Integrand size = 17, antiderivative size = 490 \[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\frac {\sqrt [8]{-a+b x^2}}{3 a x^3}+\frac {11 b \sqrt [8]{-a+b x^2}}{12 a^2 x}+\frac {11 b \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {\frac {\left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {-\frac {\sqrt [4]{a} \left (\sqrt {2}-\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{8 \sqrt {2+\sqrt {2}} a^2 x \left (\sqrt [4]{a}+\sqrt [4]{-a+b x^2}\right )}-\frac {11 b \sqrt {-\frac {b x^2}{\sqrt {a} \sqrt {-a+b x^2}}} \left (-a+b x^2\right )^{3/8} \sqrt {-\frac {\left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )^2}{\sqrt [4]{a} \sqrt [4]{-a+b x^2}}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1}{2} \sqrt {\frac {\sqrt [4]{a} \left (\sqrt {2}+\frac {2 \sqrt [4]{-a+b x^2}}{\sqrt [4]{a}}+\frac {\sqrt {2} \sqrt {-a+b x^2}}{\sqrt {a}}\right )}{\sqrt [4]{-a+b x^2}}}\right ),-2 \left (1-\sqrt {2}\right )\right )}{8 \sqrt {2+\sqrt {2}} a^2 x \left (\sqrt [4]{a}-\sqrt [4]{-a+b x^2}\right )} \] Output:
1/3*(b*x^2-a)^(1/8)/a/x^3+11/12*b*(b*x^2-a)^(1/8)/a^2/x+11/8*b*(-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^2/x/(a^(1/4)+(b*x^2-a)^(1/4))-11/8* b*(-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^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 = 0.11 \[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=-\frac {\left (1-\frac {b x^2}{a}\right )^{7/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {7}{8},-\frac {1}{2},\frac {b x^2}{a}\right )}{3 x^3 \left (-a+b x^2\right )^{7/8}} \] Input:
Integrate[1/(x^4*(-a + b*x^2)^(7/8)),x]
Output:
-1/3*((1 - (b*x^2)/a)^(7/8)*Hypergeometric2F1[-3/2, 7/8, -1/2, (b*x^2)/a]) /(x^3*(-a + b*x^2)^(7/8))
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 = 0.11, 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.
\(\displaystyle \int \frac {1}{x^4 \left (b x^2-a\right )^{7/8}} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (1-\frac {b x^2}{a}\right )^{7/8} \int \frac {1}{x^4 \left (1-\frac {b x^2}{a}\right )^{7/8}}dx}{\left (b x^2-a\right )^{7/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (1-\frac {b x^2}{a}\right )^{7/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {7}{8},-\frac {1}{2},\frac {b x^2}{a}\right )}{3 x^3 \left (b x^2-a\right )^{7/8}}\) |
Input:
Int[1/(x^4*(-a + b*x^2)^(7/8)),x]
Output:
-1/3*((1 - (b*x^2)/a)^(7/8)*Hypergeometric2F1[-3/2, 7/8, -1/2, (b*x^2)/a]) /(x^3*(-a + b*x^2)^(7/8))
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])
\[\int \frac {1}{x^{4} \left (b \,x^{2}-a \right )^{\frac {7}{8}}}d x\]
Input:
int(1/x^4/(b*x^2-a)^(7/8),x)
Output:
int(1/x^4/(b*x^2-a)^(7/8),x)
\[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {7}{8}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2-a)^(7/8),x, algorithm="fricas")
Output:
integral((b*x^2 - a)^(1/8)/(b*x^6 - a*x^4), x)
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.07 \[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\frac {e^{\frac {i \pi }{8}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {7}{8} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2}}{a}} \right )}}{3 a^{\frac {7}{8}} x^{3}} \] Input:
integrate(1/x**4/(b*x**2-a)**(7/8),x)
Output:
exp(I*pi/8)*hyper((-3/2, 7/8), (-1/2,), b*x**2/a)/(3*a**(7/8)*x**3)
\[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {7}{8}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2-a)^(7/8),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 - a)^(7/8)*x^4), x)
\[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\int { \frac {1}{{\left (b x^{2} - a\right )}^{\frac {7}{8}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2-a)^(7/8),x, algorithm="giac")
Output:
integrate(1/((b*x^2 - a)^(7/8)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2-a\right )}^{7/8}} \,d x \] Input:
int(1/(x^4*(b*x^2 - a)^(7/8)),x)
Output:
int(1/(x^4*(b*x^2 - a)^(7/8)), x)
\[ \int \frac {1}{x^4 \left (-a+b x^2\right )^{7/8}} \, dx=\frac {16 \left (b \,x^{2}-a \right )^{\frac {7}{8}} a^{2}+68 \left (b \,x^{2}-a \right )^{\frac {7}{8}} a b \,x^{2}-68 \left (b \,x^{2}-a \right )^{\frac {7}{8}} b^{2} x^{4}+17 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {\left (b \,x^{2}-a \right )^{\frac {3}{4}}}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a -\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) a \,b^{2} x^{3}-85 \left (b \,x^{2}-a \right )^{\frac {3}{4}} \left (\int \frac {\left (b \,x^{2}-a \right )^{\frac {3}{4}} x^{2}}{\left (b \,x^{2}-a \right )^{\frac {5}{8}} a -\left (b \,x^{2}-a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) b^{3} x^{3}}{48 \left (b \,x^{2}-a \right )^{\frac {3}{4}} a^{3} x^{3}} \] Input:
int(1/x^4/(b*x^2-a)^(7/8),x)
Output:
(16*( - a + b*x**2)**(7/8)*a**2 + 68*( - a + b*x**2)**(7/8)*a*b*x**2 - 68* ( - a + b*x**2)**(7/8)*b**2*x**4 + 17*( - a + b*x**2)**(3/4)*int(( - a + b *x**2)**(3/4)/(( - a + b*x**2)**(5/8)*a - ( - a + b*x**2)**(5/8)*b*x**2),x )*a*b**2*x**3 - 85*( - a + b*x**2)**(3/4)*int((( - a + b*x**2)**(3/4)*x**2 )/(( - a + b*x**2)**(5/8)*a - ( - a + b*x**2)**(5/8)*b*x**2),x)*b**3*x**3) /(48*( - a + b*x**2)**(3/4)*a**3*x**3)