Integrand size = 33, antiderivative size = 98 \[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\frac {\sqrt {8-7 x^2+2 x^4} \left (16-6 x-14 x^2+3 x^3+4 x^4\right )}{6 x^3}-\frac {\log (x)}{2 \sqrt {2}}+\frac {\log \left (-2 \sqrt {2}+\sqrt {2} x^2+\sqrt {8-7 x^2+2 x^4}\right )}{2 \sqrt {2}} \]
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Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1676, 1604, 1265, 826, 852, 212} \[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {2} \left (2-x^2\right )}{\sqrt {2 x^4-7 x^2+8}}\right )}{2 \sqrt {2}}-\frac {\left (2-x^2\right ) \sqrt {2 x^4-7 x^2+8}}{2 x^2}+\frac {\left (2 x^4-7 x^2+8\right )^{3/2}}{3 x^3} \]
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Rule 212
Rule 826
Rule 852
Rule 1265
Rule 1604
Rule 1676
Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (2+x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^3} \, dx+\int \frac {\left (-8+2 x^4\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx \\ & = \frac {\left (8-7 x^2+2 x^4\right )^{3/2}}{3 x^3}+\frac {1}{2} \text {Subst}\left (\int \frac {(2+x) \sqrt {8-7 x+2 x^2}}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (2-x^2\right ) \sqrt {8-7 x^2+2 x^4}}{2 x^2}+\frac {\left (8-7 x^2+2 x^4\right )^{3/2}}{3 x^3}-\frac {1}{4} \text {Subst}\left (\int \frac {-2-x}{x \sqrt {8-7 x+2 x^2}} \, dx,x,x^2\right ) \\ & = -\frac {\left (2-x^2\right ) \sqrt {8-7 x^2+2 x^4}}{2 x^2}+\frac {\left (8-7 x^2+2 x^4\right )^{3/2}}{3 x^3}-2 \text {Subst}\left (\int \frac {1}{32-x^2} \, dx,x,\frac {8 \left (2-x^2\right )}{\sqrt {8-7 x^2+2 x^4}}\right ) \\ & = -\frac {\left (2-x^2\right ) \sqrt {8-7 x^2+2 x^4}}{2 x^2}+\frac {\left (8-7 x^2+2 x^4\right )^{3/2}}{3 x^3}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \left (2-x^2\right )}{\sqrt {8-7 x^2+2 x^4}}\right )}{2 \sqrt {2}} \\ \end{align*}
Time = 0.51 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\frac {\sqrt {8-7 x^2+2 x^4} \left (16-6 x-14 x^2+3 x^3+4 x^4\right )}{6 x^3}-\frac {\log (x)}{2 \sqrt {2}}+\frac {\log \left (-2 \sqrt {2}+\sqrt {2} x^2+\sqrt {8-7 x^2+2 x^4}\right )}{2 \sqrt {2}} \]
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Time = 3.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {3 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {\left (x^{2}-2\right ) \sqrt {2}}{x}\right ) x^{3}+2 \sqrt {2 x^{4}-7 x^{2}+8}\, \left (4 x^{4}+3 x^{3}-14 x^{2}-6 x +16\right )}{12 x^{3}}\) | \(64\) |
trager | \(\frac {\sqrt {2 x^{4}-7 x^{2}+8}\, \left (4 x^{4}+3 x^{3}-14 x^{2}-6 x +16\right )}{6 x^{3}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x^{2}+\sqrt {2 x^{4}-7 x^{2}+8}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )}{x}\right )}{4}\) | \(87\) |
elliptic | \(\frac {\sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {15}\, \left (x^{2}-\frac {7}{4}\right )}{15}\right )}{8}+\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{2}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (-7 x^{2}+16\right ) \sqrt {2}}{8 \sqrt {2 x^{4}-7 x^{2}+8}}\right )}{8}-\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{x^{2}}+\frac {\left (2 x^{4}-7 x^{2}+8\right )^{\frac {3}{2}}}{3 x^{3}}\) | \(104\) |
risch | \(-\frac {14 x^{6}+6 x^{5}-65 x^{4}-21 x^{3}+112 x^{2}+24 x -64}{3 x^{3} \sqrt {2 x^{4}-7 x^{2}+8}}+\frac {\sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {15}\, \left (x^{2}-\frac {7}{4}\right )}{15}\right )}{8}+\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{2}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (-7 x^{2}+16\right ) \sqrt {2}}{8 \sqrt {2 x^{4}-7 x^{2}+8}}\right )}{8}+\frac {2 x \sqrt {2 x^{4}-7 x^{2}+8}}{3}\) | \(132\) |
default | \(\frac {2 x \sqrt {2 x^{4}-7 x^{2}+8}}{3}+\frac {\sqrt {2 x^{4}-7 x^{2}+8}}{16}+\frac {\sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {15}\, \left (x^{2}-\frac {7}{4}\right )}{15}\right )}{8}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (-7 x^{2}+16\right ) \sqrt {2}}{8 \sqrt {2 x^{4}-7 x^{2}+8}}\right )}{8}+\frac {8 \sqrt {2 x^{4}-7 x^{2}+8}}{3 x^{3}}-\frac {7 \sqrt {2 x^{4}-7 x^{2}+8}}{3 x}-\frac {\left (2 x^{4}-7 x^{2}+8\right )^{\frac {3}{2}}}{8 x^{2}}+\frac {\left (4 x^{2}-7\right ) \sqrt {2 x^{4}-7 x^{2}+8}}{16}\) | \(163\) |
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Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.93 \[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\frac {3 \, \sqrt {2} x^{3} \log \left (\frac {4 \, x^{4} + 2 \, \sqrt {2} \sqrt {2 \, x^{4} - 7 \, x^{2} + 8} {\left (x^{2} - 2\right )} - 15 \, x^{2} + 16}{x^{2}}\right ) + 4 \, {\left (4 \, x^{4} + 3 \, x^{3} - 14 \, x^{2} - 6 \, x + 16\right )} \sqrt {2 \, x^{4} - 7 \, x^{2} + 8}}{24 \, x^{3}} \]
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\[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\int \frac {\left (x^{2} + 2\right ) \left (2 x^{2} + x - 4\right ) \sqrt {2 x^{4} - 7 x^{2} + 8}}{x^{4}}\, dx \]
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\[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\int { \frac {\sqrt {2 \, x^{4} - 7 \, x^{2} + 8} {\left (2 \, x^{2} + x - 4\right )} {\left (x^{2} + 2\right )}}{x^{4}} \,d x } \]
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\[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\int { \frac {\sqrt {2 \, x^{4} - 7 \, x^{2} + 8} {\left (2 \, x^{2} + x - 4\right )} {\left (x^{2} + 2\right )}}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\left (2+x^2\right ) \left (-4+x+2 x^2\right ) \sqrt {8-7 x^2+2 x^4}}{x^4} \, dx=\int \frac {\left (x^2+2\right )\,\left (2\,x^2+x-4\right )\,\sqrt {2\,x^4-7\,x^2+8}}{x^4} \,d x \]
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