Integrand size = 21, antiderivative size = 49 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a (c+d x)^4 \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}}{3 d} \]
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Time = 0.03 (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} \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a (c+d x)^4 \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3} \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right )}{3 d} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = \frac {F^a (c+d x)^4 \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}}{3 d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a (c+d x)^4 \Gamma \left (-\frac {4}{3},-\frac {b \log (F)}{(c+d x)^3}\right ) \left (-\frac {b \log (F)}{(c+d x)^3}\right )^{4/3}}{3 d} \]
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\[\int F^{a +\frac {b}{\left (d x +c \right )^{3}}} \left (d x +c \right )^{3}d x\]
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Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (43) = 86\).
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 3.63 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=-\frac {3 \, F^{a} b d \left (-\frac {b \log \left (F\right )}{d^{3}}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, -\frac {b \log \left (F\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) \log \left (F\right ) - {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4} + 3 \, {\left (b d x + b c\right )} \log \left (F\right )\right )} F^{\frac {a d^{3} x^{3} + 3 \, a c d^{2} x^{2} + 3 \, a c^{2} d x + a c^{3} + b}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}}}{4 \, d} \]
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\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{3}}} \left (c + d x\right )^{3}\, dx \]
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\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]
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\[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{3}}} \,d x } \]
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Time = 0.75 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.61 \[ \int F^{a+\frac {b}{(c+d x)^3}} (c+d x)^3 \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,{\left (c+d\,x\right )}^4}{4\,d}-\frac {3\,F^a\,\Gamma \left (\frac {2}{3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}{4\,d}+\frac {3\,F^a\,\Gamma \left (\frac {2}{3},-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )\,{\left (c+d\,x\right )}^4\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^3}\right )}^{4/3}}{4\,d}+\frac {3\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^3}}\,b\,\ln \left (F\right )\,\left (c+d\,x\right )}{4\,d} \]
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