Integrand size = 17, antiderivative size = 974 \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx =\text {Too large to display} \] Output:
1/5*(b*x^2-a)^(5/8)/a/x^5+1/4*b*(b*x^2-a)^(5/8)/a^2/x^3+7/16*b^2*(b*x^2-a) ^(5/8)/a^3/x+7/32*(2+2^(1/2))^(1/2)*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)*EllipticE(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))/a^(5 /2)/x/(a^(1/4)+(b*x^2-a)^(1/4))-7/32*(2+2^(1/2))^(1/2)*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)*EllipticE(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))/a^(5/2)/x/(a^(1/4)-(b*x^2-a)^(1/4))-7/32*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^(5/2)/x/(a^(1/4)+(b*x^2-a)^(1/4))+7/32 *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^(5/2)/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 = 0.05 \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=-\frac {\left (1-\frac {b x^2}{a}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {3}{8},-\frac {3}{2},\frac {b x^2}{a}\right )}{5 x^5 \left (-a+b x^2\right )^{3/8}} \] Input:
Integrate[1/(x^6*(-a + b*x^2)^(3/8)),x]
Output:
-1/5*((1 - (b*x^2)/a)^(3/8)*Hypergeometric2F1[-5/2, 3/8, -3/2, (b*x^2)/a]) /(x^5*(-a + b*x^2)^(3/8))
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 = 0.05, 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.
\(\displaystyle \int \frac {1}{x^6 \left (b x^2-a\right )^{3/8}} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (1-\frac {b x^2}{a}\right )^{3/8} \int \frac {1}{x^6 \left (1-\frac {b x^2}{a}\right )^{3/8}}dx}{\left (b x^2-a\right )^{3/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (1-\frac {b x^2}{a}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {3}{8},-\frac {3}{2},\frac {b x^2}{a}\right )}{5 x^5 \left (b x^2-a\right )^{3/8}}\) |
Input:
Int[1/(x^6*(-a + b*x^2)^(3/8)),x]
Output:
-1/5*((1 - (b*x^2)/a)^(3/8)*Hypergeometric2F1[-5/2, 3/8, -3/2, (b*x^2)/a]) /(x^5*(-a + b*x^2)^(3/8))
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])
\[\int \frac {1}{x^{6} \left (b \,x^{2}-a \right )^{\frac {3}{8}}}d x\]
Input:
int(1/x^6/(b*x^2-a)^(3/8),x)
Output:
int(1/x^6/(b*x^2-a)^(3/8),x)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {3}{8}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2-a)^(3/8),x, algorithm="fricas")
Output:
integral((b*x^2 - a)^(5/8)/(b*x^8 - a*x^6), x)
Result contains complex when optimal does not.
Time = 0.76 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.03 \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=\frac {e^{\frac {5 i \pi }{8}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {3}{8} \\ - \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2}}{a}} \right )}}{5 a^{\frac {3}{8}} x^{5}} \] Input:
integrate(1/x**6/(b*x**2-a)**(3/8),x)
Output:
exp(5*I*pi/8)*hyper((-5/2, 3/8), (-3/2,), b*x**2/a)/(5*a**(3/8)*x**5)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {3}{8}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2-a)^(3/8),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 - a)^(3/8)*x^6), x)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {3}{8}} x^{6}} \,d x } \] Input:
integrate(1/x^6/(b*x^2-a)^(3/8),x, algorithm="giac")
Output:
integrate(1/((b*x^2 - a)^(3/8)*x^6), x)
Timed out. \[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=\int \frac {1}{x^6\,{\left (b\,x^2-a\right )}^{3/8}} \,d x \] Input:
int(1/(x^6*(b*x^2 - a)^(3/8)),x)
Output:
int(1/(x^6*(b*x^2 - a)^(3/8)), x)
\[ \int \frac {1}{x^6 \left (-a+b x^2\right )^{3/8}} \, dx=\frac {-240 \left (b \,x^{2}-a \right )^{\frac {3}{8}} a^{2}-92 \left (b \,x^{2}-a \right )^{\frac {3}{8}} a b \,x^{2}+140 \left (b \,x^{2}-a \right )^{\frac {3}{8}} b^{2} x^{4}+485 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {\left (b \,x^{2}-a \right )^{\frac {1}{4}}}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a \,x^{2}-\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{4}}d x \right ) a \,b^{2} x^{5}-245 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {\left (b \,x^{2}-a \right )^{\frac {1}{4}}}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a -\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) b^{3} x^{5}}{1200 \left (b \,x^{2}-a \right )^{\frac {3}{4}} a^{2} x^{5}} \] Input:
int(1/x^6/(b*x^2-a)^(3/8),x)
Output:
( - 240*( - a + b*x**2)**(3/8)*a**2 - 92*( - a + b*x**2)**(3/8)*a*b*x**2 + 140*( - a + b*x**2)**(3/8)*b**2*x**4 + 485*( - a + b*x**2)**(3/4)*int(( - a + b*x**2)**(1/4)/(( - a + b*x**2)**(5/8)*a*x**2 - ( - a + b*x**2)**(5/8 )*b*x**4),x)*a*b**2*x**5 - 245*( - a + b*x**2)**(3/4)*int(( - a + b*x**2)* *(1/4)/(( - a + b*x**2)**(5/8)*a - ( - a + b*x**2)**(5/8)*b*x**2),x)*b**3* x**5)/(1200*( - a + b*x**2)**(3/4)*a**2*x**5)