Integrand size = 13, antiderivative size = 100 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{15 \sqrt {x}}+\frac {8}{15} b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F) \]
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Time = 0.06 (sec) , antiderivative size = 100, 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 = {2208, 2211, 2235} \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\frac {8}{15} \sqrt {\pi } b^{5/2} F^a \log ^{\frac {5}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right )-\frac {8 b^2 \log ^2(F) F^{a+b x}}{15 \sqrt {x}}-\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b \log (F) F^{a+b x}}{15 x^{3/2}} \]
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Rule 2208
Rule 2211
Rule 2235
Rubi steps \begin{align*} \text {integral}& = -\frac {2 F^{a+b x}}{5 x^{5/2}}+\frac {1}{5} (2 b \log (F)) \int \frac {F^{a+b x}}{x^{5/2}} \, dx \\ & = -\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}+\frac {1}{15} \left (4 b^2 \log ^2(F)\right ) \int \frac {F^{a+b x}}{x^{3/2}} \, dx \\ & = -\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{15 \sqrt {x}}+\frac {1}{15} \left (8 b^3 \log ^3(F)\right ) \int \frac {F^{a+b x}}{\sqrt {x}} \, dx \\ & = -\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{15 \sqrt {x}}+\frac {1}{15} \left (16 b^3 \log ^3(F)\right ) \text {Subst}\left (\int F^{a+b x^2} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {2 F^{a+b x}}{5 x^{5/2}}-\frac {4 b F^{a+b x} \log (F)}{15 x^{3/2}}-\frac {8 b^2 F^{a+b x} \log ^2(F)}{15 \sqrt {x}}+\frac {8}{15} b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} \sqrt {x} \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F) \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.61 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 F^a \left (-4 \Gamma \left (\frac {1}{2},-b x \log (F)\right ) (-b x \log (F))^{5/2}+F^{b x} \left (3+2 b x \log (F)+4 b^2 x^2 \log ^2(F)\right )\right )}{15 x^{5/2}} \]
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Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.84
method | result | size |
meijerg | \(-\frac {F^{a} \left (-b \right )^{\frac {7}{2}} \ln \left (F \right )^{\frac {5}{2}} \left (-\frac {2 \left (\frac {4 b^{2} x^{2} \ln \left (F \right )^{2}}{3}+\frac {2 x b \ln \left (F \right )}{3}+1\right ) {\mathrm e}^{x b \ln \left (F \right )}}{5 x^{\frac {5}{2}} \left (-b \right )^{\frac {5}{2}} \ln \left (F \right )^{\frac {5}{2}}}+\frac {8 b^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\sqrt {b}\, \sqrt {x}\, \sqrt {\ln \left (F \right )}\right )}{15 \left (-b \right )^{\frac {5}{2}}}\right )}{b}\) | \(84\) |
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Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {2 \, {\left (4 \, \sqrt {\pi } \sqrt {-b \log \left (F\right )} F^{a} b^{2} x^{3} \operatorname {erf}\left (\sqrt {-b \log \left (F\right )} \sqrt {x}\right ) \log \left (F\right )^{2} + {\left (4 \, b^{2} x^{2} \log \left (F\right )^{2} + 2 \, b x \log \left (F\right ) + 3\right )} F^{b x + a} \sqrt {x}\right )}}{15 \, x^{3}} \]
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\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\int \frac {F^{a + b x}}{x^{\frac {7}{2}}}\, dx \]
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Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.24 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {\left (-b x \log \left (F\right )\right )^{\frac {5}{2}} F^{a} \Gamma \left (-\frac {5}{2}, -b x \log \left (F\right )\right )}{x^{\frac {5}{2}}} \]
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\[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=\int { \frac {F^{b x + a}}{x^{\frac {7}{2}}} \,d x } \]
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Time = 0.10 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int \frac {F^{a+b x}}{x^{7/2}} \, dx=-\frac {\frac {2\,F^{a+b\,x}}{5}+\frac {4\,F^{a+b\,x}\,b\,x\,\ln \left (F\right )}{15}+\frac {8\,F^{a+b\,x}\,b^2\,x^2\,{\ln \left (F\right )}^2}{15}-\frac {8\,F^a\,b^2\,x^2\,\mathrm {erfc}\left (\sqrt {-b\,x\,\ln \left (F\right )}\right )\,{\ln \left (F\right )}^2\,\sqrt {-\pi \,b\,x\,\ln \left (F\right )}}{15}}{x^{5/2}} \]
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