3.64 \(\int x^2 (\pi +c^2 \pi x^2)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=165 \[ \frac{1}{6} x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} \pi x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{\pi ^{3/2} x \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}-\frac{\pi ^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3}-\frac{1}{36} \pi ^{3/2} b c^3 x^6-\frac{7}{96} \pi ^{3/2} b c x^4-\frac{\pi ^{3/2} b x^2}{32 c} \]

[Out]

-(b*Pi^(3/2)*x^2)/(32*c) - (7*b*c*Pi^(3/2)*x^4)/96 - (b*c^3*Pi^(3/2)*x^6)/36 + (Pi^(3/2)*x*Sqrt[1 + c^2*x^2]*(
a + b*ArcSinh[c*x]))/(16*c^2) + (Pi*x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (x^3*(Pi + c^2*Pi*x^2)
^(3/2)*(a + b*ArcSinh[c*x]))/6 - (Pi^(3/2)*(a + b*ArcSinh[c*x])^2)/(32*b*c^3)

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Rubi [A]  time = 0.321175, antiderivative size = 254, normalized size of antiderivative = 1.54, number of steps used = 8, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {5744, 5742, 5758, 5675, 30, 14} \[ \frac{1}{6} x^3 \left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{8} \pi x^3 \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )+\frac{\pi x \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}-\frac{\pi \sqrt{\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt{c^2 x^2+1}}-\frac{\pi b c^3 x^6 \sqrt{\pi c^2 x^2+\pi }}{36 \sqrt{c^2 x^2+1}}-\frac{7 \pi b c x^4 \sqrt{\pi c^2 x^2+\pi }}{96 \sqrt{c^2 x^2+1}}-\frac{\pi b x^2 \sqrt{\pi c^2 x^2+\pi }}{32 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Pi*x^2*Sqrt[Pi + c^2*Pi*x^2])/(32*c*Sqrt[1 + c^2*x^2]) - (7*b*c*Pi*x^4*Sqrt[Pi + c^2*Pi*x^2])/(96*Sqrt[1 +
c^2*x^2]) - (b*c^3*Pi*x^6*Sqrt[Pi + c^2*Pi*x^2])/(36*Sqrt[1 + c^2*x^2]) + (Pi*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*
ArcSinh[c*x]))/(16*c^2) + (Pi*x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/8 + (x^3*(Pi + c^2*Pi*x^2)^(3/2)
*(a + b*ArcSinh[c*x]))/6 - (Pi*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x])^2)/(32*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]]

Rubi steps

\begin{align*} \int x^2 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{2} \pi \int x^2 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx-\frac{\left (b c \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int x^3 \left (1+c^2 x^2\right ) \, dx}{6 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{8} \pi x^3 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (\pi \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}{8 \sqrt{1+c^2 x^2}}-\frac{\left (b c \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int x^3 \, dx}{8 \sqrt{1+c^2 x^2}}-\frac{\left (b c \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (x^3+c^2 x^5\right ) \, dx}{6 \sqrt{1+c^2 x^2}}\\ &=-\frac{7 b c \pi x^4 \sqrt{\pi +c^2 \pi x^2}}{96 \sqrt{1+c^2 x^2}}-\frac{b c^3 \pi x^6 \sqrt{\pi +c^2 \pi x^2}}{36 \sqrt{1+c^2 x^2}}+\frac{\pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac{1}{8} \pi x^3 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\left (\pi \sqrt{\pi +c^2 \pi x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{16 c^2 \sqrt{1+c^2 x^2}}-\frac{\left (b \pi \sqrt{\pi +c^2 \pi x^2}\right ) \int x \, dx}{16 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b \pi x^2 \sqrt{\pi +c^2 \pi x^2}}{32 c \sqrt{1+c^2 x^2}}-\frac{7 b c \pi x^4 \sqrt{\pi +c^2 \pi x^2}}{96 \sqrt{1+c^2 x^2}}-\frac{b c^3 \pi x^6 \sqrt{\pi +c^2 \pi x^2}}{36 \sqrt{1+c^2 x^2}}+\frac{\pi x \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{16 c^2}+\frac{1}{8} \pi x^3 \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{1}{6} x^3 \left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{\pi \sqrt{\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c^3 \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.347191, size = 154, normalized size = 0.93 \[ \frac{\pi ^{3/2} \left (-12 \sinh ^{-1}(c x) \left (12 a+3 b \sinh \left (2 \sinh ^{-1}(c x)\right )-3 b \sinh \left (4 \sinh ^{-1}(c x)\right )-b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )+384 a c^5 x^5 \sqrt{c^2 x^2+1}+672 a c^3 x^3 \sqrt{c^2 x^2+1}+144 a c x \sqrt{c^2 x^2+1}-72 b \sinh ^{-1}(c x)^2+18 b \cosh \left (2 \sinh ^{-1}(c x)\right )-9 b \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b \cosh \left (6 \sinh ^{-1}(c x)\right )\right )}{2304 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(3/2)*(144*a*c*x*Sqrt[1 + c^2*x^2] + 672*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 384*a*c^5*x^5*Sqrt[1 + c^2*x^2] - 7
2*b*ArcSinh[c*x]^2 + 18*b*Cosh[2*ArcSinh[c*x]] - 9*b*Cosh[4*ArcSinh[c*x]] - 2*b*Cosh[6*ArcSinh[c*x]] - 12*ArcS
inh[c*x]*(12*a + 3*b*Sinh[2*ArcSinh[c*x]] - 3*b*Sinh[4*ArcSinh[c*x]] - b*Sinh[6*ArcSinh[c*x]])))/(2304*c^3)

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Maple [A]  time = 0.058, size = 240, normalized size = 1.5 \begin{align*}{\frac{ax}{6\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{5}{2}}}}-{\frac{ax}{24\,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{3}{2}}}}-{\frac{a\pi \,x}{16\,{c}^{2}}\sqrt{\pi \,{c}^{2}{x}^{2}+\pi }}-{\frac{a{\pi }^{2}}{16\,{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{3}{2}}}{c}^{2}{\it Arcsinh} \left ( cx \right ){x}^{5}}{6}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{c}^{3}{\pi }^{{\frac{3}{2}}}{x}^{6}}{36}}+{\frac{7\,b{\pi }^{3/2}{\it Arcsinh} \left ( cx \right ){x}^{3}}{24}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{7\,bc{\pi }^{3/2}{x}^{4}}{96}}+{\frac{b{\pi }^{{\frac{3}{2}}}{\it Arcsinh} \left ( cx \right ) x}{16\,{c}^{2}}\sqrt{{c}^{2}{x}^{2}+1}}-{\frac{b{\pi }^{{\frac{3}{2}}}{x}^{2}}{32\,c}}-{\frac{b{\pi }^{{\frac{3}{2}}} \left ({\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{32\,{c}^{3}}}+{\frac{b{\pi }^{{\frac{3}{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)^(3/2)*(a+b*arcsinh(c*x)),x)

[Out]

1/6*a*x*(Pi*c^2*x^2+Pi)^(5/2)/Pi/c^2-1/24*a/c^2*x*(Pi*c^2*x^2+Pi)^(3/2)-1/16*a/c^2*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)-
1/16*a/c^2*Pi^2*ln(Pi*x*c^2/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/6*b*Pi^(3/2)*c^2*arcsinh(c*
x)*(c^2*x^2+1)^(1/2)*x^5-1/36*b*c^3*Pi^(3/2)*x^6+7/24*b*Pi^(3/2)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3-7/96*b*c*P
i^(3/2)*x^4+1/16*b*Pi^(3/2)/c^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x-1/32*b*Pi^(3/2)*x^2/c-1/32*b*Pi^(3/2)/c^3*arc
sinh(c*x)^2+1/72*b*Pi^(3/2)/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)^(3/2)*(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 a c^{2} x^{4} + \pi a x^{2} +{\left (\pi b c^{2} x^{4} + \pi 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)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi*a*c^2*x^4 + pi*a*x^2 + (pi*b*c^2*x^4 + pi*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)**(3/2)*(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{3}{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)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="giac")

[Out]

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