Integrand size = 9, antiderivative size = 40 \[ \int \left (1+a^{m x}\right )^n \, dx=-\frac {\left (1+a^{m x}\right )^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+a^{m x}\right )}{m (1+n) \log (a)} \]
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Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2320, 67} \[ \int \left (1+a^{m x}\right )^n \, dx=-\frac {\left (a^{m x}+1\right )^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,a^{m x}+1\right )}{m (n+1) \log (a)} \]
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Rule 67
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(1+x)^n}{x} \, dx,x,a^{m x}\right )}{m \log (a)} \\ & = -\frac {\left (1+a^{m x}\right )^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+a^{m x}\right )}{m (1+n) \log (a)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \left (1+a^{m x}\right )^n \, dx=-\frac {\left (1+a^{m x}\right )^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+a^{m x}\right )}{m (1+n) \log (a)} \]
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\[\int \left (1+a^{m x}\right )^{n}d x\]
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\[ \int \left (1+a^{m x}\right )^n \, dx=\int { {\left (a^{m x} + 1\right )}^{n} \,d x } \]
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\[ \int \left (1+a^{m x}\right )^n \, dx=\int \left (a^{m x} + 1\right )^{n}\, dx \]
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\[ \int \left (1+a^{m x}\right )^n \, dx=\int { {\left (a^{m x} + 1\right )}^{n} \,d x } \]
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\[ \int \left (1+a^{m x}\right )^n \, dx=\int { {\left (a^{m x} + 1\right )}^{n} \,d x } \]
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Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38 \[ \int \left (1+a^{m x}\right )^n \, dx=\frac {{\left (a^{m\,x}+1\right )}^n\,{{}}_2{\mathrm {F}}_1\left (-n,-n;\ 1-n;\ -\frac {1}{a^{m\,x}}\right )}{m\,n\,\ln \left (a\right )\,{\left (\frac {1}{a^{m\,x}}+1\right )}^n} \]
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