Integrand size = 15, antiderivative size = 34 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=-\frac {2^x}{\log (2)}+\frac {2^{-1+2 x}}{\log (2)}+\frac {2 \log \left (1+2^x\right )}{\log (2)} \]
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Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2320, 711} \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {2 \log \left (2^x+1\right )}{\log (2)}-\frac {2^x}{\log (2)}+\frac {2^{2 x-1}}{\log (2)} \]
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Rule 711
Rule 2320
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1+x^2}{1+x} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \left (-1+x+\frac {2}{1+x}\right ) \, dx,x,2^x\right )}{\log (2)} \\ & = -\frac {2^x}{\log (2)}+\frac {2^{-1+2 x}}{\log (2)}+\frac {2 \log \left (1+2^x\right )}{\log (2)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.68 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {2^x \left (-2+2^x\right )+4 \log \left (1+2^x\right )}{\log (4)} \]
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Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {2^{x}}{\ln \left (2\right )}+\frac {2^{2 x}}{2 \ln \left (2\right )}+\frac {2 \ln \left (1+2^{x}\right )}{\ln \left (2\right )}\) | \(34\) |
norman | \(-\frac {{\mathrm e}^{x \ln \left (2\right )}}{\ln \left (2\right )}+\frac {{\mathrm e}^{2 x \ln \left (2\right )}}{2 \ln \left (2\right )}+\frac {2 \ln \left (1+{\mathrm e}^{x \ln \left (2\right )}\right )}{\ln \left (2\right )}\) | \(40\) |
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Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {2^{2 \, x} - 2 \cdot 2^{x} + 4 \, \log \left (2^{x} + 1\right )}{2 \, \log \left (2\right )} \]
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Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\frac {4^{x} \log {\left (2 \right )} - 2 e^{\frac {x \log {\left (4 \right )}}{2}} \log {\left (2 \right )}}{2 \log {\left (2 \right )}^{2}} + \frac {2 \log {\left (e^{\frac {x \log {\left (4 \right )}}{2}} + 1 \right )}}{\log {\left (2 \right )}} \]
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Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=2 \, x - \frac {2^{2 \, x - 1} {\left (2^{-x + 1} - 1\right )}}{\log \left (2\right )} + \frac {2 \, \log \left (\frac {1}{2^{x}} + 1\right )}{\log \left (2\right )} \]
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\[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\int { \frac {4^{x} + 1}{\frac {1}{2^{x}} + 1} \,d x } \]
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Timed out. \[ \int \frac {1+4^x}{1+2^{-x}} \, dx=\int \frac {4^x+1}{\frac {1}{2^x}+1} \,d x \]
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