Integrand size = 25, antiderivative size = 31 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\frac {b^5 F^a \Gamma \left (-5,-b (c+d x)^n \log (F)\right ) \log ^5(F)}{d n} \]
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Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2250} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\frac {b^5 F^a \log ^5(F) \Gamma \left (-5,-b (c+d x)^n \log (F)\right )}{d n} \]
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Rule 2250
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 F^a \Gamma \left (-5,-b (c+d x)^n \log (F)\right ) \log ^5(F)}{d n} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\frac {b^5 F^a \Gamma \left (-5,-b (c+d x)^n \log (F)\right ) \log ^5(F)}{d n} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(173\) vs. \(2(34)=68\).
Time = 0.64 (sec) , antiderivative size = 174, normalized size of antiderivative = 5.61
method | result | size |
risch | \(-\frac {F^{a} \left (\operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{n} \ln \left (F \right )\right ) \ln \left (F \right )^{5} b^{5} \left (d x +c \right )^{5 n}+F^{b \left (d x +c \right )^{n}} \ln \left (F \right )^{4} b^{4} \left (d x +c \right )^{4 n}+F^{b \left (d x +c \right )^{n}} \ln \left (F \right )^{3} b^{3} \left (d x +c \right )^{3 n}+2 F^{b \left (d x +c \right )^{n}} \ln \left (F \right )^{2} b^{2} \left (d x +c \right )^{2 n}+6 F^{b \left (d x +c \right )^{n}} \ln \left (F \right ) b \left (d x +c \right )^{n}+24 F^{b \left (d x +c \right )^{n}}\right ) \left (d x +c \right )^{-5 n}}{120 n d}\) | \(174\) |
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Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (36) = 72\).
Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 4.42 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\frac {{\left (d x + c\right )}^{5 \, n} F^{a} b^{5} {\rm Ei}\left ({\left (d x + c\right )}^{n} b \log \left (F\right )\right ) \log \left (F\right )^{5} - {\left ({\left (d x + c\right )}^{4 \, n} b^{4} \log \left (F\right )^{4} + {\left (d x + c\right )}^{3 \, n} b^{3} \log \left (F\right )^{3} + 2 \, {\left (d x + c\right )}^{2 \, n} b^{2} \log \left (F\right )^{2} + 6 \, {\left (d x + c\right )}^{n} b \log \left (F\right ) + 24\right )} e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{120 \, {\left (d x + c\right )}^{5 \, n} d n} \]
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\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\int F^{a + b \left (c + d x\right )^{n}} \left (c + d x\right )^{- 5 n - 1}\, dx \]
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\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\int { {\left (d x + c\right )}^{-5 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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\[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\int { {\left (d x + c\right )}^{-5 \, n - 1} F^{{\left (d x + c\right )}^{n} b + a} \,d x } \]
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Timed out. \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1-5 n} \, dx=\int \frac {F^{a+b\,{\left (c+d\,x\right )}^n}}{{\left (c+d\,x\right )}^{5\,n+1}} \,d x \]
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