3.4.32 \(\int \frac {\sinh ^{-1}(a x)^3}{(c+a^2 c x^2)^2} \, dx\) [332]

Optimal. Leaf size=294 \[ \frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {6 \sinh ^{-1}(a x) \text {ArcTan}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {\sinh ^{-1}(a x)^3 \text {ArcTan}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \text {PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sinh ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x) \text {PolyLog}\left (3,i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {PolyLog}\left (4,-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {PolyLog}\left (4,i e^{\sinh ^{-1}(a x)}\right )}{a c^2} \]

[Out]

1/2*x*arcsinh(a*x)^3/c^2/(a^2*x^2+1)-6*arcsinh(a*x)*arctan(a*x+(a^2*x^2+1)^(1/2))/a/c^2+arcsinh(a*x)^3*arctan(
a*x+(a^2*x^2+1)^(1/2))/a/c^2+3*I*polylog(2,-I*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2-3/2*I*arcsinh(a*x)^2*polylog(2,-I
*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2-3*I*polylog(2,I*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2+3/2*I*arcsinh(a*x)^2*polylog(2,
I*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2+3*I*arcsinh(a*x)*polylog(3,-I*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2-3*I*arcsinh(a*x)
*polylog(3,I*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2-3*I*polylog(4,-I*(a*x+(a^2*x^2+1)^(1/2)))/a/c^2+3*I*polylog(4,I*(a
*x+(a^2*x^2+1)^(1/2)))/a/c^2+3/2*arcsinh(a*x)^2/a/c^2/(a^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {5788, 5789, 4265, 2611, 6744, 2320, 6724, 5798, 2317, 2438} \begin {gather*} \frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (a^2 x^2+1\right )}+\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {a^2 x^2+1}}+\frac {\sinh ^{-1}(a x)^3 \text {ArcTan}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {6 \sinh ^{-1}(a x) \text {ArcTan}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {Li}_4\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_4\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^3/(c + a^2*c*x^2)^2,x]

[Out]

(3*ArcSinh[a*x]^2)/(2*a*c^2*Sqrt[1 + a^2*x^2]) + (x*ArcSinh[a*x]^3)/(2*c^2*(1 + a^2*x^2)) - (6*ArcSinh[a*x]*Ar
cTan[E^ArcSinh[a*x]])/(a*c^2) + (ArcSinh[a*x]^3*ArcTan[E^ArcSinh[a*x]])/(a*c^2) + ((3*I)*PolyLog[2, (-I)*E^Arc
Sinh[a*x]])/(a*c^2) - (((3*I)/2)*ArcSinh[a*x]^2*PolyLog[2, (-I)*E^ArcSinh[a*x]])/(a*c^2) - ((3*I)*PolyLog[2, I
*E^ArcSinh[a*x]])/(a*c^2) + (((3*I)/2)*ArcSinh[a*x]^2*PolyLog[2, I*E^ArcSinh[a*x]])/(a*c^2) + ((3*I)*ArcSinh[a
*x]*PolyLog[3, (-I)*E^ArcSinh[a*x]])/(a*c^2) - ((3*I)*ArcSinh[a*x]*PolyLog[3, I*E^ArcSinh[a*x]])/(a*c^2) - ((3
*I)*PolyLog[4, (-I)*E^ArcSinh[a*x]])/(a*c^2) + ((3*I)*PolyLog[4, I*E^ArcSinh[a*x]])/(a*c^2)

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 5788

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(
p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 1))), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a
+ b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2
)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

Rule 5789

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sech[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

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*} \int \frac {\sinh ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^2} \, dx &=\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {(3 a) \int \frac {x \sinh ^{-1}(a x)^2}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{2 c^2}+\frac {\int \frac {\sinh ^{-1}(a x)^3}{c+a^2 c x^2} \, dx}{2 c}\\ &=\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {3 \int \frac {\sinh ^{-1}(a x)}{1+a^2 x^2} \, dx}{c^2}+\frac {\text {Subst}\left (\int x^3 \text {sech}(x) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a c^2}\\ &=\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}+\frac {\sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {(3 i) \text {Subst}\left (\int x^2 \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a c^2}+\frac {(3 i) \text {Subst}\left (\int x^2 \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a c^2}-\frac {3 \text {Subst}\left (\int x \text {sech}(x) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}\\ &=\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {6 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {\sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {(3 i) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}-\frac {(3 i) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}+\frac {(3 i) \text {Subst}\left (\int x \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}-\frac {(3 i) \text {Subst}\left (\int x \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}\\ &=\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {6 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {\sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {(3 i) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {(3 i) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {(3 i) \text {Subst}\left (\int \text {Li}_3\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}+\frac {(3 i) \text {Subst}\left (\int \text {Li}_3\left (i e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2}\\ &=\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {6 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {\sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {(3 i) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )}{a c^2}\\ &=\frac {3 \sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2}}+\frac {x \sinh ^{-1}(a x)^3}{2 c^2 \left (1+a^2 x^2\right )}-\frac {6 \sinh ^{-1}(a x) \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {\sinh ^{-1}(a x)^3 \tan ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (-i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sinh ^{-1}(a x)^2 \text {Li}_2\left (i e^{\sinh ^{-1}(a x)}\right )}{2 a c^2}+\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sinh ^{-1}(a x) \text {Li}_3\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {Li}_4\left (-i e^{\sinh ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_4\left (i e^{\sinh ^{-1}(a x)}\right )}{a c^2}\\ \end {align*}

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Mathematica [A]
time = 1.60, size = 568, normalized size = 1.93 \begin {gather*} -\frac {i \left (7 \pi ^4+8 i \pi ^3 \sinh ^{-1}(a x)+24 \pi ^2 \sinh ^{-1}(a x)^2+\frac {192 i \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}}-32 i \pi \sinh ^{-1}(a x)^3+\frac {64 i a x \sinh ^{-1}(a x)^3}{1+a^2 x^2}-16 \sinh ^{-1}(a x)^4-384 \sinh ^{-1}(a x) \log \left (1-i e^{-\sinh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )+384 \sinh ^{-1}(a x) \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )+48 \pi ^2 \sinh ^{-1}(a x) \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )-96 i \pi \sinh ^{-1}(a x)^2 \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )-64 \sinh ^{-1}(a x)^3 \log \left (1+i e^{-\sinh ^{-1}(a x)}\right )-48 \pi ^2 \sinh ^{-1}(a x) \log \left (1-i e^{\sinh ^{-1}(a x)}\right )+96 i \pi \sinh ^{-1}(a x)^2 \log \left (1-i e^{\sinh ^{-1}(a x)}\right )-8 i \pi ^3 \log \left (1+i e^{\sinh ^{-1}(a x)}\right )+64 \sinh ^{-1}(a x)^3 \log \left (1+i e^{\sinh ^{-1}(a x)}\right )+8 i \pi ^3 \log \left (\tan \left (\frac {1}{4} \left (\pi +2 i \sinh ^{-1}(a x)\right )\right )\right )-48 \left (8+\pi ^2-4 i \pi \sinh ^{-1}(a x)-4 \sinh ^{-1}(a x)^2\right ) \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(a x)}\right )+384 \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(a x)}\right )+192 \sinh ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(a x)}\right )-48 \pi ^2 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )+192 i \pi \sinh ^{-1}(a x) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(a x)}\right )+192 i \pi \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(a x)}\right )+384 \sinh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(a x)}\right )-384 \sinh ^{-1}(a x) \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(a x)}\right )-192 i \pi \text {PolyLog}\left (3,i e^{\sinh ^{-1}(a x)}\right )+384 \text {PolyLog}\left (4,-i e^{-\sinh ^{-1}(a x)}\right )+384 \text {PolyLog}\left (4,-i e^{\sinh ^{-1}(a x)}\right )\right )}{128 a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^3/(c + a^2*c*x^2)^2,x]

[Out]

((-1/128*I)*(7*Pi^4 + (8*I)*Pi^3*ArcSinh[a*x] + 24*Pi^2*ArcSinh[a*x]^2 + ((192*I)*ArcSinh[a*x]^2)/Sqrt[1 + a^2
*x^2] - (32*I)*Pi*ArcSinh[a*x]^3 + ((64*I)*a*x*ArcSinh[a*x]^3)/(1 + a^2*x^2) - 16*ArcSinh[a*x]^4 - 384*ArcSinh
[a*x]*Log[1 - I/E^ArcSinh[a*x]] + (8*I)*Pi^3*Log[1 + I/E^ArcSinh[a*x]] + 384*ArcSinh[a*x]*Log[1 + I/E^ArcSinh[
a*x]] + 48*Pi^2*ArcSinh[a*x]*Log[1 + I/E^ArcSinh[a*x]] - (96*I)*Pi*ArcSinh[a*x]^2*Log[1 + I/E^ArcSinh[a*x]] -
64*ArcSinh[a*x]^3*Log[1 + I/E^ArcSinh[a*x]] - 48*Pi^2*ArcSinh[a*x]*Log[1 - I*E^ArcSinh[a*x]] + (96*I)*Pi*ArcSi
nh[a*x]^2*Log[1 - I*E^ArcSinh[a*x]] - (8*I)*Pi^3*Log[1 + I*E^ArcSinh[a*x]] + 64*ArcSinh[a*x]^3*Log[1 + I*E^Arc
Sinh[a*x]] + (8*I)*Pi^3*Log[Tan[(Pi + (2*I)*ArcSinh[a*x])/4]] - 48*(8 + Pi^2 - (4*I)*Pi*ArcSinh[a*x] - 4*ArcSi
nh[a*x]^2)*PolyLog[2, (-I)/E^ArcSinh[a*x]] + 384*PolyLog[2, I/E^ArcSinh[a*x]] + 192*ArcSinh[a*x]^2*PolyLog[2,
(-I)*E^ArcSinh[a*x]] - 48*Pi^2*PolyLog[2, I*E^ArcSinh[a*x]] + (192*I)*Pi*ArcSinh[a*x]*PolyLog[2, I*E^ArcSinh[a
*x]] + (192*I)*Pi*PolyLog[3, (-I)/E^ArcSinh[a*x]] + 384*ArcSinh[a*x]*PolyLog[3, (-I)/E^ArcSinh[a*x]] - 384*Arc
Sinh[a*x]*PolyLog[3, (-I)*E^ArcSinh[a*x]] - (192*I)*Pi*PolyLog[3, I*E^ArcSinh[a*x]] + 384*PolyLog[4, (-I)/E^Ar
cSinh[a*x]] + 384*PolyLog[4, (-I)*E^ArcSinh[a*x]]))/(a*c^2)

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Maple [F]
time = 2.05, size = 0, normalized size = 0.00 \[\int \frac {\arcsinh \left (a x \right )^{3}}{\left (a^{2} c \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^3/(a^2*c*x^2+c)^2,x)

[Out]

int(arcsinh(a*x)^3/(a^2*c*x^2+c)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^3/(a^2*c*x^2 + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

integral(arcsinh(a*x)^3/(a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**3/(a**2*c*x**2+c)**2,x)

[Out]

Integral(asinh(a*x)**3/(a**4*x**4 + 2*a**2*x**2 + 1), x)/c**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)^3/(a^2*c*x^2 + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {asinh}\left (a\,x\right )}^3}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)^3/(c + a^2*c*x^2)^2,x)

[Out]

int(asinh(a*x)^3/(c + a^2*c*x^2)^2, x)

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