Integrand size = 21, antiderivative size = 136 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac {4 b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}-\frac {4 b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {5}{2}}(F)}{15 d} \]
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Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2245, 2237, 2242, 2235} \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=-\frac {4 \sqrt {\pi } b^{5/2} F^a \log ^{\frac {5}{2}}(F) \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{15 d}+\frac {4 b^2 \log ^2(F) (c+d x) F^{a+\frac {b}{(c+d x)^2}}}{15 d}+\frac {(c+d x)^5 F^{a+\frac {b}{(c+d x)^2}}}{5 d}+\frac {2 b \log (F) (c+d x)^3 F^{a+\frac {b}{(c+d x)^2}}}{15 d} \]
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Rule 2235
Rule 2237
Rule 2242
Rule 2245
Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {1}{5} (2 b \log (F)) \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^2 \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac {1}{15} \left (4 b^2 \log ^2(F)\right ) \int F^{a+\frac {b}{(c+d x)^2}} \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac {4 b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}+\frac {1}{15} \left (8 b^3 \log ^3(F)\right ) \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac {4 b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}-\frac {\left (8 b^3 \log ^3(F)\right ) \text {Subst}\left (\int F^{a+b x^2} \, dx,x,\frac {1}{c+d x}\right )}{15 d} \\ & = \frac {F^{a+\frac {b}{(c+d x)^2}} (c+d x)^5}{5 d}+\frac {2 b F^{a+\frac {b}{(c+d x)^2}} (c+d x)^3 \log (F)}{15 d}+\frac {4 b^2 F^{a+\frac {b}{(c+d x)^2}} (c+d x) \log ^2(F)}{15 d}-\frac {4 b^{5/2} F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {5}{2}}(F)}{15 d} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.71 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {F^a \left (-4 b^{5/2} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right ) \log ^{\frac {5}{2}}(F)+F^{\frac {b}{(c+d x)^2}} (c+d x) \left (3 (c+d x)^4+2 b (c+d x)^2 \log (F)+4 b^2 \log ^2(F)\right )\right )}{15 d} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(118)=236\).
Time = 0.74 (sec) , antiderivative size = 324, normalized size of antiderivative = 2.38
method | result | size |
risch | \(\frac {F^{a} d^{4} F^{\frac {b}{\left (d x +c \right )^{2}}} x^{5}}{5}+F^{a} d^{3} F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{4}+2 F^{a} d^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x^{3}+2 F^{a} d \,F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3} x^{2}+F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{4} x +\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}} c^{5}}{5 d}+\frac {2 F^{a} d^{2} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} x^{3}}{15}+\frac {2 F^{a} d b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c \,x^{2}}{5}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{2} x}{5}+\frac {2 F^{a} b \ln \left (F \right ) F^{\frac {b}{\left (d x +c \right )^{2}}} c^{3}}{15 d}+\frac {4 F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} x}{15}+\frac {4 F^{a} b^{2} \ln \left (F \right )^{2} F^{\frac {b}{\left (d x +c \right )^{2}}} c}{15 d}-\frac {4 F^{a} b^{3} \ln \left (F \right )^{3} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{15 d \sqrt {-b \ln \left (F \right )}}\) | \(324\) |
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Time = 0.26 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.48 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {4 \, \sqrt {\pi } F^{a} b^{2} d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) \log \left (F\right )^{2} + {\left (3 \, d^{5} x^{5} + 15 \, c d^{4} x^{4} + 30 \, c^{2} d^{3} x^{3} + 30 \, c^{3} d^{2} x^{2} + 15 \, c^{4} d x + 3 \, c^{5} + 4 \, {\left (b^{2} d x + b^{2} c\right )} \log \left (F\right )^{2} + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{15 \, d} \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\int F^{a + \frac {b}{\left (c + d x\right )^{2}}} \left (c + d x\right )^{4}\, dx \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\int { {\left (d x + c\right )}^{4} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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\[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\int { {\left (d x + c\right )}^{4} F^{a + \frac {b}{{\left (d x + c\right )}^{2}}} \,d x } \]
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Time = 0.68 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.22 \[ \int F^{a+\frac {b}{(c+d x)^2}} (c+d x)^4 \, dx=\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,{\left (c+d\,x\right )}^5}{5\,d}+\frac {4\,F^a\,\sqrt {\pi }\,{\left (c+d\,x\right )}^5\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{5/2}}{15\,d}+\frac {2\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )\,{\left (c+d\,x\right )}^3}{15\,d}+\frac {4\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2\,\left (c+d\,x\right )}{15\,d}-\frac {4\,F^a\,\sqrt {\pi }\,\mathrm {erfc}\left (\sqrt {-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}}\right )\,{\left (c+d\,x\right )}^5\,{\left (-\frac {b\,\ln \left (F\right )}{{\left (c+d\,x\right )}^2}\right )}^{5/2}}{15\,d} \]
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