Integrand size = 11, antiderivative size = 251 \[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\frac {3^{3/4} \sqrt [6]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{a+b x^2}+\left (a+b x^2\right )^{2/3}}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{a}-\left (1-\sqrt {3}\right ) \sqrt [3]{a+b x^2}}{\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [3]{a} b x \sqrt {-\frac {\sqrt [3]{a+b x^2} \left (\sqrt [3]{a}-\sqrt [3]{a+b x^2}\right )}{\left (\sqrt [3]{a}-\left (1+\sqrt {3}\right ) \sqrt [3]{a+b x^2}\right )^2}}} \] Output:
1/2*3^(3/4)*(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))/a^(1/3)/b/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 = 6.74 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.18 \[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\frac {x \left (1+\frac {b x^2}{a}\right )^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},-\frac {b x^2}{a}\right )}{\left (a+b x^2\right )^{5/6}} \] Input:
Integrate[(a + b*x^2)^(-5/6),x]
Output:
(x*(1 + (b*x^2)/a)^(5/6)*Hypergeometric2F1[1/2, 5/6, 3/2, -((b*x^2)/a)])/( a + b*x^2)^(5/6)
Time = 0.23 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.32, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {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.
\(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx\) |
\(\Big \downarrow \) 236 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\frac {b x^2}{b x^2+a}\right )^{2/3}}d\frac {x}{\sqrt {b x^2+a}}}{\sqrt [3]{\frac {a}{a+b x^2}} \sqrt [3]{a+b x^2}}\) |
\(\Big \downarrow \) 234 |
\(\displaystyle -\frac {3 \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \int \frac {1}{\sqrt {\frac {x^3}{\left (b x^2+a\right )^{3/2}}-1}}d\sqrt [3]{1-\frac {b x^2}{b x^2+a}}}{2 b x \sqrt [3]{\frac {a}{a+b x^2}}}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {3^{3/4} \sqrt {2-\sqrt {3}} \sqrt {-\frac {b x^2}{a+b x^2}} \sqrt [6]{a+b x^2} \left (1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}\right ) \sqrt {\frac {\frac {x^2}{a+b x^2}+\sqrt [3]{1-\frac {b x^2}{a+b x^2}}+1}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}+\sqrt {3}+1}{-\sqrt [3]{1-\frac {b x^2}{b x^2+a}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{b x \sqrt [3]{\frac {a}{a+b x^2}} \sqrt {\frac {x^3}{\left (a+b x^2\right )^{3/2}}-1} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}}{\left (-\sqrt [3]{1-\frac {b x^2}{a+b x^2}}-\sqrt {3}+1\right )^2}}}\) |
Input:
Int[(a + b*x^2)^(-5/6),x]
Output:
(3^(3/4)*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 - Sq rt[3] - (1 - (b*x^2)/(a + b*x^2))^(1/3))], -7 + 4*Sqrt[3]])/(b*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[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]
\[\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {5}{6}}}d x\]
Input:
int(1/(b*x^2+a)^(5/6),x)
Output:
int(1/(b*x^2+a)^(5/6),x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(5/6),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(-5/6), x)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.10 \[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\frac {x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{6} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{a^{\frac {5}{6}}} \] Input:
integrate(1/(b*x**2+a)**(5/6),x)
Output:
x*hyper((1/2, 5/6), (3/2,), b*x**2*exp_polar(I*pi)/a)/a**(5/6)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(5/6),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(-5/6), x)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{6}}} \,d x } \] Input:
integrate(1/(b*x^2+a)^(5/6),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(-5/6), x)
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.15 \[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\frac {x\,{\left (\frac {b\,x^2}{a}+1\right )}^{5/6}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{6};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (b\,x^2+a\right )}^{5/6}} \] Input:
int(1/(a + b*x^2)^(5/6),x)
Output:
(x*((b*x^2)/a + 1)^(5/6)*hypergeom([1/2, 5/6], 3/2, -(b*x^2)/a))/(a + b*x^ 2)^(5/6)
\[ \int \frac {1}{\left (a+b x^2\right )^{5/6}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {2}{3}}}{\sqrt {b \,x^{2}+a}\, a +\sqrt {b \,x^{2}+a}\, b \,x^{2}}d x \] Input:
int(1/(b*x^2+a)^(5/6),x)
Output:
int((a + b*x**2)**(2/3)/(sqrt(a + b*x**2)*a + sqrt(a + b*x**2)*b*x**2),x)