Integrand size = 18, antiderivative size = 58 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {b^2 x}{a^3}+\frac {2^{-1+2 x}}{a \log (2)}+\frac {2^x b}{a^2 \log (2)}+\frac {b^2 \log \left (a-2^{-x} b\right )}{a^3 \log (2)} \]
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Time = 0.03 (sec) , antiderivative size = 58, 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, 46} \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {b^2 x}{a^3}+\frac {b^2 \log \left (a-b 2^{-x}\right )}{a^3 \log (2)}+\frac {b 2^x}{a^2 \log (2)}+\frac {2^{2 x-1}}{a \log (2)} \]
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Rule 46
Rule 2280
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{x^3 (a-b x)} \, dx,x,2^{-x}\right )}{\log (2)} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{a x^3}+\frac {b}{a^2 x^2}+\frac {b^2}{a^3 x}+\frac {b^3}{a^3 (a-b x)}\right ) \, dx,x,2^{-x}\right )}{\log (2)} \\ & = \frac {b^2 x}{a^3}+\frac {2^{-1+2 x}}{a \log (2)}+\frac {2^x b}{a^2 \log (2)}+\frac {b^2 \log \left (a-2^{-x} b\right )}{a^3 \log (2)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.66 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {2^x a \left (2^x a+2 b\right )+2 b^2 \log \left (2^x a-b\right )}{a^3 \log (4)} \]
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Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a 2^{2 x}}{2}+2^{x} b}{a^{2}}+\frac {b^{2} \ln \left (a 2^{x}-b \right )}{a^{3}}}{\ln \left (2\right )}\) | \(42\) |
default | \(\frac {\frac {\frac {a 2^{2 x}}{2}+2^{x} b}{a^{2}}+\frac {b^{2} \ln \left (a 2^{x}-b \right )}{a^{3}}}{\ln \left (2\right )}\) | \(42\) |
risch | \(\frac {2^{2 x}}{2 \ln \left (2\right ) a}+\frac {2^{x} b}{a^{2} \ln \left (2\right )}+\frac {b^{2} \ln \left (2^{x}-\frac {b}{a}\right )}{\ln \left (2\right ) a^{3}}\) | \(50\) |
norman | \(\frac {b \,{\mathrm e}^{x \ln \left (2\right )}}{\ln \left (2\right ) a^{2}}+\frac {{\mathrm e}^{2 x \ln \left (2\right )}}{2 \ln \left (2\right ) a}+\frac {b^{2} \ln \left (a \,{\mathrm e}^{x \ln \left (2\right )}-b \right )}{\ln \left (2\right ) a^{3}}\) | \(55\) |
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Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.71 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {2^{2 \, x} a^{2} + 2 \cdot 2^{x} a b + 2 \, b^{2} \log \left (2^{x} a - b\right )}{2 \, a^{3} \log \left (2\right )} \]
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Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\begin {cases} \frac {2^{2 x} a^{2} \log {\left (2 \right )} + 2 \cdot 2^{x} a b \log {\left (2 \right )}}{2 a^{3} \log {\left (2 \right )}^{2}} & \text {for}\: a^{3} \log {\left (2 \right )}^{2} \neq 0 \\\frac {x \left (a + b\right )}{a^{2}} & \text {otherwise} \end {cases} + \frac {b^{2} \log {\left (2^{x} - \frac {b}{a} \right )}}{a^{3} \log {\left (2 \right )}} \]
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Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {b^{2} x}{a^{3}} + \frac {{\left (2^{-x + 1} b + a\right )} 2^{2 \, x - 1}}{a^{2} \log \left (2\right )} + \frac {b^{2} \log \left (-a + \frac {b}{2^{x}}\right )}{a^{3} \log \left (2\right )} \]
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Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.86 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {b^{2} \log \left ({\left | 2^{x} a - b \right |}\right )}{a^{3} \log \left (2\right )} + \frac {2^{2 \, x} a \log \left (2\right ) + 2 \cdot 2^{x} b \log \left (2\right )}{2 \, a^{2} \log \left (2\right )^{2}} \]
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Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81 \[ \int \frac {2^{2 x}}{a-2^{-x} b} \, dx=\frac {2^{2\,x}}{2\,a\,\ln \left (2\right )}+\frac {2^x\,b}{a^2\,\ln \left (2\right )}+\frac {b^2\,\ln \left (b-2^x\,a\right )}{a^3\,\ln \left (2\right )} \]
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