Integrand size = 21, antiderivative size = 61 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {1+m}{2}}}{2 d} \]
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Time = 0.03 (sec) , antiderivative size = 61, 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} \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{m+1} \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),-\frac {b \log (F)}{(c+d x)^2}\right )}{2 d} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = \frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {1+m}{2}}}{2 d} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a (c+d x)^{1+m} \Gamma \left (\frac {1}{2} (-1-m),-\frac {b \log (F)}{(c+d x)^2}\right ) \left (-\frac {b \log (F)}{(c+d x)^2}\right )^{\frac {1+m}{2}}}{2 d} \]
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\[\int F^{a +\frac {b}{\left (d x +c \right )^{2}}} \left (d x +c \right )^{m}d x\]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{m}\, dx \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\int { {\left (d x + c\right )}^{m} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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Time = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^m \, dx=\frac {F^a\,{\mathrm {e}}^{\frac {b\,\ln \left (F\right )}{2\,{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^{m+1}\,{\mathrm {M}}_{\frac {m}{4}+\frac {3}{4},-\frac {m}{4}-\frac {1}{4}}\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )\,{\left (\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{\frac {m}{4}-\frac {1}{4}}}{d\,\left (m+1\right )} \]
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