Integrand size = 20, antiderivative size = 96 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-2^{-x} b}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)} \]
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Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2280, 44, 65, 214} \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-b 2^{-x}}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)}+\frac {3 b 2^{x-2} \sqrt {a-b 2^{-x}}}{a^2 \log (2)}+\frac {2^{2 x-1} \sqrt {a-b 2^{-x}}}{a \log (2)} \]
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Rule 44
Rule 65
Rule 214
Rule 2280
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{x^3 \sqrt {a-b x}} \, dx,x,2^{-x}\right )}{\log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}-\frac {(3 b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a-b x}} \, dx,x,2^{-x}\right )}{4 a \log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a-b x}} \, dx,x,2^{-x}\right )}{8 a^2 \log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {(3 b) \text {Subst}\left (\int \frac {1}{\frac {a}{b}-\frac {x^2}{b}} \, dx,x,\sqrt {a-2^{-x} b}\right )}{4 a^2 \log (2)} \\ & = \frac {2^{-1+2 x} \sqrt {a-2^{-x} b}}{a \log (2)}+\frac {3\ 2^{-2+x} b \sqrt {a-2^{-x} b}}{a^2 \log (2)}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a-2^{-x} b}}{\sqrt {a}}\right )}{4 a^{5/2} \log (2)} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.20 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {2^{-2-\frac {x}{2}} \left (2^{x/2} \sqrt {a} \left (2^{1+2 x} a^2+2^x a b-3 b^2\right )+3 \sqrt {2^x a-b} b^2 \text {arctanh}\left (\frac {2^{x/2} \sqrt {a}}{\sqrt {2^x a-b}}\right )\right )}{a^{5/2} \sqrt {a-2^{-x} b} \log (2)} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(196\) vs. \(2(84)=168\).
Time = 0.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.05
method | result | size |
derivativedivides | \(-\frac {\sqrt {\left (a 2^{x}-b \right ) 2^{-x}}\, 2^{x} \left (-4 a^{\frac {5}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, 2^{x}-4 \ln \left (\frac {2 \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) a \,b^{2}+2 a^{\frac {3}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, b -8 a^{\frac {3}{2}} \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, b +b^{2} \ln \left (\frac {2 \sqrt {a 2^{2 x}-2^{x} b}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) a \right )}{8 \ln \left (2\right ) \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, a^{\frac {7}{2}}}\) | \(197\) |
default | \(-\frac {\sqrt {\left (a 2^{x}-b \right ) 2^{-x}}\, 2^{x} \left (-4 a^{\frac {5}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, 2^{x}-4 \ln \left (\frac {2 \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) a \,b^{2}+2 a^{\frac {3}{2}} \sqrt {a 2^{2 x}-2^{x} b}\, b -8 a^{\frac {3}{2}} \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, b +b^{2} \ln \left (\frac {2 \sqrt {a 2^{2 x}-2^{x} b}\, \sqrt {a}+2 a 2^{x}-b}{2 \sqrt {a}}\right ) a \right )}{8 \ln \left (2\right ) \sqrt {\left (a 2^{x}-b \right ) 2^{x}}\, a^{\frac {7}{2}}}\) | \(197\) |
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none
Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.81 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \log \left (-2 \cdot 2^{x} a - 2 \cdot 2^{x} \sqrt {a} \sqrt {\frac {2^{x} a - b}{2^{x}}} + b\right ) + 2 \, {\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{8 \, a^{3} \log \left (2\right )}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{a}\right ) - {\left (2 \cdot 2^{2 \, x} a^{2} + 3 \cdot 2^{x} a b\right )} \sqrt {\frac {2^{x} a - b}{2^{x}}}}{4 \, a^{3} \log \left (2\right )}\right ] \]
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\[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\int \frac {2^{2 x}}{\sqrt {a - 2^{- x} b}}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.35 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=-\frac {3 \, b^{2} \log \left (\frac {\sqrt {a - \frac {b}{2^{x}}} - \sqrt {a}}{\sqrt {a - \frac {b}{2^{x}}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}} \log \left (2\right )} - \frac {3 \, {\left (a - \frac {b}{2^{x}}\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {a - \frac {b}{2^{x}}} a b^{2}}{4 \, {\left ({\left (a - \frac {b}{2^{x}}\right )}^{2} a^{2} - 2 \, {\left (a - \frac {b}{2^{x}}\right )} a^{3} + a^{4}\right )} \log \left (2\right )} \]
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Time = 0.35 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\frac {2 \, \sqrt {2^{2 \, x} a - 2^{x} b} {\left (\frac {2 \cdot 2^{x}}{a} + \frac {3 \, b}{a^{2}}\right )} - \frac {3 \, b^{2} \log \left ({\left | 2 \, {\left (2^{x} \sqrt {a} - \sqrt {2^{2 \, x} a - 2^{x} b}\right )} \sqrt {a} - b \right |}\right )}{a^{\frac {5}{2}}} + \frac {3 \, b^{2} \log \left ({\left | b \right |}\right )}{a^{\frac {5}{2}}}}{8 \, \log \left (2\right )} \]
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Timed out. \[ \int \frac {2^{2 x}}{\sqrt {a-2^{-x} b}} \, dx=\int \frac {2^{2\,x}}{\sqrt {a-\frac {b}{2^x}}} \,d x \]
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