Integrand size = 17, antiderivative size = 547 \[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=-\frac {\left (-a+b x^2\right )^{3/8}}{7 x^7}+\frac {3 b \left (-a+b x^2\right )^{3/8}}{140 a x^5}+\frac {17 b^2 \left (-a+b x^2\right )^{3/8}}{560 a^2 x^3}+\frac {153 b^3 \left (-a+b x^2\right )^{3/8}}{2240 a^3 x}-\frac {153 b^3 \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{4480 \sqrt {2+\sqrt {2}} a^{11/4} x \left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )}-\frac {153 b^3 \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{4480 \sqrt {2+\sqrt {2}} a^{11/4} x \left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )} \] Output:
-1/7*(b*x^2-a)^(3/8)/x^7+3/140*b*(b*x^2-a)^(3/8)/a/x^5+17/560*b^2*(b*x^2-a )^(3/8)/a^2/x^3+153/2240*b^3*(b*x^2-a)^(3/8)/a^3/x-153/4480*b^3*(-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)) -153/4480*b^3*(-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 \[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=-\frac {\left (-a+b x^2\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},-\frac {3}{8},-\frac {5}{2},\frac {b x^2}{a}\right )}{7 x^7 \left (1-\frac {b x^2}{a}\right )^{3/8}} \] Input:
Integrate[(-a + b*x^2)^(3/8)/x^8,x]
Output:
-1/7*((-a + b*x^2)^(3/8)*Hypergeometric2F1[-7/2, -3/8, -5/2, (b*x^2)/a])/( x^7*(1 - (b*x^2)/a)^(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 = 0.10, 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.
\(\displaystyle \int \frac {\left (b x^2-a\right )^{3/8}}{x^8} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (b x^2-a\right )^{3/8} \int \frac {\left (1-\frac {b x^2}{a}\right )^{3/8}}{x^8}dx}{\left (1-\frac {b x^2}{a}\right )^{3/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (b x^2-a\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},-\frac {3}{8},-\frac {5}{2},\frac {b x^2}{a}\right )}{7 x^7 \left (1-\frac {b x^2}{a}\right )^{3/8}}\) |
Input:
Int[(-a + b*x^2)^(3/8)/x^8,x]
Output:
-1/7*((-a + b*x^2)^(3/8)*Hypergeometric2F1[-7/2, -3/8, -5/2, (b*x^2)/a])/( x^7*(1 - (b*x^2)/a)^(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])
\[\int \frac {\left (b \,x^{2}-a \right )^{\frac {3}{8}}}{x^{8}}d x\]
Input:
int((b*x^2-a)^(3/8)/x^8,x)
Output:
int((b*x^2-a)^(3/8)/x^8,x)
\[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=\int { \frac {{\left (b x^{2} - a\right )}^{\frac {3}{8}}}{x^{8}} \,d x } \] Input:
integrate((b*x^2-a)^(3/8)/x^8,x, algorithm="fricas")
Output:
integral((b*x^2 - a)^(3/8)/x^8, x)
Result contains complex when optimal does not.
Time = 0.97 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.07 \[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=\frac {a^{\frac {3}{8}} e^{- \frac {5 i \pi }{8}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{2}, - \frac {3}{8} \\ - \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2}}{a}} \right )}}{7 x^{7}} \] Input:
integrate((b*x**2-a)**(3/8)/x**8,x)
Output:
a**(3/8)*exp(-5*I*pi/8)*hyper((-7/2, -3/8), (-5/2,), b*x**2/a)/(7*x**7)
\[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=\int { \frac {{\left (b x^{2} - a\right )}^{\frac {3}{8}}}{x^{8}} \,d x } \] Input:
integrate((b*x^2-a)^(3/8)/x^8,x, algorithm="maxima")
Output:
integrate((b*x^2 - a)^(3/8)/x^8, x)
\[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=\int { \frac {{\left (b x^{2} - a\right )}^{\frac {3}{8}}}{x^{8}} \,d x } \] Input:
integrate((b*x^2-a)^(3/8)/x^8,x, algorithm="giac")
Output:
integrate((b*x^2 - a)^(3/8)/x^8, x)
Timed out. \[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=\int \frac {{\left (b\,x^2-a\right )}^{3/8}}{x^8} \,d x \] Input:
int((b*x^2 - a)^(3/8)/x^8,x)
Output:
int((b*x^2 - a)^(3/8)/x^8, x)
\[ \int \frac {\left (-a+b x^2\right )^{3/8}}{x^8} \, dx=\frac {80 \left (b \,x^{2}-a \right )^{\frac {1}{8}} a^{2}-60 \left (b \,x^{2}-a \right )^{\frac {1}{8}} a b \,x^{2}-4 \left (b \,x^{2}-a \right )^{\frac {1}{8}} b^{2} x^{4}-45 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a \,x^{4}-\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{6}}d x \right ) a \,b^{2} x^{7}+25 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a \,x^{2}-\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{4}}d x \right ) b^{3} x^{7}}{560 \left (b \,x^{2}-a \right )^{\frac {3}{4}} a \,x^{7}} \] Input:
int((b*x^2-a)^(3/8)/x^8,x)
Output:
(80*( - a + b*x**2)**(1/8)*a**2 - 60*( - a + b*x**2)**(1/8)*a*b*x**2 - 4*( - a + b*x**2)**(1/8)*b**2*x**4 - 45*( - a + b*x**2)**(3/4)*int(1/(( - a + b*x**2)**(5/8)*a*x**4 - ( - a + b*x**2)**(5/8)*b*x**6),x)*a*b**2*x**7 + 2 5*( - 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)*b**3*x**7)/(560*( - a + b*x**2)**(3/4)*a*x**7)