Integrand size = 21, antiderivative size = 49 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=-\frac {F^a \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{4/3}}{3 d (c+d x)^4} \]
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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} \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=-\frac {F^a \left (-b \log (F) (c+d x)^3\right )^{4/3} \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right )}{3 d (c+d x)^4} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = -\frac {F^a \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{4/3}}{3 d (c+d x)^4} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=-\frac {F^a \Gamma \left (-\frac {4}{3},-b (c+d x)^3 \log (F)\right ) \left (-b (c+d x)^3 \log (F)\right )^{4/3}}{3 d (c+d x)^4} \]
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\[\int \frac {F^{a +b \left (d x +c \right )^{3}}}{\left (d x +c \right )^{5}}d x\]
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Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (43) = 86\).
Time = 0.09 (sec) , antiderivative size = 226, normalized size of antiderivative = 4.61 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\frac {3 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \left (-b d^{3} \log \left (F\right )\right )^{\frac {1}{3}} F^{a} \Gamma \left (\frac {2}{3}, -{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right ) \log \left (F\right ) - {\left (3 \, {\left (b d^{4} x^{3} + 3 \, b c d^{3} x^{2} + 3 \, b c^{2} d^{2} x + b c^{3} d\right )} \log \left (F\right ) + d\right )} F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{4 \, {\left (d^{6} x^{4} + 4 \, c d^{5} x^{3} + 6 \, c^{2} d^{4} x^{2} + 4 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} \]
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\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{3}}}{\left (c + d x\right )^{5}}\, dx \]
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\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \]
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\[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{3} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \]
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Time = 1.00 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.65 \[ \int \frac {F^{a+b (c+d x)^3}}{(c+d x)^5} \, dx=\frac {3\,F^a\,\Gamma \left (\frac {2}{3}\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}^{4/3}}{4\,d\,{\left (c+d\,x\right )}^4}-\frac {F^a\,F^{b\,{\left (c+d\,x\right )}^3}}{4\,d\,{\left (c+d\,x\right )}^4}-\frac {3\,F^a\,\Gamma \left (\frac {2}{3},-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )\,{\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3\right )}^{4/3}}{4\,d\,{\left (c+d\,x\right )}^4}-\frac {3\,F^a\,F^{b\,{\left (c+d\,x\right )}^3}\,b\,\ln \left (F\right )}{4\,d\,\left (c+d\,x\right )} \]
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