Integrand size = 17, antiderivative size = 44 \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {x^{-m (1+p)} \left (a x^m+b x^{1+m+m p}\right )^{1+p}}{b (1+p) (1+m p)} \]
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Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2004, 2025} \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {x^{-m (p+1)} \left (a x^m+b x^{m p+m+1}\right )^{p+1}}{b (p+1) (m p+1)} \]
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Rule 2004
Rule 2025
Rubi steps \begin{align*} \text {integral}& = \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx \\ & = \frac {x^{-m (1+p)} \left (a x^m+b x^{1+m+m p}\right )^{1+p}}{b (1+p) (1+m p)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {x^{-m (1+p)} \left (x^m \left (a+b x^{1+m p}\right )\right )^{1+p}}{b (1+p) (1+m p)} \]
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\[\int \left (x^{m} \left (a +b \,x^{m p +1}\right )\right )^{p}d x\]
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none
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.39 \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\frac {{\left (b x x^{m p + 1} + a x\right )} {\left (b x^{m p + 1} x^{m} + a x^{m}\right )}^{p}}{{\left (b m p^{2} + {\left (b m + b\right )} p + b\right )} x^{m p + 1}} \]
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\[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int \left (x^{m} \left (a + b x^{m p + 1}\right )\right )^{p}\, dx \]
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\[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int { \left ({\left (b x^{m p + 1} + a\right )} x^{m}\right )^{p} \,d x } \]
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\[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int { \left ({\left (b x^{m p + 1} + a\right )} x^{m}\right )^{p} \,d x } \]
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Timed out. \[ \int \left (x^m \left (a+b x^{1+m p}\right )\right )^p \, dx=\int {\left (x^m\,\left (a+b\,x^{m\,p+1}\right )\right )}^p \,d x \]
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