3.72 \(\int x^2 (\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=213 \[ \frac{1}{8} x^3 \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{48} \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{64} \pi ^2 x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 \pi ^{5/2} x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{128 c^2}-\frac{5 \pi ^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{256 b c^3}-\frac{1}{64} \pi ^{5/2} b c^5 x^8-\frac{17}{288} \pi ^{5/2} b c^3 x^6-\frac{59}{768} \pi ^{5/2} b c x^4-\frac{5 \pi ^{5/2} b x^2}{256 c} \]

[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)

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Rubi [A]  time = 0.457729, antiderivative size = 337, normalized size of antiderivative = 1.58, number of steps used = 12, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {5744, 5742, 5758, 5675, 30, 14, 266, 43} \[ \frac{1}{8} x^3 \left (\pi c^2 x^2+\pi \right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{48} \pi x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5}{64} \pi ^2 x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 \pi ^2 x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{128 c^2}-\frac{5 \pi ^2 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{256 b c^3 \sqrt{c^2 x^2+1}}-\frac{\pi ^2 b c^5 x^8 \sqrt{\pi c^2 x^2+\pi }}{64 \sqrt{c^2 x^2+1}}-\frac{17 \pi ^2 b c^3 x^6 \sqrt{\pi c^2 x^2+\pi }}{288 \sqrt{c^2 x^2+1}}-\frac{59 \pi ^2 b c x^4 \sqrt{\pi c^2 x^2+\pi }}{768 \sqrt{c^2 x^2+1}}-\frac{5 \pi ^2 b x^2 \sqrt{\pi c^2 x^2+\pi }}{256 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5744

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*d^IntPart[p]*(d + e*x^2)^FracPart[p]
)/(f*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[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]
&& (RationalQ[m] || EqQ[n, 1])

Rule 5742

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[Sqrt[d + e*x^2]/((m + 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*Sqrt[d + e*x^2])/(f*
(m + 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] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 5758

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

Rule 5675

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

Rule 30

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

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 266

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 43

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])

Rubi steps

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

Mathematica [A]  time = 0.565704, size = 196, normalized size = 0.92 \[ \frac{\pi ^{5/2} \left (-24 \sinh ^{-1}(c x) \left (120 a+48 b \sinh \left (2 \sinh ^{-1}(c x)\right )-24 b \sinh \left (4 \sinh ^{-1}(c x)\right )-16 b \sinh \left (6 \sinh ^{-1}(c x)\right )-3 b \sinh \left (8 \sinh ^{-1}(c x)\right )\right )+9216 a c^7 x^7 \sqrt{c^2 x^2+1}+26112 a c^5 x^5 \sqrt{c^2 x^2+1}+22656 a c^3 x^3 \sqrt{c^2 x^2+1}+2880 a c x \sqrt{c^2 x^2+1}-1440 b \sinh ^{-1}(c x)^2+576 b \cosh \left (2 \sinh ^{-1}(c x)\right )-144 b \cosh \left (4 \sinh ^{-1}(c x)\right )-64 b \cosh \left (6 \sinh ^{-1}(c x)\right )-9 b \cosh \left (8 \sinh ^{-1}(c x)\right )\right )}{73728 c^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.093, size = 301, normalized size = 1.4 \begin{align*}{\frac{ax}{8\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{7}{2}}}}-{\frac{ax}{48\,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}-{\frac{5\,a\pi \,x}{192\,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}-{\frac{5\,a{\pi }^{2}x}{128\,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{5\,a{\pi }^{3}}{128\,{c}^{2}}\ln \left ({\pi \,{c}^{2}x{\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+\sqrt{\pi \,{c}^{2}{x}^{2}+\pi } \right ){\frac{1}{\sqrt{\pi \,{c}^{2}}}}}+{\frac{b{\pi }^{{\frac{5}{2}}}{c}^{4}{\it Arcsinh} \left ( cx \right ){x}^{7}}{8}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{c}^{5}{\pi }^{{\frac{5}{2}}}{x}^{8}}{64}}+{\frac{17\,b{\pi }^{5/2}{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{5}}{48}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{17\,b{c}^{3}{\pi }^{5/2}{x}^{6}}{288}}+{\frac{59\,b{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{192}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{59\,bc{\pi }^{5/2}{x}^{4}}{768}}+{\frac{5\,b{\pi }^{5/2}{\it Arcsinh} \left ( cx \right ) x}{128\,{c}^{2}}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{5\,b{\pi }^{5/2}{x}^{2}}{256\,c}}-{\frac{5\,b{\pi }^{5/2} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{256\,{c}^{3}}}+{\frac{b{\pi }^{{\frac{5}{2}}}}{72\,{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(
Pi*c^2)^(1/2)+1/8*b*Pi^(5/2)*c^4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^7-1/64*b*c^5*Pi^(5/2)*x^8+17/48*b*Pi^(5/2)*c
^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^5-17/288*b*c^3*Pi^(5/2)*x^6+59/192*b*Pi^(5/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/
2)*x^3-59/768*b*c*Pi^(5/2)*x^4+5/128*b*Pi^(5/2)/c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-5/256*b*Pi^(5/2)*x^2/c-5/
256*b*Pi^(5/2)/c^3*arcsinh(c*x)^2+1/72*b*Pi^(5/2)/c^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{\pi + \pi c^{2} x^{2}}{\left (\pi ^{2} a c^{4} x^{6} + 2 \, \pi ^{2} a c^{2} x^{4} + \pi ^{2} a x^{2} +{\left (\pi ^{2} b c^{4} x^{6} + 2 \, \pi ^{2} b c^{2} x^{4} + \pi ^{2} b x^{2}\right )} \operatorname{arsinh}\left (c x\right )\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)