Integrand size = 43, antiderivative size = 17 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (12 \left (\frac {e^2}{x}+2025 x^4\right )\right )\right ) \]
[Out]
Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {1607, 6816} \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (24300 x^4+\frac {12 e^2}{x}\right )\right ) \]
[In]
[Out]
Rule 1607
Rule 6816
Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^2+8100 x^5}{x \left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx \\ & = \log \left (\log \left (\frac {12 e^2}{x}+24300 x^4\right )\right ) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (\frac {12 \left (e^2+2025 x^5\right )}{x}\right )\right ) \]
[In]
[Out]
Time = 2.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94
method | result | size |
norman | \(\ln \left (\ln \left (\frac {12 \,{\mathrm e}^{2}}{x}+24300 x^{4}\right )\right )\) | \(16\) |
risch | \(\ln \left (\ln \left (\frac {12 \,{\mathrm e}^{2}}{x}+24300 x^{4}\right )\right )\) | \(16\) |
parallelrisch | \(\ln \left (\ln \left (12 \,{\mathrm e}^{-\ln \left (x \right )+2}+24300 x^{4}\right )\right )\) | \(18\) |
default | \(\ln \left (\ln \left (3\right )+2 \ln \left (2\right )+\ln \left (\frac {{\mathrm e}^{2}+2025 x^{5}}{x}\right )\right )\) | \(22\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (\frac {12 \, {\left (2025 \, x^{5} + e^{2}\right )}}{x}\right )\right ) \]
[In]
[Out]
Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log {\left (\log {\left (24300 x^{4} + \frac {12 e^{2}}{x} \right )} \right )} \]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (3\right ) + 2 \, \log \left (2\right ) + \log \left (2025 \, x^{5} + e^{2}\right ) - \log \left (x\right )\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (12\right ) + \log \left (2025 \, x^{5} + e^{2}\right ) - \log \left (x\right )\right ) \]
[In]
[Out]
Time = 11.76 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\ln \left (\ln \left (\frac {12\,{\mathrm {e}}^2}{x}+24300\,x^4\right )\right ) \]
[In]
[Out]