Integrand size = 17, antiderivative size = 465 \[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=\frac {\left (-a+b x^2\right )^{3/8}}{a x}-\frac {\sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} a^{3/4} x \left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )}-\frac {\sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{2 \sqrt {2+\sqrt {2}} a^{3/4} x \left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )} \] Output:
(b*x^2-a)^(3/8)/a/x-1/2*(-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^ (3/4)/x/(a^(1/4)+(b*x^2-a)^(1/4))-1/2*(-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^(3/4)/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 = 8.65 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.11 \[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=-\frac {\left (1-\frac {b x^2}{a}\right )^{5/8} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{8},\frac {1}{2},\frac {b x^2}{a}\right )}{x \left (-a+b x^2\right )^{5/8}} \] Input:
Integrate[1/(x^2*(-a + b*x^2)^(5/8)),x]
Output:
-(((1 - (b*x^2)/a)^(5/8)*Hypergeometric2F1[-1/2, 5/8, 1/2, (b*x^2)/a])/(x* (-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 = 51, normalized size of antiderivative = 0.11, 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.
\(\displaystyle \int \frac {1}{x^2 \left (b x^2-a\right )^{5/8}} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (1-\frac {b x^2}{a}\right )^{5/8} \int \frac {1}{x^2 \left (1-\frac {b x^2}{a}\right )^{5/8}}dx}{\left (b x^2-a\right )^{5/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (1-\frac {b x^2}{a}\right )^{5/8} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{8},\frac {1}{2},\frac {b x^2}{a}\right )}{x \left (b x^2-a\right )^{5/8}}\) |
Input:
Int[1/(x^2*(-a + b*x^2)^(5/8)),x]
Output:
-(((1 - (b*x^2)/a)^(5/8)*Hypergeometric2F1[-1/2, 5/8, 1/2, (b*x^2)/a])/(x* (-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])
\[\int \frac {1}{x^{2} \left (b \,x^{2}-a \right )^{\frac {5}{8}}}d x\]
Input:
int(1/x^2/(b*x^2-a)^(5/8),x)
Output:
int(1/x^2/(b*x^2-a)^(5/8),x)
\[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {5}{8}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2-a)^(5/8),x, algorithm="fricas")
Output:
integral((b*x^2 - a)^(3/8)/(b*x^4 - a*x^2), x)
Result contains complex when optimal does not.
Time = 0.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.06 \[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=\frac {e^{\frac {3 i \pi }{8}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{8} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2}}{a}} \right )}}{a^{\frac {5}{8}} x} \] Input:
integrate(1/x**2/(b*x**2-a)**(5/8),x)
Output:
exp(3*I*pi/8)*hyper((-1/2, 5/8), (1/2,), b*x**2/a)/(a**(5/8)*x)
\[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {5}{8}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2-a)^(5/8),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 - a)^(5/8)*x^2), x)
\[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {5}{8}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(b*x^2-a)^(5/8),x, algorithm="giac")
Output:
integrate(1/((b*x^2 - a)^(5/8)*x^2), x)
Time = 0.46 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.09 \[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=-\frac {4\,{\left (1-\frac {a}{b\,x^2}\right )}^{5/8}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{8},\frac {9}{8};\ \frac {17}{8};\ \frac {a}{b\,x^2}\right )}{9\,x\,{\left (b\,x^2-a\right )}^{5/8}} \] Input:
int(1/(x^2*(b*x^2 - a)^(5/8)),x)
Output:
-(4*(1 - a/(b*x^2))^(5/8)*hypergeom([5/8, 9/8], 17/8, a/(b*x^2)))/(9*x*(b* x^2 - a)^(5/8))
\[ \int \frac {1}{x^2 \left (-a+b x^2\right )^{5/8}} \, dx=\frac {-36 \left (b \,x^{2}-a \right )^{\frac {1}{8}} a +20 \left (b \,x^{2}-a \right )^{\frac {1}{8}} b \,x^{2}-45 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {x^{2}}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a -\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) b^{2} x +65 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a -\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) a b x}{36 \left (b \,x^{2}-a \right )^{\frac {3}{4}} a x} \] Input:
int(1/x^2/(b*x^2-a)^(5/8),x)
Output:
( - 36*( - a + b*x**2)**(1/8)*a + 20*( - a + b*x**2)**(1/8)*b*x**2 - 45*( - a + b*x**2)**(3/4)*int(x**2/(( - a + b*x**2)**(5/8)*a - ( - a + b*x**2)* *(5/8)*b*x**2),x)*b**2*x + 65*( - a + b*x**2)**(3/4)*int(1/(( - a + b*x**2 )**(5/8)*a - ( - a + b*x**2)**(5/8)*b*x**2),x)*a*b*x)/(36*( - a + b*x**2)* *(3/4)*a*x)