f (e+f x)^2 \text{PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}-\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}-\frac{6 i a f^3 \text{PolyLog}\left (4,-i e^{c+d x}\right )}{d^4 \left (a^2+b^2\right )}+\frac{6 i a f^3 \text{PolyLog}\left (4,i e^{c+d x}\right )}{d^4 \left (a^2+b^2\right )}+\frac{6 b f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^4 \left (a^2+b^2\right )}+\frac{6 b f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^4 \left (a^2+b^2\right )}-\frac{3 b f^3 \text{PolyLog}\left (4,-e^{2 (c+d x)}\right )}{4 d^4 \left (a^2+b^2\right )}+\frac{b (e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac{b (e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac{b (e+f x)^3 \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac{2 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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]

Rule 6742

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

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 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]

Rubi steps

\begin{align*} \int \frac{(e+f x)^3 \text{sech}(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x)^3 \text{sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{a^2+b^2}+\frac{b^2 \int \frac{(e+f x)^3 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac{b (e+f x)^4}{4 \left (a^2+b^2\right ) f}+\frac{\int \left (a (e+f x)^3 \text{sech}(c+d x)-b (e+f x)^3 \tanh (c+d x)\right ) \, dx}{a^2+b^2}+\frac{b^2 \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{b^2 \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{b (e+f x)^4}{4 \left (a^2+b^2\right ) f}+\frac{b (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{b (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 \int (e+f x)^3 \text{sech}(c+d x) \, dx}{a^2+b^2}-\frac{b \int (e+f x)^3 \tanh (c+d x) \, dx}{a^2+b^2}-\frac{(3 b 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 b 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{2 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac{b (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{b (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 b 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 b 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{(2 b) \int \frac{e^{2 (c+d x)} (e+f x)^3}{1+e^{2 (c+d x)}} \, dx}{a^2+b^2}-\frac{(3 i a f) \int (e+f x)^2 \log \left (1-i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac{(3 i a f) \int (e+f x)^2 \log \left (1+i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d}-\frac{\left (6 b 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 b 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{2 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac{b (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{b (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{b (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i a f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 i a f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 b 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 b 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 b 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 b 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 b 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 f^2\right ) \int (e+f x) \text{Li}_2\left (-i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d^2}-\frac{\left (6 i a f^2\right ) \int (e+f x) \text{Li}_2\left (i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d^2}+\frac{\left (6 b 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 b 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 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac{b (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{b (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{b (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i a f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 i a f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 b 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 b 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 b 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 a f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 i a f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 b 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 b 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{\left (3 b 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 b 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 b 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 f^3\right ) \int \text{Li}_3\left (-i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d^3}+\frac{\left (6 i a f^3\right ) \int \text{Li}_3\left (i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d^3}\\ &=\frac{2 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac{b (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{b (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{b (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i a f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 i a f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 b 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 b 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 b 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 a f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 i a f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 b 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 b 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 b 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 b 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 b 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 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac{\left (6 i a f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}-\frac{\left (3 b 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 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac{b (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{b (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{b (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i a f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 i a f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 b 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 b 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 b 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 a f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 i a f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 b 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 b 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 b 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 a f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac{6 i a f^3 \text{Li}_4\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac{6 b 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 b 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 b 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 a (e+f x)^3 \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac{b (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{b (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{b (e+f x)^3 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{3 i a f (e+f x)^2 \text{Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 i a f (e+f x)^2 \text{Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac{3 b 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 b 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 b 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 a f^2 (e+f x) \text{Li}_3\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 i a f^2 (e+f x) \text{Li}_3\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{6 b 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 b 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 b 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 a f^3 \text{Li}_4\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac{6 i a f^3 \text{Li}_4\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^4}+\frac{6 b 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 b 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 b 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 = 28.6787, size = 10644, normalized size = 13.54 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.351, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}{\rm sech} \left (dx+c\right )}{a+b\sinh \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -e^{3}{\left (\frac{2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac{b \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{b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + \int \frac{4 \, f^{3} x^{3}}{{\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{12 \, e f^{2} x^{2}}{{\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{12 \, e^{2} f x}{{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^3*(2*a*arctan(e^(-d*x - c))/((a^2 + b^2)*d) - b*log(-2*a*e^(-d*x - c) + b*e^(-2*d*x - 2*c) - b)/((a^2 + b^2
)*d) + b*log(e^(-2*d*x - 2*c) + 1)/((a^2 + b^2)*d)) + integrate(4*f^3*x^3/((b*(e^(d*x + c) - e^(-d*x - c)) + 2
*a)*(e^(d*x + c) + e^(-d*x - c))) + 12*e*f^2*x^2/((b*(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*
x - c))) + 12*e^2*f*x/((b*(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c))), x)

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Fricas [C]  time = 3.03533, size = 4177, normalized size = 5.31 \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*sech(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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