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

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

[Out]

6/55*a*x*(b*x^2+a)^(1/3)/b+3/11*x^3*(b*x^2+a)^(1/3)+6/55*3^(3/4)*a^2*(a^(1/3)-(b*x^2+a)^(1/3))*EllipticF((-(b*
x^2+a)^(1/3)+a^(1/3)*(1+3^(1/2)))/(-(b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2))),2*I-I*3^(1/2))*((a^(2/3)+a^(1/3)*(b*x
^2+a)^(1/3)+(b*x^2+a)^(2/3))/(-(b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2)))^2)^(1/2)*(1/2*6^(1/2)-1/2*2^(1/2))/b^2/x/(
-a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/(-(b*x^2+a)^(1/3)+a^(1/3)*(1-3^(1/2)))^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 290, normalized size of antiderivative = 1.00, 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 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]

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 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]

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 = 0.05, 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)

________________________________________________________________________________________

fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac {1}{3}} x^{2}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{3}} x^{2}\,{d x} \]

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)

________________________________________________________________________________________

maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{2}+a \right )^{\frac {1}{3}} x^{2}\, dx \]

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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{3}} x^{2}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (b\,x^2+a\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^(1/3),x)

[Out]

int(x^2*(a + b*x^2)^(1/3), x)

________________________________________________________________________________________

sympy [A]  time = 0.84, size = 29, normalized size = 0.10 \[ \frac {\sqrt [3]{a} x^{3} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} \]

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

________________________________________________________________________________________