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

Optimal. Leaf size=436 \[ -\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {(c+d x)^3 \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}+1\right )}{f \sqrt {a^2-b^2}}-\frac {(c+d x)^3 \log \left (\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}+1\right )}{f \sqrt {a^2-b^2}}+\frac {6 d^3 \text {Li}_4\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}}-\frac {6 d^3 \text {Li}_4\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}} \]

[Out]

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

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Rubi [A]  time = 0.82, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3320, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,-\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}\right )}{f^3 \sqrt {a^2-b^2}}+\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^2 \sqrt {a^2-b^2}}-\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,-\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}\right )}{f^2 \sqrt {a^2-b^2}}+\frac {6 d^3 \text {PolyLog}\left (4,-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{f^4 \sqrt {a^2-b^2}}-\frac {6 d^3 \text {PolyLog}\left (4,-\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}\right )}{f^4 \sqrt {a^2-b^2}}+\frac {(c+d x)^3 \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}+1\right )}{f \sqrt {a^2-b^2}}-\frac {(c+d x)^3 \log \left (\frac {b e^{e+f x}}{\sqrt {a^2-b^2}+a}+1\right )}{f \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3320

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol]
:> Dist[2, Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(E^(I*Pi*(k - 1/2))*(b + (2*a*E^(-(I*e) + f*fz*x))/E^(I*Pi*(k
- 1/2)) - (b*E^(2*(-(I*e) + f*fz*x)))/E^(2*I*k*Pi))), x], x] /; FreeQ[{a, b, c, d, e, f, fz}, x] && IntegerQ[
2*k] && NeQ[a^2 - b^2, 0] && IGtQ[m, 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 {(c+d x)^3}{a+b \cosh (e+f x)} \, dx &=2 \int \frac {e^{e+f x} (c+d x)^3}{b+2 a e^{e+f x}+b e^{2 (e+f x)}} \, dx\\ &=\frac {(2 b) \int \frac {e^{e+f x} (c+d x)^3}{2 a-2 \sqrt {a^2-b^2}+2 b e^{e+f x}} \, dx}{\sqrt {a^2-b^2}}-\frac {(2 b) \int \frac {e^{e+f x} (c+d x)^3}{2 a+2 \sqrt {a^2-b^2}+2 b e^{e+f x}} \, dx}{\sqrt {a^2-b^2}}\\ &=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {(3 d) \int (c+d x)^2 \log \left (1+\frac {2 b e^{e+f x}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f}+\frac {(3 d) \int (c+d x)^2 \log \left (1+\frac {2 b e^{e+f x}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f}\\ &=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {\left (6 d^2\right ) \int (c+d x) \text {Li}_2\left (-\frac {2 b e^{e+f x}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^2}+\frac {\left (6 d^2\right ) \int (c+d x) \text {Li}_2\left (-\frac {2 b e^{e+f x}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^2}\\ &=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {\left (6 d^3\right ) \int \text {Li}_3\left (-\frac {2 b e^{e+f x}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^3}-\frac {\left (6 d^3\right ) \int \text {Li}_3\left (-\frac {2 b e^{e+f x}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\sqrt {a^2-b^2} f^3}\\ &=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {\left (6 d^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^{e+f x}\right )}{\sqrt {a^2-b^2} f^4}-\frac {\left (6 d^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^{e+f x}\right )}{\sqrt {a^2-b^2} f^4}\\ &=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}-\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f}+\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^2}-\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^3}+\frac {6 d^3 \text {Li}_4\left (-\frac {b e^{e+f x}}{a-\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4}-\frac {6 d^3 \text {Li}_4\left (-\frac {b e^{e+f x}}{a+\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} f^4}\\ \end {align*}

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Mathematica [A]  time = 1.62, size = 384, normalized size = 0.88 \[ \frac {\frac {3 d \left (f^2 (c+d x)^2 \text {Li}_2\left (\frac {b (\cosh (e+f x)+\sinh (e+f x))}{\sqrt {a^2-b^2}-a}\right )-2 d f (c+d x) \text {Li}_3\left (\frac {b (\cosh (e+f x)+\sinh (e+f x))}{\sqrt {a^2-b^2}-a}\right )+2 d^2 \text {Li}_4\left (\frac {b (\cosh (e+f x)+\sinh (e+f x))}{\sqrt {a^2-b^2}-a}\right )\right )}{f^3}-\frac {3 d \left (f^2 (c+d x)^2 \text {Li}_2\left (-\frac {b (\cosh (e+f x)+\sinh (e+f x))}{a+\sqrt {a^2-b^2}}\right )-2 d f (c+d x) \text {Li}_3\left (-\frac {b (\cosh (e+f x)+\sinh (e+f x))}{a+\sqrt {a^2-b^2}}\right )+2 d^2 \text {Li}_4\left (-\frac {b (\cosh (e+f x)+\sinh (e+f x))}{a+\sqrt {a^2-b^2}}\right )\right )}{f^3}+(c+d x)^3 \log \left (\frac {b (\sinh (e+f x)+\cosh (e+f x))}{a-\sqrt {a^2-b^2}}+1\right )-(c+d x)^3 \log \left (\frac {b (\sinh (e+f x)+\cosh (e+f x))}{\sqrt {a^2-b^2}+a}+1\right )}{f \sqrt {a^2-b^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c + d*x)^3*Log[1 + (b*(Cosh[e + f*x] + Sinh[e + f*x]))/(a - Sqrt[a^2 - b^2])] - (c + d*x)^3*Log[1 + (b*(Cosh
[e + f*x] + Sinh[e + f*x]))/(a + Sqrt[a^2 - b^2])] + (3*d*(f^2*(c + d*x)^2*PolyLog[2, (b*(Cosh[e + f*x] + Sinh
[e + f*x]))/(-a + Sqrt[a^2 - b^2])] - 2*d*f*(c + d*x)*PolyLog[3, (b*(Cosh[e + f*x] + Sinh[e + f*x]))/(-a + Sqr
t[a^2 - b^2])] + 2*d^2*PolyLog[4, (b*(Cosh[e + f*x] + Sinh[e + f*x]))/(-a + Sqrt[a^2 - b^2])]))/f^3 - (3*d*(f^
2*(c + d*x)^2*PolyLog[2, -((b*(Cosh[e + f*x] + Sinh[e + f*x]))/(a + Sqrt[a^2 - b^2]))] - 2*d*f*(c + d*x)*PolyL
og[3, -((b*(Cosh[e + f*x] + Sinh[e + f*x]))/(a + Sqrt[a^2 - b^2]))] + 2*d^2*PolyLog[4, -((b*(Cosh[e + f*x] + S
inh[e + f*x]))/(a + Sqrt[a^2 - b^2]))]))/f^3)/(Sqrt[a^2 - b^2]*f)

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fricas [C]  time = 0.62, size = 1042, normalized size = 2.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*b*d^3*sqrt((a^2 - b^2)/b^2)*polylog(4, -(a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x
+ e))*sqrt((a^2 - b^2)/b^2))/b) - 6*b*d^3*sqrt((a^2 - b^2)/b^2)*polylog(4, -(a*cosh(f*x + e) + a*sinh(f*x + e)
- (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2))/b) + 3*(b*d^3*f^2*x^2 + 2*b*c*d^2*f^2*x + b*c^2*
d*f^2)*sqrt((a^2 - b^2)/b^2)*dilog(-(a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*s
qrt((a^2 - b^2)/b^2) + b)/b + 1) - 3*(b*d^3*f^2*x^2 + 2*b*c*d^2*f^2*x + b*c^2*d*f^2)*sqrt((a^2 - b^2)/b^2)*dil
og(-(a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) + b)/b + 1)
+ (b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f^3)*sqrt((a^2 - b^2)/b^2)*log(2*b*cosh(f*x + e) + 2
*b*sinh(f*x + e) + 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) - (b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2 - b*c^3*f
^3)*sqrt((a^2 - b^2)/b^2)*log(2*b*cosh(f*x + e) + 2*b*sinh(f*x + e) - 2*b*sqrt((a^2 - b^2)/b^2) + 2*a) + (b*d^
3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2)*sqrt((a^2 - b
^2)/b^2)*log((a*cosh(f*x + e) + a*sinh(f*x + e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2) +
b)/b) - (b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2)*
sqrt((a^2 - b^2)/b^2)*log((a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 -
b^2)/b^2) + b)/b) - 6*(b*d^3*f*x + b*c*d^2*f)*sqrt((a^2 - b^2)/b^2)*polylog(3, -(a*cosh(f*x + e) + a*sinh(f*x
+ e) + (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2)/b^2))/b) + 6*(b*d^3*f*x + b*c*d^2*f)*sqrt((a^2 -
b^2)/b^2)*polylog(3, -(a*cosh(f*x + e) + a*sinh(f*x + e) - (b*cosh(f*x + e) + b*sinh(f*x + e))*sqrt((a^2 - b^2
)/b^2))/b))/((a^2 - b^2)*f^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.48, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x +c \right )^{3}}{a +b \cosh \left (f x +e \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)**3/(a + b*cosh(e + f*x)), x)

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