Integrand size = 15, antiderivative size = 265 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^2}}{a x}+\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}} \] Output:
-(b*x^2+a)^(1/3)/a/x+1/3*(1/2*6^(1/2)-1/2*2^(1/2))*(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))*3^(3/4)/a/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 = 4.81 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {2}{3},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \left (a+b x^2\right )^{2/3}} \] Input:
Integrate[1/(x^2*(a + b*x^2)^(2/3)),x]
Output:
-(((1 + (b*x^2)/a)^(2/3)*Hypergeometric2F1[-1/2, 2/3, 1/2, -((b*x^2)/a)])/ (x*(a + b*x^2)^(2/3)))
Time = 0.25 (sec) , antiderivative size = 265, 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 = {264, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {b \int \frac {1}{\left (b x^2+a\right )^{2/3}}dx}{3 a}-\frac {\sqrt [3]{a+b x^2}}{a x}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle -\frac {\sqrt {b x^2} \int \frac {1}{\sqrt {b x^2}}d\sqrt [3]{b x^2+a}}{2 a x}-\frac {\sqrt [3]{a+b x^2}}{a x}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}{\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b x^2+a}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a x \sqrt {-\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )^2}}}-\frac {\sqrt [3]{a+b x^2}}{a x}\) |
Input:
Int[1/(x^2*(a + b*x^2)^(2/3)),x]
Output:
-((a + b*x^2)^(1/3)/(a*x)) + (Sqrt[2 - Sqrt[3]]*(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 - S qrt[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]])/(3^(1/4)*a*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*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && 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]
\[\int \frac {1}{x^{2} \left (b \,x^{2}+a \right )^{\frac {2}{3}}}d x\]
Input:
int(1/x^2/(b*x^2+a)^(2/3),x)
Output:
int(1/x^2/(b*x^2+a)^(2/3),x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {2}{3}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(2/3),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/3)/(b*x^4 + a*x^2), x)
Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.10 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {2}{3} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {2}{3}} x} \] Input:
integrate(1/x**2/(b*x**2+a)**(2/3),x)
Output:
-hyper((-1/2, 2/3), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(2/3)*x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {2}{3}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(2/3),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(2/3)*x^2), x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {2}{3}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(2/3),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(2/3)*x^2), x)
Time = 0.47 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.15 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=-\frac {3\,{\left (\frac {a}{b\,x^2}+1\right )}^{2/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {2}{3},\frac {7}{6};\ \frac {13}{6};\ -\frac {a}{b\,x^2}\right )}{7\,x\,{\left (b\,x^2+a\right )}^{2/3}} \] Input:
int(1/(x^2*(a + b*x^2)^(2/3)),x)
Output:
-(3*(a/(b*x^2) + 1)^(2/3)*hypergeom([2/3, 7/6], 13/6, -a/(b*x^2)))/(7*x*(a + b*x^2)^(2/3))
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{2/3}} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} x^{2}}d x \] Input:
int(1/x^2/(b*x^2+a)^(2/3),x)
Output:
int(1/((a + b*x**2)**(2/3)*x**2),x)