Integrand size = 114, antiderivative size = 32 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=\frac {5 x}{\left (\frac {1}{x}-\frac {x}{2}\right ) \log \left (-e^x+\frac {3}{x \log (2)}\right )} \]
Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(32)=64\).
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.16 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=-\frac {10 x^2 \left (-6+3 x^2+e^x x^4 \log (2)-e^x x^2 \log (4)\right )}{\left (-2+x^2\right )^2 \left (3+e^x x^2 \log (2)\right ) \log \left (-e^x+\frac {3}{x \log (2)}\right )} \]
Integrate[(-60*x + 30*x^3 + E^x*(-20*x^3 + 10*x^5)*Log[2] + (-120*x + 40*E ^x*x^2*Log[2])*Log[(3 - E^x*x*Log[2])/(x*Log[2])])/((-12 + 12*x^2 - 3*x^4 + E^x*(4*x - 4*x^3 + x^5)*Log[2])*Log[(3 - E^x*x*Log[2])/(x*Log[2])]^2),x]
(-10*x^2*(-6 + 3*x^2 + E^x*x^4*Log[2] - E^x*x^2*Log[4]))/((-2 + x^2)^2*(3 + E^x*x^2*Log[2])*Log[-E^x + 3/(x*Log[2])])
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {30 x^3+\left (40 e^x x^2 \log (2)-120 x\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )+e^x \left (10 x^5-20 x^3\right ) \log (2)-60 x}{\left (-3 x^4+12 x^2+e^x \left (x^5-4 x^3+4 x\right ) \log (2)-12\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-30 x^3-\left (40 e^x x^2 \log (2)-120 x\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )-e^x \left (10 x^5-20 x^3\right ) \log (2)+60 x}{\left (2-x^2\right )^2 \left (3-e^x x \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {30 x (x+1)}{\left (x^2-2\right ) \left (e^x x \log (2)-3\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}+\frac {10 x \left (x^3-2 x+4 \log \left (\frac {3}{x \log (2)}-e^x\right )\right )}{\left (x^2-2\right )^2 \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 40 \int \frac {x}{\left (x^2-2\right )^2 \log \left (\frac {3}{x \log (2)}-e^x\right )}dx+10 \int \frac {1}{\log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx-5 \sqrt {2} \int \frac {1}{\left (\sqrt {2}-x\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx-5 \sqrt {2} \int \frac {1}{\left (x+\sqrt {2}\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx+30 \int \frac {1}{\left (e^x x \log (2)-3\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx-15 \sqrt {2} \int \frac {1}{\left (\sqrt {2}-x\right ) \left (e^x x \log (2)-3\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx-15 \int \frac {1}{\left (\sqrt {2}-x\right ) \left (e^x x \log (2)-3\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx-15 \sqrt {2} \int \frac {1}{\left (x+\sqrt {2}\right ) \left (e^x x \log (2)-3\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx+15 \int \frac {1}{\left (x+\sqrt {2}\right ) \left (e^x x \log (2)-3\right ) \log ^2\left (\frac {3}{x \log (2)}-e^x\right )}dx\) |
Int[(-60*x + 30*x^3 + E^x*(-20*x^3 + 10*x^5)*Log[2] + (-120*x + 40*E^x*x^2 *Log[2])*Log[(3 - E^x*x*Log[2])/(x*Log[2])])/((-12 + 12*x^2 - 3*x^4 + E^x* (4*x - 4*x^3 + x^5)*Log[2])*Log[(3 - E^x*x*Log[2])/(x*Log[2])]^2),x]
3.3.74.3.1 Defintions of rubi rules used
Time = 5.18 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(-\frac {10 x^{2}}{\left (x^{2}-2\right ) \ln \left (-\frac {x \ln \left (2\right ) {\mathrm e}^{x}-3}{x \ln \left (2\right )}\right )}\) | \(33\) |
risch | \(-\frac {20 i x^{2}}{\left (x^{2}-2\right ) \left (\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )}{x}\right )}^{2}+2 \pi {\operatorname {csgn}\left (\frac {i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )}{x}\right )}^{2}-\pi {\operatorname {csgn}\left (\frac {i \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )}{x}\right )}^{3}-2 \pi -2 i \ln \left (\ln \left (2\right )\right )-2 i \ln \left (x \right )+2 i \ln \left (x \ln \left (2\right ) {\mathrm e}^{x}-3\right )\right )}\) | \(178\) |
int(((40*x^2*ln(2)*exp(x)-120*x)*ln((-x*ln(2)*exp(x)+3)/x/ln(2))+(10*x^5-2 0*x^3)*ln(2)*exp(x)+30*x^3-60*x)/((x^5-4*x^3+4*x)*ln(2)*exp(x)-3*x^4+12*x^ 2-12)/ln((-x*ln(2)*exp(x)+3)/x/ln(2))^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=-\frac {10 \, x^{2}}{{\left (x^{2} - 2\right )} \log \left (-\frac {x e^{x} \log \left (2\right ) - 3}{x \log \left (2\right )}\right )} \]
integrate(((40*x^2*log(2)*exp(x)-120*x)*log((-x*log(2)*exp(x)+3)/x/log(2)) +(10*x^5-20*x^3)*log(2)*exp(x)+30*x^3-60*x)/((x^5-4*x^3+4*x)*log(2)*exp(x) -3*x^4+12*x^2-12)/log((-x*log(2)*exp(x)+3)/x/log(2))^2,x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=- \frac {10 x^{2}}{\left (x^{2} - 2\right ) \log {\left (\frac {- x e^{x} \log {\left (2 \right )} + 3}{x \log {\left (2 \right )}} \right )}} \]
integrate(((40*x**2*ln(2)*exp(x)-120*x)*ln((-x*ln(2)*exp(x)+3)/x/ln(2))+(1 0*x**5-20*x**3)*ln(2)*exp(x)+30*x**3-60*x)/((x**5-4*x**3+4*x)*ln(2)*exp(x) -3*x**4+12*x**2-12)/ln((-x*ln(2)*exp(x)+3)/x/ln(2))**2,x)
Time = 0.36 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=\frac {10 \, x^{2}}{x^{2} \log \left (\log \left (2\right )\right ) - {\left (x^{2} - 2\right )} \log \left (-x e^{x} \log \left (2\right ) + 3\right ) + {\left (x^{2} - 2\right )} \log \left (x\right ) - 2 \, \log \left (\log \left (2\right )\right )} \]
integrate(((40*x^2*log(2)*exp(x)-120*x)*log((-x*log(2)*exp(x)+3)/x/log(2)) +(10*x^5-20*x^3)*log(2)*exp(x)+30*x^3-60*x)/((x^5-4*x^3+4*x)*log(2)*exp(x) -3*x^4+12*x^2-12)/log((-x*log(2)*exp(x)+3)/x/log(2))^2,x, algorithm=\
10*x^2/(x^2*log(log(2)) - (x^2 - 2)*log(-x*e^x*log(2) + 3) + (x^2 - 2)*log (x) - 2*log(log(2)))
Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=-\frac {10 \, x^{2}}{x^{2} \log \left (-x e^{x} \log \left (2\right ) + 3\right ) - x^{2} \log \left (x \log \left (2\right )\right ) - 2 \, \log \left (-x e^{x} \log \left (2\right ) + 3\right ) + 2 \, \log \left (x \log \left (2\right )\right )} \]
integrate(((40*x^2*log(2)*exp(x)-120*x)*log((-x*log(2)*exp(x)+3)/x/log(2)) +(10*x^5-20*x^3)*log(2)*exp(x)+30*x^3-60*x)/((x^5-4*x^3+4*x)*log(2)*exp(x) -3*x^4+12*x^2-12)/log((-x*log(2)*exp(x)+3)/x/log(2))^2,x, algorithm=\
-10*x^2/(x^2*log(-x*e^x*log(2) + 3) - x^2*log(x*log(2)) - 2*log(-x*e^x*log (2) + 3) + 2*log(x*log(2)))
Time = 9.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-60 x+30 x^3+e^x \left (-20 x^3+10 x^5\right ) \log (2)+\left (-120 x+40 e^x x^2 \log (2)\right ) \log \left (\frac {3-e^x x \log (2)}{x \log (2)}\right )}{\left (-12+12 x^2-3 x^4+e^x \left (4 x-4 x^3+x^5\right ) \log (2)\right ) \log ^2\left (\frac {3-e^x x \log (2)}{x \log (2)}\right )} \, dx=-\frac {10\,x^2}{\left (x^2-2\right )\,\left (\ln \left (-\frac {x\,{\mathrm {e}}^x\,\ln \left (2\right )-3}{x}\right )-\ln \left (\ln \left (2\right )\right )\right )} \]