Integrand size = 13, antiderivative size = 96 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=-\frac {f^{a+b x^2}}{5 x^5}-\frac {2 b f^{a+b x^2} \log (f)}{15 x^3}-\frac {4 b^2 f^{a+b x^2} \log ^2(f)}{15 x}+\frac {4}{15} b^{5/2} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {5}{2}}(f) \]
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Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2235} \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\frac {4}{15} \sqrt {\pi } b^{5/2} f^a \log ^{\frac {5}{2}}(f) \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right )-\frac {4 b^2 \log ^2(f) f^{a+b x^2}}{15 x}-\frac {f^{a+b x^2}}{5 x^5}-\frac {2 b \log (f) f^{a+b x^2}}{15 x^3} \]
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Rule 2235
Rule 2245
Rubi steps \begin{align*} \text {integral}& = -\frac {f^{a+b x^2}}{5 x^5}+\frac {1}{5} (2 b \log (f)) \int \frac {f^{a+b x^2}}{x^4} \, dx \\ & = -\frac {f^{a+b x^2}}{5 x^5}-\frac {2 b f^{a+b x^2} \log (f)}{15 x^3}+\frac {1}{15} \left (4 b^2 \log ^2(f)\right ) \int \frac {f^{a+b x^2}}{x^2} \, dx \\ & = -\frac {f^{a+b x^2}}{5 x^5}-\frac {2 b f^{a+b x^2} \log (f)}{15 x^3}-\frac {4 b^2 f^{a+b x^2} \log ^2(f)}{15 x}+\frac {1}{15} \left (8 b^3 \log ^3(f)\right ) \int f^{a+b x^2} \, dx \\ & = -\frac {f^{a+b x^2}}{5 x^5}-\frac {2 b f^{a+b x^2} \log (f)}{15 x^3}-\frac {4 b^2 f^{a+b x^2} \log ^2(f)}{15 x}+\frac {4}{15} b^{5/2} f^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {5}{2}}(f) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.80 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\frac {f^a \left (4 b^{5/2} \sqrt {\pi } x^5 \text {erfi}\left (\sqrt {b} x \sqrt {\log (f)}\right ) \log ^{\frac {5}{2}}(f)-f^{b x^2} \left (3+2 b x^2 \log (f)+4 b^2 x^4 \log ^2(f)\right )\right )}{15 x^5} \]
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Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90
method | result | size |
meijerg | \(-\frac {f^{a} \ln \left (f \right )^{\frac {5}{2}} b^{3} \left (-\frac {2 \left (\frac {4 b^{2} x^{4} \ln \left (f \right )^{2}}{3}+\frac {2 b \,x^{2} \ln \left (f \right )}{3}+1\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{5 x^{5} \left (-b \right )^{\frac {5}{2}} \ln \left (f \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (x \sqrt {b}\, \sqrt {\ln \left (f \right )}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{2 \sqrt {-b}}\) | \(86\) |
risch | \(-\frac {f^{a} f^{b \,x^{2}}}{5 x^{5}}-\frac {2 f^{a} \ln \left (f \right ) b \,f^{b \,x^{2}}}{15 x^{3}}-\frac {4 f^{a} \ln \left (f \right )^{2} b^{2} f^{b \,x^{2}}}{15 x}+\frac {4 f^{a} \ln \left (f \right )^{3} b^{3} \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-b \ln \left (f \right )}\, x \right )}{15 \sqrt {-b \ln \left (f \right )}}\) | \(89\) |
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Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=-\frac {4 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} b^{2} f^{a} x^{5} \operatorname {erf}\left (\sqrt {-b \log \left (f\right )} x\right ) \log \left (f\right )^{2} + {\left (4 \, b^{2} x^{4} \log \left (f\right )^{2} + 2 \, b x^{2} \log \left (f\right ) + 3\right )} f^{b x^{2} + a}}{15 \, x^{5}} \]
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\[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\int \frac {f^{a + b x^{2}}}{x^{6}}\, dx \]
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Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.29 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=-\frac {\left (-b x^{2} \log \left (f\right )\right )^{\frac {5}{2}} f^{a} \Gamma \left (-\frac {5}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, x^{5}} \]
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\[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{6}} \,d x } \]
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Time = 0.21 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int \frac {f^{a+b x^2}}{x^6} \, dx=\frac {4\,f^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,x^2\,\ln \left (f\right )}\right )\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{5/2}}{15\,x^5}-\frac {4\,f^a\,\sqrt {\pi }\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{5/2}}{15\,x^5}-\frac {f^a\,f^{b\,x^2}}{5\,x^5}-\frac {4\,b^2\,f^a\,f^{b\,x^2}\,{\ln \left (f\right )}^2}{15\,x}-\frac {2\,b\,f^a\,f^{b\,x^2}\,\ln \left (f\right )}{15\,x^3} \]
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