3.339 \(\int \frac {\sinh ^{-1}(a x)^3}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=363 \[ -\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \text {Li}_3\left (-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \log \left (a^2 x^2+1\right )}{2 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2 \log \left (e^{2 \sinh ^{-1}(a x)}+1\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

[Out]

1/3*x*arcsinh(a*x)^3/c/(a^2*c*x^2+c)^(3/2)-x*arcsinh(a*x)/c^2/(a^2*c*x^2+c)^(1/2)+2/3*x*arcsinh(a*x)^3/c^2/(a^
2*c*x^2+c)^(1/2)+1/2*arcsinh(a*x)^2/a/c^2/(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+2/3*arcsinh(a*x)^3*(a^2*x^2+1)
^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)-2*arcsinh(a*x)^2*ln(1+(a*x+(a^2*x^2+1)^(1/2))^2)*(a^2*x^2+1)^(1/2)/a/c^2/(a^2
*c*x^2+c)^(1/2)+1/2*ln(a^2*x^2+1)*(a^2*x^2+1)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)-2*arcsinh(a*x)*polylog(2,-(a*x+(
a^2*x^2+1)^(1/2))^2)*(a^2*x^2+1)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)+polylog(3,-(a*x+(a^2*x^2+1)^(1/2))^2)*(a^2*x^
2+1)^(1/2)/a/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 363, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {5690, 5687, 5714, 3718, 2190, 2531, 2282, 6589, 5717, 260} \[ -\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x) \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 x^2+1} \log \left (a^2 x^2+1\right )}{2 a c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {a^2 c x^2+c}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {a^2 x^2+1} \sqrt {a^2 c x^2+c}}-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2 \log \left (e^{2 \sinh ^{-1}(a x)}+1\right )}{a c^2 \sqrt {a^2 c x^2+c}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^3/(c + a^2*c*x^2)^(5/2),x]

[Out]

-((x*ArcSinh[a*x])/(c^2*Sqrt[c + a^2*c*x^2])) + ArcSinh[a*x]^2/(2*a*c^2*Sqrt[1 + a^2*x^2]*Sqrt[c + a^2*c*x^2])
+ (x*ArcSinh[a*x]^3)/(3*c*(c + a^2*c*x^2)^(3/2)) + (2*x*ArcSinh[a*x]^3)/(3*c^2*Sqrt[c + a^2*c*x^2]) + (2*Sqrt
[1 + a^2*x^2]*ArcSinh[a*x]^3)/(3*a*c^2*Sqrt[c + a^2*c*x^2]) - (2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2*Log[1 + E^(2
*ArcSinh[a*x])])/(a*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[1 + a^2*x^2]*Log[1 + a^2*x^2])/(2*a*c^2*Sqrt[c + a^2*c*x^
2]) - (2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]*PolyLog[2, -E^(2*ArcSinh[a*x])])/(a*c^2*Sqrt[c + a^2*c*x^2]) + (Sqrt[1
+ a^2*x^2]*PolyLog[3, -E^(2*ArcSinh[a*x])])/(a*c^2*Sqrt[c + a^2*c*x^2])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 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 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 3718

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

Rule 5687

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

Rule 5690

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[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 5714

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

Rule 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 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 {\sinh ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 \int \frac {\sinh ^{-1}(a x)^3}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 c}-\frac {\left (a \sqrt {1+a^2 x^2}\right ) \int \frac {x \sinh ^{-1}(a x)^2}{\left (1+a^2 x^2\right )^2} \, dx}{c^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {1+a^2 x^2} \int \frac {\sinh ^{-1}(a x)}{\left (1+a^2 x^2\right )^{3/2}} \, dx}{c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (2 a \sqrt {1+a^2 x^2}\right ) \int \frac {x \sinh ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \tanh (x) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (a \sqrt {1+a^2 x^2}\right ) \int \frac {x}{1+a^2 x^2} \, dx}{c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \log \left (1+a^2 x^2\right )}{2 a c^2 \sqrt {c+a^2 c x^2}}-\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 x} x^2}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2 \log \left (1+e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \log \left (1+a^2 x^2\right )}{2 a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (4 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2 \log \left (1+e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \log \left (1+a^2 x^2\right )}{2 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\left (2 \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2 \log \left (1+e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \log \left (1+a^2 x^2\right )}{2 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {x \sinh ^{-1}(a x)}{c^2 \sqrt {c+a^2 c x^2}}+\frac {\sinh ^{-1}(a x)^2}{2 a c^2 \sqrt {1+a^2 x^2} \sqrt {c+a^2 c x^2}}+\frac {x \sinh ^{-1}(a x)^3}{3 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x \sinh ^{-1}(a x)^3}{3 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^3}{3 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2 \log \left (1+e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \log \left (1+a^2 x^2\right )}{2 a c^2 \sqrt {c+a^2 c x^2}}-\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x) \text {Li}_2\left (-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}+\frac {\sqrt {1+a^2 x^2} \text {Li}_3\left (-e^{2 \sinh ^{-1}(a x)}\right )}{a c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 195, normalized size = 0.54 \[ \frac {\left (a^2 x^2+1\right )^{3/2} \left (3 \log \left (a^2 x^2+1\right )+\frac {4 a x \sinh ^{-1}(a x)^3}{\sqrt {a^2 x^2+1}}+\frac {2 a x \sinh ^{-1}(a x)^3}{\left (a^2 x^2+1\right )^{3/2}}+\frac {3 \sinh ^{-1}(a x)^2}{a^2 x^2+1}-\frac {6 a x \sinh ^{-1}(a x)}{\sqrt {a^2 x^2+1}}+12 \sinh ^{-1}(a x) \text {Li}_2\left (-e^{-2 \sinh ^{-1}(a x)}\right )+6 \text {Li}_3\left (-e^{-2 \sinh ^{-1}(a x)}\right )-4 \sinh ^{-1}(a x)^3-12 \sinh ^{-1}(a x)^2 \log \left (e^{-2 \sinh ^{-1}(a x)}+1\right )\right )}{6 a c \left (a^2 c x^2+c\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]^3/(c + a^2*c*x^2)^(5/2),x]

[Out]

((1 + a^2*x^2)^(3/2)*((-6*a*x*ArcSinh[a*x])/Sqrt[1 + a^2*x^2] + (3*ArcSinh[a*x]^2)/(1 + a^2*x^2) - 4*ArcSinh[a
*x]^3 + (2*a*x*ArcSinh[a*x]^3)/(1 + a^2*x^2)^(3/2) + (4*a*x*ArcSinh[a*x]^3)/Sqrt[1 + a^2*x^2] - 12*ArcSinh[a*x
]^2*Log[1 + E^(-2*ArcSinh[a*x])] + 3*Log[1 + a^2*x^2] + 12*ArcSinh[a*x]*PolyLog[2, -E^(-2*ArcSinh[a*x])] + 6*P
olyLog[3, -E^(-2*ArcSinh[a*x])]))/(6*a*c*(c + a^2*c*x^2)^(3/2))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \operatorname {arsinh}\left (a x\right )^{3}}{a^{6} c^{3} x^{6} + 3 \, a^{4} c^{3} x^{4} + 3 \, a^{2} c^{3} x^{2} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*arcsinh(a*x)^3/(a^6*c^3*x^6 + 3*a^4*c^3*x^4 + 3*a^2*c^3*x^2 + c^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, choosing root of [1,0,%%%{-2,[2,1,2]%%%}+%%%{-2,[2,0,2]%%%}+%%%{-2,[0,1,0]%%%}+%%%{-2,[0,0,0]%%%},0,%%%{1
,[4,2,4]%%%}+%%%{-2,[4,1,4]%%%}+%%%{1,[4,0,4]%%%}+%%%{2,[2,2,2]%%%}+%%%{-4,[2,1,2]%%%}+%%%{2,[2,0,2]%%%}+%%%{1
,[0,2,0]%%%}+%%%{-2,[0,1,0]%%%}+%%%{1,[0,0,0]%%%}] at parameters values [86,-97,-82]sym2poly/r2sym(const gen &
e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.27, size = 550, normalized size = 1.52 \[ \frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \left (2 x^{3} a^{3}-2 \sqrt {a^{2} x^{2}+1}\, x^{2} a^{2}+3 a x -2 \sqrt {a^{2} x^{2}+1}\right ) \arcsinh \left (a x \right ) \left (-6 \arcsinh \left (a x \right ) x^{4} a^{4}-6 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}-6 x^{4} a^{4}-6 \sqrt {a^{2} x^{2}+1}\, x^{3} a^{3}+6 \arcsinh \left (a x \right )^{2} a^{2} x^{2}-12 \arcsinh \left (a x \right ) x^{2} a^{2}-9 \arcsinh \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x -18 a^{2} x^{2}-6 \sqrt {a^{2} x^{2}+1}\, x a +8 \arcsinh \left (a x \right )^{2}-6 \arcsinh \left (a x \right )-12\right )}{6 \left (3 x^{6} a^{6}+10 x^{4} a^{4}+11 a^{2} x^{2}+4\right ) a \,c^{3}}-\frac {2 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \ln \left (a x +\sqrt {a^{2} x^{2}+1}\right )}{\sqrt {a^{2} x^{2}+1}\, a \,c^{3}}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \ln \left (1+\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}\right )}{\sqrt {a^{2} x^{2}+1}\, a \,c^{3}}+\frac {4 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )^{3}}{3 \sqrt {a^{2} x^{2}+1}\, a \,c^{3}}-\frac {2 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right )^{2} \ln \left (1+\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}\right )}{\sqrt {a^{2} x^{2}+1}\, a \,c^{3}}-\frac {2 \sqrt {c \left (a^{2} x^{2}+1\right )}\, \arcsinh \left (a x \right ) \polylog \left (2, -\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}\right )}{\sqrt {a^{2} x^{2}+1}\, a \,c^{3}}+\frac {\sqrt {c \left (a^{2} x^{2}+1\right )}\, \polylog \left (3, -\left (a x +\sqrt {a^{2} x^{2}+1}\right )^{2}\right )}{\sqrt {a^{2} x^{2}+1}\, a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^3/(a^2*c*x^2+c)^(5/2),x)

[Out]

1/6*(c*(a^2*x^2+1))^(1/2)*(2*x^3*a^3-2*(a^2*x^2+1)^(1/2)*x^2*a^2+3*a*x-2*(a^2*x^2+1)^(1/2))*arcsinh(a*x)*(-6*a
rcsinh(a*x)*x^4*a^4-6*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a^3*x^3-6*x^4*a^4-6*(a^2*x^2+1)^(1/2)*x^3*a^3+6*arcsinh(a
*x)^2*a^2*x^2-12*arcsinh(a*x)*x^2*a^2-9*arcsinh(a*x)*(a^2*x^2+1)^(1/2)*a*x-18*a^2*x^2-6*(a^2*x^2+1)^(1/2)*x*a+
8*arcsinh(a*x)^2-6*arcsinh(a*x)-12)/(3*a^6*x^6+10*a^4*x^4+11*a^2*x^2+4)/a/c^3-2*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2
+1)^(1/2)/a/c^3*ln(a*x+(a^2*x^2+1)^(1/2))+(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a/c^3*ln(1+(a*x+(a^2*x^2+1)^
(1/2))^2)+4/3*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a/c^3*arcsinh(a*x)^3-2*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)
^(1/2)/a/c^3*arcsinh(a*x)^2*ln(1+(a*x+(a^2*x^2+1)^(1/2))^2)-2*(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a/c^3*ar
csinh(a*x)*polylog(2,-(a*x+(a^2*x^2+1)^(1/2))^2)+(c*(a^2*x^2+1))^(1/2)/(a^2*x^2+1)^(1/2)/a/c^3*polylog(3,-(a*x
+(a^2*x^2+1)^(1/2))^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^3/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^3/(a^2*c*x^2 + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {asinh}\left (a\,x\right )}^3}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)^3/(c + a^2*c*x^2)^(5/2),x)

[Out]

int(asinh(a*x)^3/(c + a^2*c*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{3}{\left (a x \right )}}{\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**3/(a**2*c*x**2+c)**(5/2),x)

[Out]

Integral(asinh(a*x)**3/(c*(a**2*x**2 + 1))**(5/2), x)

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