Integrand size = 21, antiderivative size = 87 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b F^{a+b (c+d x)^2} \log (F)}{4 d (c+d x)^2}+\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^2(F)}{4 d} \]
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Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2245, 2241} \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\frac {b^2 F^a \log ^2(F) \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right )}{4 d}-\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b \log (F) F^{a+b (c+d x)^2}}{4 d (c+d x)^2} \]
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Rule 2241
Rule 2245
Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}+\frac {1}{2} (b \log (F)) \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^3} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b F^{a+b (c+d x)^2} \log (F)}{4 d (c+d x)^2}+\frac {1}{2} \left (b^2 \log ^2(F)\right ) \int \frac {F^{a+b (c+d x)^2}}{c+d x} \, dx \\ & = -\frac {F^{a+b (c+d x)^2}}{4 d (c+d x)^4}-\frac {b F^{a+b (c+d x)^2} \log (F)}{4 d (c+d x)^2}+\frac {b^2 F^a \text {Ei}\left (b (c+d x)^2 \log (F)\right ) \log ^2(F)}{4 d} \\ \end{align*}
Time = 0.21 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\frac {F^a \left (b^2 \operatorname {ExpIntegralEi}\left (b (c+d x)^2 \log (F)\right ) \log ^2(F)-\frac {F^{b (c+d x)^2} \left (1+b (c+d x)^2 \log (F)\right )}{(c+d x)^4}\right )}{4 d} \]
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Time = 0.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.99
method | result | size |
risch | \(-\frac {F^{b \left (d x +c \right )^{2}} F^{a}}{4 d \left (d x +c \right )^{4}}-\frac {b \ln \left (F \right ) F^{b \left (d x +c \right )^{2}} F^{a}}{4 d \left (d x +c \right )^{2}}-\frac {b^{2} \ln \left (F \right )^{2} F^{a} \operatorname {Ei}_{1}\left (-b \left (d x +c \right )^{2} \ln \left (F \right )\right )}{4 d}\) | \(86\) |
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Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (81) = 162\).
Time = 0.27 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.10 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\frac {{\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} F^{a} {\rm Ei}\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right ) \log \left (F\right )^{2} - {\left ({\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right ) + 1\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{4 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 6 \, c^{2} d^{3} x^{2} + 4 \, c^{3} d^{2} x + c^{4} d\right )}} \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\int \frac {F^{a + b \left (c + d x\right )^{2}}}{\left (c + d x\right )^{5}}\, dx \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \]
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\[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=\int { \frac {F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{5}} \,d x } \]
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Time = 2.18 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int \frac {F^{a+b (c+d x)^2}}{(c+d x)^5} \, dx=-\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2\right )}{2}+F^{b\,{\left (c+d\,x\right )}^2}\,\left (\frac {1}{2\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^2}+\frac {1}{2\,b^2\,{\ln \left (F\right )}^2\,{\left (c+d\,x\right )}^4}\right )\right )}{2\,d} \]
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