Integrand size = 15, antiderivative size = 269 \[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 x \sqrt [3]{a+b x^2}}{5 b}+\frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
3/5*x*(b*x^2+a)^(1/3)/b+3/5*3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a*(a^(1/3)-( b*x^2+a)^(1/3))*((a^(2/3)+a^(1/3)*(b*x^2+a)^(1/3)+(b*x^2+a)^(2/3))/((1-3^( 1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2)^(1/2)*EllipticF(((1+3^(1/2))*a^(1/3)-(b* x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3)),2*I-I*3^(1/2))/b^2/x/( -a^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/((1-3^(1/2))*a^(1/3)-(b*x^2+a)^(1/3))^2 )^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.41 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.23 \[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {3 x \left (a+b x^2-a \left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{5 b \left (a+b x^2\right )^{2/3}} \] Input:
Integrate[x^2/(a + b*x^2)^(2/3),x]
Output:
(3*x*(a + b*x^2 - a*(1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[1/2, 2/3, 3/2, -((b*x^2)/a)]))/(5*b*(a + b*x^2)^(2/3))
Time = 0.25 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {262, 234, 760}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {3 x \sqrt [3]{a+b x^2}}{5 b}-\frac {3 a \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx}{5 b}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {3 x \sqrt [3]{a+b x^2}}{5 b}-\frac {9 a \sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{10 b^2 x}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {3\ 3^{3/4} \sqrt {2-\sqrt {3}} a \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{5 b^2 x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}+\frac {3 x \sqrt [3]{a+b x^2}}{5 b}\) |
Input:
Int[x^2/(a + b*x^2)^(2/3),x]
Output:
(3*x*(a + b*x^2)^(1/3))/(5*b) + (3*3^(3/4)*Sqrt[2 - Sqrt[3]]*a*(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]])/(5*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)])
Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((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]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
\[\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x\]
Input:
int(x^2/(b*x^2+a)^(2/3),x)
Output:
int(x^2/(b*x^2+a)^(2/3),x)
\[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(x^2/(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
integral(x^2/(b*x^2 + a)^(2/3), x)
Time = 0.46 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.10 \[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\frac {x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {2}{3}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {2}{3}}} \] Input:
integrate(x**2/(b*x**2+a)**(2/3),x)
Output:
x**3*hyper((2/3, 3/2), (5/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(2/3))
\[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(x^2/(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
integrate(x^2/(b*x^2 + a)^(2/3), x)
\[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {x^{2}}{{\left (b x^{2} + a\right )}^{\frac {2}{3}}} \,d x } \] Input:
integrate(x^2/(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
integrate(x^2/(b*x^2 + a)^(2/3), x)
Timed out. \[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\int \frac {x^2}{{\left (b\,x^2+a\right )}^{2/3}} \,d x \] Input:
int(x^2/(a + b*x^2)^(2/3),x)
Output:
int(x^2/(a + b*x^2)^(2/3), x)
\[ \int \frac {x^2}{\left (a+b x^2\right )^{2/3}} \, dx=\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x \] Input:
int(x^2/(b*x^2+a)^(2/3),x)
Output:
int(x**2/(a + b*x**2)**(2/3),x)