3.18 \(\int \frac{(e+f x)^2 \cosh (c+d x)}{a+b \text{csch}(c+d x)} \, dx\)

Optimal. Leaf size=330 \[ -\frac{2 b f (e+f x) \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}\right )}{a^2 d^2}-\frac{2 b f (e+f x) \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}\right )}{a^2 d^2}+\frac{2 b f^2 \text{PolyLog}\left (3,-\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}\right )}{a^2 d^3}+\frac{2 b f^2 \text{PolyLog}\left (3,-\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}\right )}{a^2 d^3}-\frac{b (e+f x)^2 \log \left (\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}+1\right )}{a^2 d}-\frac{b (e+f x)^2 \log \left (\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}+1\right )}{a^2 d}+\frac{b (e+f x)^3}{3 a^2 f}-\frac{2 f (e+f x) \cosh (c+d x)}{a d^2}+\frac{2 f^2 \sinh (c+d x)}{a d^3}+\frac{(e+f x)^2 \sinh (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.598721, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {5594, 5579, 3296, 2637, 5561, 2190, 2531, 2282, 6589} \[ -\frac{2 b f (e+f x) \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}\right )}{a^2 d^2}-\frac{2 b f (e+f x) \text{PolyLog}\left (2,-\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}\right )}{a^2 d^2}+\frac{2 b f^2 \text{PolyLog}\left (3,-\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}\right )}{a^2 d^3}+\frac{2 b f^2 \text{PolyLog}\left (3,-\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}\right )}{a^2 d^3}-\frac{b (e+f x)^2 \log \left (\frac{a e^{c+d x}}{b-\sqrt{a^2+b^2}}+1\right )}{a^2 d}-\frac{b (e+f x)^2 \log \left (\frac{a e^{c+d x}}{\sqrt{a^2+b^2}+b}+1\right )}{a^2 d}+\frac{b (e+f x)^3}{3 a^2 f}-\frac{2 f (e+f x) \cosh (c+d x)}{a d^2}+\frac{2 f^2 \sinh (c+d x)}{a d^3}+\frac{(e+f x)^2 \sinh (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5594

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.))/(Csch[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Sym
bol] :> Int[((e + f*x)^m*Sinh[c + d*x]*F[c + d*x]^n)/(b + a*Sinh[c + d*x]), x] /; FreeQ[{a, b, c, d, e, f}, x]
&& HyperbolicQ[F] && IntegersQ[m, n]

Rule 5579

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 5561

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

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

Rubi steps

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

Mathematica [C]  time = 10.2523, size = 1167, normalized size = 3.54 \[ -\frac{\text{csch}(c+d x) \left (\frac{b \log (b+a \sinh (c+d x))}{a^2}-\frac{\sinh (c+d x)}{a}\right ) (b+a \sinh (c+d x)) e^2}{d (a+b \text{csch}(c+d x))}+\frac{2 f \text{csch}(c+d x) (b+a \sinh (c+d x)) \left (-a \cosh (c+d x)-b (c+d x) \log (b+a \sinh (c+d x))+b c \log \left (\frac{a \sinh (c+d x)}{b}+1\right )+i b \left (-\frac{1}{8} i (2 c+2 d x+i \pi )^2-4 i \sin ^{-1}\left (\frac{\sqrt{\frac{i b}{a}+1}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(i a+b) \cot \left (\frac{1}{4} (2 i c+2 i d x+\pi )\right )}{\sqrt{a^2+b^2}}\right )-\frac{1}{2} \left (-2 i c-2 i d x+4 \sin ^{-1}\left (\frac{\sqrt{\frac{i b}{a}+1}}{\sqrt{2}}\right )+\pi \right ) \log \left (\frac{e^{c+d x} \left (\sqrt{a^2+b^2}-b\right )}{a}+1\right )-\frac{1}{2} \left (-2 i c-2 i d x-4 \sin ^{-1}\left (\frac{\sqrt{\frac{i b}{a}+1}}{\sqrt{2}}\right )+\pi \right ) \log \left (1-\frac{\left (b+\sqrt{a^2+b^2}\right ) e^{c+d x}}{a}\right )+\left (\frac{\pi }{2}-i (c+d x)\right ) \log (b+a \sinh (c+d x))+i \left (\text{PolyLog}\left (2,\frac{\left (b-\sqrt{a^2+b^2}\right ) e^{c+d x}}{a}\right )+\text{PolyLog}\left (2,\frac{\left (b+\sqrt{a^2+b^2}\right ) e^{c+d x}}{a}\right )\right )\right )+a d x \sinh (c+d x)\right ) e}{a^2 d^2 (a+b \text{csch}(c+d x))}+\frac{f^2 \text{csch}(c+d x) \left (2 b (\coth (c)-1) x^3-2 b \coth (c) x^3-\frac{6 a^2 b \left (d^2 \log \left (\frac{\left (b-\sqrt{a^2+b^2}\right ) (\cosh (c+d x)-\sinh (c+d x))}{a}+1\right ) x^2-2 d \text{PolyLog}\left (2,\frac{\left (\sqrt{a^2+b^2}-b\right ) (\cosh (c+d x)-\sinh (c+d x))}{a}\right ) x-2 \text{PolyLog}\left (3,\frac{\left (\sqrt{a^2+b^2}-b\right ) (\cosh (c+d x)-\sinh (c+d x))}{a}\right )\right )}{\sqrt{a^2+b^2} \left (\sqrt{a^2+b^2}-b\right ) d^3}-\frac{6 a^2 b \left (d^2 \log \left (\frac{\left (b+\sqrt{a^2+b^2}\right ) (\cosh (c+d x)-\sinh (c+d x))}{a}+1\right ) x^2-2 d \text{PolyLog}\left (2,\frac{\left (b+\sqrt{a^2+b^2}\right ) (\sinh (c+d x)-\cosh (c+d x))}{a}\right ) x-2 \text{PolyLog}\left (3,\frac{\left (b+\sqrt{a^2+b^2}\right ) (\sinh (c+d x)-\cosh (c+d x))}{a}\right )\right )}{\sqrt{a^2+b^2} \left (b+\sqrt{a^2+b^2}\right ) d^3}+\frac{6 b^2 \left (d^2 \log \left (\frac{a (\cosh (c+d x)+\sinh (c+d x))}{b-\sqrt{a^2+b^2}}+1\right ) x^2+2 d \text{PolyLog}\left (2,\frac{a (\cosh (c+d x)+\sinh (c+d x))}{\sqrt{a^2+b^2}-b}\right ) x-2 \text{PolyLog}\left (3,\frac{a (\cosh (c+d x)+\sinh (c+d x))}{\sqrt{a^2+b^2}-b}\right )\right )}{\sqrt{a^2+b^2} d^3}-\frac{6 b^2 \left (d^2 \log \left (\frac{a (\cosh (c+d x)+\sinh (c+d x))}{b+\sqrt{a^2+b^2}}+1\right ) x^2+2 d \text{PolyLog}\left (2,-\frac{a (\cosh (c+d x)+\sinh (c+d x))}{b+\sqrt{a^2+b^2}}\right ) x-2 \text{PolyLog}\left (3,-\frac{a (\cosh (c+d x)+\sinh (c+d x))}{b+\sqrt{a^2+b^2}}\right )\right )}{\sqrt{a^2+b^2} d^3}+\frac{6 a \cosh (d x) \left (\left (d^2 x^2+2\right ) \sinh (c)-2 d x \cosh (c)\right )}{d^3}+\frac{6 a \left (\left (d^2 x^2+2\right ) \cosh (c)-2 d x \sinh (c)\right ) \sinh (d x)}{d^3}\right ) (b+a \sinh (c+d x))}{6 a^2 (a+b \text{csch}(c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(f^2*Csch[c + d*x]*(2*b*x^3*(-1 + Coth[c]) - 2*b*x^3*Coth[c] - (6*a^2*b*(d^2*x^2*Log[1 + ((b - Sqrt[a^2 + b^2]
)*(Cosh[c + d*x] - Sinh[c + d*x]))/a] - 2*d*x*PolyLog[2, ((-b + Sqrt[a^2 + b^2])*(Cosh[c + d*x] - Sinh[c + d*x
]))/a] - 2*PolyLog[3, ((-b + Sqrt[a^2 + b^2])*(Cosh[c + d*x] - Sinh[c + d*x]))/a]))/(Sqrt[a^2 + b^2]*(-b + Sqr
t[a^2 + b^2])*d^3) - (6*a^2*b*(d^2*x^2*Log[1 + ((b + Sqrt[a^2 + b^2])*(Cosh[c + d*x] - Sinh[c + d*x]))/a] - 2*
d*x*PolyLog[2, ((b + Sqrt[a^2 + b^2])*(-Cosh[c + d*x] + Sinh[c + d*x]))/a] - 2*PolyLog[3, ((b + Sqrt[a^2 + b^2
])*(-Cosh[c + d*x] + Sinh[c + d*x]))/a]))/(Sqrt[a^2 + b^2]*(b + Sqrt[a^2 + b^2])*d^3) + (6*b^2*(d^2*x^2*Log[1
+ (a*(Cosh[c + d*x] + Sinh[c + d*x]))/(b - Sqrt[a^2 + b^2])] + 2*d*x*PolyLog[2, (a*(Cosh[c + d*x] + Sinh[c + d
*x]))/(-b + Sqrt[a^2 + b^2])] - 2*PolyLog[3, (a*(Cosh[c + d*x] + Sinh[c + d*x]))/(-b + Sqrt[a^2 + b^2])]))/(Sq
rt[a^2 + b^2]*d^3) - (6*b^2*(d^2*x^2*Log[1 + (a*(Cosh[c + d*x] + Sinh[c + d*x]))/(b + Sqrt[a^2 + b^2])] + 2*d*
x*PolyLog[2, -((a*(Cosh[c + d*x] + Sinh[c + d*x]))/(b + Sqrt[a^2 + b^2]))] - 2*PolyLog[3, -((a*(Cosh[c + d*x]
+ Sinh[c + d*x]))/(b + Sqrt[a^2 + b^2]))]))/(Sqrt[a^2 + b^2]*d^3) + (6*a*Cosh[d*x]*(-2*d*x*Cosh[c] + (2 + d^2*
x^2)*Sinh[c]))/d^3 + (6*a*((2 + d^2*x^2)*Cosh[c] - 2*d*x*Sinh[c])*Sinh[d*x])/d^3)*(b + a*Sinh[c + d*x]))/(6*a^
2*(a + b*Csch[c + d*x])) - (e^2*Csch[c + d*x]*((b*Log[b + a*Sinh[c + d*x]])/a^2 - Sinh[c + d*x]/a)*(b + a*Sinh
[c + d*x]))/(d*(a + b*Csch[c + d*x])) + (2*e*f*Csch[c + d*x]*(b + a*Sinh[c + d*x])*(-(a*Cosh[c + d*x]) - b*(c
+ d*x)*Log[b + a*Sinh[c + d*x]] + b*c*Log[1 + (a*Sinh[c + d*x])/b] + I*b*((-I/8)*(2*c + I*Pi + 2*d*x)^2 - (4*I
)*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]]*ArcTan[((I*a + b)*Cot[((2*I)*c + Pi + (2*I)*d*x)/4])/Sqrt[a^2 + b^2]] - ((
(-2*I)*c + Pi - (2*I)*d*x + 4*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]])*Log[1 + ((-b + Sqrt[a^2 + b^2])*E^(c + d*x))/
a])/2 - (((-2*I)*c + Pi - (2*I)*d*x - 4*ArcSin[Sqrt[1 + (I*b)/a]/Sqrt[2]])*Log[1 - ((b + Sqrt[a^2 + b^2])*E^(c
+ d*x))/a])/2 + (Pi/2 - I*(c + d*x))*Log[b + a*Sinh[c + d*x]] + I*(PolyLog[2, ((b - Sqrt[a^2 + b^2])*E^(c + d
*x))/a] + PolyLog[2, ((b + Sqrt[a^2 + b^2])*E^(c + d*x))/a])) + a*d*x*Sinh[c + d*x]))/(a^2*d^2*(a + b*Csch[c +
d*x]))

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Maple [F]  time = 0.283, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2}\cosh \left ( dx+c \right ) }{a+b{\rm csch} \left (dx+c\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, e^{2}{\left (\frac{2 \,{\left (d x + c\right )} b}{a^{2} d} - \frac{e^{\left (d x + c\right )}}{a d} + \frac{e^{\left (-d x - c\right )}}{a d} + \frac{2 \, b \log \left (-2 \, b e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )}{a^{2} d}\right )} - \frac{{\left (2 \, b d^{3} f^{2} x^{3} e^{c} + 6 \, b d^{3} e f x^{2} e^{c} - 3 \,{\left (a d^{2} f^{2} x^{2} e^{\left (2 \, c\right )} + 2 \,{\left (d^{2} e f - d f^{2}\right )} a x e^{\left (2 \, c\right )} - 2 \,{\left (d e f - f^{2}\right )} a e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} + 3 \,{\left (a d^{2} f^{2} x^{2} + 2 \,{\left (d^{2} e f + d f^{2}\right )} a x + 2 \,{\left (d e f + f^{2}\right )} a\right )} e^{\left (-d x\right )}\right )} e^{\left (-c\right )}}{6 \, a^{2} d^{3}} + \int -\frac{2 \,{\left (a b f^{2} x^{2} + 2 \, a b e f x -{\left (b^{2} f^{2} x^{2} e^{c} + 2 \, b^{2} e f x e^{c}\right )} e^{\left (d x\right )}\right )}}{a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} b e^{\left (d x + c\right )} - a^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 1.69167, size = 3051, normalized size = 9.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*a*d^2*f^2*x^2 + 3*a*d^2*e^2 + 6*a*d*e*f + 6*a*f^2 - 3*(a*d^2*f^2*x^2 + a*d^2*e^2 - 2*a*d*e*f + 2*a*f^2
+ 2*(a*d^2*e*f - a*d*f^2)*x)*cosh(d*x + c)^2 - 3*(a*d^2*f^2*x^2 + a*d^2*e^2 - 2*a*d*e*f + 2*a*f^2 + 2*(a*d^2*
e*f - a*d*f^2)*x)*sinh(d*x + c)^2 + 6*(a*d^2*e*f + a*d*f^2)*x - 2*(b*d^3*f^2*x^3 + 3*b*d^3*e*f*x^2 + 3*b*d^3*e
^2*x + 6*b*c*d^2*e^2 - 6*b*c^2*d*e*f + 2*b*c^3*f^2)*cosh(d*x + c) + 12*((b*d*f^2*x + b*d*e*f)*cosh(d*x + c) +
(b*d*f^2*x + b*d*e*f)*sinh(d*x + c))*dilog((b*cosh(d*x + c) + b*sinh(d*x + c) + (a*cosh(d*x + c) + a*sinh(d*x
+ c))*sqrt((a^2 + b^2)/a^2) - a)/a + 1) + 12*((b*d*f^2*x + b*d*e*f)*cosh(d*x + c) + (b*d*f^2*x + b*d*e*f)*sinh
(d*x + c))*dilog((b*cosh(d*x + c) + b*sinh(d*x + c) - (a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt((a^2 + b^2)/a^2
) - a)/a + 1) + 6*((b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*cosh(d*x + c) + (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)
*sinh(d*x + c))*log(2*a*cosh(d*x + c) + 2*a*sinh(d*x + c) + 2*a*sqrt((a^2 + b^2)/a^2) + 2*b) + 6*((b*d^2*e^2 -
2*b*c*d*e*f + b*c^2*f^2)*cosh(d*x + c) + (b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*sinh(d*x + c))*log(2*a*cosh(d*
x + c) + 2*a*sinh(d*x + c) - 2*a*sqrt((a^2 + b^2)/a^2) + 2*b) + 6*((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)*cosh(d*x + c) + (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)*sinh(d*x + c))*log(-(
b*cosh(d*x + c) + b*sinh(d*x + c) + (a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt((a^2 + b^2)/a^2) - a)/a) + 6*((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)*cosh(d*x + c) + (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)*sinh(d*x + c))*log(-(b*cosh(d*x + c) + b*sinh(d*x + c) - (a*cosh(d*x + c) + a*sinh(d*x + c)
)*sqrt((a^2 + b^2)/a^2) - a)/a) - 12*(b*f^2*cosh(d*x + c) + b*f^2*sinh(d*x + c))*polylog(3, (b*cosh(d*x + c) +
b*sinh(d*x + c) + (a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt((a^2 + b^2)/a^2))/a) - 12*(b*f^2*cosh(d*x + c) + b
*f^2*sinh(d*x + c))*polylog(3, (b*cosh(d*x + c) + b*sinh(d*x + c) - (a*cosh(d*x + c) + a*sinh(d*x + c))*sqrt((
a^2 + b^2)/a^2))/a) - 2*(b*d^3*f^2*x^3 + 3*b*d^3*e*f*x^2 + 3*b*d^3*e^2*x + 6*b*c*d^2*e^2 - 6*b*c^2*d*e*f + 2*b
*c^3*f^2 + 3*(a*d^2*f^2*x^2 + a*d^2*e^2 - 2*a*d*e*f + 2*a*f^2 + 2*(a*d^2*e*f - a*d*f^2)*x)*cosh(d*x + c))*sinh
(d*x + c))/(a^2*d^3*cosh(d*x + c) + a^2*d^3*sinh(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*cosh(d*x+c)/(a+b*csch(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2} \cosh \left (d x + c\right )}{b \operatorname{csch}\left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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