3.323 \(\int \frac {(e+f x)^3 \cosh (c+d x)}{(a+b \sinh (c+d x))^2} \, dx\)

Optimal. Leaf size=348 \[ -\frac {6 f^3 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 f^3 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 f^2 (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {6 f^2 (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}+\frac {3 f (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {3 f (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.74, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5464, 3322, 2264, 2190, 2531, 2282, 6589} \[ \frac {6 f^2 (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {6 f^2 (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^3 \sqrt {a^2+b^2}}-\frac {6 f^3 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {6 f^3 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{b d^4 \sqrt {a^2+b^2}}+\frac {3 f (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {3 f (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{b d^2 \sqrt {a^2+b^2}}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Cosh[c + d*x])/(a + b*Sinh[c + d*x])^2,x]

[Out]

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

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 5464

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

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

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Mathematica [A]  time = 2.28, size = 368, normalized size = 1.06 \[ \frac {3 f \left (-2 d^2 e^2 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )+2 d^2 e f x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-2 d^2 e f x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )-d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+2 d f (e+f x) \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-2 d f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )-2 f^2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{b d^4 \sqrt {a^2+b^2}}-\frac {(e+f x)^3}{b d (a+b \sinh (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Cosh[c + d*x])/(a + b*Sinh[c + d*x])^2,x]

[Out]

(3*f*(-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 - 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 + Sqrt[a^2 + b^2])] + 2*d*f*(e + f*x)*PolyLo
g[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 + Sqr
t[a^2 + b^2]))]))/(b*Sqrt[a^2 + b^2]*d^4) - (e + f*x)^3/(b*d*(a + b*Sinh[c + d*x]))

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fricas [C]  time = 0.60, size = 2420, normalized size = 6.95 \[ \text {result too large to display} \]

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))^2,x, algorithm="fricas")

[Out]

-(6*(b^2*d*f^3*x + b^2*d*e*f^2 - (b^2*d*f^3*x + b^2*d*e*f^2)*cosh(d*x + c)^2 - (b^2*d*f^3*x + b^2*d*e*f^2)*sin
h(d*x + c)^2 - 2*(a*b*d*f^3*x + a*b*d*e*f^2)*cosh(d*x + c) - 2*(a*b*d*f^3*x + a*b*d*e*f^2 + (b^2*d*f^3*x + b^2
*d*e*f^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*dilog((a*cosh(d*x + c) + a*sinh(d*x + c) + (b*co
sh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b + 1) - 6*(b^2*d*f^3*x + b^2*d*e*f^2 - (b^2*d*f^3*x
+ b^2*d*e*f^2)*cosh(d*x + c)^2 - (b^2*d*f^3*x + b^2*d*e*f^2)*sinh(d*x + c)^2 - 2*(a*b*d*f^3*x + a*b*d*e*f^2)*
cosh(d*x + c) - 2*(a*b*d*f^3*x + a*b*d*e*f^2 + (b^2*d*f^3*x + b^2*d*e*f^2)*cosh(d*x + c))*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) - 3*(b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3 - (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^
2*c^2*f^3)*cosh(d*x + c)^2 - (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3)*sinh(d*x + c)^2 - 2*(a*b*d^2*e^2*
f - 2*a*b*c*d*e*f^2 + a*b*c^2*f^3)*cosh(d*x + c) - 2*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 + a*b*c^2*f^3 + (b^2*d^2
*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3)*cosh(d*x + c))*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) + 3*(b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3
- (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3)*cosh(d*x + c)^2 - (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2
*f^3)*sinh(d*x + c)^2 - 2*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 + a*b*c^2*f^3)*cosh(d*x + c) - 2*(a*b*d^2*e^2*f - 2
*a*b*c*d*e*f^2 + a*b*c^2*f^3 + (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^2 + b^2*c^2*f^3)*cosh(d*x + c))*sinh(d*x + c))*s
qrt((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) + 3*(b^2*d^2
*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2*b^2*c*d*e*f^2 - b^2*c^2*f^3 - (b^2*d^2*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2*b^2*c*
d*e*f^2 - b^2*c^2*f^3)*cosh(d*x + c)^2 - (b^2*d^2*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2*b^2*c*d*e*f^2 - b^2*c^2*f^3)
*sinh(d*x + c)^2 - 2*(a*b*d^2*f^3*x^2 + 2*a*b*d^2*e*f^2*x + 2*a*b*c*d*e*f^2 - a*b*c^2*f^3)*cosh(d*x + c) - 2*(
a*b*d^2*f^3*x^2 + 2*a*b*d^2*e*f^2*x + 2*a*b*c*d*e*f^2 - a*b*c^2*f^3 + (b^2*d^2*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2
*b^2*c*d*e*f^2 - b^2*c^2*f^3)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(d*x + c) + a*si
nh(d*x + c) + (b*cosh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) - 3*(b^2*d^2*f^3*x^2 + 2*b^2*d
^2*e*f^2*x + 2*b^2*c*d*e*f^2 - b^2*c^2*f^3 - (b^2*d^2*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2*b^2*c*d*e*f^2 - b^2*c^2*
f^3)*cosh(d*x + c)^2 - (b^2*d^2*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2*b^2*c*d*e*f^2 - b^2*c^2*f^3)*sinh(d*x + c)^2 -
2*(a*b*d^2*f^3*x^2 + 2*a*b*d^2*e*f^2*x + 2*a*b*c*d*e*f^2 - a*b*c^2*f^3)*cosh(d*x + c) - 2*(a*b*d^2*f^3*x^2 +
2*a*b*d^2*e*f^2*x + 2*a*b*c*d*e*f^2 - a*b*c^2*f^3 + (b^2*d^2*f^3*x^2 + 2*b^2*d^2*e*f^2*x + 2*b^2*c*d*e*f^2 - b
^2*c^2*f^3)*cosh(d*x + c))*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2)*log(-(a*cosh(d*x + c) + a*sinh(d*x + c) - (b*c
osh(d*x + c) + b*sinh(d*x + c))*sqrt((a^2 + b^2)/b^2) - b)/b) + 6*(b^2*f^3*cosh(d*x + c)^2 + b^2*f^3*sinh(d*x
+ c)^2 + 2*a*b*f^3*cosh(d*x + c) - b^2*f^3 + 2*(b^2*f^3*cosh(d*x + c) + a*b*f^3)*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^2*f^3*cosh(d*x + c)^2 + b^2*f^3*sinh(d*x + c)^2 + 2*a*b*f^3*cosh(d*x + c) - b^2*f^3 + 2*(b^2*f^
3*cosh(d*x + c) + a*b*f^3)*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) + 2*((a^2 + b^2)*d^3*f^3*x^3 + 3*(a^2 + b^2)*d
^3*e*f^2*x^2 + 3*(a^2 + b^2)*d^3*e^2*f*x + (a^2 + b^2)*d^3*e^3)*cosh(d*x + c) + 2*((a^2 + b^2)*d^3*f^3*x^3 + 3
*(a^2 + b^2)*d^3*e*f^2*x^2 + 3*(a^2 + b^2)*d^3*e^2*f*x + (a^2 + b^2)*d^3*e^3)*sinh(d*x + c))/((a^2*b^2 + b^4)*
d^4*cosh(d*x + c)^2 + (a^2*b^2 + b^4)*d^4*sinh(d*x + c)^2 + 2*(a^3*b + a*b^3)*d^4*cosh(d*x + c) - (a^2*b^2 + b
^4)*d^4 + 2*((a^2*b^2 + b^4)*d^4*cosh(d*x + c) + (a^3*b + a*b^3)*d^4)*sinh(d*x + c))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \cosh \left (d x + c\right )}{{\left (b \sinh \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

integrate((f*x + e)^3*cosh(d*x + c)/(b*sinh(d*x + c) + a)^2, x)

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maple [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \cosh \left (d x +c \right )}{\left (a +b \sinh \left (d x +c \right )\right )^{2}}\, dx \]

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))^2,x)

[Out]

int((f*x+e)^3*cosh(d*x+c)/(a+b*sinh(d*x+c))^2,x)

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

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {cosh}\left (c+d\,x\right )\,{\left (e+f\,x\right )}^3}{{\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(c + d*x)*(e + f*x)^3)/(a + b*sinh(c + d*x))^2,x)

[Out]

int((cosh(c + d*x)*(e + f*x)^3)/(a + b*sinh(c + d*x))^2, x)

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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*cosh(d*x+c)/(a+b*sinh(d*x+c))**2,x)

[Out]

Timed out

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