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

Optimal. Leaf size=356 \[ -\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^3}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^3}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^2}+\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^4}+\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^4}+\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b d}+\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b d}-\frac{(e+f x)^4}{4 b f} \]

[Out]

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

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Rubi [A]  time = 0.482998, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5561, 2190, 2531, 6609, 2282, 6589} \[ -\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^3}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^3}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{3 f (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^2}+\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^4}+\frac{6 f^3 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{b d^4}+\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{b d}+\frac{(e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{b d}-\frac{(e+f x)^4}{4 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)}{a+b \sinh (c+d x)} \, dx &=-\frac{(e+f x)^4}{4 b f}+\int \frac{e^{c+d x} (e+f x)^3}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx+\int \frac{e^{c+d x} (e+f x)^3}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx\\ &=-\frac{(e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d}-\frac{(3 f) \int (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{b d}-\frac{(3 f) \int (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{b d}\\ &=-\frac{(e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^2}-\frac{\left (6 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 d^2}-\frac{\left (6 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 d^2}\\ &=-\frac{(e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^3}+\frac{\left (6 f^3\right ) \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{b d^3}+\frac{\left (6 f^3\right ) \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{b d^3}\\ &=-\frac{(e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^3}+\frac{\left (6 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 d^4}+\frac{\left (6 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 d^4}\\ &=-\frac{(e+f x)^4}{4 b f}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d}+\frac{(e+f x)^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^2}+\frac{3 f (e+f x)^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^3}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^3}+\frac{6 f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{b d^4}+\frac{6 f^3 \text{Li}_4\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{b d^4}\\ \end{align*}

Mathematica [A]  time = 0.149016, size = 329, normalized size = 0.92 \[ \frac{\frac{12 f \left (d^2 (e+f x)^2 \text{PolyLog}\left (2,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )-2 d f (e+f x) \text{PolyLog}\left (3,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )+2 f^2 \text{PolyLog}\left (4,\frac{b e^{c+d x}}{\sqrt{a^2+b^2}-a}\right )\right )}{d^4}+\frac{12 f \left (d^2 (e+f x)^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )-2 d f (e+f x) \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )+2 f^2 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )\right )}{d^4}+\frac{4 (e+f x)^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d}+\frac{4 (e+f x)^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d}-\frac{(e+f x)^4}{f}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((e + f*x)^4/f) + (4*(e + f*x)^3*Log[1 + (b*E^(c + d*x))/(a - Sqrt[a^2 + b^2])])/d + (4*(e + f*x)^3*Log[1 +
(b*E^(c + d*x))/(a + Sqrt[a^2 + b^2])])/d + (12*f*(d^2*(e + f*x)^2*PolyLog[2, (b*E^(c + d*x))/(-a + Sqrt[a^2 +
b^2])] - 2*d*f*(e + f*x)*PolyLog[3, (b*E^(c + d*x))/(-a + Sqrt[a^2 + b^2])] + 2*f^2*PolyLog[4, (b*E^(c + d*x)
)/(-a + Sqrt[a^2 + b^2])]))/d^4 + (12*f*(d^2*(e + f*x)^2*PolyLog[2, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))]
- 2*d*f*(e + f*x)*PolyLog[3, -((b*E^(c + d*x))/(a + Sqrt[a^2 + b^2]))] + 2*f^2*PolyLog[4, -((b*E^(c + d*x))/(a
+ Sqrt[a^2 + b^2]))]))/d^4)/(4*b)

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Maple [F]  time = 0.259, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}\cosh \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)/(a+b*sinh(d*x+c)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.47518, size = 2128, normalized size = 5.98 \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)/(a+b*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

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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)**3*cosh(d*x+c)/(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 )}^{3} \cosh \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)/(a+b*sinh(d*x+c)),x, algorithm="giac")

[Out]

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