Integrand size = 19, antiderivative size = 67 \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\frac {\sqrt {1+2 x^2}}{2 \sqrt {2}}-\frac {5}{8} \text {arcsinh}\left (\sqrt {2} x\right )-\frac {1}{8} \sqrt {33} \text {arctanh}\left (\frac {\sqrt {\frac {2}{33}} (2-5 x)}{\sqrt {1+2 x^2}}\right ) \]
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Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {749, 858, 221, 739, 212} \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=-\frac {5}{8} \text {arcsinh}\left (\sqrt {2} x\right )-\frac {1}{8} \sqrt {33} \text {arctanh}\left (\frac {\sqrt {\frac {2}{33}} (2-5 x)}{\sqrt {2 x^2+1}}\right )+\frac {\sqrt {2 x^2+1}}{2 \sqrt {2}} \]
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Rule 212
Rule 221
Rule 739
Rule 749
Rule 858
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+2 x^2}}{2 \sqrt {2}}+\frac {1}{4} \int \frac {8-20 x}{(5+4 x) \sqrt {2+4 x^2}} \, dx \\ & = \frac {\sqrt {1+2 x^2}}{2 \sqrt {2}}-\frac {5}{4} \int \frac {1}{\sqrt {2+4 x^2}} \, dx+\frac {33}{4} \int \frac {1}{(5+4 x) \sqrt {2+4 x^2}} \, dx \\ & = \frac {\sqrt {1+2 x^2}}{2 \sqrt {2}}-\frac {5}{8} \sinh ^{-1}\left (\sqrt {2} x\right )-\frac {33}{4} \text {Subst}\left (\int \frac {1}{132-x^2} \, dx,x,\frac {8-20 x}{\sqrt {2+4 x^2}}\right ) \\ & = \frac {\sqrt {1+2 x^2}}{2 \sqrt {2}}-\frac {5}{8} \sinh ^{-1}\left (\sqrt {2} x\right )-\frac {1}{8} \sqrt {33} \tanh ^{-1}\left (\frac {\sqrt {\frac {2}{33}} (2-5 x)}{\sqrt {1+2 x^2}}\right ) \\ \end{align*}
Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\frac {1}{8} \left (2 \sqrt {2+4 x^2}+2 \sqrt {33} \text {arctanh}\left (\frac {5+4 x-2 \sqrt {2+4 x^2}}{\sqrt {33}}\right )+5 \log \left (-2 x+\sqrt {2+4 x^2}\right )\right ) \]
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Time = 1.97 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\sqrt {16 \left (x +\frac {5}{4}\right )^{2}-40 x -17}}{8}-\frac {5 \,\operatorname {arcsinh}\left (\sqrt {2}\, x \right )}{8}-\frac {\sqrt {33}\, \operatorname {arctanh}\left (\frac {2 \left (4-10 x \right ) \sqrt {33}}{33 \sqrt {16 \left (x +\frac {5}{4}\right )^{2}-40 x -17}}\right )}{8}\) | \(56\) |
risch | \(\frac {2 x^{2}+1}{2 \sqrt {4 x^{2}+2}}-\frac {5 \,\operatorname {arcsinh}\left (\sqrt {2}\, x \right )}{8}-\frac {\sqrt {33}\, \operatorname {arctanh}\left (\frac {2 \left (4-10 x \right ) \sqrt {33}}{33 \sqrt {16 \left (x +\frac {5}{4}\right )^{2}-40 x -17}}\right )}{8}\) | \(58\) |
trager | \(\frac {\sqrt {4 x^{2}+2}}{4}-\frac {5 \ln \left (2 x +\sqrt {4 x^{2}+2}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-33\right ) \ln \left (-\frac {10 \operatorname {RootOf}\left (\textit {\_Z}^{2}-33\right ) x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-33\right )-33 \sqrt {4 x^{2}+2}}{5+4 x}\right )}{8}\) | \(76\) |
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Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\frac {1}{8} \, \sqrt {33} \log \left (-\frac {2 \, \sqrt {33} {\left (5 \, x - 2\right )} + \sqrt {4 \, x^{2} + 2} {\left (5 \, \sqrt {33} + 33\right )} + 50 \, x - 20}{4 \, x + 5}\right ) + \frac {1}{4} \, \sqrt {4 \, x^{2} + 2} + \frac {5}{8} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 2}\right ) \]
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\[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\sqrt {2} \int \frac {\sqrt {2 x^{2} + 1}}{4 x + 5}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\frac {1}{8} \, \sqrt {33} \operatorname {arsinh}\left (\frac {5 \, \sqrt {2} x}{{\left | 4 \, x + 5 \right |}} - \frac {2 \, \sqrt {2}}{{\left | 4 \, x + 5 \right |}}\right ) + \frac {1}{4} \, \sqrt {4 \, x^{2} + 2} - \frac {5}{8} \, \operatorname {arsinh}\left (\sqrt {2} x\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (48) = 96\).
Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.57 \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\frac {1}{16} \, \sqrt {2} {\left (5 \, \sqrt {2} \log \left (-\sqrt {2} x + \sqrt {2 \, x^{2} + 1}\right ) + \sqrt {66} \log \left (-\frac {{\left | -4 \, \sqrt {2} x - \sqrt {66} - 5 \, \sqrt {2} + 4 \, \sqrt {2 \, x^{2} + 1} \right |}}{4 \, \sqrt {2} x - \sqrt {66} + 5 \, \sqrt {2} - 4 \, \sqrt {2 \, x^{2} + 1}}\right ) + 4 \, \sqrt {2 \, x^{2} + 1}\right )} \]
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Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {2+4 x^2}}{5+4 x} \, dx=\frac {\sqrt {x^2+\frac {1}{2}}}{2}-\frac {5\,\mathrm {asinh}\left (\sqrt {2}\,x\right )}{8}+\frac {\sqrt {33}\,\left (132\,\ln \left (x+\frac {5}{4}\right )-132\,\ln \left (x-\frac {\sqrt {33}\,\sqrt {x^2+\frac {1}{2}}}{5}-\frac {2}{5}\right )\right )}{1056} \]
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