Integrand size = 13, antiderivative size = 109 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {15 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}-\frac {15 f^{a+\frac {b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)} \]
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Time = 0.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2243, 2242, 2235} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {15 \sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}-\frac {15 f^{a+\frac {b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)} \]
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Rule 2235
Rule 2242
Rule 2243
Rubi steps \begin{align*} \text {integral}& = -\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}-\frac {5 \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx}{2 b \log (f)} \\ & = \frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}+\frac {15 \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx}{4 b^2 \log ^2(f)} \\ & = -\frac {15 f^{a+\frac {b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}-\frac {15 \int \frac {f^{a+\frac {b}{x^2}}}{x^2} \, dx}{8 b^3 \log ^3(f)} \\ & = -\frac {15 f^{a+\frac {b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)}+\frac {15 \text {Subst}\left (\int f^{a+b x^2} \, dx,x,\frac {1}{x}\right )}{8 b^3 \log ^3(f)} \\ & = \frac {15 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}-\frac {15 f^{a+\frac {b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac {5 f^{a+\frac {b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^5 \log (f)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.79 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {15 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac {7}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}} \left (15 x^4-10 b x^2 \log (f)+4 b^2 \log ^2(f)\right )}{8 b^3 x^5 \log ^3(f)} \]
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Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83
method | result | size |
meijerg | \(-\frac {f^{a} \sqrt {-b}\, \left (\frac {\left (-b \right )^{\frac {7}{2}} \sqrt {\ln \left (f \right )}\, \left (\frac {28 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {70 b \ln \left (f \right )}{x^{2}}+105\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{28 x \,b^{3}}-\frac {15 \left (-b \right )^{\frac {7}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{8 b^{\frac {7}{2}}}\right )}{2 \ln \left (f \right )^{\frac {7}{2}} b^{4}}\) | \(91\) |
risch | \(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{5} b \ln \left (f \right )}+\frac {5 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x^{3}}-\frac {15 f^{a} f^{\frac {b}{x^{2}}}}{8 \ln \left (f \right )^{3} b^{3} x}+\frac {15 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{16 \ln \left (f \right )^{3} b^{3} \sqrt {-b \ln \left (f \right )}}\) | \(102\) |
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Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=-\frac {15 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{5} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (15 \, b x^{4} \log \left (f\right ) - 10 \, b^{2} x^{2} \log \left (f\right )^{2} + 4 \, b^{3} \log \left (f\right )^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{16 \, b^{4} x^{5} \log \left (f\right )^{4}} \]
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Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\text {Timed out} \]
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Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.26 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {f^{a} \Gamma \left (\frac {7}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{7} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {7}{2}}} \]
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\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{8}} \,d x } \]
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Time = 0.34 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^8} \, dx=\frac {5\,f^a\,f^{\frac {b}{x^2}}}{4\,b^2\,x^3\,{\ln \left (f\right )}^2}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x^5\,\ln \left (f\right )}-\frac {15\,f^a\,f^{\frac {b}{x^2}}}{8\,b^3\,x\,{\ln \left (f\right )}^3}+\frac {15\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )}{16\,b^3\,{\ln \left (f\right )}^3\,\sqrt {b\,\ln \left (f\right )}} \]
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