Optimal. Leaf size=86 \[ \frac {a (e x)^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{4 e^2 (m+2)}+\frac {(e x)^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{4 e (m+1)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {82, 73, 364} \[ \frac {a (e x)^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};a^2 x^2\right )}{4 e^2 (m+2)}+\frac {(e x)^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )}{4 e (m+1)} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 73
Rule 82
Rule 364
Rubi steps
\begin {align*} \int \frac {(e x)^m}{(2-2 a x)^2 (1+a x)} \, dx &=\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*}
________________________________________________________________________________________
Mathematica [A] time = 0.04, size = 77, normalized size = 0.90 \[ \frac {x (e x)^m \left (a (m+1) x \, _2F_1\left (2,\frac {m}{2}+1;\frac {m}{2}+2;a^2 x^2\right )+(m+2) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};a^2 x^2\right )\right )}{4 (m+1) (m+2)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e x\right )^{m}}{4 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x\right )^{m}}{4 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [F] time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{m}}{\left (-2 a x +2\right )^{2} \left (a x +1\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, \int \frac {\left (e x\right )^{m}}{{\left (a x + 1\right )} {\left (a x - 1\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e\,x\right )}^m}{\left (a\,x+1\right )\,{\left (2\,a\,x-2\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [C] time = 3.25, size = 337, normalized size = 3.92 \[ \frac {2 a e^{m} m^{2} x x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} - \frac {a e^{m} m x x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac {a e^{m} m x x^{m} \Phi \left (\frac {e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac {2 a e^{m} m x x^{m} \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} - \frac {2 e^{m} m^{2} x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} + \frac {e^{m} m x^{m} \Phi \left (\frac {1}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} - \frac {e^{m} m x^{m} \Phi \left (\frac {e^{i \pi }}{a x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{16 a^{2} x \Gamma \left (1 - m\right ) - 16 a \Gamma \left (1 - m\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________