Integrand size = 13, antiderivative size = 62 \[ \int f^{a+b x^2} x^5 \, dx=\frac {f^{a+b x^2}}{b^3 \log ^3(f)}-\frac {f^{a+b x^2} x^2}{b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^4}{2 b \log (f)} \]
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Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int f^{a+b x^2} x^5 \, dx=\frac {f^{a+b x^2}}{b^3 \log ^3(f)}-\frac {x^2 f^{a+b x^2}}{b^2 \log ^2(f)}+\frac {x^4 f^{a+b x^2}}{2 b \log (f)} \]
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Rule 2240
Rule 2243
Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^2} x^4}{2 b \log (f)}-\frac {2 \int f^{a+b x^2} x^3 \, dx}{b \log (f)} \\ & = -\frac {f^{a+b x^2} x^2}{b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^4}{2 b \log (f)}+\frac {2 \int f^{a+b x^2} x \, dx}{b^2 \log ^2(f)} \\ & = \frac {f^{a+b x^2}}{b^3 \log ^3(f)}-\frac {f^{a+b x^2} x^2}{b^2 \log ^2(f)}+\frac {f^{a+b x^2} x^4}{2 b \log (f)} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.66 \[ \int f^{a+b x^2} x^5 \, dx=\frac {f^{a+b x^2} \left (2-2 b x^2 \log (f)+b^2 x^4 \log ^2(f)\right )}{2 b^3 \log ^3(f)} \]
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Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.65
method | result | size |
gosper | \(\frac {\left (b^{2} x^{4} \ln \left (f \right )^{2}-2 b \,x^{2} \ln \left (f \right )+2\right ) f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{3} b^{3}}\) | \(40\) |
risch | \(\frac {\left (b^{2} x^{4} \ln \left (f \right )^{2}-2 b \,x^{2} \ln \left (f \right )+2\right ) f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{3} b^{3}}\) | \(40\) |
meijerg | \(-\frac {f^{a} \left (2-\frac {\left (3 b^{2} x^{4} \ln \left (f \right )^{2}-6 b \,x^{2} \ln \left (f \right )+6\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{3}\right )}{2 b^{3} \ln \left (f \right )^{3}}\) | \(47\) |
parallelrisch | \(\frac {f^{b \,x^{2}+a} x^{4} \ln \left (f \right )^{2} b^{2}-2 f^{b \,x^{2}+a} x^{2} \ln \left (f \right ) b +2 f^{b \,x^{2}+a}}{2 \ln \left (f \right )^{3} b^{3}}\) | \(59\) |
norman | \(\frac {{\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{3} b^{3}}+\frac {x^{4} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}-\frac {x^{2} {\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{\ln \left (f \right )^{2} b^{2}}\) | \(67\) |
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Time = 0.28 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.63 \[ \int f^{a+b x^2} x^5 \, dx=\frac {{\left (b^{2} x^{4} \log \left (f\right )^{2} - 2 \, b x^{2} \log \left (f\right ) + 2\right )} f^{b x^{2} + a}}{2 \, b^{3} \log \left (f\right )^{3}} \]
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Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85 \[ \int f^{a+b x^2} x^5 \, dx=\begin {cases} \frac {f^{a + b x^{2}} \left (b^{2} x^{4} \log {\left (f \right )}^{2} - 2 b x^{2} \log {\left (f \right )} + 2\right )}{2 b^{3} \log {\left (f \right )}^{3}} & \text {for}\: b^{3} \log {\left (f \right )}^{3} \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases} \]
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Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76 \[ \int f^{a+b x^2} x^5 \, dx=\frac {{\left (b^{2} f^{a} x^{4} \log \left (f\right )^{2} - 2 \, b f^{a} x^{2} \log \left (f\right ) + 2 \, f^{a}\right )} f^{b x^{2}}}{2 \, b^{3} \log \left (f\right )^{3}} \]
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Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69 \[ \int f^{a+b x^2} x^5 \, dx=\frac {{\left (b^{2} x^{4} \log \left (f\right )^{2} - 2 \, b x^{2} \log \left (f\right ) + 2\right )} e^{\left (b x^{2} \log \left (f\right ) + a \log \left (f\right )\right )}}{2 \, b^{3} \log \left (f\right )^{3}} \]
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Time = 0.17 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.63 \[ \int f^{a+b x^2} x^5 \, dx=\frac {f^{b\,x^2+a}\,\left (\frac {b^2\,x^4\,{\ln \left (f\right )}^2}{2}-b\,x^2\,\ln \left (f\right )+1\right )}{b^3\,{\ln \left (f\right )}^3} \]
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