Integrand size = 15, antiderivative size = 43 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\frac {2^x}{a \log (2)}-\frac {\sqrt {b} \arctan \left (\frac {2^x \sqrt {a}}{\sqrt {b}}\right )}{a^{3/2} \log (2)} \]
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Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2281, 199, 327, 211} \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\frac {2^x}{a \log (2)}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a} 2^x}{\sqrt {b}}\right )}{a^{3/2} \log (2)} \]
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Rule 199
Rule 211
Rule 327
Rule 2281
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{a+\frac {b}{x^2}} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{b+a x^2} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {2^x}{a \log (2)}-\frac {b \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,2^x\right )}{a \log (2)} \\ & = \frac {2^x}{a \log (2)}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {2^x \sqrt {a}}{\sqrt {b}}\right )}{a^{3/2} \log (2)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\frac {\frac {2^x}{a}-\frac {\sqrt {b} \arctan \left (\frac {2^x \sqrt {a}}{\sqrt {b}}\right )}{a^{3/2}}}{\log (2)} \]
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Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84
method | result | size |
derivativedivides | \(\frac {\frac {2^{x}}{a}-\frac {b \arctan \left (\frac {a 2^{x}}{\sqrt {a b}}\right )}{a \sqrt {a b}}}{\ln \left (2\right )}\) | \(36\) |
default | \(\frac {\frac {2^{x}}{a}-\frac {b \arctan \left (\frac {a 2^{x}}{\sqrt {a b}}\right )}{a \sqrt {a b}}}{\ln \left (2\right )}\) | \(36\) |
risch | \(\frac {2^{x}}{a \ln \left (2\right )}+\frac {\sqrt {-a b}\, \ln \left (2^{x}-\frac {\sqrt {-a b}}{a}\right )}{2 a^{2} \ln \left (2\right )}-\frac {\sqrt {-a b}\, \ln \left (2^{x}+\frac {\sqrt {-a b}}{a}\right )}{2 a^{2} \ln \left (2\right )}\) | \(74\) |
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Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.37 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\left [\frac {\sqrt {-\frac {b}{a}} \log \left (-\frac {2 \cdot 2^{x} a \sqrt {-\frac {b}{a}} - 2^{2 \, x} a + b}{2^{2 \, x} a + b}\right ) + 2 \cdot 2^{x}}{2 \, a \log \left (2\right )}, -\frac {\sqrt {\frac {b}{a}} \arctan \left (\frac {2^{x} a \sqrt {\frac {b}{a}}}{b}\right ) - 2^{x}}{a \log \left (2\right )}\right ] \]
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Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\begin {cases} \frac {2^{x}}{a \log {\left (2 \right )}} & \text {for}\: a \log {\left (2 \right )} \neq 0 \\\frac {x}{a} & \text {otherwise} \end {cases} + \frac {\operatorname {RootSum} {\left (4 z^{2} a^{3} + b, \left ( i \mapsto i \log {\left (2^{x} - 2 i a \right )} \right )\right )}}{\log {\left (2 \right )}} \]
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Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\frac {b \arctan \left (\frac {b}{\sqrt {a b} 2^{x}}\right )}{\sqrt {a b} a \log \left (2\right )} + \frac {2^{x}}{a \log \left (2\right )} \]
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Time = 0.47 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.88 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=-\frac {b \arctan \left (\frac {2^{x} a}{\sqrt {a b}}\right )}{\sqrt {a b} a \log \left (2\right )} + \frac {2^{x}}{a \log \left (2\right )} \]
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Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {2^x}{a+2^{-2 x} b} \, dx=\frac {2^x}{a\,\ln \left (2\right )}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {2^x\,\sqrt {a}}{\sqrt {b}}\right )}{a^{3/2}\,\ln \left (2\right )} \]
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