Integrand size = 13, antiderivative size = 35 \[ \int (b x)^m (2+d x)^n \, dx=\frac {2^n (b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d x}{2}\right )}{b (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {66} \[ \int (b x)^m (2+d x)^n \, dx=\frac {2^n (b x)^{m+1} \operatorname {Hypergeometric2F1}\left (m+1,-n,m+2,-\frac {d x}{2}\right )}{b (m+1)} \]
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Rule 66
Rubi steps \begin{align*} \text {integral}& = \frac {2^n (b x)^{1+m} \, _2F_1\left (1+m,-n;2+m;-\frac {d x}{2}\right )}{b (1+m)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int (b x)^m (2+d x)^n \, dx=\frac {2^n x (b x)^m \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d x}{2}\right )}{1+m} \]
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Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91
method | result | size |
meijerg | \(\frac {2^{n} \left (b x \right )^{m} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-n ,1+m ;2+m ;-\frac {d x}{2}\right )}{1+m}\) | \(32\) |
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\[ \int (b x)^m (2+d x)^n \, dx=\int { \left (b x\right )^{m} {\left (d x + 2\right )}^{n} \,d x } \]
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Result contains complex when optimal does not.
Time = 1.60 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int (b x)^m (2+d x)^n \, dx=\frac {2^{n} b^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - n, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{2}} \right )}}{\Gamma \left (m + 2\right )} \]
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\[ \int (b x)^m (2+d x)^n \, dx=\int { \left (b x\right )^{m} {\left (d x + 2\right )}^{n} \,d x } \]
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\[ \int (b x)^m (2+d x)^n \, dx=\int { \left (b x\right )^{m} {\left (d x + 2\right )}^{n} \,d x } \]
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Timed out. \[ \int (b x)^m (2+d x)^n \, dx=\int {\left (b\,x\right )}^m\,{\left (d\,x+2\right )}^n \,d x \]
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