Integrand size = 17, antiderivative size = 24 \[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\frac {2^x \sqrt {a+2^{-2 x} b}}{a \log (2)} \]
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Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2281, 197} \[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\frac {2^x \sqrt {a+b 2^{-2 x}}}{a \log (2)} \]
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Rule 197
Rule 2281
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b}{x^2}}} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {2^x \sqrt {a+2^{-2 x} b}}{a \log (2)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\frac {2^{-x} \left (2^{2 x} a+b\right )}{a \sqrt {a+2^{-2 x} b} \log (2)} \]
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Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67
method | result | size |
risch | \(\frac {\left (a 2^{2 x}+b \right ) 2^{-x}}{\sqrt {\left (a 2^{2 x}+b \right ) 2^{-2 x}}\, a \ln \left (2\right )}\) | \(40\) |
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none
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\frac {2^{x} \sqrt {\frac {2^{2 \, x} a + b}{2^{2 \, x}}}}{a \log \left (2\right )} \]
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\[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\int \frac {2^{x}}{\sqrt {a + 4^{- x} b}}\, dx \]
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none
Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\frac {\sqrt {2^{2 \, x} a + b}}{a \log \left (2\right )} \]
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\[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\int { \frac {2^{x}}{\sqrt {a + \frac {b}{4^{x}}}} \,d x } \]
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Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2^x}{\sqrt {a+4^{-x} b}} \, dx=\frac {2^x\,\sqrt {a+\frac {b}{2^{2\,x}}}}{a\,\ln \left (2\right )} \]
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