Integrand size = 97, antiderivative size = 35 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {4}{-4+x+\frac {3+x}{x}}+\left (2-\frac {1}{3} e^5 (2-\log (2))\right ) \log \left (x^2\right ) \]
[Out]
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {2099, 652, 632, 210} \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {4 x}{x^2-3 x+3}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x) \]
[In]
[Out]
Rule 210
Rule 632
Rule 652
Rule 2099
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {12 (-2+x)}{\left (3-3 x+x^2\right )^2}-\frac {4}{3-3 x+x^2}+\frac {2 \left (6-e^5 (2-\log (2))\right )}{3 x}\right ) \, dx \\ & = \frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)-4 \int \frac {1}{3-3 x+x^2} \, dx-12 \int \frac {-2+x}{\left (3-3 x+x^2\right )^2} \, dx \\ & = \frac {4 x}{3-3 x+x^2}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)+4 \int \frac {1}{3-3 x+x^2} \, dx+8 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-3+2 x\right ) \\ & = \frac {4 x}{3-3 x+x^2}+\frac {8 \tan ^{-1}\left (\frac {3-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x)-8 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-3+2 x\right ) \\ & = \frac {4 x}{3-3 x+x^2}+\frac {2}{3} \left (6-e^5 (2-\log (2))\right ) \log (x) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2}{3} \left (\frac {6 x}{3-3 x+x^2}+\left (6+e^5 (-2+\log (2))\right ) \log (x)\right ) \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86
method | result | size |
default | \(\frac {4 x}{x^{2}-3 x +3}+\frac {2 \left ({\mathrm e}^{5} \ln \left (2\right )-2 \,{\mathrm e}^{5}+6\right ) \ln \left (x \right )}{3}\) | \(30\) |
norman | \(\frac {4 x}{x^{2}-3 x +3}+\left (\frac {2 \,{\mathrm e}^{5} \ln \left (2\right )}{3}-\frac {4 \,{\mathrm e}^{5}}{3}+4\right ) \ln \left (x \right )\) | \(30\) |
risch | \(\frac {4 x}{x^{2}-3 x +3}+\frac {2 \,{\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )}{3}-\frac {4 \,{\mathrm e}^{5} \ln \left (x \right )}{3}+4 \ln \left (x \right )\) | \(33\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{5} \ln \left (x \right ) \ln \left (2\right ) x^{2}-6 \,{\mathrm e}^{5} \ln \left (x \right ) \ln \left (2\right ) x -4 x^{2} {\mathrm e}^{5} \ln \left (x \right )+6 \,{\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )+12 x \,{\mathrm e}^{5} \ln \left (x \right )+12 x^{2} \ln \left (x \right )-12 \,{\mathrm e}^{5} \ln \left (x \right )-36 x \ln \left (x \right )+36 \ln \left (x \right )+12 x}{3 x^{2}-9 x +9}\) | \(83\) |
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.54 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2 \, {\left ({\left ({\left (x^{2} - 3 \, x + 3\right )} e^{5} \log \left (2\right ) + 6 \, x^{2} - 2 \, {\left (x^{2} - 3 \, x + 3\right )} e^{5} - 18 \, x + 18\right )} \log \left (x\right ) + 6 \, x\right )}}{3 \, {\left (x^{2} - 3 \, x + 3\right )}} \]
[In]
[Out]
Time = 0.60 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {4 x}{x^{2} - 3 x + 3} + \frac {2 \left (- 2 e^{5} + 6 + e^{5} \log {\left (2 \right )}\right ) \log {\left (x \right )}}{3} \]
[In]
[Out]
none
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2}{3} \, {\left (e^{5} \log \left (2\right ) - 2 \, e^{5} + 6\right )} \log \left (x\right ) + \frac {4 \, x}{x^{2} - 3 \, x + 3} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\frac {2}{3} \, {\left (e^{5} \log \left (2\right ) - 2 \, e^{5} + 6\right )} \log \left ({\left | x \right |}\right ) + \frac {4 \, x}{x^{2} - 3 \, x + 3} \]
[In]
[Out]
Time = 18.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {108-180 x+180 x^2-84 x^3+12 x^4+e^5 \left (-36+72 x-60 x^2+24 x^3-4 x^4\right )+e^5 \left (18-36 x+30 x^2-12 x^3+2 x^4\right ) \log (2)}{27 x-54 x^2+45 x^3-18 x^4+3 x^5} \, dx=\ln \left (x\right )\,\left (\frac {2\,{\mathrm {e}}^5\,\ln \left (2\right )}{3}-\frac {4\,{\mathrm {e}}^5}{3}+4\right )+\frac {12\,x}{3\,x^2-9\,x+9} \]
[In]
[Out]