Integrand size = 22, antiderivative size = 295 \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,1,2+m,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,1,2+m,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,2,2+m,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} \operatorname {AppellF1}\left (1+m,-n,2,2+m,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)} \]
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Time = 0.28 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {975, 140, 138, 926} \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,1,m+2,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,1,m+2,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,2,m+2,-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)}+\frac {(g x)^{m+1} (d+e x)^n \left (\frac {e x}{d}+1\right )^{-n} \operatorname {AppellF1}\left (m+1,-n,2,m+2,-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (m+1)} \]
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Rule 138
Rule 140
Rule 926
Rule 975
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {c (g x)^m (d+e x)^n}{4 a \left (\sqrt {-a} \sqrt {c}-c x\right )^2}-\frac {c (g x)^m (d+e x)^n}{4 a \left (\sqrt {-a} \sqrt {c}+c x\right )^2}-\frac {c (g x)^m (d+e x)^n}{2 a \left (-a c-c^2 x^2\right )}\right ) \, dx \\ & = -\frac {c \int \frac {(g x)^m (d+e x)^n}{\left (\sqrt {-a} \sqrt {c}-c x\right )^2} \, dx}{4 a}-\frac {c \int \frac {(g x)^m (d+e x)^n}{\left (\sqrt {-a} \sqrt {c}+c x\right )^2} \, dx}{4 a}-\frac {c \int \frac {(g x)^m (d+e x)^n}{-a c-c^2 x^2} \, dx}{2 a} \\ & = -\frac {c \int \left (-\frac {\sqrt {-a} (g x)^m (d+e x)^n}{2 a c \left (\sqrt {-a}-\sqrt {c} x\right )}-\frac {\sqrt {-a} (g x)^m (d+e x)^n}{2 a c \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a}-\frac {\left (c (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n}\right ) \int \frac {(g x)^m \left (1+\frac {e x}{d}\right )^n}{\left (\sqrt {-a} \sqrt {c}-c x\right )^2} \, dx}{4 a}-\frac {\left (c (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n}\right ) \int \frac {(g x)^m \left (1+\frac {e x}{d}\right )^n}{\left (\sqrt {-a} \sqrt {c}+c x\right )^2} \, dx}{4 a} \\ & = \frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {\int \frac {(g x)^m (d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 (-a)^{3/2}}+\frac {\int \frac {(g x)^m (d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 (-a)^{3/2}} \\ & = \frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {\left ((d+e x)^n \left (1+\frac {e x}{d}\right )^{-n}\right ) \int \frac {(g x)^m \left (1+\frac {e x}{d}\right )^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 (-a)^{3/2}}+\frac {\left ((d+e x)^n \left (1+\frac {e x}{d}\right )^{-n}\right ) \int \frac {(g x)^m \left (1+\frac {e x}{d}\right )^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 (-a)^{3/2}} \\ & = \frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,1;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},-\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)}+\frac {(g x)^{1+m} (d+e x)^n \left (1+\frac {e x}{d}\right )^{-n} F_1\left (1+m;-n,2;2+m;-\frac {e x}{d},\frac {\sqrt {c} x}{\sqrt {-a}}\right )}{4 a^2 g (1+m)} \\ \end{align*}
\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx \]
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\[\int \frac {\left (g x \right )^{m} \left (e x +d \right )^{n}}{\left (c \,x^{2}+a \right )^{2}}d x\]
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\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\text {Timed out} \]
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\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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\[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n} \left (g x\right )^{m}}{{\left (c x^{2} + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(g x)^m (d+e x)^n}{\left (a+c x^2\right )^2} \, dx=\int \frac {{\left (g\,x\right )}^m\,{\left (d+e\,x\right )}^n}{{\left (c\,x^2+a\right )}^2} \,d x \]
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