Integrand size = 19, antiderivative size = 200 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {16 b^{7/2} c^{7/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right ) \log ^{\frac {7}{2}}(F)}{105 e^{9/2}} \]
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Time = 0.14 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2208, 2211, 2235} \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\frac {16 \sqrt {\pi } b^{7/2} c^{7/2} \log ^{\frac {7}{2}}(F) F^{c \left (a-\frac {b d}{e}\right )} \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {\log (F)} \sqrt {d+e x}}{\sqrt {e}}\right )}{105 e^{9/2}}-\frac {16 b^3 c^3 \log ^3(F) F^{c (a+b x)}}{105 e^4 \sqrt {d+e x}}-\frac {8 b^2 c^2 \log ^2(F) F^{c (a+b x)}}{105 e^3 (d+e x)^{3/2}}-\frac {4 b c \log (F) F^{c (a+b x)}}{35 e^2 (d+e x)^{5/2}}-\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}} \]
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Rule 2208
Rule 2211
Rule 2235
Rubi steps \begin{align*} \text {integral}& = -\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}+\frac {(2 b c \log (F)) \int \frac {F^{c (a+b x)}}{(d+e x)^{7/2}} \, dx}{7 e} \\ & = -\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}+\frac {\left (4 b^2 c^2 \log ^2(F)\right ) \int \frac {F^{c (a+b x)}}{(d+e x)^{5/2}} \, dx}{35 e^2} \\ & = -\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}+\frac {\left (8 b^3 c^3 \log ^3(F)\right ) \int \frac {F^{c (a+b x)}}{(d+e x)^{3/2}} \, dx}{105 e^3} \\ & = -\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {\left (16 b^4 c^4 \log ^4(F)\right ) \int \frac {F^{c (a+b x)}}{\sqrt {d+e x}} \, dx}{105 e^4} \\ & = -\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {\left (32 b^4 c^4 \log ^4(F)\right ) \text {Subst}\left (\int F^{c \left (a-\frac {b d}{e}\right )+\frac {b c x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{105 e^5} \\ & = -\frac {2 F^{c (a+b x)}}{7 e (d+e x)^{7/2}}-\frac {4 b c F^{c (a+b x)} \log (F)}{35 e^2 (d+e x)^{5/2}}-\frac {8 b^2 c^2 F^{c (a+b x)} \log ^2(F)}{105 e^3 (d+e x)^{3/2}}-\frac {16 b^3 c^3 F^{c (a+b x)} \log ^3(F)}{105 e^4 \sqrt {d+e x}}+\frac {16 b^{7/2} c^{7/2} F^{c \left (a-\frac {b d}{e}\right )} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {c} \sqrt {d+e x} \sqrt {\log (F)}}{\sqrt {e}}\right ) \log ^{\frac {7}{2}}(F)}{105 e^{9/2}} \\ \end{align*}
Time = 0.33 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.72 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\frac {2 \left (-15 e^3 F^{c (a+b x)}+2 b c (d+e x) \log (F) \left (-3 e^2 F^{c (a+b x)}-2 b c (d+e x) \log (F) \left (2 e F^{c \left (a-\frac {b d}{e}\right )} \Gamma \left (\frac {1}{2},-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{3/2}+F^{c (a+b x)} (e+2 b c (d+e x) \log (F))\right )\right )\right )}{105 e^4 (d+e x)^{7/2}} \]
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\[\int \frac {F^{c \left (b x +a \right )}}{\left (e x +d \right )^{\frac {9}{2}}}d x\]
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Time = 0.26 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.60 \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=-\frac {2 \, {\left (\frac {8 \, \sqrt {\pi } {\left (b^{3} c^{3} e^{4} x^{4} + 4 \, b^{3} c^{3} d e^{3} x^{3} + 6 \, b^{3} c^{3} d^{2} e^{2} x^{2} + 4 \, b^{3} c^{3} d^{3} e x + b^{3} c^{3} d^{4}\right )} \sqrt {-\frac {b c \log \left (F\right )}{e}} \operatorname {erf}\left (\sqrt {e x + d} \sqrt {-\frac {b c \log \left (F\right )}{e}}\right ) \log \left (F\right )^{3}}{F^{\frac {b c d - a c e}{e}}} + {\left (8 \, {\left (b^{3} c^{3} e^{3} x^{3} + 3 \, b^{3} c^{3} d e^{2} x^{2} + 3 \, b^{3} c^{3} d^{2} e x + b^{3} c^{3} d^{3}\right )} \log \left (F\right )^{3} + 15 \, e^{3} + 4 \, {\left (b^{2} c^{2} e^{3} x^{2} + 2 \, b^{2} c^{2} d e^{2} x + b^{2} c^{2} d^{2} e\right )} \log \left (F\right )^{2} + 6 \, {\left (b c e^{3} x + b c d e^{2}\right )} \log \left (F\right )\right )} \sqrt {e x + d} F^{b c x + a c}\right )}}{105 \, {\left (e^{8} x^{4} + 4 \, d e^{7} x^{3} + 6 \, d^{2} e^{6} x^{2} + 4 \, d^{3} e^{5} x + d^{4} e^{4}\right )}} \]
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Timed out. \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {9}{2}}} \,d x } \]
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\[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (e x + d\right )}^{\frac {9}{2}}} \,d x } \]
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Timed out. \[ \int \frac {F^{c (a+b x)}}{(d+e x)^{9/2}} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (d+e\,x\right )}^{9/2}} \,d x \]
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