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

Optimal. Leaf size=861 \[ \frac {2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d}-\frac {2 i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^2}+\frac {2 i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^2}+\frac {2 i f^2 \text {Li}_3\left (-i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^3}-\frac {2 i f^2 \text {Li}_3\left (i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^3}+\frac {(e+f x)^2 \log \left (\frac {e^{c+d x} b}{a-\sqrt {a^2+b^2}}+1\right ) a^2}{b \left (a^2+b^2\right ) d}+\frac {(e+f x)^2 \log \left (\frac {e^{c+d x} b}{a+\sqrt {a^2+b^2}}+1\right ) a^2}{b \left (a^2+b^2\right ) d}-\frac {(e+f x)^2 \log \left (1+e^{2 (c+d x)}\right ) a^2}{b \left (a^2+b^2\right ) d}+\frac {2 f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^2}+\frac {2 f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^2}-\frac {f (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right ) a^2}{b \left (a^2+b^2\right ) d^2}-\frac {2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^3}-\frac {2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) a^2}{b \left (a^2+b^2\right ) d^3}+\frac {f^2 \text {Li}_3\left (-e^{2 (c+d x)}\right ) a^2}{2 b \left (a^2+b^2\right ) d^3}-\frac {2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right ) a}{b^2 d}+\frac {2 i f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) a}{b^2 d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) a}{b^2 d^2}-\frac {2 i f^2 \text {Li}_3\left (-i e^{c+d x}\right ) a}{b^2 d^3}+\frac {2 i f^2 \text {Li}_3\left (i e^{c+d x}\right ) a}{b^2 d^3}-\frac {(e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{b d}+\frac {f (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right )}{b d^2}-\frac {f^2 \text {Li}_3\left (-e^{2 (c+d x)}\right )}{2 b d^3} \]

[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*f^2*polylog(3,I*exp(d*x+c))/b^2/d^3-2*I*a*f^2*p
olylog(3,-I*exp(d*x+c))/b^2/d^3-2*I*a^3*f*(f*x+e)*polylog(2,-I*exp(d*x+c))/b^2/(a^2+b^2)/d^2+f*(f*x+e)*polylog
(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)*polylog(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*f*(f*x+e)*polylog(2,-I*exp(d*x+c))/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^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
^2-1/2*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 = 1.37, 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 = {5581, 3718, 2190, 2531, 2282, 6589, 5567, 4180, 5573, 5561, 6742} \[ \frac {2 (e+f x)^2 \tan ^{-1}\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 {PolyLog}\left (2,-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 {PolyLog}\left (2,i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^2}+\frac {2 i f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right ) a^3}{b^2 \left (a^2+b^2\right ) d^3}-\frac {2 i f^2 \text {PolyLog}\left (3,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 {PolyLog}\left (2,-\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 {PolyLog}\left (2,-\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 {PolyLog}\left (2,-e^{2 (c+d x)}\right ) a^2}{b \left (a^2+b^2\right ) d^2}-\frac {2 f^2 \text {PolyLog}\left (3,-\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 {PolyLog}\left (3,-\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 {PolyLog}\left (3,-e^{2 (c+d x)}\right ) a^2}{2 b \left (a^2+b^2\right ) d^3}-\frac {2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right ) a}{b^2 d}+\frac {2 i f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right ) a}{b^2 d^2}-\frac {2 i f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right ) a}{b^2 d^2}-\frac {2 i f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right ) a}{b^2 d^3}+\frac {2 i f^2 \text {PolyLog}\left (3,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 {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{b d^2}-\frac {f^2 \text {PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 b d^3} \]

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]

-(e + f*x)^3/(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)*PolyLo
g[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 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 5581

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 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]]

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 ) \operatorname {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 ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^3}-\frac {f^2 \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 = 10.66, size = 997, normalized size = 1.16 \[ \frac {-2 a^2 f^2 x^3 d^3-2 b^2 f^2 x^3 d^3-6 a^2 e f x^2 d^3-6 b^2 e f x^2 d^3-6 a^2 e^2 x d^3-6 b^2 e^2 x d^3-12 a b e^2 \tan ^{-1}\left (e^{c+d x}\right ) d^2-6 i a b f^2 x^2 \log \left (1-i e^{c+d x}\right ) d^2-12 i a b e f x \log \left (1-i e^{c+d x}\right ) d^2+6 i a b f^2 x^2 \log \left (1+i e^{c+d x}\right ) d^2+12 i a b e f x \log \left (1+i e^{c+d x}\right ) d^2+6 b^2 e^2 \log \left (1+e^{2 (c+d x)}\right ) d^2+6 b^2 f^2 x^2 \log \left (1+e^{2 (c+d x)}\right ) d^2+12 b^2 e f x \log \left (1+e^{2 (c+d x)}\right ) d^2+6 a^2 e^2 \log \left (-2 e^{c+d x} a-b e^{2 (c+d x)}+b\right ) d^2+6 a^2 f^2 x^2 \log \left (\frac {e^{2 c+d x} b}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}+1\right ) d^2+12 a^2 e f x \log \left (\frac {e^{2 c+d x} b}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}+1\right ) d^2+6 a^2 f^2 x^2 \log \left (\frac {e^{2 c+d x} b}{e^c a+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}+1\right ) d^2+12 a^2 e f x \log \left (\frac {e^{2 c+d x} b}{e^c a+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}+1\right ) d^2+12 i a b f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right ) d-12 i a b f (e+f x) \text {Li}_2\left (i e^{c+d x}\right ) d+6 b^2 e f \text {Li}_2\left (-e^{2 (c+d x)}\right ) d+6 b^2 f^2 x \text {Li}_2\left (-e^{2 (c+d x)}\right ) d+12 a^2 e f \text {Li}_2\left (-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right ) d+12 a^2 f^2 x \text {Li}_2\left (-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right ) d+12 a^2 e f \text {Li}_2\left (-\frac {b e^{2 c+d x}}{e^c a+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right ) d+12 a^2 f^2 x \text {Li}_2\left (-\frac {b e^{2 c+d x}}{e^c a+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right ) d-12 i a b f^2 \text {Li}_3\left (-i e^{c+d x}\right )+12 i a b f^2 \text {Li}_3\left (i e^{c+d x}\right )-3 b^2 f^2 \text {Li}_3\left (-e^{2 (c+d x)}\right )-12 a^2 f^2 \text {Li}_3\left (-\frac {b e^{2 c+d x}}{a e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-12 a^2 f^2 \text {Li}_3\left (-\frac {b e^{2 c+d x}}{e^c a+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )}{6 b \left (a^2+b^2\right ) d^3} \]

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]

(-6*a^2*d^3*e^2*x - 6*b^2*d^3*e^2*x - 6*a^2*d^3*e*f*x^2 - 6*b^2*d^3*e*f*x^2 - 2*a^2*d^3*f^2*x^3 - 2*b^2*d^3*f^
2*x^3 - 12*a*b*d^2*e^2*ArcTan[E^(c + d*x)] - (12*I)*a*b*d^2*e*f*x*Log[1 - I*E^(c + d*x)] - (6*I)*a*b*d^2*f^2*x
^2*Log[1 - I*E^(c + d*x)] + (12*I)*a*b*d^2*e*f*x*Log[1 + I*E^(c + d*x)] + (6*I)*a*b*d^2*f^2*x^2*Log[1 + I*E^(c
+ d*x)] + 6*b^2*d^2*e^2*Log[1 + E^(2*(c + d*x))] + 12*b^2*d^2*e*f*x*Log[1 + E^(2*(c + d*x))] + 6*b^2*d^2*f^2*
x^2*Log[1 + E^(2*(c + d*x))] + 6*a^2*d^2*e^2*Log[b - 2*a*E^(c + d*x) - b*E^(2*(c + d*x))] + 12*a^2*d^2*e*f*x*L
og[1 + (b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)])] + 6*a^2*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)])] + 12*a^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)])] + 6*a^2*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)])] + (12*I)*a*b*d*f*(e
+ f*x)*PolyLog[2, (-I)*E^(c + d*x)] - (12*I)*a*b*d*f*(e + f*x)*PolyLog[2, I*E^(c + d*x)] + 6*b^2*d*e*f*PolyLog
[2, -E^(2*(c + d*x))] + 6*b^2*d*f^2*x*PolyLog[2, -E^(2*(c + d*x))] + 12*a^2*d*e*f*PolyLog[2, -((b*E^(2*c + d*x
))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] + 12*a^2*d*f^2*x*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 +
b^2)*E^(2*c)]))] + 12*a^2*d*e*f*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] + 12*a^2*
d*f^2*x*PolyLog[2, -((b*E^(2*c + d*x))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] - (12*I)*a*b*f^2*PolyLog[3, (-I)*
E^(c + d*x)] + (12*I)*a*b*f^2*PolyLog[3, I*E^(c + d*x)] - 3*b^2*f^2*PolyLog[3, -E^(2*(c + d*x))] - 12*a^2*f^2*
PolyLog[3, -((b*E^(2*c + d*x))/(a*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 12*a^2*f^2*PolyLog[3, -((b*E^(2*c + d*x
))/(a*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/(6*b*(a^2 + b^2)*d^3)

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fricas [C]  time = 0.93, size = 1250, normalized size = 1.45 \[ \text {result too large to display} \]

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*e*f*x^2 + 3*(a^2 + b^2)*d^3*e^2*x + 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*po
lylog(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*e*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqr
t((a^2 + b^2)/b^2) - b)/b + 1) - 6*(a^2*d*f^2*x + a^2*d*e*f)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cos
h(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - (-6*I*a*b*d*f^2*x + 6*b^2*d*f^2*x - 6*I*a*b*
d*e*f + 6*b^2*d*e*f)*dilog(I*cosh(d*x + c) + I*sinh(d*x + c)) - (6*I*a*b*d*f^2*x + 6*b^2*d*f^2*x + 6*I*a*b*d*e
*f + 6*b^2*d*e*f)*dilog(-I*cosh(d*x + c) - I*sinh(d*x + c)) - 3*(a^2*d^2*e^2 - 2*a^2*c*d*e*f + a^2*c^2*f^2)*lo
g(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*e^2 - 2*a^2*c*d*e*f +
a^2*c^2*f^2)*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
+ 2*a^2*d^2*e*f*x + 2*a^2*c*d*e*f - a^2*c^2*f^2)*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 + 2*a^2*d^2*e*f*x + 2*a^2*c*d*e*f - a^2*c
^2*f^2)*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*d^2*e^2 + 3*b^2*d^2*e^2 + 6*I*a*b*c*d*e*f - 6*b^2*c*d*e*f - 3*I*a*b*c^2*f^2 + 3*b^2*c^2*f^2)
*log(cosh(d*x + c) + sinh(d*x + c) + I) - (3*I*a*b*d^2*e^2 + 3*b^2*d^2*e^2 - 6*I*a*b*c*d*e*f - 6*b^2*c*d*e*f +
3*I*a*b*c^2*f^2 + 3*b^2*c^2*f^2)*log(cosh(d*x + c) + sinh(d*x + c) - I) - (3*I*a*b*d^2*f^2*x^2 + 3*b^2*d^2*f^
2*x^2 + 6*I*a*b*d^2*e*f*x + 6*b^2*d^2*e*f*x + 6*I*a*b*c*d*e*f + 6*b^2*c*d*e*f - 3*I*a*b*c^2*f^2 - 3*b^2*c^2*f^
2)*log(I*cosh(d*x + c) + I*sinh(d*x + c) + 1) - (-3*I*a*b*d^2*f^2*x^2 + 3*b^2*d^2*f^2*x^2 - 6*I*a*b*d^2*e*f*x
+ 6*b^2*d^2*e*f*x - 6*I*a*b*c*d*e*f + 6*b^2*c*d*e*f + 3*I*a*b*c^2*f^2 - 3*b^2*c^2*f^2)*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)))/((a^2*b + b^3)*d^3)

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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)^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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maple [F]  time = 1.10, 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 \[ e^{2} {\left (\frac {a^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} b + b^{3}\right )} d} + \frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {d x + c}{b d}\right )} + \frac {f^{2} x^{3} + 3 \, e f x^{2}}{3 \, b} - \int \frac {2 \, {\left (a^{2} b f^{2} x^{2} + 2 \, a^{2} b e f x - {\left (a^{3} f^{2} x^{2} e^{c} + 2 \, a^{3} e f x e^{c}\right )} e^{\left (d x\right )}\right )}}{a^{2} b^{2} + b^{4} - {\left (a^{2} b^{2} e^{\left (2 \, c\right )} + b^{4} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - 2 \, {\left (a^{3} b e^{c} + a b^{3} e^{c}\right )} e^{\left (d x\right )}}\,{d x} - \int \frac {2 \, {\left (b f^{2} x^{2} + 2 \, b e f x + {\left (a f^{2} x^{2} e^{c} + 2 \, a e f x e^{c}\right )} e^{\left (d x\right )}\right )}}{a^{2} + b^{2} + {\left (a^{2} e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}}\,{d x} \]

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]

e^2*(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)) + 1/3*(f^2*x^3 + 3*e*f*x^2)/b - integ
rate(2*(a^2*b*f^2*x^2 + 2*a^2*b*e*f*x - (a^3*f^2*x^2*e^c + 2*a^3*e*f*x*e^c)*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*e*f*
x + (a*f^2*x^2*e^c + 2*a*e*f*x*e^c)*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {sinh}\left (c+d\,x\right )\,\mathrm {tanh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

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

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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