3.4.78 \(\int \frac {(e+f x)^2 \sinh (c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [378]

Optimal. Leaf size=861 \[ -\frac {(e+f x)^3}{3 b f}-\frac {2 a (e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right )}{b^2 d}+\frac {2 a^3 (e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d}+\frac {a^2 (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right ) d}+\frac {a^2 (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right ) d}+\frac {(e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{b d}-\frac {a^2 (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{b \left (a^2+b^2\right ) d}+\frac {2 i a f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{b^2 d^2}-\frac {2 i a^3 f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d^2}-\frac {2 i a f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )}{b^2 d^2}+\frac {2 i a^3 f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d^2}+\frac {2 a^2 f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {2 a^2 f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right ) d^2}+\frac {f (e+f x) \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{b d^2}-\frac {a^2 f (e+f x) \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{b \left (a^2+b^2\right ) d^2}-\frac {2 i a f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{b^2 d^3}+\frac {2 i a^3 f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d^3}+\frac {2 i a f^2 \text {PolyLog}\left (3,i e^{c+d x}\right )}{b^2 d^3}-\frac {2 i a^3 f^2 \text {PolyLog}\left (3,i e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d^3}-\frac {2 a^2 f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right ) d^3}-\frac {2 a^2 f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b \left (a^2+b^2\right ) d^3}-\frac {f^2 \text {PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 b d^3}+\frac {a^2 f^2 \text {PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 b \left (a^2+b^2\right ) d^3} \]

[Out]

-1/3*(f*x+e)^3/b/f-2*a*(f*x+e)^2*arctan(exp(d*x+c))/b^2/d+2*a^3*(f*x+e)^2*arctan(exp(d*x+c))/b^2/(a^2+b^2)/d+(
f*x+e)^2*ln(1+exp(2*d*x+2*c))/b/d-a^2*(f*x+e)^2*ln(1+exp(2*d*x+2*c))/b/(a^2+b^2)/d+a^2*(f*x+e)^2*ln(1+b*exp(d*
x+c)/(a-(a^2+b^2)^(1/2)))/b/(a^2+b^2)/d+a^2*(f*x+e)^2*ln(1+b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/(a^2+b^2)/d+2*I
*a^3*f*(f*x+e)*polylog(2,I*exp(d*x+c))/b^2/(a^2+b^2)/d^2-2*I*a^3*f^2*polylog(3,I*exp(d*x+c))/b^2/(a^2+b^2)/d^3
+2*I*a*f*(f*x+e)*polylog(2,-I*exp(d*x+c))/b^2/d^2-2*I*a*f*(f*x+e)*polylog(2,I*exp(d*x+c))/b^2/d^2+f*(f*x+e)*po
lylog(2,-exp(2*d*x+2*c))/b/d^2-a^2*f*(f*x+e)*polylog(2,-exp(2*d*x+2*c))/b/(a^2+b^2)/d^2+2*a^2*f*(f*x+e)*polylo
g(2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/(a^2+b^2)/d^2+2*a^2*f*(f*x+e)*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1
/2)))/b/(a^2+b^2)/d^2+2*I*a^3*f^2*polylog(3,-I*exp(d*x+c))/b^2/(a^2+b^2)/d^3-2*I*a^3*f*(f*x+e)*polylog(2,-I*ex
p(d*x+c))/b^2/(a^2+b^2)/d^2-2*I*a*f^2*polylog(3,-I*exp(d*x+c))/b^2/d^3+2*I*a*f^2*polylog(3,I*exp(d*x+c))/b^2/d
^3-1/2*f^2*polylog(3,-exp(2*d*x+2*c))/b/d^3+1/2*a^2*f^2*polylog(3,-exp(2*d*x+2*c))/b/(a^2+b^2)/d^3-2*a^2*f^2*p
olylog(3,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/(a^2+b^2)/d^3-2*a^2*f^2*polylog(3,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/
2)))/b/(a^2+b^2)/d^3

________________________________________________________________________________________

Rubi [A]
time = 0.99, antiderivative size = 861, normalized size of antiderivative = 1.00, number of steps used = 38, number of rules used = 11, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {5700, 3799, 2221, 2611, 2320, 6724, 5686, 4265, 5692, 5680, 6874} \begin {gather*} \frac {2 (e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d}-\frac {2 i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^2}+\frac {2 i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^2}+\frac {2 i f^2 \text {Li}_3\left (-i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^3}-\frac {2 i f^2 \text {Li}_3\left (i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^3}+\frac {(e+f x)^2 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right ) a^2}{b \left (a^2+b^2\right ) d}+\frac {(e+f x)^2 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right ) a^2}{b \left (a^2+b^2\right ) d}-\frac {(e+f x)^2 \log \left (1+e^{2 (c+d x)}\right ) a^2}{b \left (a^2+b^2\right ) d}+\frac {2 f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^2}+\frac {2 f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^2}-\frac {f (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right ) a^2}{b \left (a^2+b^2\right ) d^2}-\frac {2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^3}-\frac {2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^3}+\frac {f^2 \text {Li}_3\left (-e^{2 (c+d x)}\right ) a^2}{2 b \left (a^2+b^2\right ) d^3}-\frac {2 (e+f x)^2 \text {ArcTan}\left (e^{c+d x}\right ) a}{b^2 d}+\frac {2 i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) a}{b^2 d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) a}{b^2 d^2}-\frac {2 i f^2 \text {Li}_3\left (-i e^{c+d x}\right ) a}{b^2 d^3}+\frac {2 i f^2 \text {Li}_3\left (i e^{c+d x}\right ) a}{b^2 d^3}-\frac {(e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{b d}+\frac {f (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right )}{b d^2}-\frac {f^2 \text {Li}_3\left (-e^{2 (c+d x)}\right )}{2 b d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^2*Sinh[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3799

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 4265

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 5680

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 5686

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*Sech
[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 5692

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 5700

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

Rule 6724

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 6874

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

Rubi steps

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

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Mathematica [A]
time = 15.24, size = 1202, normalized size = 1.40 \begin {gather*} -\frac {12 b d^3 e^2 e^{2 c} x+12 b d^3 e e^{2 c} f x^2+4 b d^3 e^{2 c} f^2 x^3+12 a d^2 e^2 \left (1+e^{2 c}\right ) \text {ArcTan}\left (e^{c+d x}\right )-6 b d^2 e^2 \left (1+e^{2 c}\right ) \log \left (1+e^{2 (c+d x)}\right )+12 i a d e \left (1+e^{2 c}\right ) f \left (d x \left (\log \left (1-i e^{c+d x}\right )-\log \left (1+i e^{c+d x}\right )\right )-\text {PolyLog}\left (2,-i e^{c+d x}\right )+\text {PolyLog}\left (2,i e^{c+d x}\right )\right )-6 b d e \left (1+e^{2 c}\right ) f \left (2 d x \log \left (1+e^{2 (c+d x)}\right )+\text {PolyLog}\left (2,-e^{2 (c+d x)}\right )\right )+6 i a \left (1+e^{2 c}\right ) f^2 \left (d^2 x^2 \log \left (1-i e^{c+d x}\right )-d^2 x^2 \log \left (1+i e^{c+d x}\right )-2 d x \text {PolyLog}\left (2,-i e^{c+d x}\right )+2 d x \text {PolyLog}\left (2,i e^{c+d x}\right )+2 \text {PolyLog}\left (3,-i e^{c+d x}\right )-2 \text {PolyLog}\left (3,i e^{c+d x}\right )\right )-3 b \left (1+e^{2 c}\right ) f^2 \left (2 d^2 x^2 \log \left (1+e^{2 (c+d x)}\right )+2 d x \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )-\text {PolyLog}\left (3,-e^{2 (c+d x)}\right )\right )}{6 \left (a^2+b^2\right ) d^3 \left (1+e^{2 c}\right )}+\frac {a^2 \left (-\frac {2 e^{2 c} x \left (3 e^2+3 e f x+f^2 x^2\right )}{-1+e^{2 c}}+\frac {3 \left (\frac {2 a \sqrt {a^2+b^2} d^2 e^2 \text {ArcTan}\left (\frac {a+b e^{c+d x}}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-\left (a^2+b^2\right )^2}}+\frac {2 a \sqrt {-a^2-b^2} d^2 e^2 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )}{\sqrt {-\left (a^2+b^2\right )^2}}+d^2 e^2 \log \left (2 a e^{c+d x}+b \left (-1+e^{2 (c+d x)}\right )\right )+2 d^2 e f x \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+d^2 f^2 x^2 \log \left (1+\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+2 d^2 e f x \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+d^2 f^2 x^2 \log \left (1+\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+2 d f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+2 d f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-2 f^2 \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-2 f^2 \text {PolyLog}\left (3,-\frac {b e^{2 c+d x}}{a e^c+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )\right )}{d^3}\right )}{3 b \left (a^2+b^2\right )}+\frac {\left (3 a^2 e^2 x-3 b^2 e^2 x+3 a^2 e f x^2-3 b^2 e f x^2+a^2 f^2 x^3-b^2 f^2 x^3+3 a^2 e^2 x \cosh (2 c)+3 b^2 e^2 x \cosh (2 c)+3 a^2 e f x^2 \cosh (2 c)+3 b^2 e f x^2 \cosh (2 c)+a^2 f^2 x^3 \cosh (2 c)+b^2 f^2 x^3 \cosh (2 c)\right ) \text {csch}(c) \text {sech}(c)}{6 b \left (a^2+b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^2*Sinh[c + d*x]*Tanh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

-1/6*(12*b*d^3*e^2*E^(2*c)*x + 12*b*d^3*e*E^(2*c)*f*x^2 + 4*b*d^3*E^(2*c)*f^2*x^3 + 12*a*d^2*e^2*(1 + E^(2*c))
*ArcTan[E^(c + d*x)] - 6*b*d^2*e^2*(1 + E^(2*c))*Log[1 + E^(2*(c + d*x))] + (12*I)*a*d*e*(1 + E^(2*c))*f*(d*x*
(Log[1 - I*E^(c + d*x)] - Log[1 + I*E^(c + d*x)]) - PolyLog[2, (-I)*E^(c + d*x)] + PolyLog[2, I*E^(c + d*x)])
- 6*b*d*e*(1 + E^(2*c))*f*(2*d*x*Log[1 + E^(2*(c + d*x))] + PolyLog[2, -E^(2*(c + d*x))]) + (6*I)*a*(1 + E^(2*
c))*f^2*(d^2*x^2*Log[1 - I*E^(c + d*x)] - d^2*x^2*Log[1 + I*E^(c + d*x)] - 2*d*x*PolyLog[2, (-I)*E^(c + d*x)]
+ 2*d*x*PolyLog[2, I*E^(c + d*x)] + 2*PolyLog[3, (-I)*E^(c + d*x)] - 2*PolyLog[3, I*E^(c + d*x)]) - 3*b*(1 + E
^(2*c))*f^2*(2*d^2*x^2*Log[1 + E^(2*(c + d*x))] + 2*d*x*PolyLog[2, -E^(2*(c + d*x))] - PolyLog[3, -E^(2*(c + d
*x))]))/((a^2 + b^2)*d^3*(1 + E^(2*c))) + (a^2*((-2*E^(2*c)*x*(3*e^2 + 3*e*f*x + f^2*x^2))/(-1 + E^(2*c)) + (3
*((2*a*Sqrt[a^2 + b^2]*d^2*e^2*ArcTan[(a + b*E^(c + d*x))/Sqrt[-a^2 - b^2]])/Sqrt[-(a^2 + b^2)^2] + (2*a*Sqrt[
-a^2 - b^2]*d^2*e^2*ArcTanh[(a + b*E^(c + d*x))/Sqrt[a^2 + b^2]])/Sqrt[-(a^2 + b^2)^2] + d^2*e^2*Log[2*a*E^(c
+ d*x) + b*(-1 + E^(2*(c + d*x)))] + 2*d^2*e*f*x*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])
] + d^2*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])] + 2*d^2*e*f*x*Log[1 + (b*E^(2*c
+ d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])] + d^2*f^2*x^2*Log[1 + (b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2
)*E^(2*c)])] + 2*d*f*(e + f*x)*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] + 2*d*f*(e
+ f*x)*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] - 2*f^2*PolyLog[3, -((b*E^(2*c +
d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 2*f^2*PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E
^(2*c)]))]))/d^3))/(3*b*(a^2 + b^2)) + ((3*a^2*e^2*x - 3*b^2*e^2*x + 3*a^2*e*f*x^2 - 3*b^2*e*f*x^2 + a^2*f^2*x
^3 - b^2*f^2*x^3 + 3*a^2*e^2*x*Cosh[2*c] + 3*b^2*e^2*x*Cosh[2*c] + 3*a^2*e*f*x^2*Cosh[2*c] + 3*b^2*e*f*x^2*Cos
h[2*c] + a^2*f^2*x^3*Cosh[2*c] + b^2*f^2*x^3*Cosh[2*c])*Csch[c]*Sech[c])/(6*b*(a^2 + b^2))

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Maple [F]
time = 2.52, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \sinh \left (d x +c \right ) \tanh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^2*sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1694 vs. \(2 (802) = 1604\).
time = 0.41, size = 1694, normalized size = 1.97 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-1/3*((a^2 + b^2)*d^3*f^2*x^3 + 3*(a^2 + b^2)*d^3*f*x^2*cosh(1) + 3*(a^2 + b^2)*d^3*x*cosh(1)^2 + 3*(a^2 + b^2
)*d^3*x*sinh(1)^2 + 6*a^2*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^2*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^2*d*f^2*x + a^2*d*f*cosh(1) + a^2*d*f*sinh(1))*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) - 6*(a^2*d*
f^2*x + a^2*d*f*cosh(1) + a^2*d*f*sinh(1))*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d*x + c) + b*sin
h(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) + 6*(I*a*b*d*f^2*x - b^2*d*f^2*x + I*a*b*d*f*cosh(1) - b^2*d*f*c
osh(1) + I*a*b*d*f*sinh(1) - b^2*d*f*sinh(1))*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) + 6*(-I*a*b*d*f^2*x - b
^2*d*f^2*x - I*a*b*d*f*cosh(1) - b^2*d*f*cosh(1) - I*a*b*d*f*sinh(1) - b^2*d*f*sinh(1))*dilog(-I*cosh(d*x + c)
- I*sinh(d*x + c)) - 3*(a^2*c^2*f^2 - 2*a^2*c*d*f*cosh(1) + a^2*d^2*cosh(1)^2 + a^2*d^2*sinh(1)^2 - 2*(a^2*c*
d*f - a^2*d^2*cosh(1))*sinh(1))*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) -
3*(a^2*c^2*f^2 - 2*a^2*c*d*f*cosh(1) + a^2*d^2*cosh(1)^2 + a^2*d^2*sinh(1)^2 - 2*(a^2*c*d*f - a^2*d^2*cosh(1)
)*sinh(1))*log(2*b*cosh(d*x + c) + 2*b*sinh(d*x + c) - 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) - 3*(a^2*d^2*f^2*x^2 -
a^2*c^2*f^2 + 2*(a^2*d^2*f*x + a^2*c*d*f)*cosh(1) + 2*(a^2*d^2*f*x + a^2*c*d*f)*sinh(1))*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) - 3*(a^2*d^2*f^2*x^2 -
a^2*c^2*f^2 + 2*(a^2*d^2*f*x + a^2*c*d*f)*cosh(1) + 2*(a^2*d^2*f*x + a^2*c*d*f)*sinh(1))*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) + 3*(I*a*b*c^2*f^2 - b
^2*c^2*f^2 - 2*I*a*b*c*d*f*cosh(1) + 2*b^2*c*d*f*cosh(1) + I*a*b*d^2*cosh(1)^2 - b^2*d^2*cosh(1)^2 + I*a*b*d^2
*sinh(1)^2 - b^2*d^2*sinh(1)^2 - 2*I*(a*b*c*d*f - a*b*d^2*cosh(1))*sinh(1) + 2*(b^2*c*d*f - b^2*d^2*cosh(1))*s
inh(1))*log(cosh(d*x + c) + sinh(d*x + c) + I) + 3*(-I*a*b*c^2*f^2 - b^2*c^2*f^2 + 2*I*a*b*c*d*f*cosh(1) + 2*b
^2*c*d*f*cosh(1) - I*a*b*d^2*cosh(1)^2 - b^2*d^2*cosh(1)^2 - I*a*b*d^2*sinh(1)^2 - b^2*d^2*sinh(1)^2 + 2*I*(a*
b*c*d*f - a*b*d^2*cosh(1))*sinh(1) + 2*(b^2*c*d*f - b^2*d^2*cosh(1))*sinh(1))*log(cosh(d*x + c) + sinh(d*x + c
) - I) + 3*(-I*a*b*d^2*f^2*x^2 - b^2*d^2*f^2*x^2 + I*a*b*c^2*f^2 + b^2*c^2*f^2 - 2*I*(a*b*d^2*f*x + a*b*c*d*f)
*cosh(1) - 2*(b^2*d^2*f*x + b^2*c*d*f)*cosh(1) - 2*I*(a*b*d^2*f*x + a*b*c*d*f)*sinh(1) - 2*(b^2*d^2*f*x + b^2*
c*d*f)*sinh(1))*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1) + 3*(I*a*b*d^2*f^2*x^2 - b^2*d^2*f^2*x^2 - I*a*b*c^
2*f^2 + b^2*c^2*f^2 + 2*I*(a*b*d^2*f*x + a*b*c*d*f)*cosh(1) - 2*(b^2*d^2*f*x + b^2*c*d*f)*cosh(1) + 2*I*(a*b*d
^2*f*x + a*b*c*d*f)*sinh(1) - 2*(b^2*d^2*f*x + b^2*c*d*f)*sinh(1))*log(-I*cosh(d*x + c) - I*sinh(d*x + c) + 1)
+ 6*(-I*a*b*f^2 + b^2*f^2)*polylog(3, I*cosh(d*x + c) + I*sinh(d*x + c)) + 6*(I*a*b*f^2 + b^2*f^2)*polylog(3,
-I*cosh(d*x + c) - I*sinh(d*x + c)) + 3*((a^2 + b^2)*d^3*f*x^2 + 2*(a^2 + b^2)*d^3*x*cosh(1))*sinh(1))/((a^2*
b + b^3)*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e + f x\right )^{2} \sinh {\left (c + d x \right )} \tanh {\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral((e + f*x)**2*sinh(c + d*x)*tanh(c + d*x)/(a + b*sinh(c + d*x)), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sinh(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {sinh}\left (c+d\,x\right )\,\mathrm {tanh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinh(c + d*x)*tanh(c + d*x)*(e + f*x)^2)/(a + b*sinh(c + d*x)),x)

[Out]

int((sinh(c + d*x)*tanh(c + d*x)*(e + f*x)^2)/(a + b*sinh(c + d*x)), x)

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