Optimal. Leaf size=149 \[ \frac {15 \sqrt {\pi } F^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^3}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5} \]
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Rubi [A] time = 0.21, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2212, 2211, 2204} \[ \frac {15 \sqrt {\pi } F^a \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d \log ^2(F) (c+d x)^3}-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d \log ^3(F) (c+d x)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d \log (F) (c+d x)^5} \]
Antiderivative was successfully verified.
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Rule 2204
Rule 2211
Rule 2212
Rubi steps
\begin {align*} \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^8} \, dx &=-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}-\frac {5 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^6} \, dx}{2 b \log (F)}\\ &=\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}+\frac {15 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^4} \, dx}{4 b^2 \log ^2(F)}\\ &=-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}-\frac {15 \int \frac {F^{a+\frac {b}{(c+d x)^2}}}{(c+d x)^2} \, dx}{8 b^3 \log ^3(F)}\\ &=-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}+\frac {15 \operatorname {Subst}\left (\int F^{a+b x^2} \, dx,x,\frac {1}{c+d x}\right )}{8 b^3 d \log ^3(F)}\\ &=\frac {15 F^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)}-\frac {15 F^{a+\frac {b}{(c+d x)^2}}}{8 b^3 d (c+d x) \log ^3(F)}+\frac {5 F^{a+\frac {b}{(c+d x)^2}}}{4 b^2 d (c+d x)^3 \log ^2(F)}-\frac {F^{a+\frac {b}{(c+d x)^2}}}{2 b d (c+d x)^5 \log (F)}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 111, normalized size = 0.74 \[ \frac {F^a \left (15 \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (F)}}{c+d x}\right )-\frac {2 \sqrt {b} \sqrt {\log (F)} F^{\frac {b}{(c+d x)^2}} \left (4 b^2 \log ^2(F)-10 b \log (F) (c+d x)^2+15 (c+d x)^4\right )}{(c+d x)^5}\right )}{16 b^{7/2} d \log ^{\frac {7}{2}}(F)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.43, size = 305, normalized size = 2.05 \[ -\frac {15 \, \sqrt {\pi } {\left (d^{6} x^{5} + 5 \, c d^{5} x^{4} + 10 \, c^{2} d^{4} x^{3} + 10 \, c^{3} d^{3} x^{2} + 5 \, c^{4} d^{2} x + c^{5} d\right )} F^{a} \sqrt {-\frac {b \log \relax (F)}{d^{2}}} \operatorname {erf}\left (\frac {d \sqrt {-\frac {b \log \relax (F)}{d^{2}}}}{d x + c}\right ) + 2 \, {\left (4 \, b^{3} \log \relax (F)^{3} - 10 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \log \relax (F)^{2} + 15 \, {\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + 4 \, b c^{3} d x + b c^{4}\right )} \log \relax (F)\right )} F^{\frac {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{16 \, {\left (b^{4} d^{6} x^{5} + 5 \, b^{4} c d^{5} x^{4} + 10 \, b^{4} c^{2} d^{4} x^{3} + 10 \, b^{4} c^{3} d^{3} x^{2} + 5 \, b^{4} c^{4} d^{2} x + b^{4} c^{5} d\right )} \log \relax (F)^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{8}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 142, normalized size = 0.95 \[ -\frac {F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{2 \left (d x +c \right )^{5} b d \ln \relax (F )}+\frac {5 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{4 \left (d x +c \right )^{3} b^{2} d \ln \relax (F )^{2}}-\frac {15 F^{a} F^{\frac {b}{\left (d x +c \right )^{2}}}}{8 \left (d x +c \right ) b^{3} d \ln \relax (F )^{3}}+\frac {15 \sqrt {\pi }\, F^{a} \erf \left (\frac {\sqrt {-b \ln \relax (F )}}{d x +c}\right )}{16 \sqrt {-b \ln \relax (F )}\, b^{3} d \ln \relax (F )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {F^{a + \frac {b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{8}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.37, size = 142, normalized size = 0.95 \[ \frac {5\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{4\,b^2\,d\,{\ln \relax (F)}^2\,{\left (c+d\,x\right )}^3}-\frac {F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{2\,b\,d\,\ln \relax (F)\,{\left (c+d\,x\right )}^5}-\frac {15\,F^a\,F^{\frac {b}{{\left (c+d\,x\right )}^2}}}{8\,b^3\,d\,{\ln \relax (F)}^3\,\left (c+d\,x\right )}+\frac {15\,F^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (F)}{\sqrt {b\,\ln \relax (F)}\,\left (c+d\,x\right )}\right )}{16\,b^3\,d\,{\ln \relax (F)}^3\,\sqrt {b\,\ln \relax (F)}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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