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

Optimal. Leaf size=558 \[ \frac {2 i a f^2 \text {Li}_3\left (-i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac {2 i a f^2 \text {Li}_3\left (i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac {2 b f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}-\frac {2 b f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}+\frac {b f^2 \text {Li}_3\left (-e^{2 (c+d x)}\right )}{2 d^3 \left (a^2+b^2\right )}-\frac {2 i a f (e+f x) \text {Li}_2\left (-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 i a f (e+f x) \text {Li}_2\left (i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac {b f (e+f x) \text {Li}_2\left (-e^{2 (c+d x)}\right )}{d^2 \left (a^2+b^2\right )}+\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac {b (e+f x)^2 \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac {2 a (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.03, antiderivative size = 558, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5573, 5561, 2190, 2531, 2282, 6589, 6742, 4180, 3718} \[ -\frac {2 i a f (e+f x) \text {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 i a f (e+f x) \text {PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 b f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}-\frac {b f (e+f x) \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{d^2 \left (a^2+b^2\right )}+\frac {2 i a f^2 \text {PolyLog}\left (3,-i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac {2 i a f^2 \text {PolyLog}\left (3,i e^{c+d x}\right )}{d^3 \left (a^2+b^2\right )}-\frac {2 b f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}-\frac {2 b f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{d^3 \left (a^2+b^2\right )}+\frac {b f^2 \text {PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 d^3 \left (a^2+b^2\right )}+\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac {b (e+f x)^2 \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac {2 a (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(e + f*x)^2*ArcTan[E^(c + d*x)])/((a^2 + b^2)*d) + (b*(e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 +
b^2])])/((a^2 + b^2)*d) + (b*(e + f*x)^2*Log[1 + (b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/((a^2 + b^2)*d) - (b
*(e + f*x)^2*Log[1 + E^(2*(c + d*x))])/((a^2 + b^2)*d) - ((2*I)*a*f*(e + f*x)*PolyLog[2, (-I)*E^(c + d*x)])/((
a^2 + b^2)*d^2) + ((2*I)*a*f*(e + f*x)*PolyLog[2, I*E^(c + d*x)])/((a^2 + b^2)*d^2) + (2*b*f*(e + f*x)*PolyLog
[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^2) + (2*b*f*(e + f*x)*PolyLog[2, -((b*E^(c + d*x
))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^2) - (b*f*(e + f*x)*PolyLog[2, -E^(2*(c + d*x))])/((a^2 + b^2)*d^2)
+ ((2*I)*a*f^2*PolyLog[3, (-I)*E^(c + d*x)])/((a^2 + b^2)*d^3) - ((2*I)*a*f^2*PolyLog[3, I*E^(c + d*x)])/((a^
2 + b^2)*d^3) - (2*b*f^2*PolyLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^3) - (2*b*f^2*Po
lyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/((a^2 + b^2)*d^3) + (b*f^2*PolyLog[3, -E^(2*(c + d*x))])/(
2*(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 5573

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

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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

________________________________________________________________________________________

Mathematica [B]  time = 16.55, size = 1639, normalized size = 2.94 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.48, size = 1098, normalized size = 1.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \mathrm {sech}\left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -e^{2} {\left (\frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + \int \frac {4 \, f^{2} x^{2}}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} + \frac {8 \, e f x}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^2*(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^2*x^2/((b*(e^(d*x + c) - e^(-d*x - c)) + 2
*a)*(e^(d*x + c) + e^(-d*x - c))) + 8*e*f*x/((b*(e^(d*x + c) - e^(-d*x - c)) + 2*a)*(e^(d*x + c) + e^(-d*x - c
))), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________