Integrand size = 21, antiderivative size = 87 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4}{4 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log (F)}{4 d}-\frac {b^2 F^a \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log ^2(F)}{4 d} \]
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Time = 0.08 (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} \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=-\frac {b^2 F^a \log ^2(F) \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right )}{4 d}+\frac {(c+d x)^4 F^{a+\frac {b}{(c+d x)^2}}}{4 d}+\frac {b \log (F) (c+d x)^2 F^{a+\frac {b}{(c+d x)^2}}}{4 d} \]
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Rule 2241
Rule 2245
Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4}{4 d}+\frac {1}{2} (b \log (F)) \int F^{a+\frac {b}{(c+d x)^2}} (c+d x) \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4}{4 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log (F)}{4 d}+\frac {1}{2} \left (b^2 \log ^2(F)\right ) \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{c+d x} \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4}{4 d}+\frac {b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \log (F)}{4 d}-\frac {b^2 F^a \text {Ei}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log ^2(F)}{4 d} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.82 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\frac {F^a \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^4+b \log (F) \left (F^{\frac {b}{(c+d x)^2}} (c+d x)^2-b \operatorname {ExpIntegralEi}\left (\frac {b \log (F)}{(c+d x)^2}\right ) \log (F)\right )\right )}{4 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(207\) vs. \(2(81)=162\).
Time = 0.57 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.39
method | result | size |
risch | \(\frac {F^{a} d^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{4}}{4}+F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{3}+\frac {3 F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{2}}{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4}}{4 d}+\frac {F^{a} d b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x^{2}}{4}+\frac {F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c x}{2}+\frac {F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2}}{4 d}+\frac {F^{a} b^{2} \ln \left (F \right )^{2} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (F \right )}{\left (d x +c \right )^{2}}\right )}{4 d}\) | \(208\) |
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none
Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.67 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=-\frac {F^{a} b^{2} {\rm Ei}\left (\frac {b \log \left (F\right )}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) \log \left (F\right )^{2} - {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4} + {\left (b d^{2} x^{2} + 2 \, b c d x + b c^{2}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{4 \, d} \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{3}\, dx \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\int { {\left (d x + c\right )}^{3} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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Time = 0.36 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \, dx=\frac {F^a\,b^2\,{\ln \left (F\right )}^2\,\left (\frac {\mathrm {expint}\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}{2}+F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\left (\frac {{\left (c+d\,x\right )}^2}{2\,b\,\ln \left (F\right )}+\frac {{\left (c+d\,x\right )}^4}{2\,b^2\,{\ln \left (F\right )}^2}\right )\right )}{2\,d} \]
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