Integrand size = 21, antiderivative size = 86 \[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )}{4 e (1+m)}+\frac {a (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},a^2 x^2\right )}{4 e^2 (2+m)} \]
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Time = 0.02 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {83, 74, 371} \[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\frac {a (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (2,\frac {m+2}{2},\frac {m+4}{2},a^2 x^2\right )}{4 e^2 (m+2)}+\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},a^2 x^2\right )}{4 e (m+1)} \]
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Rule 74
Rule 83
Rule 371
Rubi steps \begin{align*} \text {integral}& = \frac {a \int \frac {(e x)^{1+m}}{(2-2 a x)^2 (1+a x)^2} \, dx}{e}+\int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)^2} \, dx \\ & = \frac {a \int \frac {(e x)^{1+m}}{\left (2-2 a^2 x^2\right )^2} \, dx}{e}+\int \frac {(e x)^m}{\left (2-2 a^2 x^2\right )^2} \, dx \\ & = \frac {(e x)^{1+m} \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};a^2 x^2\right )}{4 e (1+m)}+\frac {a (e x)^{2+m} \, _2F_1\left (2,\frac {2+m}{2};\frac {4+m}{2};a^2 x^2\right )}{4 e^2 (2+m)} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\frac {x (e x)^m \left (a (1+m) x \operatorname {Hypergeometric2F1}\left (2,1+\frac {m}{2},2+\frac {m}{2},a^2 x^2\right )+(2+m) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},a^2 x^2\right )\right )}{4 (1+m) (2+m)} \]
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\[\int \frac {\left (e x \right )^{m}}{\left (-2 a x +2\right )^{2} \left (a x +1\right )}d x\]
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\[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\int { \frac {\left (e x\right )^{m}}{4 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2}} \,d x } \]
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Result contains complex when optimal does not.
Time = 1.28 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.92 \[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\frac {2 a e^{m} m^{2} x x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} - \frac {a e^{m} m x x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac {a e^{m} m x x^{m} \Phi \left (\frac {e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac {2 a e^{m} m x x^{m} \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} - \frac {2 e^{m} m^{2} x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac {e^{m} m x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} - \frac {e^{m} m x^{m} \Phi \left (\frac {e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} \]
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\[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\int { \frac {\left (e x\right )^{m}}{4 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2}} \,d x } \]
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\[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\int { \frac {\left (e x\right )^{m}}{4 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx=\int \frac {{\left (e\,x\right )}^m}{\left (a\,x+1\right )\,{\left (2\,a\,x-2\right )}^2} \,d x \]
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