Integrand size = 21, antiderivative size = 136 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac {4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac {4 b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F)}{15 d} \]
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Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2235} \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=\frac {4 \sqrt {\pi } b^{5/2} F^a \log ^{\frac {5}{2}}(F) \text {erfi}\left (\sqrt {b} \sqrt {\log (F)} (c+d x)\right )}{15 d}-\frac {4 b^2 \log ^2(F) F^{a+b (c+d x)^2}}{15 d (c+d x)}-\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b \log (F) F^{a+b (c+d x)^2}}{15 d (c+d x)^3} \]
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Rule 2235
Rule 2245
Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}+\frac {1}{5} (2 b \log (F)) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^4} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}+\frac {1}{15} \left (4 b^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^2} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac {4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac {1}{15} \left (8 b^3 \log ^3(F)\right ) \int F^{a+b (c+d x)^2} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{5 d (c+d x)^5}-\frac {2 b F^{a+b (c+d x)^2} \log (F)}{15 d (c+d x)^3}-\frac {4 b^2 F^{a+b (c+d x)^2} \log ^2(F)}{15 d (c+d x)}+\frac {4 b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F)}{15 d} \\ \end{align*}
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=\frac {F^a \left (4 b^{5/2} \sqrt {\pi } \text {erfi}\left (\sqrt {b} (c+d x) \sqrt {\log (F)}\right ) \log ^{\frac {5}{2}}(F)-\frac {F^{b (c+d x)^2} \left (3+2 b (c+d x)^2 \log (F)+4 b^2 (c+d x)^4 \log ^2(F)\right )}{(c+d x)^5}\right )}{15 d} \]
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Time = 0.47 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.95
method | result | size |
risch | \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{5 d \left (d x +c \right )^{5}}-\frac {2 b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{15 d \left (d x +c \right )^{3}}-\frac {4 b^{2} \ln \left (F \right )^{2} F^{b \left (d x +c \right )^{2}} F^{a}}{15 d \left (d x +c \right )}+\frac {4 b^{3} \ln \left (F \right )^{3} \sqrt {\pi }\, F^{a} \operatorname {erf}\left (\sqrt {-b \ln \left (F \right )}\, \left (d x +c \right )\right )}{15 d \sqrt {-b \ln \left (F \right )}}\) | \(129\) |
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Leaf count of result is larger than twice the leaf count of optimal. 288 vs. \(2 (118) = 236\).
Time = 0.32 (sec) , antiderivative size = 288, normalized size of antiderivative = 2.12 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=-\frac {4 \, \sqrt {\pi } {\left (b^{2} d^{5} x^{5} + 5 \, b^{2} c d^{4} x^{4} + 10 \, b^{2} c^{2} d^{3} x^{3} + 10 \, b^{2} c^{3} d^{2} x^{2} + 5 \, b^{2} c^{4} d x + b^{2} c^{5}\right )} \sqrt {-b d^{2} \log \left (F\right )} F^{a} \operatorname {erf}\left (\frac {\sqrt {-b d^{2} \log \left (F\right )} {\left (d x + c\right )}}{d}\right ) \log \left (F\right )^{2} + {\left (4 \, {\left (b^{2} d^{5} x^{4} + 4 \, b^{2} c d^{4} x^{3} + 6 \, b^{2} c^{2} d^{3} x^{2} + 4 \, b^{2} c^{3} d^{2} x + b^{2} c^{4} d\right )} \log \left (F\right )^{2} + 2 \, {\left (b d^{3} x^{2} + 2 \, b c d^{2} x + b c^{2} d\right )} \log \left (F\right ) + 3 \, d\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{15 \, {\left (d^{7} x^{5} + 5 \, c d^{6} x^{4} + 10 \, c^{2} d^{5} x^{3} + 10 \, c^{3} d^{4} x^{2} + 5 \, c^{4} d^{3} x + c^{5} d^{2}\right )}} \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{6}}\, dx \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{6}} \,d x } \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{6}} \,d x } \]
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Time = 1.47 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.24 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^6} \, dx=\frac {4\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{5/2}}{15\,d\,{\left (c+d\,x\right )}^5}-\frac {4\,F^a\,\sqrt {\pi }\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}^{5/2}}{15\,d\,{\left (c+d\,x\right )}^5}-\frac {4\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b^2\,{\ln \left (F\right )}^2}{15\,d\,\left (c+d\,x\right )}-\frac {2\,F^a\,F^{b\,{\left (c+d\,x\right )}^2}\,b\,\ln \left (F\right )}{15\,d\,{\left (c+d\,x\right )}^3}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^2}}{5\,d\,{\left (c+d\,x\right )}^5} \]
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