\(\int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx\) [169]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 404 \[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f}-\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^2}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^2}-\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^3}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^3}+\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^4}-\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^4} \]

[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)

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3403, 2296, 2221, 2611, 6744, 2320, 6724} \[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=-\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{f^3 \sqrt {a^2+b^2}}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{f^3 \sqrt {a^2+b^2}}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{f^2 \sqrt {a^2+b^2}}-\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{f^2 \sqrt {a^2+b^2}}+\frac {(c+d x)^3 \log \left (\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}+1\right )}{f \sqrt {a^2+b^2}}-\frac {(c+d x)^3 \log \left (\frac {b e^{e+f x}}{\sqrt {a^2+b^2}+a}+1\right )}{f \sqrt {a^2+b^2}}+\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{f^4 \sqrt {a^2+b^2}}-\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{f^4 \sqrt {a^2+b^2}} \]

[In]

Int[(c + d*x)^3/(a + b*Sinh[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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 3403

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 6724

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 6744

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*} \text {integral}& = 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 \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,-\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) \operatorname {PolyLog}\left (2,-\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) \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,-\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) \operatorname {PolyLog}\left (3,-\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) \operatorname {PolyLog}\left (3,-\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 \operatorname {PolyLog}\left (3,-\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 \operatorname {PolyLog}\left (3,-\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 \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,-\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) \operatorname {PolyLog}\left (3,-\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) \operatorname {PolyLog}\left (3,-\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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\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 ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,-\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 \operatorname {PolyLog}\left (2,-\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 \operatorname {PolyLog}\left (2,-\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) \operatorname {PolyLog}\left (3,-\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) \operatorname {PolyLog}\left (3,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^3}+\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^4}-\frac {6 d^3 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2} f^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.79 \[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\frac {(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a-\sqrt {a^2+b^2}}\right )-(c+d x)^3 \log \left (1+\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )+\frac {3 d \left (f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,\frac {b e^{e+f x}}{-a+\sqrt {a^2+b^2}}\right )-2 d f (c+d x) \operatorname {PolyLog}\left (3,\frac {b e^{e+f x}}{-a+\sqrt {a^2+b^2}}\right )+2 d^2 \operatorname {PolyLog}\left (4,\frac {b e^{e+f x}}{-a+\sqrt {a^2+b^2}}\right )\right )}{f^3}-\frac {3 d \left (f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )-2 d f (c+d x) \operatorname {PolyLog}\left (3,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )+2 d^2 \operatorname {PolyLog}\left (4,-\frac {b e^{e+f x}}{a+\sqrt {a^2+b^2}}\right )\right )}{f^3}}{\sqrt {a^2+b^2} f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (d x +c \right )^{3}}{a +b \sinh \left (f x +e \right )}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1004 vs. \(2 (362) = 724\).

Time = 0.28 (sec) , antiderivative size = 1004, normalized size of antiderivative = 2.49 \[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^3/(a+b*sinh(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))*sqrt
((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)*dilog(
(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*s
inh(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)*sqr
t((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)

Sympy [F]

\[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\int \frac {\left (c + d x\right )^{3}}{a + b \sinh {\left (e + f x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{b \sinh \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{b \sinh \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^3}{a+b \sinh (e+f x)} \, dx=\int \frac {{\left (c+d\,x\right )}^3}{a+b\,\mathrm {sinh}\left (e+f\,x\right )} \,d x \]

[In]

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

[Out]

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