3.673 \(\int x^2 \sqrt [3]{a+b x^2} \, dx\)

Optimal. Leaf size=290 \[ \frac{6\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),4 \sqrt{3}-7\right )}{55 b^2 x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}+\frac{3}{11} x^3 \sqrt [3]{a+b x^2}+\frac{6 a x \sqrt [3]{a+b x^2}}{55 b} \]

[Out]

(6*a*x*(a + b*x^2)^(1/3))/(55*b) + (3*x^3*(a + b*x^2)^(1/3))/11 + (6*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(a^(1/3) -
(a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a
+ b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a +
b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(55*b^2*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/
3) - (a + b*x^2)^(1/3))^2)])

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Rubi [A]  time = 0.161355, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {279, 321, 236, 219} \[ \frac{6\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right )|-7+4 \sqrt{3}\right )}{55 b^2 x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}+\frac{3}{11} x^3 \sqrt [3]{a+b x^2}+\frac{6 a x \sqrt [3]{a+b x^2}}{55 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^(1/3),x]

[Out]

(6*a*x*(a + b*x^2)^(1/3))/(55*b) + (3*x^3*(a + b*x^2)^(1/3))/11 + (6*3^(3/4)*Sqrt[2 - Sqrt[3]]*a^2*(a^(1/3) -
(a + b*x^2)^(1/3))*Sqrt[(a^(2/3) + a^(1/3)*(a + b*x^2)^(1/3) + (a + b*x^2)^(2/3))/((1 - Sqrt[3])*a^(1/3) - (a
+ b*x^2)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*a^(1/3) - (a + b*x^2)^(1/3))/((1 - Sqrt[3])*a^(1/3) - (a +
b*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(55*b^2*x*Sqrt[-((a^(1/3)*(a^(1/3) - (a + b*x^2)^(1/3)))/((1 - Sqrt[3])*a^(1/
3) - (a + b*x^2)^(1/3))^2)])

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 236

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[(3*Sqrt[b*x^2])/(2*b*x), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

\begin{align*} \int x^2 \sqrt [3]{a+b x^2} \, dx &=\frac{3}{11} x^3 \sqrt [3]{a+b x^2}+\frac{1}{11} (2 a) \int \frac{x^2}{\left (a+b x^2\right )^{2/3}} \, dx\\ &=\frac{6 a x \sqrt [3]{a+b x^2}}{55 b}+\frac{3}{11} x^3 \sqrt [3]{a+b x^2}-\frac{\left (6 a^2\right ) \int \frac{1}{\left (a+b x^2\right )^{2/3}} \, dx}{55 b}\\ &=\frac{6 a x \sqrt [3]{a+b x^2}}{55 b}+\frac{3}{11} x^3 \sqrt [3]{a+b x^2}-\frac{\left (9 a^2 \sqrt{b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a+x^3}} \, dx,x,\sqrt [3]{a+b x^2}\right )}{55 b^2 x}\\ &=\frac{6 a x \sqrt [3]{a+b x^2}}{55 b}+\frac{3}{11} x^3 \sqrt [3]{a+b x^2}+\frac{6\ 3^{3/4} \sqrt{2-\sqrt{3}} a^2 \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt{\frac{a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right )|-7+4 \sqrt{3}\right )}{55 b^2 x \sqrt{-\frac{\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0452349, size = 62, normalized size = 0.21 \[ \frac{3 x \sqrt [3]{a+b x^2} \left (-\frac{a \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )}{\sqrt [3]{\frac{b x^2}{a}+1}}+a+b x^2\right )}{11 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^(1/3),x]

[Out]

(3*x*(a + b*x^2)^(1/3)*(a + b*x^2 - (a*Hypergeometric2F1[-1/3, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(1/3))
)/(11*b)

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Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\sqrt [3]{b{x}^{2}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*x^2+a)^(1/3),x)

[Out]

int(x^2*(b*x^2+a)^(1/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{3}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(1/3),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/3)*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{1}{3}} x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(1/3),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/3)*x^2, x)

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Sympy [A]  time = 0.709148, size = 29, normalized size = 0.1 \begin{align*} \frac{\sqrt [3]{a} x^{3}{{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{3}, \frac{3}{2} \\ \frac{5}{2} \end{matrix}\middle |{\frac{b x^{2} e^{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)**(1/3),x)

[Out]

a**(1/3)*x**3*hyper((-1/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{\frac{1}{3}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(1/3),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/3)*x^2, x)