Integrand size = 21, antiderivative size = 183 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=-\frac {105 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)} \]
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Time = 0.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2243, 2242, 2235} \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=-\frac {105 \sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d \log ^4(F) (c+d x)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)^3}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^5}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^7} \]
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Rule 2235
Rule 2242
Rule 2243
Rubi steps \begin{align*} \text {integral}& = -\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}-\frac {7 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx}{2 b \log (F)} \\ & = \frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}+\frac {35 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx}{4 b^2 \log ^2(F)} \\ & = -\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}-\frac {105 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx}{8 b^3 \log ^3(F)} \\ & = \frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}+\frac {105 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx}{16 b^4 \log ^4(F)} \\ & = \frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)}-\frac {105 \text {Subst}\left (\int F^{a+b x^2} \, dx,x,\frac {1}{c+d x}\right )}{16 b^4 d \log ^4(F)} \\ & = -\frac {105 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)}+\frac {105 F^{a+\frac {b}{(c+d x)^2}}}{16 b^4 d (c+d x) \log ^4(F)}-\frac {35 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x)^3 \log ^3(F)}+\frac {7 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^5 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^7 \log (F)} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.69 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\frac {F^a \left (-105 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )+\frac {2 \sqrt {b} F^{\frac {b}{(c+d x)^2}} \sqrt {\log (F)} \left (105 (c+d x)^6-70 b (c+d x)^4 \log (F)+28 b^2 (c+d x)^2 \log ^2(F)-8 b^3 \log ^3(F)\right )}{(c+d x)^7}\right )}{32 b^{9/2} d \log ^{\frac {9}{2}}(F)} \]
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Time = 4.23 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.96
method | result | size |
risch | \(-\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 d \left (d x +c \right )^{7} b \ln \left (F \right )}+\frac {7 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 d \,b^{2} \ln \left (F \right )^{2} \left (d x +c \right )^{5}}-\frac {35 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{8 d \,b^{3} \ln \left (F \right )^{3} \left (d x +c \right )^{3}}+\frac {105 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{16 d \,b^{4} \ln \left (F \right )^{4} \left (d x +c \right )}-\frac {105 F^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (F \right )}}{d x +c}\right )}{32 d \,b^{4} \ln \left (F \right )^{4} \sqrt {-b \ln \left (F \right )}}\) | \(175\) |
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Leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (163) = 326\).
Time = 0.28 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.40 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\frac {105 \, \sqrt {\pi } {\left (d^{8} x^{7} + 7 \, c d^{7} x^{6} + 21 \, c^{2} d^{6} x^{5} + 35 \, c^{3} d^{5} x^{4} + 35 \, c^{4} d^{4} x^{3} + 21 \, c^{5} d^{3} x^{2} + 7 \, c^{6} d^{2} x + c^{7} d\right )} F^{a} \sqrt {-\frac {b \log \left (F\right )}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) - 2 \, {\left (8 \, b^{4} \log \left (F\right )^{4} - 28 \, {\left (b^{3} d^{2} x^{2} + 2 \, b^{3} c d x + b^{3} c^{2}\right )} \log \left (F\right )^{3} + 70 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )} \log \left (F\right )^{2} - 105 \, {\left (b d^{6} x^{6} + 6 \, b c d^{5} x^{5} + 15 \, b c^{2} d^{4} x^{4} + 20 \, b c^{3} d^{3} x^{3} + 15 \, b c^{4} d^{2} x^{2} + 6 \, b c^{5} d x + b c^{6}\right )} \log \left (F\right )\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{32 \, {\left (b^{5} d^{8} x^{7} + 7 \, b^{5} c d^{7} x^{6} + 21 \, b^{5} c^{2} d^{6} x^{5} + 35 \, b^{5} c^{3} d^{5} x^{4} + 35 \, b^{5} c^{4} d^{4} x^{3} + 21 \, b^{5} c^{5} d^{3} x^{2} + 7 \, b^{5} c^{6} d^{2} x + b^{5} c^{7} d\right )} \log \left (F\right )^{5}} \]
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Timed out. \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\text {Timed out} \]
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\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{10}} \,d x } \]
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\[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=\int { \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{10}} \,d x } \]
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Time = 1.25 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.87 \[ \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^{10}} \, dx=-\frac {\frac {F^a\,\left (105\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (F\right )}{\sqrt {b\,\ln \left (F\right )}\,\left (c+d\,x\right )}\right )-\frac {210\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,\sqrt {b\,\ln \left (F\right )}}{c+d\,x}\right )}{32\,\sqrt {b\,\ln \left (F\right )}}-\frac {7\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^2\,{\ln \left (F\right )}^2}{4\,{\left (c+d\,x\right )}^5}+\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b^3\,{\ln \left (F\right )}^3}{2\,{\left (c+d\,x\right )}^7}+\frac {35\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}\,b\,\ln \left (F\right )}{8\,{\left (c+d\,x\right )}^3}}{b^4\,d\,{\ln \left (F\right )}^4} \]
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