Integrand size = 22, antiderivative size = 84 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=-\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (-\frac {3}{2},-p,-q,-\frac {1}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 x^3} \]
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Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {525, 524} \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=-\frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (c+d x^2\right )^q \left (\frac {d x^2}{c}+1\right )^{-q} \operatorname {AppellF1}\left (-\frac {3}{2},-p,-q,-\frac {1}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 x^3} \]
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Rule 524
Rule 525
Rubi steps \begin{align*} \text {integral}& = \left (\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p \left (c+d x^2\right )^q}{x^4} \, dx \\ & = \left (\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q}\right ) \int \frac {\left (1+\frac {b x^2}{a}\right )^p \left (1+\frac {d x^2}{c}\right )^q}{x^4} \, dx \\ & = -\frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (1+\frac {d x^2}{c}\right )^{-q} F_1\left (-\frac {3}{2};-p,-q;-\frac {1}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 x^3} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=-\frac {\left (a+b x^2\right )^p \left (\frac {a+b x^2}{a}\right )^{-p} \left (c+d x^2\right )^q \left (\frac {c+d x^2}{c}\right )^{-q} \operatorname {AppellF1}\left (-\frac {3}{2},-p,-q,-\frac {1}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{3 x^3} \]
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\[\int \frac {\left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{q}}{x^{4}}d x\]
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\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{x^{4}} \,d x } \]
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\[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{p} {\left (d x^{2} + c\right )}^{q}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x^4} \, dx=\int \frac {{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q}{x^4} \,d x \]
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