Integrand size = 9, antiderivative size = 65 \[ \int \left (1+a^{m x}\right )^4 \, dx=x+\frac {4 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{m \log (a)}+\frac {4 a^{3 m x}}{3 m \log (a)}+\frac {a^{4 m x}}{4 m \log (a)} \]
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Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2320, 45} \[ \int \left (1+a^{m x}\right )^4 \, dx=\frac {4 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{m \log (a)}+\frac {4 a^{3 m x}}{3 m \log (a)}+\frac {a^{4 m x}}{4 m \log (a)}+x \]
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Rule 45
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(1+x)^4}{x} \, dx,x,a^{m x}\right )}{m \log (a)} \\ & = \frac {\text {Subst}\left (\int \left (4+\frac {1}{x}+6 x+4 x^2+x^3\right ) \, dx,x,a^{m x}\right )}{m \log (a)} \\ & = x+\frac {4 a^{m x}}{m \log (a)}+\frac {3 a^{2 m x}}{m \log (a)}+\frac {4 a^{3 m x}}{3 m \log (a)}+\frac {a^{4 m x}}{4 m \log (a)} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int \left (1+a^{m x}\right )^4 \, dx=\frac {\frac {a^{m x} \left (48+36 a^{m x}+16 a^{2 m x}+3 a^{3 m x}\right )}{12 m}+\frac {\log \left (a^{m x}\right )}{m}}{\log (a)} \]
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Time = 0.04 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {\frac {a^{4 m x}}{4}+\frac {4 a^{3 m x}}{3}+3 a^{2 m x}+4 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \left (a \right )}\) | \(50\) |
default | \(\frac {\frac {a^{4 m x}}{4}+\frac {4 a^{3 m x}}{3}+3 a^{2 m x}+4 a^{m x}+\ln \left (a^{m x}\right )}{m \ln \left (a \right )}\) | \(50\) |
parallelrisch | \(\frac {3 a^{4 m x}+16 a^{3 m x}+12 m x \ln \left (a \right )+36 a^{2 m x}+48 a^{m x}}{12 \ln \left (a \right ) m}\) | \(51\) |
risch | \(x +\frac {4 a^{m x}}{m \ln \left (a \right )}+\frac {3 a^{2 m x}}{m \ln \left (a \right )}+\frac {4 a^{3 m x}}{3 m \ln \left (a \right )}+\frac {a^{4 m x}}{4 m \ln \left (a \right )}\) | \(65\) |
norman | \(x +\frac {4 \,{\mathrm e}^{m x \ln \left (a \right )}}{m \ln \left (a \right )}+\frac {3 \,{\mathrm e}^{2 m x \ln \left (a \right )}}{m \ln \left (a \right )}+\frac {4 \,{\mathrm e}^{3 m x \ln \left (a \right )}}{3 m \ln \left (a \right )}+\frac {{\mathrm e}^{4 m x \ln \left (a \right )}}{4 m \ln \left (a \right )}\) | \(69\) |
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none
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72 \[ \int \left (1+a^{m x}\right )^4 \, dx=\frac {12 \, m x \log \left (a\right ) + 3 \, a^{4 \, m x} + 16 \, a^{3 \, m x} + 36 \, a^{2 \, m x} + 48 \, a^{m x}}{12 \, m \log \left (a\right )} \]
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Time = 0.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.34 \[ \int \left (1+a^{m x}\right )^4 \, dx=x + \begin {cases} \frac {3 a^{4 m x} m^{3} \log {\left (a \right )}^{3} + 16 a^{3 m x} m^{3} \log {\left (a \right )}^{3} + 36 a^{2 m x} m^{3} \log {\left (a \right )}^{3} + 48 a^{m x} m^{3} \log {\left (a \right )}^{3}}{12 m^{4} \log {\left (a \right )}^{4}} & \text {for}\: m^{4} \log {\left (a \right )}^{4} \neq 0 \\15 x & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \left (1+a^{m x}\right )^4 \, dx=x + \frac {a^{4 \, m x}}{4 \, m \log \left (a\right )} + \frac {4 \, a^{3 \, m x}}{3 \, m \log \left (a\right )} + \frac {3 \, a^{2 \, m x}}{m \log \left (a\right )} + \frac {4 \, a^{m x}}{m \log \left (a\right )} \]
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Time = 0.33 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.74 \[ \int \left (1+a^{m x}\right )^4 \, dx=\frac {12 \, m x \log \left ({\left | a \right |}\right ) + 3 \, a^{4 \, m x} + 16 \, a^{3 \, m x} + 36 \, a^{2 \, m x} + 48 \, a^{m x}}{12 \, m \log \left (a\right )} \]
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Time = 0.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \left (1+a^{m x}\right )^4 \, dx=x+\frac {4\,a^{m\,x}+3\,a^{2\,m\,x}+\frac {4\,a^{3\,m\,x}}{3}+\frac {a^{4\,m\,x}}{4}}{m\,\ln \left (a\right )} \]
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