\(\int x^2 (\pi +c^2 \pi x^2)^{5/2} (a+b \text {arcsinh}(c x)) \, dx\) [72]

Optimal result

Integrand size = 26, antiderivative size = 213 \[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {5 b \pi ^{5/2} x^2}{256 c}-\frac {59}{768} b c \pi ^{5/2} x^4-\frac {17}{288} b c^3 \pi ^{5/2} x^6-\frac {1}{64} b c^5 \pi ^{5/2} x^8+\frac {5 \pi ^{5/2} x \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{128 c^2}+\frac {5}{64} \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {5}{48} \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {1}{8} x^3 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))-\frac {5 \pi ^{5/2} (a+b \text {arcsinh}(c x))^2}{256 b c^3} \]

[Out]

-5/256*b*Pi^(5/2)*x^2/c-59/768*b*c*Pi^(5/2)*x^4-17/288*b*c^3*Pi^(5/2)*x^6-1/64*b*c^5*Pi^(5/2)*x^8+5/48*Pi*x^3*
(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x))+1/8*x^3*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x))-5/256*Pi^(5/2)*(a+b*
arcsinh(c*x))^2/b/c^3+5/128*Pi^(5/2)*x*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^2+5/64*Pi^2*x^3*(a+b*arcsinh(c*x
))*(Pi*c^2*x^2+Pi)^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5808, 5806, 5812, 5783, 30, 14, 272, 45} \[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=-\frac {5 \pi ^{5/2} (a+b \text {arcsinh}(c x))^2}{256 b c^3}+\frac {5 \pi ^{5/2} x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{128 c^2}+\frac {1}{8} x^3 \left (\pi c^2 x^2+\pi \right )^{5/2} (a+b \text {arcsinh}(c x))+\frac {5}{48} \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {5}{64} \pi ^2 x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))-\frac {1}{64} \pi ^{5/2} b c^5 x^8-\frac {17}{288} \pi ^{5/2} b c^3 x^6-\frac {59}{768} \pi ^{5/2} b c x^4-\frac {5 \pi ^{5/2} b x^2}{256 c} \]

[In]

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

[Out]

(-5*b*Pi^(5/2)*x^2)/(256*c) - (59*b*c*Pi^(5/2)*x^4)/768 - (17*b*c^3*Pi^(5/2)*x^6)/288 - (b*c^5*Pi^(5/2)*x^8)/6
4 + (5*Pi^(5/2)*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(128*c^2) + (5*Pi^2*x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b
*ArcSinh[c*x]))/64 + (5*Pi*x^3*(Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/48 + (x^3*(Pi + c^2*Pi*x^2)^(5/2)
*(a + b*ArcSinh[c*x]))/8 - (5*Pi^(5/2)*(a + b*ArcSinh[c*x])^2)/(256*b*c^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5783

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

Rule 5806

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

Rule 5808

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} x^3 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))+\frac {1}{8} (5 \pi ) \int x^2 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \, dx-\frac {1}{8} \left (b c \pi ^{5/2}\right ) \int x^3 \left (1+c^2 x^2\right )^2 \, dx \\ & = \frac {5}{48} \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {1}{8} x^3 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))+\frac {1}{16} \left (5 \pi ^2\right ) \int x^2 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x)) \, dx-\frac {1}{16} \left (b c \pi ^{5/2}\right ) \text {Subst}\left (\int x \left (1+c^2 x\right )^2 \, dx,x,x^2\right )-\frac {1}{48} \left (5 b c \pi ^{5/2}\right ) \int x^3 \left (1+c^2 x^2\right ) \, dx \\ & = \frac {5}{64} \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {5}{48} \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {1}{8} x^3 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))+\frac {1}{64} \left (5 \pi ^{5/2}\right ) \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {1+c^2 x^2}} \, dx-\frac {1}{16} \left (b c \pi ^{5/2}\right ) \text {Subst}\left (\int \left (x+2 c^2 x^2+c^4 x^3\right ) \, dx,x,x^2\right )-\frac {1}{64} \left (5 b c \pi ^{5/2}\right ) \int x^3 \, dx-\frac {1}{48} \left (5 b c \pi ^{5/2}\right ) \int \left (x^3+c^2 x^5\right ) \, dx \\ & = -\frac {59}{768} b c \pi ^{5/2} x^4-\frac {17}{288} b c^3 \pi ^{5/2} x^6-\frac {1}{64} b c^5 \pi ^{5/2} x^8+\frac {5 \pi ^{5/2} x \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{128 c^2}+\frac {5}{64} \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {5}{48} \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {1}{8} x^3 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))-\frac {\left (5 \pi ^{5/2}\right ) \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{128 c^2}-\frac {\left (5 b \pi ^{5/2}\right ) \int x \, dx}{128 c} \\ & = -\frac {5 b \pi ^{5/2} x^2}{256 c}-\frac {59}{768} b c \pi ^{5/2} x^4-\frac {17}{288} b c^3 \pi ^{5/2} x^6-\frac {1}{64} b c^5 \pi ^{5/2} x^8+\frac {5 \pi ^{5/2} x \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{128 c^2}+\frac {5}{64} \pi ^2 x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))+\frac {5}{48} \pi x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} (a+b \text {arcsinh}(c x))+\frac {1}{8} x^3 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x))-\frac {5 \pi ^{5/2} (a+b \text {arcsinh}(c x))^2}{256 b c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.92 \[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {\pi ^{5/2} \left (2880 a c x \sqrt {1+c^2 x^2}+22656 a c^3 x^3 \sqrt {1+c^2 x^2}+26112 a c^5 x^5 \sqrt {1+c^2 x^2}+9216 a c^7 x^7 \sqrt {1+c^2 x^2}-1440 b \text {arcsinh}(c x)^2+576 b \cosh (2 \text {arcsinh}(c x))-144 b \cosh (4 \text {arcsinh}(c x))-64 b \cosh (6 \text {arcsinh}(c x))-9 b \cosh (8 \text {arcsinh}(c x))-24 \text {arcsinh}(c x) (120 a+48 b \sinh (2 \text {arcsinh}(c x))-24 b \sinh (4 \text {arcsinh}(c x))-16 b \sinh (6 \text {arcsinh}(c x))-3 b \sinh (8 \text {arcsinh}(c x)))\right )}{73728 c^3} \]

[In]

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

[Out]

(Pi^(5/2)*(2880*a*c*x*Sqrt[1 + c^2*x^2] + 22656*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 26112*a*c^5*x^5*Sqrt[1 + c^2*x^2
] + 9216*a*c^7*x^7*Sqrt[1 + c^2*x^2] - 1440*b*ArcSinh[c*x]^2 + 576*b*Cosh[2*ArcSinh[c*x]] - 144*b*Cosh[4*ArcSi
nh[c*x]] - 64*b*Cosh[6*ArcSinh[c*x]] - 9*b*Cosh[8*ArcSinh[c*x]] - 24*ArcSinh[c*x]*(120*a + 48*b*Sinh[2*ArcSinh
[c*x]] - 24*b*Sinh[4*ArcSinh[c*x]] - 16*b*Sinh[6*ArcSinh[c*x]] - 3*b*Sinh[8*ArcSinh[c*x]])))/(73728*c^3)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {7}{2}}}{8 \pi \,c^{2}}-\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{48 c^{2}}-\frac {5 a \pi x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{192 c^{2}}-\frac {5 a \,\pi ^{2} x \sqrt {\pi \,c^{2} x^{2}+\pi }}{128 c^{2}}-\frac {5 a \,\pi ^{3} \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{128 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \,\pi ^{\frac {5}{2}} \left (-288 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{7} c^{7}+36 c^{8} x^{8}-816 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+136 c^{6} x^{6}-708 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+177 c^{4} x^{4}-90 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+45 c^{2} x^{2}+45 \operatorname {arcsinh}\left (c x \right )^{2}-32\right )}{2304 c^{3}}\)	\(267\)
parts	\(\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {7}{2}}}{8 \pi \,c^{2}}-\frac {a x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{48 c^{2}}-\frac {5 a \pi x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{192 c^{2}}-\frac {5 a \,\pi ^{2} x \sqrt {\pi \,c^{2} x^{2}+\pi }}{128 c^{2}}-\frac {5 a \,\pi ^{3} \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{128 c^{2} \sqrt {\pi \,c^{2}}}-\frac {b \,\pi ^{\frac {5}{2}} \left (-288 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{7} c^{7}+36 c^{8} x^{8}-816 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{5} c^{5}+136 c^{6} x^{6}-708 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}+177 c^{4} x^{4}-90 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+45 c^{2} x^{2}+45 \operatorname {arcsinh}\left (c x \right )^{2}-32\right )}{2304 c^{3}}\)	\(267\)

[In]

int(x^2*(Pi*c^2*x^2+Pi)^(5/2)*(a+b*arcsinh(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/8*a*x*(Pi*c^2*x^2+Pi)^(7/2)/Pi/c^2-1/48*a/c^2*x*(Pi*c^2*x^2+Pi)^(5/2)-5/192*a/c^2*Pi*x*(Pi*c^2*x^2+Pi)^(3/2)
-5/128*a/c^2*Pi^2*x*(Pi*c^2*x^2+Pi)^(1/2)-5/128*a/c^2*Pi^3*ln(Pi*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(
Pi*c^2)^(1/2)-1/2304*b*Pi^(5/2)*(-288*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^7*c^7+36*c^8*x^8-816*arcsinh(c*x)*(c^2*
x^2+1)^(1/2)*x^5*c^5+136*c^6*x^6-708*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3*c^3+177*c^4*x^4-90*arcsinh(c*x)*c*x*(c
^2*x^2+1)^(1/2)+45*c^2*x^2+45*arcsinh(c*x)^2-32)/c^3

Fricas [F]

\[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2} \,d x } \]

[In]

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

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi^2*a*c^4*x^6 + 2*pi^2*a*c^2*x^4 + pi^2*a*x^2 + (pi^2*b*c^4*x^6 + 2*pi^2*b*c^
2*x^4 + pi^2*b*x^2)*arcsinh(c*x)), x)

Sympy [A] (verification not implemented)

Time = 49.52 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.64 \[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\begin {cases} \frac {\pi ^{\frac {5}{2}} a c^{4} x^{7} \sqrt {c^{2} x^{2} + 1}}{8} + \frac {17 \pi ^{\frac {5}{2}} a c^{2} x^{5} \sqrt {c^{2} x^{2} + 1}}{48} + \frac {59 \pi ^{\frac {5}{2}} a x^{3} \sqrt {c^{2} x^{2} + 1}}{192} + \frac {5 \pi ^{\frac {5}{2}} a x \sqrt {c^{2} x^{2} + 1}}{128 c^{2}} - \frac {5 \pi ^{\frac {5}{2}} a \operatorname {asinh}{\left (c x \right )}}{128 c^{3}} - \frac {\pi ^{\frac {5}{2}} b c^{5} x^{8}}{64} + \frac {\pi ^{\frac {5}{2}} b c^{4} x^{7} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{8} - \frac {17 \pi ^{\frac {5}{2}} b c^{3} x^{6}}{288} + \frac {17 \pi ^{\frac {5}{2}} b c^{2} x^{5} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{48} - \frac {59 \pi ^{\frac {5}{2}} b c x^{4}}{768} + \frac {59 \pi ^{\frac {5}{2}} b x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{192} - \frac {5 \pi ^{\frac {5}{2}} b x^{2}}{256 c} + \frac {5 \pi ^{\frac {5}{2}} b x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{128 c^{2}} - \frac {5 \pi ^{\frac {5}{2}} b \operatorname {asinh}^{2}{\left (c x \right )}}{256 c^{3}} & \text {for}\: c \neq 0 \\\frac {\pi ^{\frac {5}{2}} a x^{3}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((pi**(5/2)*a*c**4*x**7*sqrt(c**2*x**2 + 1)/8 + 17*pi**(5/2)*a*c**2*x**5*sqrt(c**2*x**2 + 1)/48 + 59*
pi**(5/2)*a*x**3*sqrt(c**2*x**2 + 1)/192 + 5*pi**(5/2)*a*x*sqrt(c**2*x**2 + 1)/(128*c**2) - 5*pi**(5/2)*a*asin
h(c*x)/(128*c**3) - pi**(5/2)*b*c**5*x**8/64 + pi**(5/2)*b*c**4*x**7*sqrt(c**2*x**2 + 1)*asinh(c*x)/8 - 17*pi*
*(5/2)*b*c**3*x**6/288 + 17*pi**(5/2)*b*c**2*x**5*sqrt(c**2*x**2 + 1)*asinh(c*x)/48 - 59*pi**(5/2)*b*c*x**4/76
8 + 59*pi**(5/2)*b*x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/192 - 5*pi**(5/2)*b*x**2/(256*c) + 5*pi**(5/2)*b*x*sqrt
(c**2*x**2 + 1)*asinh(c*x)/(128*c**2) - 5*pi**(5/2)*b*asinh(c*x)**2/(256*c**3), Ne(c, 0)), (pi**(5/2)*a*x**3/3
, True))

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

\[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (\pi +c^2 \pi x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \, dx=\int x^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2} \,d x \]

[In]

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

[Out]

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