Integrand size = 118, antiderivative size = 28 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=-x^2+\left (4 x-4 x \log \left (-e^x+\frac {2}{\log (x)}\right )\right )^2 \]
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=2 \left (\frac {15 x^2}{2}-16 x^2 \log \left (-e^x+\frac {2}{\log (x)}\right )+8 x^2 \log ^2\left (-e^x+\frac {2}{\log (x)}\right )\right ) \]
Integrate[(-64*x - 60*x*Log[x] + E^x*(30*x - 32*x^2)*Log[x]^2 + (64*x + 12 8*x*Log[x] + E^x*(-64*x + 32*x^2)*Log[x]^2)*Log[(2 - E^x*Log[x])/Log[x]] + (-64*x*Log[x] + 32*E^x*x*Log[x]^2)*Log[(2 - E^x*Log[x])/Log[x]]^2)/(-2*Lo g[x] + E^x*Log[x]^2),x]
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.
\(\displaystyle \int \frac {e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (e^x \left (32 x^2-64 x\right ) \log ^2(x)+64 x+128 x \log (x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )-64 x+\left (32 e^x x \log ^2(x)-64 x \log (x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )-60 x \log (x)}{e^x \log ^2(x)-2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^x \left (30 x-32 x^2\right ) \log ^2(x)-\left (e^x \left (32 x^2-64 x\right ) \log ^2(x)+64 x+128 x \log (x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+64 x-\left (32 e^x x \log ^2(x)-64 x \log (x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )+60 x \log (x)}{\log (x) \left (2-e^x \log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 x \left (-16 x+16 \log ^2\left (\frac {2}{\log (x)}-e^x\right )+16 x \log \left (\frac {2}{\log (x)}-e^x\right )-32 \log \left (\frac {2}{\log (x)}-e^x\right )+15\right )+\frac {64 x (x \log (x)+1) \left (\log \left (\frac {2}{\log (x)}-e^x\right )-1\right )}{\log (x) \left (e^x \log (x)-2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {64}{3} \int \frac {x^3}{e^x \log (x)-2}dx-\frac {64}{3} \int \frac {x^2}{\log (x) \left (e^x \log (x)-2\right )}dx+64 \int \frac {x^2 \log \left (-\frac {e^x \log (x)-2}{\log (x)}\right )}{e^x \log (x)-2}dx+32 \int x \log ^2\left (-\frac {e^x \log (x)-2}{\log (x)}\right )dx+64 \int \frac {x \log \left (-\frac {e^x \log (x)-2}{\log (x)}\right )}{\log (x) \left (e^x \log (x)-2\right )}dx-\frac {8 x^4}{3}+\frac {32}{3} x^3 \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+15 x^2-32 x^2 \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )\) |
Int[(-64*x - 60*x*Log[x] + E^x*(30*x - 32*x^2)*Log[x]^2 + (64*x + 128*x*Lo g[x] + E^x*(-64*x + 32*x^2)*Log[x]^2)*Log[(2 - E^x*Log[x])/Log[x]] + (-64* x*Log[x] + 32*E^x*x*Log[x]^2)*Log[(2 - E^x*Log[x])/Log[x]]^2)/(-2*Log[x] + E^x*Log[x]^2),x]
3.26.1.3.1 Defintions of rubi rules used
Time = 2.44 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68
method | result | size |
parallelrisch | \(16 \ln \left (-\frac {{\mathrm e}^{x} \ln \left (x \right )-2}{\ln \left (x \right )}\right )^{2} x^{2}-32 \ln \left (-\frac {{\mathrm e}^{x} \ln \left (x \right )-2}{\ln \left (x \right )}\right ) x^{2}+15 x^{2}\) | \(47\) |
risch | \(\text {Expression too large to display}\) | \(1209\) |
int(((32*x*exp(x)*ln(x)^2-64*x*ln(x))*ln((-exp(x)*ln(x)+2)/ln(x))^2+((32*x ^2-64*x)*exp(x)*ln(x)^2+128*x*ln(x)+64*x)*ln((-exp(x)*ln(x)+2)/ln(x))+(-32 *x^2+30*x)*exp(x)*ln(x)^2-60*x*ln(x)-64*x)/(exp(x)*ln(x)^2-2*ln(x)),x,meth od=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=16 \, x^{2} \log \left (-\frac {e^{x} \log \left (x\right ) - 2}{\log \left (x\right )}\right )^{2} - 32 \, x^{2} \log \left (-\frac {e^{x} \log \left (x\right ) - 2}{\log \left (x\right )}\right ) + 15 \, x^{2} \]
integrate(((32*x*exp(x)*log(x)^2-64*x*log(x))*log((-exp(x)*log(x)+2)/log(x ))^2+((32*x^2-64*x)*exp(x)*log(x)^2+128*x*log(x)+64*x)*log((-exp(x)*log(x) +2)/log(x))+(-32*x^2+30*x)*exp(x)*log(x)^2-60*x*log(x)-64*x)/(exp(x)*log(x )^2-2*log(x)),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.40 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=16 x^{2} \log {\left (\frac {- e^{x} \log {\left (x \right )} + 2}{\log {\left (x \right )}} \right )}^{2} - 32 x^{2} \log {\left (\frac {- e^{x} \log {\left (x \right )} + 2}{\log {\left (x \right )}} \right )} + 15 x^{2} \]
integrate(((32*x*exp(x)*ln(x)**2-64*x*ln(x))*ln((-exp(x)*ln(x)+2)/ln(x))** 2+((32*x**2-64*x)*exp(x)*ln(x)**2+128*x*ln(x)+64*x)*ln((-exp(x)*ln(x)+2)/l n(x))+(-32*x**2+30*x)*exp(x)*ln(x)**2-60*x*ln(x)-64*x)/(exp(x)*ln(x)**2-2* ln(x)),x)
16*x**2*log((-exp(x)*log(x) + 2)/log(x))**2 - 32*x**2*log((-exp(x)*log(x) + 2)/log(x)) + 15*x**2
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (28) = 56\).
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.21 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=16 \, x^{2} \log \left (-e^{x} \log \left (x\right ) + 2\right )^{2} + 16 \, x^{2} \log \left (\log \left (x\right )\right )^{2} + 32 \, x^{2} \log \left (\log \left (x\right )\right ) + 15 \, x^{2} - 32 \, {\left (x^{2} \log \left (\log \left (x\right )\right ) + x^{2}\right )} \log \left (-e^{x} \log \left (x\right ) + 2\right ) \]
integrate(((32*x*exp(x)*log(x)^2-64*x*log(x))*log((-exp(x)*log(x)+2)/log(x ))^2+((32*x^2-64*x)*exp(x)*log(x)^2+128*x*log(x)+64*x)*log((-exp(x)*log(x) +2)/log(x))+(-32*x^2+30*x)*exp(x)*log(x)^2-60*x*log(x)-64*x)/(exp(x)*log(x )^2-2*log(x)),x, algorithm=\
16*x^2*log(-e^x*log(x) + 2)^2 + 16*x^2*log(log(x))^2 + 32*x^2*log(log(x)) + 15*x^2 - 32*(x^2*log(log(x)) + x^2)*log(-e^x*log(x) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (28) = 56\).
Time = 0.39 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.54 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=16 \, x^{2} \log \left (-e^{x} \log \left (x\right ) + 2\right )^{2} - 32 \, x^{2} \log \left (-e^{x} \log \left (x\right ) + 2\right ) \log \left (\log \left (x\right )\right ) + 16 \, x^{2} \log \left (\log \left (x\right )\right )^{2} - 32 \, x^{2} \log \left (-e^{x} \log \left (x\right ) + 2\right ) + 32 \, x^{2} \log \left (\log \left (x\right )\right ) + 15 \, x^{2} \]
integrate(((32*x*exp(x)*log(x)^2-64*x*log(x))*log((-exp(x)*log(x)+2)/log(x ))^2+((32*x^2-64*x)*exp(x)*log(x)^2+128*x*log(x)+64*x)*log((-exp(x)*log(x) +2)/log(x))+(-32*x^2+30*x)*exp(x)*log(x)^2-60*x*log(x)-64*x)/(exp(x)*log(x )^2-2*log(x)),x, algorithm=\
16*x^2*log(-e^x*log(x) + 2)^2 - 32*x^2*log(-e^x*log(x) + 2)*log(log(x)) + 16*x^2*log(log(x))^2 - 32*x^2*log(-e^x*log(x) + 2) + 32*x^2*log(log(x)) + 15*x^2
Time = 13.59 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-64 x-60 x \log (x)+e^x \left (30 x-32 x^2\right ) \log ^2(x)+\left (64 x+128 x \log (x)+e^x \left (-64 x+32 x^2\right ) \log ^2(x)\right ) \log \left (\frac {2-e^x \log (x)}{\log (x)}\right )+\left (-64 x \log (x)+32 e^x x \log ^2(x)\right ) \log ^2\left (\frac {2-e^x \log (x)}{\log (x)}\right )}{-2 \log (x)+e^x \log ^2(x)} \, dx=16\,x^2\,{\ln \left (-\frac {{\mathrm {e}}^x\,\ln \left (x\right )-2}{\ln \left (x\right )}\right )}^2-32\,x^2\,\ln \left (-\frac {{\mathrm {e}}^x\,\ln \left (x\right )-2}{\ln \left (x\right )}\right )+15\,x^2 \]
int((64*x + log(-(exp(x)*log(x) - 2)/log(x))^2*(64*x*log(x) - 32*x*exp(x)* log(x)^2) + 60*x*log(x) - log(-(exp(x)*log(x) - 2)/log(x))*(64*x + 128*x*l og(x) - exp(x)*log(x)^2*(64*x - 32*x^2)) - exp(x)*log(x)^2*(30*x - 32*x^2) )/(2*log(x) - exp(x)*log(x)^2),x)