Integrand size = 18, antiderivative size = 44 \[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\frac {x^{-m (1+p)} \left (a x^m+b x^{1+m+m p}\right )^{1+p}}{b (1+p) (1+m p)} \]
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Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2025} \[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\frac {x^{-m (p+1)} \left (a x^m+b x^{m p+m+1}\right )^{p+1}}{b (p+1) (m p+1)} \]
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Rule 2025
Rubi steps \begin{align*} \text {integral}& = \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 = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\frac {x^{-m (1+p)} \left (x^m \left (a+b x^{1+m p}\right )\right )^{1+p}}{b (1+p) (1+m p)} \]
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\[\int \left (x^{m} a +b \,x^{m p +m +1}\right )^{p}d x\]
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none
Time = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.45 \[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\frac {{\left (b x x^{m p + m + 1} + a x x^{m}\right )} {\left (b x^{m p + m + 1} + a x^{m}\right )}^{p}}{{\left (b m p^{2} + {\left (b m + b\right )} p + b\right )} x^{m p + m + 1}} \]
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\[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\int \left (a x^{m} + b x^{m p + m + 1}\right )^{p}\, dx \]
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\[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\int { {\left (b x^{m p + m + 1} + a x^{m}\right )}^{p} \,d x } \]
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\[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\int { {\left (b x^{m p + m + 1} + a x^{m}\right )}^{p} \,d x } \]
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Time = 9.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.73 \[ \int \left (a x^m+b x^{1+m+m p}\right )^p \, dx=\frac {a\,{\left (a\,x^m+b\,x^{m+m\,p+1}\right )}^p\,\left (\frac {b\,x^{m\,p+1}}{a}-\frac {1}{{\left (\frac {b\,x^{m\,p+1}}{a}+1\right )}^p}+1\right )}{b\,x^{m\,p}\,\left (m\,p+1\right )\,\left (p+1\right )} \]
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