Integrand size = 17, antiderivative size = 524 \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\frac {\left (-a+b x^2\right )^{3/8}}{5 a x^5}+\frac {17 b \left (-a+b x^2\right )^{3/8}}{60 a^2 x^3}+\frac {51 b^2 \left (-a+b x^2\right )^{3/8}}{80 a^3 x}-\frac {51 b^2 \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{160 \sqrt {2+\sqrt {2}} a^{11/4} x \left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )}-\frac {51 b^2 \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{160 \sqrt {2+\sqrt {2}} a^{11/4} x \left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )} \] Output:
1/5*(b*x^2-a)^(3/8)/a/x^5+17/60*b*(b*x^2-a)^(3/8)/a^2/x^3+51/80*b^2*(b*x^2 -a)^(3/8)/a^3/x-51/160*b^2*(-b*x^2/a^(1/2)/(b*x^2-a)^(1/2))^(1/2)*(b*x^2-a )^(3/8)*((a^(1/4)+(b*x^2-a)^(1/4))^2/a^(1/4)/(b*x^2-a)^(1/4))^(1/2)*Ellipt icF(1/2*(-a^(1/4)*(2^(1/2)-2*(b*x^2-a)^(1/4)/a^(1/4)+2^(1/2)*(b*x^2-a)^(1/ 2)/a^(1/2))/(b*x^2-a)^(1/4))^(1/2),(-2+2*2^(1/2))^(1/2))/(2+2^(1/2))^(1/2) /a^(11/4)/x/(a^(1/4)+(b*x^2-a)^(1/4))-51/160*b^2*(-b*x^2/a^(1/2)/(b*x^2-a) ^(1/2))^(1/2)*(b*x^2-a)^(3/8)*(-(a^(1/4)-(b*x^2-a)^(1/4))^2/a^(1/4)/(b*x^2 -a)^(1/4))^(1/2)*EllipticF(1/2*(a^(1/4)*(2^(1/2)+2*(b*x^2-a)^(1/4)/a^(1/4) +2^(1/2)*(b*x^2-a)^(1/2)/a^(1/2))/(b*x^2-a)^(1/4))^(1/2),(-2+2*2^(1/2))^(1 /2))/(2+2^(1/2))^(1/2)/a^(11/4)/x/(a^(1/4)-(b*x^2-a)^(1/4))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.10 \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=-\frac {\left (1-\frac {b x^2}{a}\right )^{5/8} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {5}{8},-\frac {3}{2},\frac {b x^2}{a}\right )}{5 x^5 \left (-a+b x^2\right )^{5/8}} \] Input:
Integrate[1/(x^6*(-a + b*x^2)^(5/8)),x]
Output:
-1/5*((1 - (b*x^2)/a)^(5/8)*Hypergeometric2F1[-5/2, 5/8, -3/2, (b*x^2)/a]) /(x^5*(-a + b*x^2)^(5/8))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {279, 278}
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.
\(\displaystyle \int \frac {1}{x^6 \left (b x^2-a\right )^{5/8}} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (1-\frac {b x^2}{a}\right )^{5/8} \int \frac {1}{x^6 \left (1-\frac {b x^2}{a}\right )^{5/8}}dx}{\left (b x^2-a\right )^{5/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (1-\frac {b x^2}{a}\right )^{5/8} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {5}{8},-\frac {3}{2},\frac {b x^2}{a}\right )}{5 x^5 \left (b x^2-a\right )^{5/8}}\) |
Input:
Int[1/(x^6*(-a + b*x^2)^(5/8)),x]
Output:
-1/5*((1 - (b*x^2)/a)^(5/8)*Hypergeometric2F1[-5/2, 5/8, -3/2, (b*x^2)/a]) /(x^5*(-a + b*x^2)^(5/8))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[(c*x)^m* (1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && !(ILtQ[p, 0] || GtQ[a, 0])
\[\int \frac {1}{x^{6} \left (b \,x^{2}-a \right )^{\frac {5}{8}}}d x\]
Input:
int(1/x^6/(b*x^2-a)^(5/8),x)
Output:
int(1/x^6/(b*x^2-a)^(5/8),x)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {5}{8}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2-a)^(5/8),x, algorithm="fricas")
Output:
integral((b*x^2 - a)^(3/8)/(b*x^8 - a*x^6), x)
Result contains complex when optimal does not.
Time = 0.74 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.06 \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\frac {e^{\frac {3 i \pi }{8}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {5}{8} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2}}{a}} \right )}}{5 a^{\frac {5}{8}} x^{5}} \] Input:
integrate(1/x**6/(b*x**2-a)**(5/8),x)
Output:
exp(3*I*pi/8)*hyper((-5/2, 5/8), (-3/2,), b*x**2/a)/(5*a**(5/8)*x**5)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {5}{8}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2-a)^(5/8),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 - a)^(5/8)*x^6), x)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {5}{8}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2-a)^(5/8),x, algorithm="giac")
Output:
integrate(1/((b*x^2 - a)^(5/8)*x^6), x)
Timed out. \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^2-a\right )}^{5/8}} \,d x \] Input:
int(1/(x^6*(b*x^2 - a)^(5/8)),x)
Output:
int(1/(x^6*(b*x^2 - a)^(5/8)), x)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{5/8}} \, dx=\frac {-1360 \left (b \,x^{2}-a \right )^{\frac {1}{8}} a^{2}+1892 \left (b \,x^{2}-a \right )^{\frac {1}{8}} a b \,x^{2}+1440 \left (b \,x^{2}-a \right )^{\frac {1}{8}} b^{2} x^{4}+7920 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a \,x^{4}-\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{6}}d x \right ) a^{2} b \,x^{5}-10385 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a \,x^{2}-\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{4}}d x \right ) a \,b^{2} x^{5}}{6800 \left (b \,x^{2}-a \right )^{\frac {3}{4}} a^{2} x^{5}} \] Input:
int(1/x^6/(b*x^2-a)^(5/8),x)
Output:
( - 1360*( - a + b*x**2)**(1/8)*a**2 + 1892*( - a + b*x**2)**(1/8)*a*b*x** 2 + 1440*( - a + b*x**2)**(1/8)*b**2*x**4 + 7920*( - a + b*x**2)**(3/4)*in t(1/(( - a + b*x**2)**(5/8)*a*x**4 - ( - a + b*x**2)**(5/8)*b*x**6),x)*a** 2*b*x**5 - 10385*( - a + b*x**2)**(3/4)*int(1/(( - a + b*x**2)**(5/8)*a*x* *2 - ( - a + b*x**2)**(5/8)*b*x**4),x)*a*b**2*x**5)/(6800*( - a + b*x**2)* *(3/4)*a**2*x**5)