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

Optimal. Leaf size=453 \[ -\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 a f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {6 a f^2 (e+f x) \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}+\frac {3 a f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {a (e+f x)^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {a (e+f x)^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {(e+f x)^4}{4 b f} \]

[Out]

1/4*(f*x+e)^4/b/f-a*(f*x+e)^3*ln(1+b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/d/(a^2+b^2)^(1/2)+a*(f*x+e)^3*ln(1+b*ex
p(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d/(a^2+b^2)^(1/2)-3*a*f*(f*x+e)^2*polylog(2,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))
/b/d^2/(a^2+b^2)^(1/2)+3*a*f*(f*x+e)^2*polylog(2,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d^2/(a^2+b^2)^(1/2)+6*a*
f^2*(f*x+e)*polylog(3,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)-6*a*f^2*(f*x+e)*polylog(3,-b*ex
p(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d^3/(a^2+b^2)^(1/2)-6*a*f^3*polylog(4,-b*exp(d*x+c)/(a-(a^2+b^2)^(1/2)))/b/d^4
/(a^2+b^2)^(1/2)+6*a*f^3*polylog(4,-b*exp(d*x+c)/(a+(a^2+b^2)^(1/2)))/b/d^4/(a^2+b^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.79, antiderivative size = 453, normalized size of antiderivative = 1.00, 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 = {5557, 32, 3322, 2264, 2190, 2531, 6609, 2282, 6589} \[ \frac {6 a f^2 (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {6 a f^2 (e+f x) \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {3 a f (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^2 \sqrt {a^2+b^2}}+\frac {3 a f (e+f x)^2 \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {6 a f^3 \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 a f^3 \text {PolyLog}\left (4,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^4 \sqrt {a^2+b^2}}-\frac {a (e+f x)^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {a (e+f x)^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d \sqrt {a^2+b^2}}+\frac {(e+f x)^4}{4 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 5557

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*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]

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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 2.28, size = 607, normalized size = 1.34 \[ \frac {a \left (2 d^3 e^3 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-3 d^3 e^2 f x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+3 d^3 e^2 f x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-3 d^3 e f^2 x^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+3 d^3 e f^2 x^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-d^3 f^3 x^3 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+d^3 f^3 x^3 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-3 d^2 f (e+f x)^2 \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+3 d^2 f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+6 d e f^2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-6 d e f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+6 d f^3 x \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-6 d f^3 x \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-6 f^3 \text {Li}_4\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+6 f^3 \text {Li}_4\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 1.01, size = 1112, normalized size = 2.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \sinh \left (d x + c\right )}{b \sinh \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.38, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \sinh \left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -e^{3} {\left (\frac {a \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} b d} - \frac {d x + c}{b d}\right )} + \frac {f^{3} x^{4} + 4 \, e f^{2} x^{3} + 6 \, e^{2} f x^{2}}{4 \, b} - \int \frac {2 \, {\left (a f^{3} x^{3} e^{c} + 3 \, a e f^{2} x^{2} e^{c} + 3 \, a e^{2} f x e^{c}\right )} e^{\left (d x\right )}}{b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} - b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {sinh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{a+b\,\mathrm {sinh}\left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________