Integrand size = 15, antiderivative size = 51 \[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=-\frac {\left (a+b x^2\right )^{7/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {7}{8},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \left (1+\frac {b x^2}{a}\right )^{7/8}} \] Output:
-1/3*(b*x^2+a)^(7/8)*hypergeom([-3/2, -7/8],[-1/2],-b*x^2/a)/x^3/(1+b*x^2/ a)^(7/8)
Time = 10.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=-\frac {\left (a+b x^2\right )^{7/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {7}{8},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \left (1+\frac {b x^2}{a}\right )^{7/8}} \] Input:
Integrate[(a + b*x^2)^(7/8)/x^4,x]
Output:
-1/3*((a + b*x^2)^(7/8)*Hypergeometric2F1[-3/2, -7/8, -1/2, -((b*x^2)/a)]) /(x^3*(1 + (b*x^2)/a)^(7/8))
Time = 0.16 (sec) , antiderivative size = 51, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx\) |
\(\Big \downarrow \) 279 |
\(\displaystyle \frac {\left (a+b x^2\right )^{7/8} \int \frac {\left (\frac {b x^2}{a}+1\right )^{7/8}}{x^4}dx}{\left (\frac {b x^2}{a}+1\right )^{7/8}}\) |
\(\Big \downarrow \) 278 |
\(\displaystyle -\frac {\left (a+b x^2\right )^{7/8} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {7}{8},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \left (\frac {b x^2}{a}+1\right )^{7/8}}\) |
Input:
Int[(a + b*x^2)^(7/8)/x^4,x]
Output:
-1/3*((a + b*x^2)^(7/8)*Hypergeometric2F1[-3/2, -7/8, -1/2, -((b*x^2)/a)]) /(x^3*(1 + (b*x^2)/a)^(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])
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {7}{8}}}{x^{4}}d x\]
Input:
int((b*x^2+a)^(7/8)/x^4,x)
Output:
int((b*x^2+a)^(7/8)/x^4,x)
\[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {7}{8}}}{x^{4}} \,d x } \] Input:
integrate((b*x^2+a)^(7/8)/x^4,x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(7/8)/x^4, x)
Result contains complex when optimal does not.
Time = 1.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=- \frac {a^{\frac {7}{8}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {7}{8} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 x^{3}} \] Input:
integrate((b*x**2+a)**(7/8)/x**4,x)
Output:
-a**(7/8)*hyper((-3/2, -7/8), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*x**3)
\[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {7}{8}}}{x^{4}} \,d x } \] Input:
integrate((b*x^2+a)^(7/8)/x^4,x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(7/8)/x^4, x)
\[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {7}{8}}}{x^{4}} \,d x } \] Input:
integrate((b*x^2+a)^(7/8)/x^4,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(7/8)/x^4, x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{7/8}}{x^4} \,d x \] Input:
int((a + b*x^2)^(7/8)/x^4,x)
Output:
int((a + b*x^2)^(7/8)/x^4, x)
\[ \int \frac {\left (a+b x^2\right )^{7/8}}{x^4} \, dx=\frac {-612 \left (b \,x^{2}+a \right )^{\frac {5}{8}} a^{2}-892 \left (b \,x^{2}+a \right )^{\frac {5}{8}} a b \,x^{2}-280 \left (b \,x^{2}+a \right )^{\frac {5}{8}} b^{2} x^{4}-865 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (\int \frac {\sqrt {b \,x^{2}+a}}{\left (b \,x^{2}+a \right )^{\frac {5}{8}} a \,x^{2}+\left (b \,x^{2}+a \right )^{\frac {5}{8}} b \,x^{4}}d x \right ) a^{2} b \,x^{3}-1215 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (\int \frac {\sqrt {b \,x^{2}+a}}{\left (b \,x^{2}+a \right )^{\frac {5}{8}} a +\left (b \,x^{2}+a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) a \,b^{2} x^{3}-350 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (\int \frac {\sqrt {b \,x^{2}+a}\, x^{2}}{\left (b \,x^{2}+a \right )^{\frac {5}{8}} a +\left (b \,x^{2}+a \right )^{\frac {5}{8}} b \,x^{2}}d x \right ) b^{3} x^{3}}{1836 \left (b \,x^{2}+a \right )^{\frac {3}{4}} a \,x^{3}} \] Input:
int((b*x^2+a)^(7/8)/x^4,x)
Output:
( - 612*(a + b*x**2)**(5/8)*a**2 - 892*(a + b*x**2)**(5/8)*a*b*x**2 - 280* (a + b*x**2)**(5/8)*b**2*x**4 - 865*(a + b*x**2)**(3/4)*int(sqrt(a + b*x** 2)/((a + b*x**2)**(5/8)*a*x**2 + (a + b*x**2)**(5/8)*b*x**4),x)*a**2*b*x** 3 - 1215*(a + b*x**2)**(3/4)*int(sqrt(a + b*x**2)/((a + b*x**2)**(5/8)*a + (a + b*x**2)**(5/8)*b*x**2),x)*a*b**2*x**3 - 350*(a + b*x**2)**(3/4)*int( (sqrt(a + b*x**2)*x**2)/((a + b*x**2)**(5/8)*a + (a + b*x**2)**(5/8)*b*x** 2),x)*b**3*x**3)/(1836*(a + b*x**2)**(3/4)*a*x**3)