Integrand size = 13, antiderivative size = 65 \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^{a+b x^2} \left (24-24 b x^2 \log (f)+12 b^2 x^4 \log ^2(f)-4 b^3 x^6 \log ^3(f)+b^4 x^8 \log ^4(f)\right )}{2 b^5 \log ^5(f)} \]
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Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2249} \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^{a+b x^2} \left (b^4 x^8 \log ^4(f)-4 b^3 x^6 \log ^3(f)+12 b^2 x^4 \log ^2(f)-24 b x^2 \log (f)+24\right )}{2 b^5 \log ^5(f)} \]
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Rule 2249
Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^2} \left (24-24 b x^2 \log (f)+12 b^2 x^4 \log ^2(f)-4 b^3 x^6 \log ^3(f)+b^4 x^8 \log ^4(f)\right )}{2 b^5 \log ^5(f)} \\ \end{align*}
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.37 \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^a \Gamma \left (5,-b x^2 \log (f)\right )}{2 b^5 \log ^5(f)} \]
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Time = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98
method | result | size |
gosper | \(\frac {f^{b \,x^{2}+a} \left (24-24 b \,x^{2} \ln \left (f \right )+12 b^{2} x^{4} \ln \left (f \right )^{2}-4 b^{3} x^{6} \ln \left (f \right )^{3}+b^{4} x^{8} \ln \left (f \right )^{4}\right )}{2 b^{5} \ln \left (f \right )^{5}}\) | \(64\) |
risch | \(\frac {f^{b \,x^{2}+a} \left (24-24 b \,x^{2} \ln \left (f \right )+12 b^{2} x^{4} \ln \left (f \right )^{2}-4 b^{3} x^{6} \ln \left (f \right )^{3}+b^{4} x^{8} \ln \left (f \right )^{4}\right )}{2 b^{5} \ln \left (f \right )^{5}}\) | \(64\) |
meijerg | \(-\frac {f^{a} \left (24-\frac {\left (5 b^{4} x^{8} \ln \left (f \right )^{4}-20 b^{3} x^{6} \ln \left (f \right )^{3}+60 b^{2} x^{4} \ln \left (f \right )^{2}-120 b \,x^{2} \ln \left (f \right )+120\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{5}\right )}{2 b^{5} \ln \left (f \right )^{5}}\) | \(71\) |
parallelrisch | \(\frac {f^{b \,x^{2}+a} x^{8} \ln \left (f \right )^{4} b^{4}-4 f^{b \,x^{2}+a} x^{6} \ln \left (f \right )^{3} b^{3}+12 f^{b \,x^{2}+a} x^{4} \ln \left (f \right )^{2} b^{2}-24 f^{b \,x^{2}+a} x^{2} \ln \left (f \right ) b +24 f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{5} b^{5}}\) | \(101\) |
norman | \(\frac {12 \,{\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{b^{5} \ln \left (f \right )^{5}}+\frac {x^{8} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}-\frac {12 x^{2} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{4} b^{4}}+\frac {6 x^{4} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{3} b^{3}}-\frac {2 x^{6} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{2} b^{2}}\) | \(114\) |
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Time = 0.25 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^2} x^9 \, dx=\frac {{\left (b^{4} x^{8} \log \left (f\right )^{4} - 4 \, b^{3} x^{6} \log \left (f\right )^{3} + 12 \, b^{2} x^{4} \log \left (f\right )^{2} - 24 \, b x^{2} \log \left (f\right ) + 24\right )} f^{b x^{2} + a}}{2 \, b^{5} \log \left (f\right )^{5}} \]
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Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.23 \[ \int f^{a+b x^2} x^9 \, dx=\begin {cases} \frac {f^{a + b x^{2}} \left (b^{4} x^{8} \log {\left (f \right )}^{4} - 4 b^{3} x^{6} \log {\left (f \right )}^{3} + 12 b^{2} x^{4} \log {\left (f \right )}^{2} - 24 b x^{2} \log {\left (f \right )} + 24\right )}{2 b^{5} \log {\left (f \right )}^{5}} & \text {for}\: b^{5} \log {\left (f \right )}^{5} \neq 0 \\\frac {x^{10}}{10} & \text {otherwise} \end {cases} \]
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Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int f^{a+b x^2} x^9 \, dx=\frac {{\left (b^{4} f^{a} x^{8} \log \left (f\right )^{4} - 4 \, b^{3} f^{a} x^{6} \log \left (f\right )^{3} + 12 \, b^{2} f^{a} x^{4} \log \left (f\right )^{2} - 24 \, b f^{a} x^{2} \log \left (f\right ) + 24 \, f^{a}\right )} f^{b x^{2}}}{2 \, b^{5} \log \left (f\right )^{5}} \]
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Time = 0.32 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int f^{a+b x^2} x^9 \, dx=\frac {{\left (b^{4} x^{8} \log \left (f\right )^{4} - 4 \, b^{3} x^{6} \log \left (f\right )^{3} + 12 \, b^{2} x^{4} \log \left (f\right )^{2} - 24 \, b x^{2} \log \left (f\right ) + 24\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{2 \, b^{5} \log \left (f\right )^{5}} \]
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Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int f^{a+b x^2} x^9 \, dx=\frac {f^{b\,x^2+a}\,\left (\frac {b^4\,x^8\,{\ln \left (f\right )}^4}{2}-2\,b^3\,x^6\,{\ln \left (f\right )}^3+6\,b^2\,x^4\,{\ln \left (f\right )}^2-12\,b\,x^2\,\ln \left (f\right )+12\right )}{b^5\,{\ln \left (f\right )}^5} \]
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