Integrand size = 15, antiderivative size = 266 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {\sqrt [6]{a+b x^2}}{a x}-\frac {\sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} a^{4/3} x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}} \] Output:
-(b*x^2+a)^(1/6)/a/x-1/3*(b*x^2+a)^(1/6)*(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))/(a^(1/3)-(1+3^(1/2))*(b*x^2+a) ^(1/3))^2)^(1/2)*InverseJacobiAM(arccos((a^(1/3)-(1-3^(1/2))*(b*x^2+a)^(1/ 3))/(a^(1/3)-(1+3^(1/2))*(b*x^2+a)^(1/3))),1/4*6^(1/2)+1/4*2^(1/2))*3^(3/4 )/a^(4/3)/x/(-(b*x^2+a)^(1/3)*(a^(1/3)-(b*x^2+a)^(1/3))/(a^(1/3)-(1+3^(1/2 ))*(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 = 7.50 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.18 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{6},\frac {1}{2},-\frac {b x^2}{a}\right )}{x \left (a+b x^2\right )^{5/6}} \] Input:
Integrate[1/(x^2*(a + b*x^2)^(5/6)),x]
Output:
-(((1 + (b*x^2)/a)^(5/6)*Hypergeometric2F1[-1/2, 5/6, 1/2, -((b*x^2)/a)])/ (x*(a + b*x^2)^(5/6)))
Time = 0.27 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.33, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {264, 236, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {2 b \int \frac {1}{\left (b x^2+a\right )^{5/6}}dx}{3 a}-\frac {\sqrt [6]{a+b x^2}}{a x}\) |
\(\Big \downarrow \) 236 |
\(\displaystyle -\frac {2 b \int \frac {1}{\left (1-\frac {b x^2}{b x^2+a}\right )^{2/3}}d\frac {x}{\sqrt {b x^2+a}}}{3 a \sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}-\frac {\sqrt [6]{a+b x^2}}{a x}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle \frac {\sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{a x \sqrt [3]{\frac {a}{a+b x^2}}}-\frac {\sqrt [6]{a+b x^2}}{a x}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle -\frac {2 \sqrt {2-\sqrt {3}} \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} a x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}-\frac {\sqrt [6]{a+b x^2}}{a x}\) |
Input:
Int[1/(x^2*(a + b*x^2)^(5/6)),x]
Output:
-((a + b*x^2)^(1/6)/(a*x)) - (2*Sqrt[2 - Sqrt[3]]*Sqrt[-((b*x^2)/(a + b*x^ 2))]*(a + b*x^2)^(1/6)*(1 - (1 - (b*x^2)/(a + b*x^2))^(1/3))*Sqrt[(1 + x^2 /(a + b*x^2) + (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2 )/(a + b*x^2))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))], -7 + 4*S qrt[3]])/(3^(1/4)*a*x*(a/(a + b*x^2))^(1/3)*Sqrt[-1 + x^3/(a + b*x^2)^(3/2 )]*Sqrt[-((1 - (1 - (b*x^2)/(a + b*x^2))^(1/3))/(1 - Sqrt[3] - (1 - (b*x^2 )/(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[((a_) + (b_.)*(x_)^2)^(-5/6), x_Symbol] :> Simp[1/((a/(a + b*x^2))^(1/3 )*(a + b*x^2)^(1/3)) Subst[Int[1/(1 - b*x^2)^(2/3), x], x, x/Sqrt[a + b*x ^2]], 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]
\[\int \frac {1}{x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{6}}}d x\]
Input:
int(1/x^2/(b*x^2+a)^(5/6),x)
Output:
int(1/x^2/(b*x^2+a)^(5/6),x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(5/6),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/6)/(b*x^4 + a*x^2), x)
Result contains complex when optimal does not.
Time = 0.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.10 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{6} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{6}} x} \] Input:
integrate(1/x**2/(b*x**2+a)**(5/6),x)
Output:
-hyper((-1/2, 5/6), (1/2,), b*x**2*exp_polar(I*pi)/a)/(a**(5/6)*x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(5/6),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(5/6)*x^2), x)
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2+a)^(5/6),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(5/6)*x^2), x)
Time = 0.49 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.15 \[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=-\frac {3\,{\left (\frac {a}{b\,x^2}+1\right )}^{5/6}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{6},\frac {4}{3};\ \frac {7}{3};\ -\frac {a}{b\,x^2}\right )}{8\,x\,{\left (b\,x^2+a\right )}^{5/6}} \] Input:
int(1/(x^2*(a + b*x^2)^(5/6)),x)
Output:
-(3*(a/(b*x^2) + 1)^(5/6)*hypergeom([5/6, 4/3], 7/3, -a/(b*x^2)))/(8*x*(a + b*x^2)^(5/6))
\[ \int \frac {1}{x^2 \left (a+b x^2\right )^{5/6}} \, dx=\frac {-\left (b \,x^{2}+a \right )^{\frac {5}{6}}-2 \left (b \,x^{2}+a \right )^{\frac {2}{3}} \left (\int \frac {\left (b \,x^{2}+a \right )^{\frac {7}{6}}}{b^{2} x^{4}+2 a b \,x^{2}+a^{2}}d x \right ) b x}{\left (b \,x^{2}+a \right )^{\frac {2}{3}} a x} \] Input:
int(1/x^2/(b*x^2+a)^(5/6),x)
Output:
( - (a + b*x**2)**(5/6) - 2*(a + b*x**2)**(2/3)*int((a + b*x**2)**(7/6)/(a **2 + 2*a*b*x**2 + b**2*x**4),x)*b*x)/((a + b*x**2)**(2/3)*a*x)