Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {F^a \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{9/2}}{2 d (c+d x)^9} \]
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Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2250} \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {F^a \left (-b \log (F) (c+d x)^2\right )^{9/2} \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right )}{2 d (c+d x)^9} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = -\frac {F^a \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{9/2}}{2 d (c+d x)^9} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {F^a \Gamma \left (-\frac {9}{2},-b (c+d x)^2 \log (F)\right ) \left (-b (c+d x)^2 \log (F)\right )^{9/2}}{2 d (c+d x)^9} \]
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Time = 0.84 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.98
method | result | size |
risch | \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{9 d \left (d x +c \right )^{9}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{63 d \left (d x +c \right )^{7}}-\frac {4 b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{315 d \left (d x +c \right )^{5}}-\frac {8 b^{3} \ln \left (F \right )^{3} F^{b \left (d x +c \right )^{2}} F^{a}}{945 d \left (d x +c \right )^{3}}-\frac {16 b^{4} \ln \left (F \right )^{4} F^{b \left (d x +c \right )^{2}} F^{a}}{945 d \left (d x +c \right )}+\frac {16 b^{5} \ln \left (F \right )^{5} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{945 d \sqrt {-b \ln \left (F \right )}}\) | \(195\) |
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Leaf count of result is larger than twice the leaf count of optimal. 598 vs. \(2 (185) = 370\).
Time = 0.27 (sec) , antiderivative size = 598, normalized size of antiderivative = 12.20 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=-\frac {16 \, \sqrt {\pi } {\left (b^{4} d^{9} x^{9} + 9 \, b^{4} c d^{8} x^{8} + 36 \, b^{4} c^{2} d^{7} x^{7} + 84 \, b^{4} c^{3} d^{6} x^{6} + 126 \, b^{4} c^{4} d^{5} x^{5} + 126 \, b^{4} c^{5} d^{4} x^{4} + 84 \, b^{4} c^{6} d^{3} x^{3} + 36 \, b^{4} c^{7} d^{2} x^{2} + 9 \, b^{4} c^{8} d x + b^{4} c^{9}\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) \log \left (F\right )^{4} + {\left (16 \, {\left (b^{4} d^{9} x^{8} + 8 \, b^{4} c d^{8} x^{7} + 28 \, b^{4} c^{2} d^{7} x^{6} + 56 \, b^{4} c^{3} d^{6} x^{5} + 70 \, b^{4} c^{4} d^{5} x^{4} + 56 \, b^{4} c^{5} d^{4} x^{3} + 28 \, b^{4} c^{6} d^{3} x^{2} + 8 \, b^{4} c^{7} d^{2} x + b^{4} c^{8} d\right )} \log \left (F\right )^{4} + 8 \, {\left (b^{3} d^{7} x^{6} + 6 \, b^{3} c d^{6} x^{5} + 15 \, b^{3} c^{2} d^{5} x^{4} + 20 \, b^{3} c^{3} d^{4} x^{3} + 15 \, b^{3} c^{4} d^{3} x^{2} + 6 \, b^{3} c^{5} d^{2} x + b^{3} c^{6} d\right )} \log \left (F\right )^{3} + 12 \, {\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \left (F\right )^{2} + 30 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + 105 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{945 \, {\left (d^{11} x^{9} + 9 \, c d^{10} x^{8} + 36 \, c^{2} d^{9} x^{7} + 84 \, c^{3} d^{8} x^{6} + 126 \, c^{4} d^{7} x^{5} + 126 \, c^{5} d^{6} x^{4} + 84 \, c^{6} d^{5} x^{3} + 36 \, c^{7} d^{4} x^{2} + 9 \, c^{8} d^{3} x + c^{9} d^{2}\right )}} \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{10}}\, dx \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{10}} \,d x } \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{10}} \,d x } \]
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Time = 0.81 (sec) , antiderivative size = 234, normalized size of antiderivative = 4.78 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^{10}} \, dx=\frac {16\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{9/2}}{945\,d\,{\left (c+d\,x\right )}^9}-\frac {16\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{9/2}}{945\,d\,{\left (c+d\,x\right )}^9}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \left (F\right )}^2}{315\,d\,{\left (c+d\,x\right )}^5}-\frac {8\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^3\,{\ln \left (F\right )}^3}{945\,d\,{\left (c+d\,x\right )}^3}-\frac {16\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^4\,{\ln \left (F\right )}^4}{945\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \left (F\right )}{63\,d\,{\left (c+d\,x\right )}^7}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{9\,d\,{\left (c+d\,x\right )}^9} \]
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