3.333 \(\int \frac{(e+f x)^3 \cosh (c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=448 \[ \frac{6 a f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 a f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^3}-\frac{3 a f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{3 a f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^2}-\frac{6 a f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^4}-\frac{6 a f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^4}-\frac{a (e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b^2 d}+\frac{a (e+f x)^4}{4 b^2 f}+\frac{6 f^2 (e+f x) \sinh (c+d x)}{b d^3}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{6 f^3 \cosh (c+d x)}{b d^4}+\frac{(e+f x)^3 \sinh (c+d x)}{b d} \]

[Out]

(a*(e + f*x)^4)/(4*b^2*f) - (6*f^3*Cosh[c + d*x])/(b*d^4) - (3*f*(e + f*x)^2*Cosh[c + d*x])/(b*d^2) - (a*(e +
f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(b^2*d) - (a*(e + f*x)^3*Log[1 + (b*E^(c + d*x))/(a + S
qrt[a^2 + b^2])])/(b^2*d) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^2)
- (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(b^2*d^2) + (6*a*f^2*(e + f*x)*Pol
yLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))
/(a + Sqrt[a^2 + b^2]))])/(b^2*d^3) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^4)
- (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(b^2*d^4) + (6*f^2*(e + f*x)*Sinh[c + d*x])/
(b*d^3) + ((e + f*x)^3*Sinh[c + d*x])/(b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.646703, antiderivative size = 448, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 9, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.281, Rules used = {5579, 3296, 2638, 5561, 2190, 2531, 6609, 2282, 6589} \[ \frac{6 a f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 a f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^3}-\frac{3 a f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{3 a f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^2}-\frac{6 a f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^4}-\frac{6 a f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b^2 d^4}-\frac{a (e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b^2 d}+\frac{a (e+f x)^4}{4 b^2 f}+\frac{6 f^2 (e+f x) \sinh (c+d x)}{b d^3}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{6 f^3 \cosh (c+d x)}{b d^4}+\frac{(e+f x)^3 \sinh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Cosh[c + d*x]*Sinh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

(a*(e + f*x)^4)/(4*b^2*f) - (6*f^3*Cosh[c + d*x])/(b*d^4) - (3*f*(e + f*x)^2*Cosh[c + d*x])/(b*d^2) - (a*(e +
f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/(b^2*d) - (a*(e + f*x)^3*Log[1 + (b*E^(c + d*x))/(a + S
qrt[a^2 + b^2])])/(b^2*d) - (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^2)
- (3*a*f*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(b^2*d^2) + (6*a*f^2*(e + f*x)*Pol
yLog[3, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^3) + (6*a*f^2*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))
/(a + Sqrt[a^2 + b^2]))])/(b^2*d^3) - (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a - Sqrt[a^2 + b^2]))])/(b^2*d^4)
- (6*a*f^3*PolyLog[4, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))])/(b^2*d^4) + (6*f^2*(e + f*x)*Sinh[c + d*x])/
(b*d^3) + ((e + f*x)^3*Sinh[c + d*x])/(b*d)

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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)^3 \cosh (c+d x) \sinh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x)^3 \cosh (c+d x) \, dx}{b}-\frac{a \int \frac{(e+f x)^3 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{b}\\ &=\frac{a (e+f x)^4}{4 b^2 f}+\frac{(e+f x)^3 \sinh (c+d x)}{b d}-\frac{a \int \frac{e^{c+d x} (e+f x)^3}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{b}-\frac{a \int \frac{e^{c+d x} (e+f x)^3}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{b}-\frac{(3 f) \int (e+f x)^2 \sinh (c+d x) \, dx}{b d}\\ &=\frac{a (e+f x)^4}{4 b^2 f}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}+\frac{(e+f x)^3 \sinh (c+d x)}{b d}+\frac{(3 a f) \int (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{b^2 d}+\frac{(3 a f) \int (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{b^2 d}+\frac{\left (6 f^2\right ) \int (e+f x) \cosh (c+d x) \, dx}{b d^2}\\ &=\frac{a (e+f x)^4}{4 b^2 f}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}+\frac{6 f^2 (e+f x) \sinh (c+d x)}{b d^3}+\frac{(e+f x)^3 \sinh (c+d x)}{b d}+\frac{\left (6 a f^2\right ) \int (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{b^2 d^2}+\frac{\left (6 a f^2\right ) \int (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{b^2 d^2}-\frac{\left (6 f^3\right ) \int \sinh (c+d x) \, dx}{b d^3}\\ &=\frac{a (e+f x)^4}{4 b^2 f}-\frac{6 f^3 \cosh (c+d x)}{b d^4}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 f^2 (e+f x) \sinh (c+d x)}{b d^3}+\frac{(e+f x)^3 \sinh (c+d x)}{b d}-\frac{\left (6 a f^3\right ) \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{b^2 d^3}-\frac{\left (6 a f^3\right ) \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{b^2 d^3}\\ &=\frac{a (e+f x)^4}{4 b^2 f}-\frac{6 f^3 \cosh (c+d x)}{b d^4}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 f^2 (e+f x) \sinh (c+d x)}{b d^3}+\frac{(e+f x)^3 \sinh (c+d x)}{b d}-\frac{\left (6 a f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{b x}{-a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^4}-\frac{\left (6 a f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{b x}{a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b^2 d^4}\\ &=\frac{a (e+f x)^4}{4 b^2 f}-\frac{6 f^3 \cosh (c+d x)}{b d^4}-\frac{3 f (e+f x)^2 \cosh (c+d x)}{b d^2}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{a (e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^2}-\frac{3 a f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^2}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^3}+\frac{6 a f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^3}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b^2 d^4}-\frac{6 a f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b^2 d^4}+\frac{6 f^2 (e+f x) \sinh (c+d x)}{b d^3}+\frac{(e+f x)^3 \sinh (c+d x)}{b d}\\ \end{align*}

Mathematica [B]  time = 20.9593, size = 10378, normalized size = 23.17 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Cosh[c + d*x]*Sinh[c + d*x])/(a + b*Sinh[c + d*x]),x]

[Out]

Result too large to show

________________________________________________________________________________________

Maple [F]  time = 0.198, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{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)^3*cosh(d*x+c)*sinh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^3*cosh(d*x+c)*sinh(d*x+c)/(a+b*sinh(d*x+c)),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, e^{3}{\left (\frac{2 \,{\left (d x + c\right )} a}{b^{2} d} - \frac{e^{\left (d x + c\right )}}{b d} + \frac{e^{\left (-d x - c\right )}}{b d} + \frac{2 \, a \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{b^{2} d}\right )} - \frac{{\left (a d^{4} f^{3} x^{4} e^{c} + 4 \, a d^{4} e f^{2} x^{3} e^{c} + 6 \, a d^{4} e^{2} f x^{2} e^{c} - 2 \,{\left (b d^{3} f^{3} x^{3} e^{\left (2 \, c\right )} + 3 \,{\left (d^{3} e f^{2} - d^{2} f^{3}\right )} b x^{2} e^{\left (2 \, c\right )} + 3 \,{\left (d^{3} e^{2} f - 2 \, d^{2} e f^{2} + 2 \, d f^{3}\right )} b x e^{\left (2 \, c\right )} - 3 \,{\left (d^{2} e^{2} f - 2 \, d e f^{2} + 2 \, f^{3}\right )} b e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} + 2 \,{\left (b d^{3} f^{3} x^{3} + 3 \,{\left (d^{3} e f^{2} + d^{2} f^{3}\right )} b x^{2} + 3 \,{\left (d^{3} e^{2} f + 2 \, d^{2} e f^{2} + 2 \, d f^{3}\right )} b x + 3 \,{\left (d^{2} e^{2} f + 2 \, d e f^{2} + 2 \, f^{3}\right )} b\right )} e^{\left (-d x\right )}\right )} e^{\left (-c\right )}}{4 \, b^{2} d^{4}} + \int -\frac{2 \,{\left (a b f^{3} x^{3} + 3 \, a b e f^{2} x^{2} + 3 \, a b e^{2} f x -{\left (a^{2} f^{3} x^{3} e^{c} + 3 \, a^{2} e f^{2} x^{2} e^{c} + 3 \, a^{2} e^{2} f x e^{c}\right )} e^{\left (d x\right )}\right )}}{b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b^{2} e^{\left (d x + c\right )} - b^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cosh(d*x+c)*sinh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*e^3*(2*(d*x + c)*a/(b^2*d) - e^(d*x + c)/(b*d) + e^(-d*x - c)/(b*d) + 2*a*log(-2*a*e^(-d*x - c) + b*e^(-2
*d*x - 2*c) - b)/(b^2*d)) - 1/4*(a*d^4*f^3*x^4*e^c + 4*a*d^4*e*f^2*x^3*e^c + 6*a*d^4*e^2*f*x^2*e^c - 2*(b*d^3*
f^3*x^3*e^(2*c) + 3*(d^3*e*f^2 - d^2*f^3)*b*x^2*e^(2*c) + 3*(d^3*e^2*f - 2*d^2*e*f^2 + 2*d*f^3)*b*x*e^(2*c) -
3*(d^2*e^2*f - 2*d*e*f^2 + 2*f^3)*b*e^(2*c))*e^(d*x) + 2*(b*d^3*f^3*x^3 + 3*(d^3*e*f^2 + d^2*f^3)*b*x^2 + 3*(d
^3*e^2*f + 2*d^2*e*f^2 + 2*d*f^3)*b*x + 3*(d^2*e^2*f + 2*d*e*f^2 + 2*f^3)*b)*e^(-d*x))*e^(-c)/(b^2*d^4) + inte
grate(-2*(a*b*f^3*x^3 + 3*a*b*e*f^2*x^2 + 3*a*b*e^2*f*x - (a^2*f^3*x^3*e^c + 3*a^2*e*f^2*x^2*e^c + 3*a^2*e^2*f
*x*e^c)*e^(d*x))/(b^3*e^(2*d*x + 2*c) + 2*a*b^2*e^(d*x + c) - b^3), x)

________________________________________________________________________________________

Fricas [C]  time = 2.77144, size = 4582, normalized size = 10.23 \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)^3*cosh(d*x+c)*sinh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(2*b*d^3*f^3*x^3 + 2*b*d^3*e^3 + 6*b*d^2*e^2*f + 12*b*d*e*f^2 + 12*b*f^3 + 6*(b*d^3*e*f^2 + b*d^2*f^3)*x^
2 - 2*(b*d^3*f^3*x^3 + b*d^3*e^3 - 3*b*d^2*e^2*f + 6*b*d*e*f^2 - 6*b*f^3 + 3*(b*d^3*e*f^2 - b*d^2*f^3)*x^2 + 3
*(b*d^3*e^2*f - 2*b*d^2*e*f^2 + 2*b*d*f^3)*x)*cosh(d*x + c)^2 - 2*(b*d^3*f^3*x^3 + b*d^3*e^3 - 3*b*d^2*e^2*f +
6*b*d*e*f^2 - 6*b*f^3 + 3*(b*d^3*e*f^2 - b*d^2*f^3)*x^2 + 3*(b*d^3*e^2*f - 2*b*d^2*e*f^2 + 2*b*d*f^3)*x)*sinh
(d*x + c)^2 + 6*(b*d^3*e^2*f + 2*b*d^2*e*f^2 + 2*b*d*f^3)*x - (a*d^4*f^3*x^4 + 4*a*d^4*e*f^2*x^3 + 6*a*d^4*e^2
*f*x^2 + 4*a*d^4*e^3*x + 8*a*c*d^3*e^3 - 12*a*c^2*d^2*e^2*f + 8*a*c^3*d*e*f^2 - 2*a*c^4*f^3)*cosh(d*x + c) + 1
2*((a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*x + a*d^2*e^2*f)*cosh(d*x + c) + (a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*x + a*d^2*e^
2*f)*sinh(d*x + c))*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*((a*d^2*f^3*x^2 + 2*a*d^2*e*f^2*x + a*d^2*e^2*f)*cosh(d*x + c) + (a*d^2*f^3*x^2 +
2*a*d^2*e*f^2*x + a*d^2*e^2*f)*sinh(d*x + c))*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) + 4*((a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f
^3)*cosh(d*x + c) + (a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*sinh(d*x + c))*log(2*b*cosh(d*
x + c) + 2*b*sinh(d*x + c) + 2*b*sqrt((a^2 + b^2)/b^2) + 2*a) + 4*((a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*
f^2 - a*c^3*f^3)*cosh(d*x + c) + (a*d^3*e^3 - 3*a*c*d^2*e^2*f + 3*a*c^2*d*e*f^2 - a*c^3*f^3)*sinh(d*x + c))*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) + 4*((a*d^3*f^3*x^3 + 3*a*d^3*e*f^2
*x^2 + 3*a*d^3*e^2*f*x + 3*a*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*cosh(d*x + c) + (a*d^3*f^3*x^3 + 3*a*d
^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*a*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*sinh(d*x + c))*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) + 4*((a*d^3*f^3*x
^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*a*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*cosh(d*x + c) + (a*d
^3*f^3*x^3 + 3*a*d^3*e*f^2*x^2 + 3*a*d^3*e^2*f*x + 3*a*c*d^2*e^2*f - 3*a*c^2*d*e*f^2 + a*c^3*f^3)*sinh(d*x + c
))*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)
+ 24*(a*f^3*cosh(d*x + c) + a*f^3*sinh(d*x + c))*polylog(4, (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) + 24*(a*f^3*cosh(d*x + c) + a*f^3*sinh(d*x + c))*polylog(4,
(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) - 24*((a*d*
f^3*x + a*d*e*f^2)*cosh(d*x + c) + (a*d*f^3*x + a*d*e*f^2)*sinh(d*x + c))*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) - 24*((a*d*f^3*x + a*d*e*f^2)*cosh(d
*x + c) + (a*d*f^3*x + a*d*e*f^2)*sinh(d*x + c))*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) - (a*d^4*f^3*x^4 + 4*a*d^4*e*f^2*x^3 + 6*a*d^4*e^2*f*x^2 + 4*
a*d^4*e^3*x + 8*a*c*d^3*e^3 - 12*a*c^2*d^2*e^2*f + 8*a*c^3*d*e*f^2 - 2*a*c^4*f^3 + 4*(b*d^3*f^3*x^3 + b*d^3*e^
3 - 3*b*d^2*e^2*f + 6*b*d*e*f^2 - 6*b*f^3 + 3*(b*d^3*e*f^2 - b*d^2*f^3)*x^2 + 3*(b*d^3*e^2*f - 2*b*d^2*e*f^2 +
2*b*d*f^3)*x)*cosh(d*x + c))*sinh(d*x + c))/(b^2*d^4*cosh(d*x + c) + b^2*d^4*sinh(d*x + c))

________________________________________________________________________________________

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)**3*cosh(d*x+c)*sinh(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )}{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)^3*cosh(d*x+c)*sinh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*cosh(d*x + c)*sinh(d*x + c)/(b*sinh(d*x + c) + a), x)