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

Optimal. Leaf size=389 \[ \frac{2 f \sqrt{a^2+b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 f \sqrt{a^2+b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^2}-\frac{2 f^2 \sqrt{a^2+b^2} \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{2 f^2 \sqrt{a^2+b^2} \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^3}+\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b^2 d}-\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b^2 d}-\frac{a (e+f x)^3}{3 b^2 f}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}+\frac{2 f^2 \cosh (c+d x)}{b d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.785279, antiderivative size = 389, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5565, 32, 3296, 2638, 3322, 2264, 2190, 2531, 2282, 6589} \[ \frac{2 f \sqrt{a^2+b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 f \sqrt{a^2+b^2} (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^2}-\frac{2 f^2 \sqrt{a^2+b^2} \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{2 f^2 \sqrt{a^2+b^2} \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^3}+\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b^2 d}-\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b^2 d}-\frac{a (e+f x)^3}{3 b^2 f}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}+\frac{2 f^2 \cosh (c+d x)}{b d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5565

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2638

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

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 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 ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac{a \int (e+f x)^2 \, dx}{b^2}+\frac{\int (e+f x)^2 \sinh (c+d x) \, dx}{b}+\frac{\left (a^2+b^2\right ) \int \frac{(e+f x)^2}{a+b \sinh (c+d x)} \, dx}{b^2}\\ &=-\frac{a (e+f x)^3}{3 b^2 f}+\frac{(e+f x)^2 \cosh (c+d x)}{b d}+\frac{\left (2 \left (a^2+b^2\right )\right ) \int \frac{e^{c+d x} (e+f x)^2}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{b^2}-\frac{(2 f) \int (e+f x) \cosh (c+d x) \, dx}{b d}\\ &=-\frac{a (e+f x)^3}{3 b^2 f}+\frac{(e+f x)^2 \cosh (c+d x)}{b d}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}+\frac{\left (2 \sqrt{a^2+b^2}\right ) \int \frac{e^{c+d x} (e+f x)^2}{2 a-2 \sqrt{a^2+b^2}+2 b e^{c+d x}} \, dx}{b}-\frac{\left (2 \sqrt{a^2+b^2}\right ) \int \frac{e^{c+d x} (e+f x)^2}{2 a+2 \sqrt{a^2+b^2}+2 b e^{c+d x}} \, dx}{b}+\frac{\left (2 f^2\right ) \int \sinh (c+d x) \, dx}{b d^2}\\ &=-\frac{a (e+f x)^3}{3 b^2 f}+\frac{2 f^2 \cosh (c+d x)}{b d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{b d}+\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}-\frac{\left (2 \sqrt{a^2+b^2} f\right ) \int (e+f x) \log \left (1+\frac{2 b e^{c+d x}}{2 a-2 \sqrt{a^2+b^2}}\right ) \, dx}{b^2 d}+\frac{\left (2 \sqrt{a^2+b^2} f\right ) \int (e+f x) \log \left (1+\frac{2 b e^{c+d x}}{2 a+2 \sqrt{a^2+b^2}}\right ) \, dx}{b^2 d}\\ &=-\frac{a (e+f x)^3}{3 b^2 f}+\frac{2 f^2 \cosh (c+d x)}{b d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{b d}+\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}+\frac{2 \sqrt{a^2+b^2} f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 \sqrt{a^2+b^2} f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}-\frac{\left (2 \sqrt{a^2+b^2} f^2\right ) \int \text{Li}_2\left (-\frac{2 b e^{c+d x}}{2 a-2 \sqrt{a^2+b^2}}\right ) \, dx}{b^2 d^2}+\frac{\left (2 \sqrt{a^2+b^2} f^2\right ) \int \text{Li}_2\left (-\frac{2 b e^{c+d x}}{2 a+2 \sqrt{a^2+b^2}}\right ) \, dx}{b^2 d^2}\\ &=-\frac{a (e+f x)^3}{3 b^2 f}+\frac{2 f^2 \cosh (c+d x)}{b d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{b d}+\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}+\frac{2 \sqrt{a^2+b^2} f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 \sqrt{a^2+b^2} f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}-\frac{\left (2 \sqrt{a^2+b^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^2 d^3}+\frac{\left (2 \sqrt{a^2+b^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^2 d^3}\\ &=-\frac{a (e+f x)^3}{3 b^2 f}+\frac{2 f^2 \cosh (c+d x)}{b d^3}+\frac{(e+f x)^2 \cosh (c+d x)}{b d}+\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{\sqrt{a^2+b^2} (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}+\frac{2 \sqrt{a^2+b^2} f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 \sqrt{a^2+b^2} f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{2 \sqrt{a^2+b^2} f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{2 \sqrt{a^2+b^2} f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^3}-\frac{2 f (e+f x) \sinh (c+d x)}{b d^2}\\ \end{align*}

Mathematica [A]  time = 2.91874, size = 447, normalized size = 1.15 \[ -\frac{3 \sqrt{a^2+b^2} \left (-2 d f (e+f x) \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+2 d f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+2 f^2 \text{PolyLog}\left (3,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )-2 f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+2 d^2 e^2 \tanh ^{-1}\left (\frac{a+b e^{c+d x}}{\sqrt{a^2+b^2}}\right )-2 d^2 e f x \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+2 d^2 e f x \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )-d^2 f^2 x^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )+d^2 f^2 x^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )\right )+a d^3 x \left (3 e^2+3 e f x+f^2 x^2\right )-3 b \cosh (d x) \left (\cosh (c) \left (d^2 (e+f x)^2+2 f^2\right )-2 d f \sinh (c) (e+f x)\right )+3 b \sinh (d x) \left (2 d f \cosh (c) (e+f x)-\sinh (c) \left (d^2 (e+f x)^2+2 f^2\right )\right )}{3 b^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.23, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2} \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{a+b\sinh \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 2.75005, size = 3163, normalized size = 8.13 \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)^2/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*b*d^2*f^2*x^2 + 3*b*d^2*e^2 + 6*b*d*e*f + 6*b*f^2 + 3*(b*d^2*f^2*x^2 + b*d^2*e^2 - 2*b*d*e*f + 2*b*f^2
+ 2*(b*d^2*e*f - b*d*f^2)*x)*cosh(d*x + c)^2 + 3*(b*d^2*f^2*x^2 + b*d^2*e^2 - 2*b*d*e*f + 2*b*f^2 + 2*(b*d^2*e
*f - b*d*f^2)*x)*sinh(d*x + c)^2 + 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))*sqrt((a^2 + b^2)/b^2)*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) - 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))*
sqrt((a^2 + b^2)/b^2)*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*((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))*sqrt((a^2 + b^2)/b^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) + 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))*sqrt((a^2 + b^2)/b^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) + 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))*sqrt((a^2 + b^2)/b^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) - 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))
*sqrt((a^2 + b^2)/b^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) - 12*(b*f^2*cosh(d*x + c) + b*f^2*sinh(d*x + c))*sqrt((a^2 + b^2)/b^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) + 12*(b*f^2*cosh(d
*x + c) + b*f^2*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*polylog(3, (a*cosh(d*x + c) + a*sinh(d*x + c) - (b*cosh(d
*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2))/b) + 6*(b*d^2*e*f + b*d*f^2)*x - 2*(a*d^3*f^2*x^3 + 3*a*d^3*
e*f*x^2 + 3*a*d^3*e^2*x)*cosh(d*x + c) - 2*(a*d^3*f^2*x^3 + 3*a*d^3*e*f*x^2 + 3*a*d^3*e^2*x - 3*(b*d^2*f^2*x^2
+ b*d^2*e^2 - 2*b*d*e*f + 2*b*f^2 + 2*(b*d^2*e*f - b*d*f^2)*x)*cosh(d*x + c))*sinh(d*x + c))/(b^2*d^3*cosh(d*
x + c) + b^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)**2/(a+b*sinh(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 )^{2}}{b \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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