Integrand size = 20, antiderivative size = 28 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {1}{4 \sqrt {1+x^8}}-\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^8}\right ) \]
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Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {457, 79, 65, 213} \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {1}{4} \text {arctanh}\left (\sqrt {x^8+1}\right )-\frac {1}{4 \sqrt {x^8+1}} \]
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Rule 65
Rule 79
Rule 213
Rule 457
Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \text {Subst}\left (\int \frac {1+2 x}{x (1+x)^{3/2}} \, dx,x,x^8\right ) \\ & = -\frac {1}{4 \sqrt {1+x^8}}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^8\right ) \\ & = -\frac {1}{4 \sqrt {1+x^8}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^8}\right ) \\ & = -\frac {1}{4 \sqrt {1+x^8}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {1+x^8}\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {1}{4 \sqrt {1+x^8}}-\frac {1}{4} \text {arctanh}\left (\sqrt {1+x^8}\right ) \]
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Time = 1.54 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
pseudoelliptic | \(-\frac {1}{4 \sqrt {x^{8}+1}}-\frac {\operatorname {arctanh}\left (\frac {1}{\sqrt {x^{8}+1}}\right )}{4}\) | \(21\) |
trager | \(-\frac {1}{4 \sqrt {x^{8}+1}}-\frac {\ln \left (\frac {\sqrt {x^{8}+1}+1}{x^{4}}\right )}{4}\) | \(27\) |
risch | \(-\frac {1}{4 \sqrt {x^{8}+1}}+\frac {-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{8}+1}}{2}\right )+\left (-2 \ln \left (2\right )+8 \ln \left (x \right )\right ) \sqrt {\pi }}{8 \sqrt {\pi }}\) | \(47\) |
meijerg | \(\frac {-\sqrt {\pi }+\frac {\sqrt {\pi }}{\sqrt {x^{8}+1}}-\sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{8}+1}}{2}\right )+\frac {\left (2-2 \ln \left (2\right )+8 \ln \left (x \right )\right ) \sqrt {\pi }}{2}}{4 \sqrt {\pi }}+\frac {\sqrt {\pi }-\frac {\sqrt {\pi }}{\sqrt {x^{8}+1}}}{2 \sqrt {\pi }}\) | \(77\) |
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Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {{\left (x^{8} + 1\right )} \log \left (\sqrt {x^{8} + 1} + 1\right ) - {\left (x^{8} + 1\right )} \log \left (\sqrt {x^{8} + 1} - 1\right ) + 2 \, \sqrt {x^{8} + 1}}{8 \, {\left (x^{8} + 1\right )}} \]
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Time = 5.91 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=\frac {\log {\left (\sqrt {x^{8} + 1} - 1 \right )}}{8} - \frac {\log {\left (\sqrt {x^{8} + 1} + 1 \right )}}{8} - \frac {1}{4 \sqrt {x^{8} + 1}} \]
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none
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {1}{4 \, \sqrt {x^{8} + 1}} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]
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none
Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {1}{4 \, \sqrt {x^{8} + 1}} - \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {x^{8} + 1} - 1\right ) \]
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Time = 21.79 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {1+2 x^8}{x \left (1+x^8\right )^{3/2}} \, dx=-\frac {\mathrm {atanh}\left (\sqrt {x^8+1}\right )}{4}-\frac {1}{4\,\sqrt {x^8+1}} \]
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