3.354 \(\int x^2 (a+b \sin ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=61 \[ -\frac{2 b \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{c} x\right ),-1\right )}{9 c^{3/2}}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{2 b x \sqrt{1-c^2 x^4}}{9 c} \]

[Out]

(2*b*x*Sqrt[1 - c^2*x^4])/(9*c) + (x^3*(a + b*ArcSin[c*x^2]))/3 - (2*b*EllipticF[ArcSin[Sqrt[c]*x], -1])/(9*c^
(3/2))

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Rubi [A]  time = 0.0345188, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4842, 12, 321, 221} \[ \frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^2\right )\right )+\frac{2 b x \sqrt{1-c^2 x^4}}{9 c}-\frac{2 b F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{9 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcSin[c*x^2]),x]

[Out]

(2*b*x*Sqrt[1 - c^2*x^4])/(9*c) + (x^3*(a + b*ArcSin[c*x^2]))/3 - (2*b*EllipticF[ArcSin[Sqrt[c]*x], -1])/(9*c^
(3/2))

Rule 4842

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin[
u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 - u^2], x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 321

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

\begin{align*} \int x^2 \left (a+b \sin ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{3} b \int \frac{2 c x^4}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{1}{3} (2 b c) \int \frac{x^4}{\sqrt{1-c^2 x^4}} \, dx\\ &=\frac{2 b x \sqrt{1-c^2 x^4}}{9 c}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{(2 b) \int \frac{1}{\sqrt{1-c^2 x^4}} \, dx}{9 c}\\ &=\frac{2 b x \sqrt{1-c^2 x^4}}{9 c}+\frac{1}{3} x^3 \left (a+b \sin ^{-1}\left (c x^2\right )\right )-\frac{2 b F\left (\left .\sin ^{-1}\left (\sqrt{c} x\right )\right |-1\right )}{9 c^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.144431, size = 72, normalized size = 1.18 \[ \frac{1}{9} \left (-\frac{2 i b \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt{-c} x\right ),-1\right )}{(-c)^{3/2}}+3 a x^3+\frac{2 b x \sqrt{1-c^2 x^4}}{c}+3 b x^3 \sin ^{-1}\left (c x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcSin[c*x^2]),x]

[Out]

(3*a*x^3 + (2*b*x*Sqrt[1 - c^2*x^4])/c + 3*b*x^3*ArcSin[c*x^2] - ((2*I)*b*EllipticF[I*ArcSinh[Sqrt[-c]*x], -1]
)/(-c)^(3/2))/9

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Maple [A]  time = 0.008, size = 88, normalized size = 1.4 \begin{align*}{\frac{{x}^{3}a}{3}}+b \left ({\frac{{x}^{3}\arcsin \left ( c{x}^{2} \right ) }{3}}-{\frac{2\,c}{3} \left ( -{\frac{x}{3\,{c}^{2}}\sqrt{-{c}^{2}{x}^{4}+1}}+{\frac{1}{3}\sqrt{-c{x}^{2}+1}\sqrt{c{x}^{2}+1}{\it EllipticF} \left ( x\sqrt{c},i \right ){c}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{-{c}^{2}{x}^{4}+1}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsin(c*x^2)),x)

[Out]

1/3*x^3*a+b*(1/3*x^3*arcsin(c*x^2)-2/3*c*(-1/3/c^2*x*(-c^2*x^4+1)^(1/2)+1/3/c^(5/2)*(-c*x^2+1)^(1/2)*(c*x^2+1)
^(1/2)/(-c^2*x^4+1)^(1/2)*EllipticF(x*c^(1/2),I)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsin(c*x^2)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*x^2*arcsin(c*x^2) + a*x^2, x)

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Sympy [A]  time = 1.91266, size = 58, normalized size = 0.95 \begin{align*} \frac{a x^{3}}{3} - \frac{b c x^{5} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{c^{2} x^{4} e^{2 i \pi }} \right )}}{6 \Gamma \left (\frac{9}{4}\right )} + \frac{b x^{3} \operatorname{asin}{\left (c x^{2} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asin(c*x**2)),x)

[Out]

a*x**3/3 - b*c*x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c**2*x**4*exp_polar(2*I*pi))/(6*gamma(9/4)) + b*x**3*
asin(c*x**2)/3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arcsin \left (c x^{2}\right ) + a\right )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c*x^2) + a)*x^2, x)