Integrand size = 132, antiderivative size = 29 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=\left (2-x+\log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )\right )^2 \] Output:
(2+ln(-ln(exp(x)^2/exp(exp(3)+2-x)^2)+x-ln(x))-x)^2
Leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(29)=58\).
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.03 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=2 \left (-2 x+\frac {x^2}{2}-x \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )+\frac {1}{2} \log ^2\left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )+2 \log \left (-x+\log \left (e^{-4-2 e^3+4 x}\right )+\log (x)\right )\right ) \] Input:
Integrate[(4 + 10*x - 2*x^2 - 2*x^3 + (-4*x + 2*x^2)*Log[E^(-4 - 2*E^3 + 4 *x)] + (-4*x + 2*x^2)*Log[x] + (2 + 6*x + 2*x^2 - 2*x*Log[E^(-4 - 2*E^3 + 4*x)] - 2*x*Log[x])*Log[x - Log[E^(-4 - 2*E^3 + 4*x)] - Log[x]])/(-x^2 + x *Log[E^(-4 - 2*E^3 + 4*x)] + x*Log[x]),x]
Output:
2*(-2*x + x^2/2 - x*Log[x - Log[E^(-4 - 2*E^3 + 4*x)] - Log[x]] + Log[x - Log[E^(-4 - 2*E^3 + 4*x)] - Log[x]]^2/2 + 2*Log[-x + Log[E^(-4 - 2*E^3 + 4 *x)] + Log[x]])
Time = 0.59 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {7292, 27, 25, 7237}
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.
| \(\displaystyle \int \frac {-2 x^3-2 x^2+\left (2 x^2-4 x\right ) \log \left (e^{4 x-2 e^3-4}\right )+\left (2 x^2-4 x\right ) \log (x)+\left (2 x^2+6 x-2 x \log \left (e^{4 x-2 e^3-4}\right )-2 x \log (x)+2\right ) \log \left (x-\log \left (e^{4 x-2 e^3-4}\right )-\log (x)\right )+10 x+4}{-x^2+x \log \left (e^{4 x-2 e^3-4}\right )+x \log (x)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {2 \left (x^2+3 x-x \log \left (e^{4 x-2 e^3-4}\right )-x \log (x)+1\right ) \left (x-\log \left (x-\log \left (e^{4 x-2 e^3-4}\right )-\log (x)\right )-2\right )}{x^2-x \log \left (e^{4 x-2 e^3-4}\right )-x \log (x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \int -\frac {\left (x^2-\log \left (e^{4 x-2 \left (2+e^3\right )}\right ) x-\log (x) x+3 x+1\right ) \left (-x+\log \left (x-\log \left (e^{4 x-2 \left (2+e^3\right )}\right )-\log (x)\right )+2\right )}{x^2-\log \left (e^{4 x-2 \left (2+e^3\right )}\right ) x-\log (x) x}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -2 \int \frac {\left (x^2-\log \left (e^{4 x-2 \left (2+e^3\right )}\right ) x-\log (x) x+3 x+1\right ) \left (-x+\log \left (x-\log \left (e^{4 x-2 \left (2+e^3\right )}\right )-\log (x)\right )+2\right )}{x^2-\log \left (e^{4 x-2 \left (2+e^3\right )}\right ) x-\log (x) x}dx\) |
| \(\Big \downarrow \) 7237 |
| \(\displaystyle \left (-x+\log \left (x-\log \left (e^{4 x-2 \left (2+e^3\right )}\right )-\log (x)\right )+2\right )^2\) |
Input:
Int[(4 + 10*x - 2*x^2 - 2*x^3 + (-4*x + 2*x^2)*Log[E^(-4 - 2*E^3 + 4*x)] + (-4*x + 2*x^2)*Log[x] + (2 + 6*x + 2*x^2 - 2*x*Log[E^(-4 - 2*E^3 + 4*x)] - 2*x*Log[x])*Log[x - Log[E^(-4 - 2*E^3 + 4*x)] - Log[x]])/(-x^2 + x*Log[E ^(-4 - 2*E^3 + 4*x)] + x*Log[x]),x]
Output:
(2 - x + Log[x - Log[E^(-2*(2 + E^3) + 4*x)] - Log[x]])^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(89\) vs. \(2(32)=64\).
Time = 4.54 (sec) , antiderivative size = 90, normalized size of antiderivative = 3.10
| method | result | size |
| parallelrisch | \(x^{2}-2 \ln \left (-\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{-2 \,{\mathrm e}^{3}-4+2 x}\right )+x -\ln \left (x \right )\right ) x +{\ln \left (-\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{-2 \,{\mathrm e}^{3}-4+2 x}\right )+x -\ln \left (x \right )\right )}^{2}+4 \ln \left (-\ln \left ({\mathrm e}^{2 x} {\mathrm e}^{-2 \,{\mathrm e}^{3}-4+2 x}\right )+x -\ln \left (x \right )\right )-4 x\) | \(90\) |
| risch | \(\ln \left (4+2 \,{\mathrm e}^{3}-4 \ln \left ({\mathrm e}^{x}\right )+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) {\left (-\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}^{2}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{4 x}\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}+x -\ln \left (x \right )\right )^{2}-2 x \ln \left (4+2 \,{\mathrm e}^{3}-4 \ln \left ({\mathrm e}^{x}\right )+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) {\left (-\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}^{2}}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{4 x}\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )+\operatorname {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}+x -\ln \left (x \right )\right )+x^{2}-4 x +4 \ln \left (\ln \left ({\mathrm e}^{x}\right )-\frac {i \left (\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3} \pi -2 \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \pi +\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2} \pi +\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )-\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}-\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )^{2}+\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )-\operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{3 x}\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{3 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{4 x}\right )^{3}-2 i x -4 i {\mathrm e}^{3}+2 i \ln \left (x \right )-8 i\right )}{8}\right )\) | \(529\) |
Input:
int(((-2*x*ln(exp(x)^2/exp(exp(3)+2-x)^2)-2*x*ln(x)+2*x^2+6*x+2)*ln(-ln(ex p(x)^2/exp(exp(3)+2-x)^2)+x-ln(x))+(2*x^2-4*x)*ln(exp(x)^2/exp(exp(3)+2-x) ^2)+(2*x^2-4*x)*ln(x)-2*x^3-2*x^2+10*x+4)/(x*ln(exp(x)^2/exp(exp(3)+2-x)^2 )+x*ln(x)-x^2),x,method=_RETURNVERBOSE)
Output:
x^2-2*ln(-ln(exp(x)^2/exp(exp(3)+2-x)^2)+x-ln(x))*x+ln(-ln(exp(x)^2/exp(ex p(3)+2-x)^2)+x-ln(x))^2+4*ln(-ln(exp(x)^2/exp(exp(3)+2-x)^2)+x-ln(x))-4*x
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=x^{2} - 2 \, {\left (x - 2\right )} \log \left (-3 \, x + 2 \, e^{3} - \log \left (x\right ) + 4\right ) + \log \left (-3 \, x + 2 \, e^{3} - \log \left (x\right ) + 4\right )^{2} - 4 \, x \] Input:
integrate(((-2*x*log(exp(x)^2/exp(exp(3)+2-x)^2)-2*x*log(x)+2*x^2+6*x+2)*l og(-log(exp(x)^2/exp(exp(3)+2-x)^2)+x-log(x))+(2*x^2-4*x)*log(exp(x)^2/exp (exp(3)+2-x)^2)+(2*x^2-4*x)*log(x)-2*x^3-2*x^2+10*x+4)/(x*log(exp(x)^2/exp (exp(3)+2-x)^2)+x*log(x)-x^2),x, algorithm="fricas")
Output:
x^2 - 2*(x - 2)*log(-3*x + 2*e^3 - log(x) + 4) + log(-3*x + 2*e^3 - log(x) + 4)^2 - 4*x
Timed out. \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=\text {Timed out} \] Input:
integrate(((-2*x*ln(exp(x)**2/exp(exp(3)+2-x)**2)-2*x*ln(x)+2*x**2+6*x+2)* ln(-ln(exp(x)**2/exp(exp(3)+2-x)**2)+x-ln(x))+(2*x**2-4*x)*ln(exp(x)**2/ex p(exp(3)+2-x)**2)+(2*x**2-4*x)*ln(x)-2*x**3-2*x**2+10*x+4)/(x*ln(exp(x)**2 /exp(exp(3)+2-x)**2)+x*ln(x)-x**2),x)
Output:
Timed out
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=x^{2} - 2 \, {\left (x - 2\right )} \log \left (-3 \, x + 2 \, e^{3} - \log \left (x\right ) + 4\right ) + \log \left (-3 \, x + 2 \, e^{3} - \log \left (x\right ) + 4\right )^{2} - 4 \, x \] Input:
integrate(((-2*x*log(exp(x)^2/exp(exp(3)+2-x)^2)-2*x*log(x)+2*x^2+6*x+2)*l og(-log(exp(x)^2/exp(exp(3)+2-x)^2)+x-log(x))+(2*x^2-4*x)*log(exp(x)^2/exp (exp(3)+2-x)^2)+(2*x^2-4*x)*log(x)-2*x^3-2*x^2+10*x+4)/(x*log(exp(x)^2/exp (exp(3)+2-x)^2)+x*log(x)-x^2),x, algorithm="maxima")
Output:
x^2 - 2*(x - 2)*log(-3*x + 2*e^3 - log(x) + 4) + log(-3*x + 2*e^3 - log(x) + 4)^2 - 4*x
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=x^{2} - 2 \, x \log \left (-3 \, x + 2 \, e^{3} - \log \left (x\right ) + 4\right ) + \log \left (-3 \, x + 2 \, e^{3} - \log \left (x\right ) + 4\right )^{2} - 4 \, x + 4 \, \log \left (3 \, x - 2 \, e^{3} + \log \left (x\right ) - 4\right ) \] Input:
integrate(((-2*x*log(exp(x)^2/exp(exp(3)+2-x)^2)-2*x*log(x)+2*x^2+6*x+2)*l og(-log(exp(x)^2/exp(exp(3)+2-x)^2)+x-log(x))+(2*x^2-4*x)*log(exp(x)^2/exp (exp(3)+2-x)^2)+(2*x^2-4*x)*log(x)-2*x^3-2*x^2+10*x+4)/(x*log(exp(x)^2/exp (exp(3)+2-x)^2)+x*log(x)-x^2),x, algorithm="giac")
Output:
x^2 - 2*x*log(-3*x + 2*e^3 - log(x) + 4) + log(-3*x + 2*e^3 - log(x) + 4)^ 2 - 4*x + 4*log(3*x - 2*e^3 + log(x) - 4)
Time = 1.83 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=4\,\ln \left (3\,x-2\,{\mathrm {e}}^3+\ln \left (x\right )-4\right )-4\,x+{\ln \left (2\,{\mathrm {e}}^3-3\,x-\ln \left (x\right )+4\right )}^2-2\,x\,\ln \left (2\,{\mathrm {e}}^3-3\,x-\ln \left (x\right )+4\right )+x^2 \] Input:
int(-(log(exp(2*x)*exp(2*x - 2*exp(3) - 4))*(4*x - 2*x^2) - log(x - log(ex p(2*x)*exp(2*x - 2*exp(3) - 4)) - log(x))*(6*x - 2*x*log(exp(2*x)*exp(2*x - 2*exp(3) - 4)) - 2*x*log(x) + 2*x^2 + 2) - 10*x + log(x)*(4*x - 2*x^2) + 2*x^2 + 2*x^3 - 4)/(x*log(exp(2*x)*exp(2*x - 2*exp(3) - 4)) + x*log(x) - x^2),x)
Output:
4*log(3*x - 2*exp(3) + log(x) - 4) - 4*x + log(2*exp(3) - 3*x - log(x) + 4 )^2 - 2*x*log(2*exp(3) - 3*x - log(x) + 4) + x^2
Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 6.41 \[ \int \frac {4+10 x-2 x^2-2 x^3+\left (-4 x+2 x^2\right ) \log \left (e^{-4-2 e^3+4 x}\right )+\left (-4 x+2 x^2\right ) \log (x)+\left (2+6 x+2 x^2-2 x \log \left (e^{-4-2 e^3+4 x}\right )-2 x \log (x)\right ) \log \left (x-\log \left (e^{-4-2 e^3+4 x}\right )-\log (x)\right )}{-x^2+x \log \left (e^{-4-2 e^3+4 x}\right )+x \log (x)} \, dx=\frac {10 \,\mathrm {log}\left (\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right )+\mathrm {log}\left (x \right )-x \right )}{3}+{\mathrm {log}\left (-\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right )-\mathrm {log}\left (x \right )+x \right )}^{2}-2 \,\mathrm {log}\left (-\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right )-\mathrm {log}\left (x \right )+x \right ) x +\frac {2 \,\mathrm {log}\left (-\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right )-\mathrm {log}\left (x \right )+x \right )}{3}-\frac {\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right )^{2}}{16}+\frac {\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right ) x}{2}-\mathrm {log}\left (\frac {e^{4 x}}{e^{2 e^{3}} e^{4}}\right ) \] Input:
int(((-2*x*log(exp(x)^2/exp(exp(3)+2-x)^2)-2*x*log(x)+2*x^2+6*x+2)*log(-lo g(exp(x)^2/exp(exp(3)+2-x)^2)+x-log(x))+(2*x^2-4*x)*log(exp(x)^2/exp(exp(3 )+2-x)^2)+(2*x^2-4*x)*log(x)-2*x^3-2*x^2+10*x+4)/(x*log(exp(x)^2/exp(exp(3 )+2-x)^2)+x*log(x)-x^2),x)
Output:
(160*log(log(e**(4*x)/(e**(2*e**3)*e**4)) + log(x) - x) + 48*log( - log(e* *(4*x)/(e**(2*e**3)*e**4)) - log(x) + x)**2 - 96*log( - log(e**(4*x)/(e**( 2*e**3)*e**4)) - log(x) + x)*x + 32*log( - log(e**(4*x)/(e**(2*e**3)*e**4) ) - log(x) + x) - 3*log(e**(4*x)/(e**(2*e**3)*e**4))**2 + 24*log(e**(4*x)/ (e**(2*e**3)*e**4))*x - 48*log(e**(4*x)/(e**(2*e**3)*e**4)))/48