Integrand size = 15, antiderivative size = 54 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {11}{8},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a x^3 \left (a+b x^2\right )^{3/8}} \] Output:
-1/3*(1+b*x^2/a)^(3/8)*hypergeom([-3/2, 11/8],[-1/2],-b*x^2/a)/a/x^3/(b*x^ 2+a)^(3/8)
Time = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=-\frac {\left (1+\frac {b x^2}{a}\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {11}{8},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a x^3 \left (a+b x^2\right )^{3/8}} \] Input:
Integrate[1/(x^4*(a + b*x^2)^(11/8)),x]
Output:
-1/3*((1 + (b*x^2)/a)^(3/8)*Hypergeometric2F1[-3/2, 11/8, -1/2, -((b*x^2)/ a)])/(a*x^3*(a + b*x^2)^(3/8))
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
\(\displaystyle \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/8} \int \frac {1}{x^4 \left (\frac {b x^2}{a}+1\right )^{11/8}}dx}{a \left (a+b x^2\right )^{3/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (\frac {b x^2}{a}+1\right )^{3/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {11}{8},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 a x^3 \left (a+b x^2\right )^{3/8}}\) |
Input:
Int[1/(x^4*(a + b*x^2)^(11/8)),x]
Output:
-1/3*((1 + (b*x^2)/a)^(3/8)*Hypergeometric2F1[-3/2, 11/8, -1/2, -((b*x^2)/ a)])/(a*x^3*(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])
\[\int \frac {1}{x^{4} \left (b \,x^{2}+a \right )^{\frac {11}{8}}}d x\]
Input:
int(1/x^4/(b*x^2+a)^(11/8),x)
Output:
int(1/x^4/(b*x^2+a)^(11/8),x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{8}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2+a)^(11/8),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(5/8)/(b^2*x^8 + 2*a*b*x^6 + a^2*x^4), x)
Result contains complex when optimal does not.
Time = 0.91 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {11}{8} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 a^{\frac {11}{8}} x^{3}} \] Input:
integrate(1/x**4/(b*x**2+a)**(11/8),x)
Output:
-hyper((-3/2, 11/8), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(11/8)*x**3)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{8}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2+a)^(11/8),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(11/8)*x^4), x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {11}{8}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2+a)^(11/8),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(11/8)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^{11/8}} \,d x \] Input:
int(1/(x^4*(a + b*x^2)^(11/8)),x)
Output:
int(1/(x^4*(a + b*x^2)^(11/8)), x)
\[ \int \frac {1}{x^4 \left (a+b x^2\right )^{11/8}} \, dx=\frac {-4 \left (b \,x^{2}+a \right )^{\frac {3}{8}} b +2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{8}} a \,x^{4}+\left (b \,x^{2}+a \right )^{\frac {3}{8}} b \,x^{6}}d x \right ) a^{2} x -4 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{8}} a \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {3}{8}} b \,x^{4}}d x \right ) a b x -9 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{8}} a +\left (b \,x^{2}+a \right )^{\frac {3}{8}} b \,x^{2}}d x \right ) b^{2} x}{2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a^{2} x} \] Input:
int(1/x^4/(b*x^2+a)^(11/8),x)
Output:
( - 4*(a + b*x**2)**(3/8)*b + 2*(a + b*x**2)**(3/4)*int(1/((a + b*x**2)**( 3/8)*a*x**4 + (a + b*x**2)**(3/8)*b*x**6),x)*a**2*x - 4*(a + b*x**2)**(3/4 )*int(1/((a + b*x**2)**(3/8)*a*x**2 + (a + b*x**2)**(3/8)*b*x**4),x)*a*b*x - 9*(a + b*x**2)**(3/4)*int(1/((a + b*x**2)**(3/8)*a + (a + b*x**2)**(3/8 )*b*x**2),x)*b**2*x)/(2*(a + b*x**2)**(3/4)*a**2*x)