Integrand size = 21, antiderivative size = 42 \[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\frac {2^n x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-n,\frac {3+m}{2},\frac {a^2 x^2}{4}\right )}{1+m} \]
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Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {126, 371} \[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\frac {2^n x^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-n,\frac {m+3}{2},\frac {a^2 x^2}{4}\right )}{m+1} \]
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Rule 126
Rule 371
Rubi steps \begin{align*} \text {integral}& = \int x^m \left (2-\frac {a^2 x^2}{2}\right )^n \, dx \\ & = \frac {2^n x^{1+m} \, _2F_1\left (\frac {1+m}{2},-n;\frac {3+m}{2};\frac {a^2 x^2}{4}\right )}{1+m} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\frac {2^n x^{1+m} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-n,\frac {3+m}{2},\frac {a^2 x^2}{4}\right )}{1+m} \]
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\[\int x^{m} \left (1-\frac {a x}{2}\right )^{n} \left (a x +2\right )^{n}d x\]
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\[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\int { {\left (a x + 2\right )}^{n} {\left (-\frac {1}{2} \, a x + 1\right )}^{n} x^{m} \,d x } \]
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Timed out. \[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\text {Timed out} \]
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\[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\int { {\left (a x + 2\right )}^{n} {\left (-\frac {1}{2} \, a x + 1\right )}^{n} x^{m} \,d x } \]
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\[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\int { {\left (a x + 2\right )}^{n} {\left (-\frac {1}{2} \, a x + 1\right )}^{n} x^{m} \,d x } \]
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Timed out. \[ \int x^m \left (1-\frac {a x}{2}\right )^n (2+a x)^n \, dx=\int x^m\,{\left (a\,x+2\right )}^n\,{\left (1-\frac {a\,x}{2}\right )}^n \,d x \]
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