3.806 \(\int x^2 (a-b x^2)^{3/4} \, dx\)

Optimal. Leaf size=101 \[ \frac{4 a^{5/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a-b x^2}}+\frac{2}{9} x^3 \left (a-b x^2\right )^{3/4}-\frac{2 a x \left (a-b x^2\right )^{3/4}}{15 b} \]

[Out]

(-2*a*x*(a - b*x^2)^(3/4))/(15*b) + (2*x^3*(a - b*x^2)^(3/4))/9 + (4*a^(5/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[A
rcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(15*b^(3/2)*(a - b*x^2)^(1/4))

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Rubi [A]  time = 0.0329896, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {279, 321, 229, 228} \[ \frac{4 a^{5/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a-b x^2}}+\frac{2}{9} x^3 \left (a-b x^2\right )^{3/4}-\frac{2 a x \left (a-b x^2\right )^{3/4}}{15 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a - b*x^2)^(3/4),x]

[Out]

(-2*a*x*(a - b*x^2)^(3/4))/(15*b) + (2*x^3*(a - b*x^2)^(3/4))/9 + (4*a^(5/2)*(1 - (b*x^2)/a)^(1/4)*EllipticE[A
rcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(15*b^(3/2)*(a - b*x^2)^(1/4))

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, 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 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int x^2 \left (a-b x^2\right )^{3/4} \, dx &=\frac{2}{9} x^3 \left (a-b x^2\right )^{3/4}+\frac{1}{3} a \int \frac{x^2}{\sqrt [4]{a-b x^2}} \, dx\\ &=-\frac{2 a x \left (a-b x^2\right )^{3/4}}{15 b}+\frac{2}{9} x^3 \left (a-b x^2\right )^{3/4}+\frac{\left (2 a^2\right ) \int \frac{1}{\sqrt [4]{a-b x^2}} \, dx}{15 b}\\ &=-\frac{2 a x \left (a-b x^2\right )^{3/4}}{15 b}+\frac{2}{9} x^3 \left (a-b x^2\right )^{3/4}+\frac{\left (2 a^2 \sqrt [4]{1-\frac{b x^2}{a}}\right ) \int \frac{1}{\sqrt [4]{1-\frac{b x^2}{a}}} \, dx}{15 b \sqrt [4]{a-b x^2}}\\ &=-\frac{2 a x \left (a-b x^2\right )^{3/4}}{15 b}+\frac{2}{9} x^3 \left (a-b x^2\right )^{3/4}+\frac{4 a^{5/2} \sqrt [4]{1-\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \sin ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 b^{3/2} \sqrt [4]{a-b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0519357, size = 64, normalized size = 0.63 \[ \frac{2 x \left (a-b x^2\right )^{3/4} \left (\frac{a \, _2F_1\left (-\frac{3}{4},\frac{1}{2};\frac{3}{2};\frac{b x^2}{a}\right )}{\left (1-\frac{b x^2}{a}\right )^{3/4}}-a+b x^2\right )}{9 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a - b*x^2)^(3/4),x]

[Out]

(2*x*(a - b*x^2)^(3/4)*(-a + b*x^2 + (a*Hypergeometric2F1[-3/4, 1/2, 3/2, (b*x^2)/a])/(1 - (b*x^2)/a)^(3/4)))/
(9*b)

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( -b{x}^{2}+a \right ) ^{{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(-b*x^2+a)^(3/4),x)

[Out]

int(x^2*(-b*x^2+a)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-b x^{2} + a\right )}^{\frac{3}{4}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((-b*x^2 + a)^(3/4)*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-b x^{2} + a\right )}^{\frac{3}{4}} x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

integral((-b*x^2 + a)^(3/4)*x^2, x)

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Sympy [C]  time = 1.11439, size = 31, normalized size = 0.31 \begin{align*} \frac{a^{\frac{3}{4}} x^{3}{{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{2 i \pi }}{a}} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(-b*x**2+a)**(3/4),x)

[Out]

a**(3/4)*x**3*hyper((-3/4, 3/2), (5/2,), b*x**2*exp_polar(2*I*pi)/a)/3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(-b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

Exception raised: TypeError