Integrand size = 170, antiderivative size = 29 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\left (4+x-x^2\right ) \left (-4+\frac {2 x \log \left (\log \left (\frac {x}{e}+x^2\right )\right )}{\log (x)}\right ) \] Output:
(2*ln(ln(x^2+x/exp(1)))/ln(x)*x-4)*(-x^2+x+4)
Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=-4 x+4 x^2-\frac {2 x \left (-4-x+x^2\right ) \log \left (\log \left (x \left (\frac {1}{e}+x\right )\right )\right )}{\log (x)} \] Input:
Integrate[((8 + 2*x - 2*x^2 + E*(16*x + 4*x^2 - 4*x^3))*Log[x] + (-4 + 8*x + E*(-4*x + 8*x^2))*Log[x]^2*Log[(x + E*x^2)/E] + (-8 - 2*x + 2*x^2 + E*( -8*x - 2*x^2 + 2*x^3) + (8 + 4*x - 6*x^2 + E*(8*x + 4*x^2 - 6*x^3))*Log[x] )*Log[(x + E*x^2)/E]*Log[Log[(x + E*x^2)/E]])/((1 + E*x)*Log[x]^2*Log[(x + E*x^2)/E]),x]
Output:
-4*x + 4*x^2 - (2*x*(-4 - x + x^2)*Log[Log[x*(E^(-1) + x)]])/Log[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.
| \(\displaystyle \int \frac {\left (e \left (8 x^2-4 x\right )+8 x-4\right ) \log \left (\frac {e x^2+x}{e}\right ) \log ^2(x)+\left (-2 x^2+e \left (-4 x^3+4 x^2+16 x\right )+2 x+8\right ) \log (x)+\left (2 x^2+e \left (2 x^3-2 x^2-8 x\right )+\left (-6 x^2+e \left (-6 x^3+4 x^2+8 x\right )+4 x+8\right ) \log (x)-2 x-8\right ) \log \left (\frac {e x^2+x}{e}\right ) \log \left (\log \left (\frac {e x^2+x}{e}\right )\right )}{(e x+1) \log ^2(x) \log \left (\frac {e x^2+x}{e}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-\frac {2 (e x+1) \left (-x^2+\left (3 x^2-2 x-4\right ) \log (x)+x+4\right ) \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}-\frac {2 (2 e x+1) \left (x^2-x-4\right )}{\log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}+4 e (2 x-1) x+8 x-4}{e x+1}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 \left (-2 e x^3-(1-2 e) x^2+4 e x^2 \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )+(1+8 e) x+4 \left (1-\frac {e}{2}\right ) x \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )-2 \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )+4\right )}{(e x+1) \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}-\frac {2 \left (-x^2+3 x^2 \log (x)+x-2 x \log (x)-4 \log (x)+4\right ) \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int \frac {x^2 \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}dx-4 \int \frac {x^2}{\log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}dx-6 \int \frac {x^2 \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log (x)}dx-8 \int \frac {\log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}dx-2 \int \frac {x \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log ^2(x)}dx-\frac {2 \left (1+e-8 e^2\right ) \int \frac {1}{\log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}dx}{e^2}+\frac {2 \left (1+e-4 e^2\right ) \int \frac {1}{(e x+1) \log (x) \log \left (x \left (x+\frac {1}{e}\right )\right )}dx}{e^2}-\frac {2 (1+2 e) \int \frac {x}{\log (x) (1-\log (x (e x+1)))}dx}{e}+8 \int \frac {\log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log (x)}dx+4 \int \frac {x \log \left (\log \left (x \left (x+\frac {1}{e}\right )\right )\right )}{\log (x)}dx+4 x^2-4 x\) |
Input:
Int[((8 + 2*x - 2*x^2 + E*(16*x + 4*x^2 - 4*x^3))*Log[x] + (-4 + 8*x + E*( -4*x + 8*x^2))*Log[x]^2*Log[(x + E*x^2)/E] + (-8 - 2*x + 2*x^2 + E*(-8*x - 2*x^2 + 2*x^3) + (8 + 4*x - 6*x^2 + E*(8*x + 4*x^2 - 6*x^3))*Log[x])*Log[ (x + E*x^2)/E]*Log[Log[(x + E*x^2)/E]])/((1 + E*x)*Log[x]^2*Log[(x + E*x^2 )/E]),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(76\) vs. \(2(30)=60\).
Time = 18.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.66
| method | result | size |
| parallelrisch | \(\frac {-2 \ln \left (\ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{-1}\right )\right ) x^{3}+4 x^{2} \ln \left (x \right )+2 \ln \left (\ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{-1}\right )\right ) x^{2}-4 x \ln \left (x \right )+8 \ln \left (\ln \left (\left (x^{2} {\mathrm e}+x \right ) {\mathrm e}^{-1}\right )\right ) x}{\ln \left (x \right )}\) | \(77\) |
| risch | \(-\frac {2 x \left (x^{2}-x -4\right ) \ln \left (-1+\ln \left (x \right )+\ln \left (x \,{\mathrm e}+1\right )-\frac {i \pi \,\operatorname {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right ) \left (-\operatorname {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i x \left (x \,{\mathrm e}+1\right )\right )+\operatorname {csgn}\left (i \left (x \,{\mathrm e}+1\right )\right )\right )}{2}\right )}{\ln \left (x \right )}+4 x^{2}-4 x\) | \(95\) |
Input:
int(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*ln(x)+(2*x^3-2*x^2-8*x)*exp( 1)+2*x^2-2*x-8)*ln((x^2*exp(1)+x)/exp(1))*ln(ln((x^2*exp(1)+x)/exp(1)))+(( 8*x^2-4*x)*exp(1)+8*x-4)*ln(x)^2*ln((x^2*exp(1)+x)/exp(1))+((-4*x^3+4*x^2+ 16*x)*exp(1)-2*x^2+2*x+8)*ln(x))/(x*exp(1)+1)/ln(x)^2/ln((x^2*exp(1)+x)/ex p(1)),x,method=_RETURNVERBOSE)
Output:
(-2*ln(ln((x^2*exp(1)+x)/exp(1)))*x^3+4*x^2*ln(x)+2*ln(ln((x^2*exp(1)+x)/e xp(1)))*x^2-4*x*ln(x)+8*ln(ln((x^2*exp(1)+x)/exp(1)))*x)/ln(x)
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.55 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\frac {2 \, {\left (2 \, {\left (x^{2} - x\right )} \log \left (x\right ) - {\left (x^{3} - x^{2} - 4 \, x\right )} \log \left (\log \left ({\left (x^{2} e + x\right )} e^{\left (-1\right )}\right )\right )\right )}}{\log \left (x\right )} \] Input:
integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8* x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/e xp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((- 4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(exp(1)*x+1)/log(x)^2/log((x ^2*exp(1)+x)/exp(1)),x, algorithm="fricas")
Output:
2*(2*(x^2 - x)*log(x) - (x^3 - x^2 - 4*x)*log(log((x^2*e + x)*e^(-1))))/lo g(x)
Exception generated. \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((((-6*x**3+4*x**2+8*x)*exp(1)-6*x**2+4*x+8)*ln(x)+(2*x**3-2*x** 2-8*x)*exp(1)+2*x**2-2*x-8)*ln((x**2*exp(1)+x)/exp(1))*ln(ln((x**2*exp(1)+ x)/exp(1)))+((8*x**2-4*x)*exp(1)+8*x-4)*ln(x)**2*ln((x**2*exp(1)+x)/exp(1) )+((-4*x**3+4*x**2+16*x)*exp(1)-2*x**2+2*x+8)*ln(x))/(exp(1)*x+1)/ln(x)**2 /ln((x**2*exp(1)+x)/exp(1)),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\frac {2 \, {\left (2 \, {\left (x^{2} - x\right )} \log \left (x\right ) - {\left (x^{3} - x^{2} - 4 \, x\right )} \log \left (\log \left (x e + 1\right ) + \log \left (x\right ) - 1\right )\right )}}{\log \left (x\right )} \] Input:
integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8* x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/e xp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((- 4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(exp(1)*x+1)/log(x)^2/log((x ^2*exp(1)+x)/exp(1)),x, algorithm="maxima")
Output:
2*(2*(x^2 - x)*log(x) - (x^3 - x^2 - 4*x)*log(log(x*e + 1) + log(x) - 1))/ log(x)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=-\frac {2 \, {\left (x^{3} \log \left (\log \left (x^{2} e + x\right ) - 1\right ) - 2 \, x^{2} \log \left (x\right ) - x^{2} \log \left (\log \left (x^{2} e + x\right ) - 1\right ) + 2 \, x \log \left (x\right ) - 4 \, x \log \left (\log \left (x^{2} e + x\right ) - 1\right )\right )}}{\log \left (x\right )} \] Input:
integrate(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8* x)*exp(1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/e xp(1)))+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((- 4*x^3+4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(exp(1)*x+1)/log(x)^2/log((x ^2*exp(1)+x)/exp(1)),x, algorithm="giac")
Output:
-2*(x^3*log(log(x^2*e + x) - 1) - 2*x^2*log(x) - x^2*log(log(x^2*e + x) - 1) + 2*x*log(x) - 4*x*log(log(x^2*e + x) - 1))/log(x)
Time = 4.53 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=4\,x^2-4\,x+\frac {2\,x^2\,\ln \left (\ln \left ({\mathrm {e}}^{-1}\,\left (\mathrm {e}\,x^2+x\right )\right )\right )\,\left (x\,\mathrm {e}+1\right )\,\left (-x^2+x+4\right )}{\ln \left (x\right )\,\left (\mathrm {e}\,x^2+x\right )} \] Input:
int(-(log(log(exp(-1)*(x + x^2*exp(1))))*log(exp(-1)*(x + x^2*exp(1)))*(2* x - log(x)*(4*x + exp(1)*(8*x + 4*x^2 - 6*x^3) - 6*x^2 + 8) + exp(1)*(8*x + 2*x^2 - 2*x^3) - 2*x^2 + 8) - log(x)*(2*x + exp(1)*(16*x + 4*x^2 - 4*x^3 ) - 2*x^2 + 8) + log(exp(-1)*(x + x^2*exp(1)))*log(x)^2*(exp(1)*(4*x - 8*x ^2) - 8*x + 4))/(log(exp(-1)*(x + x^2*exp(1)))*log(x)^2*(x*exp(1) + 1)),x)
Output:
4*x^2 - 4*x + (2*x^2*log(log(exp(-1)*(x + x^2*exp(1))))*(x*exp(1) + 1)*(x - x^2 + 4))/(log(x)*(x + x^2*exp(1)))
Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {\left (8+2 x-2 x^2+e \left (16 x+4 x^2-4 x^3\right )\right ) \log (x)+\left (-4+8 x+e \left (-4 x+8 x^2\right )\right ) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )+\left (-8-2 x+2 x^2+e \left (-8 x-2 x^2+2 x^3\right )+\left (8+4 x-6 x^2+e \left (8 x+4 x^2-6 x^3\right )\right ) \log (x)\right ) \log \left (\frac {x+e x^2}{e}\right ) \log \left (\log \left (\frac {x+e x^2}{e}\right )\right )}{(1+e x) \log ^2(x) \log \left (\frac {x+e x^2}{e}\right )} \, dx=\frac {2 x \left (-\mathrm {log}\left (\mathrm {log}\left (\frac {e \,x^{2}+x}{e}\right )\right ) x^{2}+\mathrm {log}\left (\mathrm {log}\left (\frac {e \,x^{2}+x}{e}\right )\right ) x +4 \,\mathrm {log}\left (\mathrm {log}\left (\frac {e \,x^{2}+x}{e}\right )\right )+2 \,\mathrm {log}\left (x \right ) x -2 \,\mathrm {log}\left (x \right )\right )}{\mathrm {log}\left (x \right )} \] Input:
int(((((-6*x^3+4*x^2+8*x)*exp(1)-6*x^2+4*x+8)*log(x)+(2*x^3-2*x^2-8*x)*exp (1)+2*x^2-2*x-8)*log((x^2*exp(1)+x)/exp(1))*log(log((x^2*exp(1)+x)/exp(1)) )+((8*x^2-4*x)*exp(1)+8*x-4)*log(x)^2*log((x^2*exp(1)+x)/exp(1))+((-4*x^3+ 4*x^2+16*x)*exp(1)-2*x^2+2*x+8)*log(x))/(exp(1)*x+1)/log(x)^2/log((x^2*exp (1)+x)/exp(1)),x)
Output:
(2*x*( - log(log((e*x**2 + x)/e))*x**2 + log(log((e*x**2 + x)/e))*x + 4*lo g(log((e*x**2 + x)/e)) + 2*log(x)*x - 2*log(x)))/log(x)