Integrand size = 16, antiderivative size = 32 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {2^x}{b \log (2)}-\frac {a \log \left (a-2^x b\right )}{b^2 \log (2)} \]
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Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2280, 45} \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {a \log \left (a-b 2^x\right )}{b^2 \log (2)}-\frac {2^x}{b \log (2)} \]
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Rule 45
Rule 2280
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x}{a-b x} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{b}-\frac {a}{b (-a+b x)}\right ) \, dx,x,2^x\right )}{\log (2)} \\ & = -\frac {2^x}{b \log (2)}-\frac {a \log \left (a-2^x b\right )}{b^2 \log (2)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {2^x b+a \log \left (a-2^x b\right )}{b^2 \log (2)} \]
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Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {2^{x}}{b}-\frac {a \ln \left (a -2^{x} b \right )}{b^{2}}}{\ln \left (2\right )}\) | \(30\) |
default | \(\frac {-\frac {2^{x}}{b}-\frac {a \ln \left (a -2^{x} b \right )}{b^{2}}}{\ln \left (2\right )}\) | \(30\) |
risch | \(-\frac {2^{x}}{b \ln \left (2\right )}-\frac {a \ln \left (2^{x}-\frac {a}{b}\right )}{\ln \left (2\right ) b^{2}}\) | \(35\) |
norman | \(-\frac {{\mathrm e}^{x \ln \left (2\right )}}{\ln \left (2\right ) b}-\frac {a \ln \left (a -{\mathrm e}^{x \ln \left (2\right )} b \right )}{\ln \left (2\right ) b^{2}}\) | \(37\) |
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Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {2^{x} b + a \log \left (2^{x} b - a\right )}{b^{2} \log \left (2\right )} \]
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Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=- \frac {a \log {\left (2^{x} - \frac {a}{b} \right )}}{b^{2} \log {\left (2 \right )}} + \begin {cases} - \frac {2^{x}}{b \log {\left (2 \right )}} & \text {for}\: b \log {\left (2 \right )} \neq 0 \\- \frac {x}{b} & \text {otherwise} \end {cases} \]
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Time = 0.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {2^{x}}{b \log \left (2\right )} - \frac {a \log \left (2^{x} b - a\right )}{b^{2} \log \left (2\right )} \]
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Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {2^{x}}{b \log \left (2\right )} - \frac {a \log \left ({\left | 2^{x} b - a \right |}\right )}{b^{2} \log \left (2\right )} \]
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Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {2^{2 x}}{a-2^x b} \, dx=-\frac {2^x\,b+a\,\ln \left (2^x\,b-a\right )}{b^2\,\ln \left (2\right )} \]
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