3.2.42 \(\int x^2 (a+b \sinh ^{-1}(c x))^{5/2} \, dx\) [142]

Optimal. Leaf size=327 \[ -\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^3}+\frac {5 b^{5/2} e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{576 c^3}+\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^3}-\frac {5 b^{5/2} e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{576 c^3} \]

[Out]

1/3*x^3*(a+b*arcsinh(c*x))^(5/2)+5/1728*b^(5/2)*exp(3*a/b)*erf(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/
2)*Pi^(1/2)/c^3-5/1728*b^(5/2)*erfi(3^(1/2)*(a+b*arcsinh(c*x))^(1/2)/b^(1/2))*3^(1/2)*Pi^(1/2)/c^3/exp(3*a/b)-
15/64*b^(5/2)*exp(a/b)*erf((a+b*arcsinh(c*x))^(1/2)/b^(1/2))*Pi^(1/2)/c^3+15/64*b^(5/2)*erfi((a+b*arcsinh(c*x)
)^(1/2)/b^(1/2))*Pi^(1/2)/c^3/exp(a/b)+5/9*b*(a+b*arcsinh(c*x))^(3/2)*(c^2*x^2+1)^(1/2)/c^3-5/18*b*x^2*(a+b*ar
csinh(c*x))^(3/2)*(c^2*x^2+1)^(1/2)/c-5/6*b^2*x*(a+b*arcsinh(c*x))^(1/2)/c^2+5/36*b^2*x^3*(a+b*arcsinh(c*x))^(
1/2)

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Rubi [A]
time = 0.77, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5777, 5812, 5798, 5772, 5819, 3389, 2211, 2236, 2235, 3393} \begin {gather*} -\frac {15 \sqrt {\pi } b^{5/2} e^{a/b} \text {Erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^3}+\frac {5 \sqrt {\frac {\pi }{3}} b^{5/2} e^{\frac {3 a}{b}} \text {Erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{576 c^3}+\frac {15 \sqrt {\pi } b^{5/2} e^{-\frac {a}{b}} \text {Erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^3}-\frac {5 \sqrt {\frac {\pi }{3}} b^{5/2} e^{-\frac {3 a}{b}} \text {Erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{576 c^3}-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}-\frac {5 b x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {5 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSinh[c*x])^(5/2),x]

[Out]

(-5*b^2*x*Sqrt[a + b*ArcSinh[c*x]])/(6*c^2) + (5*b^2*x^3*Sqrt[a + b*ArcSinh[c*x]])/36 + (5*b*Sqrt[1 + c^2*x^2]
*(a + b*ArcSinh[c*x])^(3/2))/(9*c^3) - (5*b*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(3/2))/(18*c) + (x^3*(a
+ b*ArcSinh[c*x])^(5/2))/3 - (15*b^(5/2)*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(64*c^3) + (
5*b^(5/2)*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(576*c^3) + (15*b^(5/2)*Sqrt
[Pi]*Erfi[Sqrt[a + b*ArcSinh[c*x]]/Sqrt[b]])/(64*c^3*E^(a/b)) - (5*b^(5/2)*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b
*ArcSinh[c*x]])/Sqrt[b]])/(576*c^3*E^((3*a)/b))

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 5772

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

Rule 5777

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcSinh[c*x])^n/(
m + 1)), x] - Dist[b*c*(n/(m + 1)), Int[x^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /;
FreeQ[{a, b, c}, x] && IGtQ[m, 0] && GtQ[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 5812

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

Rule 5819

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

Rubi steps

\begin {align*} \int x^2 \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \, dx &=\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {1}{6} (5 b c) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{\sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}+\frac {1}{12} \left (5 b^2\right ) \int x^2 \sqrt {a+b \sinh ^{-1}(c x)} \, dx+\frac {(5 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{\sqrt {1+c^2 x^2}} \, dx}{9 c}\\ &=\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (5 b^2\right ) \int \sqrt {a+b \sinh ^{-1}(c x)} \, dx}{6 c^2}-\frac {1}{72} \left (5 b^3 c\right ) \int \frac {x^3}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx\\ &=-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {\sinh ^3(x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{72 c^3}+\frac {\left (5 b^3\right ) \int \frac {x}{\sqrt {1+c^2 x^2} \sqrt {a+b \sinh ^{-1}(c x)}} \, dx}{12 c}\\ &=-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (5 i b^3\right ) \text {Subst}\left (\int \left (\frac {3 i \sinh (x)}{4 \sqrt {a+b x}}-\frac {i \sinh (3 x)}{4 \sqrt {a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{72 c^3}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{12 c^3}\\ &=-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {\sinh (3 x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{288 c^3}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {\sinh (x)}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{96 c^3}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{24 c^3}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{24 c^3}\\ &=-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{12 c^3}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{12 c^3}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {e^{-3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{576 c^3}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {e^{3 x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{576 c^3}-\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {e^{-x}}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{192 c^3}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {e^x}{\sqrt {a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{192 c^3}\\ &=-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {5 b^{5/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{24 c^3}+\frac {5 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{24 c^3}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int e^{\frac {3 a}{b}-\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{288 c^3}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int e^{-\frac {3 a}{b}+\frac {3 x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{288 c^3}-\frac {\left (5 b^2\right ) \text {Subst}\left (\int e^{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{96 c^3}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int e^{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sinh ^{-1}(c x)}\right )}{96 c^3}\\ &=-\frac {5 b^2 x \sqrt {a+b \sinh ^{-1}(c x)}}{6 c^2}+\frac {5}{36} b^2 x^3 \sqrt {a+b \sinh ^{-1}(c x)}+\frac {5 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{9 c^3}-\frac {5 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^{3/2}}{18 c}+\frac {1}{3} x^3 \left (a+b \sinh ^{-1}(c x)\right )^{5/2}-\frac {15 b^{5/2} e^{a/b} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^3}+\frac {5 b^{5/2} e^{\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erf}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{576 c^3}+\frac {15 b^{5/2} e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{64 c^3}-\frac {5 b^{5/2} e^{-\frac {3 a}{b}} \sqrt {\frac {\pi }{3}} \text {erfi}\left (\frac {\sqrt {3} \sqrt {a+b \sinh ^{-1}(c x)}}{\sqrt {b}}\right )}{576 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.32, size = 215, normalized size = 0.66 \begin {gather*} -\frac {e^{-\frac {3 a}{b}} \left (a+b \sinh ^{-1}(c x)\right )^{5/2} \left (81 e^{\frac {4 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {7}{2},\frac {a}{b}+\sinh ^{-1}(c x)\right )+\sqrt {3} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {7}{2},-\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-81 e^{\frac {2 a}{b}} \sqrt {\frac {a}{b}+\sinh ^{-1}(c x)} \Gamma \left (\frac {7}{2},-\frac {a+b \sinh ^{-1}(c x)}{b}\right )-\sqrt {3} e^{\frac {6 a}{b}} \sqrt {-\frac {a+b \sinh ^{-1}(c x)}{b}} \Gamma \left (\frac {7}{2},\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )\right )}{648 c^3 \left (-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{b^2}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/648*((a + b*ArcSinh[c*x])^(5/2)*(81*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[7/2, a/b + ArcSinh[c*
x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c*x]]*Gamma[7/2, (-3*(a + b*ArcSinh[c*x]))/b] - 81*E^((2*a)/b)*Sqrt[a/b + Arc
Sinh[c*x]]*Gamma[7/2, -((a + b*ArcSinh[c*x])/b)] - Sqrt[3]*E^((6*a)/b)*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[7
/2, (3*(a + b*ArcSinh[c*x]))/b]))/(c^3*E^((3*a)/b)*(-((a + b*ArcSinh[c*x])^2/b^2))^(3/2))

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \arcsinh \left (c x \right )\right )^{\frac {5}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asinh(c*x))^(5/2),x)

[Out]

int(x^2*(a + b*asinh(c*x))^(5/2), x)

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