Integrand size = 18, antiderivative size = 46 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2 a \sqrt {a-2^x b}}{b^2 \log (2)}+\frac {2 \left (a-2^x b\right )^{3/2}}{3 b^2 \log (2)} \]
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Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2280, 45} \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\frac {2 \left (a-b 2^x\right )^{3/2}}{3 b^2 \log (2)}-\frac {2 a \sqrt {a-b 2^x}}{b^2 \log (2)} \]
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Rule 45
Rule 2280
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x}{\sqrt {a-b x}} \, dx,x,2^x\right )}{\log (2)} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a}{b \sqrt {a-b x}}-\frac {\sqrt {a-b x}}{b}\right ) \, dx,x,2^x\right )}{\log (2)} \\ & = -\frac {2 a \sqrt {a-2^x b}}{b^2 \log (2)}+\frac {2 \left (a-2^x b\right )^{3/2}}{3 b^2 \log (2)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2 \sqrt {a-2^x b} \left (2 a+2^x b\right )}{b^2 \log (8)} \]
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Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.63
method | result | size |
risch | \(-\frac {2 \left (2^{x} b +2 a \right ) \sqrt {a -2^{x} b}}{3 b^{2} \ln \left (2\right )}\) | \(29\) |
derivativedivides | \(\frac {\frac {2 \left (a -2^{x} b \right )^{\frac {3}{2}}}{3}-2 \sqrt {a -2^{x} b}\, a}{b^{2} \ln \left (2\right )}\) | \(36\) |
default | \(\frac {\frac {2 \left (a -2^{x} b \right )^{\frac {3}{2}}}{3}-2 \sqrt {a -2^{x} b}\, a}{b^{2} \ln \left (2\right )}\) | \(36\) |
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Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.61 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2 \, {\left (2^{x} b + 2 \, a\right )} \sqrt {-2^{x} b + a}}{3 \, b^{2} \log \left (2\right )} \]
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Time = 0.33 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\begin {cases} - \frac {2 \cdot 2^{x} \sqrt {- 2^{x} b + a}}{3 b \log {\left (2 \right )}} - \frac {4 a \sqrt {- 2^{x} b + a}}{3 b^{2} \log {\left (2 \right )}} & \text {for}\: b \neq 0 \\\frac {2^{2 x}}{2 \sqrt {a} \log {\left (2 \right )}} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\frac {2 \, {\left (-2^{x} b + a\right )}^{\frac {3}{2}}}{3 \, b^{2} \log \left (2\right )} - \frac {2 \, \sqrt {-2^{x} b + a} a}{b^{2} \log \left (2\right )} \]
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Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=\frac {2 \, {\left ({\left (-2^{x} b + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {-2^{x} b + a} a\right )}}{3 \, b^{2} \log \left (2\right )} \]
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Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.61 \[ \int \frac {2^{2 x}}{\sqrt {a-2^x b}} \, dx=-\frac {2\,\sqrt {a-2^x\,b}\,\left (2\,a+2^x\,b\right )}{3\,b^2\,\ln \left (2\right )} \]
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