Integrand size = 143, antiderivative size = 35 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {2 \left (-x+(-3+(-4+x) x) \left (-2+\frac {e^{2 x}}{\log (x)}\right )\right )}{e^{2 x}+x} \] Output:
2/(exp(x)^2+x)*((exp(x)^2/ln(x)-2)*((-4+x)*x-3)-x)
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.23 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {2 \left (e^{2 x} \left (-3-4 x+x^2\right )+\left (6+7 x-2 x^2\right ) \log (x)\right )}{\left (e^{2 x}+x\right ) \log (x)} \] Input:
Integrate[(E^(4*x)*(6 + 8*x - 2*x^2) + E^(2*x)*(6*x + 8*x^2 - 2*x^3) + (E^ (4*x)*(-8*x + 4*x^2) + E^(2*x)*(6*x - 12*x^2 - 14*x^3 + 4*x^4))*Log[x] + ( -12*x - 4*x^3 + E^(2*x)*(-10*x - 36*x^2 + 8*x^3))*Log[x]^2)/((E^(4*x)*x + 2*E^(2*x)*x^2 + x^3)*Log[x]^2),x]
Output:
(2*(E^(2*x)*(-3 - 4*x + x^2) + (6 + 7*x - 2*x^2)*Log[x]))/((E^(2*x) + x)*L og[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.
\(\displaystyle \int \frac {e^{4 x} \left (-2 x^2+8 x+6\right )+e^{2 x} \left (-2 x^3+8 x^2+6 x\right )+\left (-4 x^3+e^{2 x} \left (8 x^3-36 x^2-10 x\right )-12 x\right ) \log ^2(x)+\left (e^{4 x} \left (4 x^2-8 x\right )+e^{2 x} \left (4 x^4-14 x^3-12 x^2+6 x\right )\right ) \log (x)}{\left (x^3+2 e^{2 x} x^2+e^{4 x} x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{4 x} \left (-2 x^2+8 x+6\right )+e^{2 x} \left (-2 x^3+8 x^2+6 x\right )+\left (-4 x^3+e^{2 x} \left (8 x^3-36 x^2-10 x\right )-12 x\right ) \log ^2(x)+\left (e^{4 x} \left (4 x^2-8 x\right )+e^{2 x} \left (4 x^4-14 x^3-12 x^2+6 x\right )\right ) \log (x)}{x \left (x+e^{2 x}\right )^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (-x^2+2 x^2 \log (x)+4 x-4 x \log (x)+3\right )}{x \log ^2(x)}+\frac {2 \left (2 x^3 \log (x)+x^2+4 x^2 \log ^2(x)-11 x^2 \log (x)-4 x-18 x \log ^2(x)-5 \log ^2(x)+2 x \log (x)+3 \log (x)-3\right )}{\left (x+e^{2 x}\right ) \log ^2(x)}-\frac {2 (2 x-1) \left (x^3-4 x^2+2 x^2 \log (x)-3 x-7 x \log (x)-6 \log (x)\right )}{\left (x+e^{2 x}\right )^2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {x^4}{\left (x+e^{2 x}\right )^2 \log (x)}dx-8 \int \frac {x^3}{\left (x+e^{2 x}\right )^2}dx+18 \int \frac {x^3}{\left (x+e^{2 x}\right )^2 \log (x)}dx+4 \int \frac {x^3}{\left (x+e^{2 x}\right ) \log (x)}dx+32 \int \frac {x^2}{\left (x+e^{2 x}\right )^2}dx+8 \int \frac {x^2}{x+e^{2 x}}dx+2 \int \frac {x^2}{\left (x+e^{2 x}\right ) \log ^2(x)}dx+2 \int \frac {-x^2+4 x+3}{x \log ^2(x)}dx+4 \int \frac {x^2}{\left (x+e^{2 x}\right )^2 \log (x)}dx-22 \int \frac {x^2}{\left (x+e^{2 x}\right ) \log (x)}dx-12 \int \frac {1}{\left (x+e^{2 x}\right )^2}dx+10 \int \frac {x}{\left (x+e^{2 x}\right )^2}dx-10 \int \frac {1}{x+e^{2 x}}dx-36 \int \frac {x}{x+e^{2 x}}dx-6 \int \frac {1}{\left (x+e^{2 x}\right ) \log ^2(x)}dx-8 \int \frac {x}{\left (x+e^{2 x}\right ) \log ^2(x)}dx-6 \int \frac {x}{\left (x+e^{2 x}\right )^2 \log (x)}dx+6 \int \frac {1}{\left (x+e^{2 x}\right ) \log (x)}dx+4 \int \frac {x}{\left (x+e^{2 x}\right ) \log (x)}dx+4 \operatorname {ExpIntegralEi}(2 \log (x))-8 \operatorname {LogIntegral}(x)\) |
Input:
Int[(E^(4*x)*(6 + 8*x - 2*x^2) + E^(2*x)*(6*x + 8*x^2 - 2*x^3) + (E^(4*x)* (-8*x + 4*x^2) + E^(2*x)*(6*x - 12*x^2 - 14*x^3 + 4*x^4))*Log[x] + (-12*x - 4*x^3 + E^(2*x)*(-10*x - 36*x^2 + 8*x^3))*Log[x]^2)/((E^(4*x)*x + 2*E^(2 *x)*x^2 + x^3)*Log[x]^2),x]
Output:
$Aborted
Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37
\[-\frac {2 \left (2 x^{2}-7 x -6\right )}{{\mathrm e}^{2 x}+x}+\frac {2 \left (x^{2}-4 x -3\right ) {\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}+x \right ) \ln \left (x \right )}\]
Input:
int((((8*x^3-36*x^2-10*x)*exp(x)^2-4*x^3-12*x)*ln(x)^2+((4*x^2-8*x)*exp(x) ^4+(4*x^4-14*x^3-12*x^2+6*x)*exp(x)^2)*ln(x)+(-2*x^2+8*x+6)*exp(x)^4+(-2*x ^3+8*x^2+6*x)*exp(x)^2)/(x*exp(x)^4+2*exp(x)^2*x^2+x^3)/ln(x)^2,x)
Output:
-2*(2*x^2-7*x-6)/(exp(2*x)+x)+2*(x^2-4*x-3)*exp(2*x)/(exp(2*x)+x)/ln(x)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {2 \, {\left ({\left (x^{2} - 4 \, x - 3\right )} e^{\left (2 \, x\right )} - {\left (2 \, x^{2} - 7 \, x - 6\right )} \log \left (x\right )\right )}}{{\left (x + e^{\left (2 \, x\right )}\right )} \log \left (x\right )} \] Input:
integrate((((8*x^3-36*x^2-10*x)*exp(x)^2-4*x^3-12*x)*log(x)^2+((4*x^2-8*x) *exp(x)^4+(4*x^4-14*x^3-12*x^2+6*x)*exp(x)^2)*log(x)+(-2*x^2+8*x+6)*exp(x) ^4+(-2*x^3+8*x^2+6*x)*exp(x)^2)/(x*exp(x)^4+2*exp(x)^2*x^2+x^3)/log(x)^2,x , algorithm="fricas")
Output:
2*((x^2 - 4*x - 3)*e^(2*x) - (2*x^2 - 7*x - 6)*log(x))/((x + e^(2*x))*log( x))
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (27) = 54\).
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {2 x^{2} - 8 x - 6}{\log {\left (x \right )}} + \frac {- 2 x^{3} - 4 x^{2} \log {\left (x \right )} + 8 x^{2} + 14 x \log {\left (x \right )} + 6 x + 12 \log {\left (x \right )}}{x \log {\left (x \right )} + e^{2 x} \log {\left (x \right )}} \] Input:
integrate((((8*x**3-36*x**2-10*x)*exp(x)**2-4*x**3-12*x)*ln(x)**2+((4*x**2 -8*x)*exp(x)**4+(4*x**4-14*x**3-12*x**2+6*x)*exp(x)**2)*ln(x)+(-2*x**2+8*x +6)*exp(x)**4+(-2*x**3+8*x**2+6*x)*exp(x)**2)/(x*exp(x)**4+2*exp(x)**2*x** 2+x**3)/ln(x)**2,x)
Output:
(2*x**2 - 8*x - 6)/log(x) + (-2*x**3 - 4*x**2*log(x) + 8*x**2 + 14*x*log(x ) + 6*x + 12*log(x))/(x*log(x) + exp(2*x)*log(x))
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.26 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {2 \, {\left ({\left (x^{2} - 4 \, x - 3\right )} e^{\left (2 \, x\right )} - {\left (2 \, x^{2} - 7 \, x - 6\right )} \log \left (x\right )\right )}}{x \log \left (x\right ) + e^{\left (2 \, x\right )} \log \left (x\right )} \] Input:
integrate((((8*x^3-36*x^2-10*x)*exp(x)^2-4*x^3-12*x)*log(x)^2+((4*x^2-8*x) *exp(x)^4+(4*x^4-14*x^3-12*x^2+6*x)*exp(x)^2)*log(x)+(-2*x^2+8*x+6)*exp(x) ^4+(-2*x^3+8*x^2+6*x)*exp(x)^2)/(x*exp(x)^4+2*exp(x)^2*x^2+x^3)/log(x)^2,x , algorithm="maxima")
Output:
2*((x^2 - 4*x - 3)*e^(2*x) - (2*x^2 - 7*x - 6)*log(x))/(x*log(x) + e^(2*x) *log(x))
Time = 0.13 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.54 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {2 \, {\left (x^{2} e^{\left (2 \, x\right )} - 2 \, x^{2} \log \left (x\right ) - 4 \, x e^{\left (2 \, x\right )} + 7 \, x \log \left (x\right ) - 3 \, e^{\left (2 \, x\right )} + 6 \, \log \left (x\right )\right )}}{x \log \left (x\right ) + e^{\left (2 \, x\right )} \log \left (x\right )} \] Input:
integrate((((8*x^3-36*x^2-10*x)*exp(x)^2-4*x^3-12*x)*log(x)^2+((4*x^2-8*x) *exp(x)^4+(4*x^4-14*x^3-12*x^2+6*x)*exp(x)^2)*log(x)+(-2*x^2+8*x+6)*exp(x) ^4+(-2*x^3+8*x^2+6*x)*exp(x)^2)/(x*exp(x)^4+2*exp(x)^2*x^2+x^3)/log(x)^2,x , algorithm="giac")
Output:
2*(x^2*e^(2*x) - 2*x^2*log(x) - 4*x*e^(2*x) + 7*x*log(x) - 3*e^(2*x) + 6*l og(x))/(x*log(x) + e^(2*x)*log(x))
Time = 3.11 (sec) , antiderivative size = 179, normalized size of antiderivative = 5.11 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=4\,x^2-\frac {\frac {2\,{\mathrm {e}}^{2\,x}\,\left (-x^2+4\,x+3\right )}{x+{\mathrm {e}}^{2\,x}}-\frac {2\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (x\right )\,\left (6\,x+4\,{\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^{2\,x}+7\,x^2-2\,x^3-3\right )}{{\left (x+{\mathrm {e}}^{2\,x}\right )}^2}}{\ln \left (x\right )}-8\,x-\frac {2\,\left (4\,x^6-20\,x^5+5\,x^4+8\,x^3-3\,x^2\right )}{\left (2\,x-1\right )\,\left ({\mathrm {e}}^{4\,x}+2\,x\,{\mathrm {e}}^{2\,x}+x^2\right )}+\frac {2\,\left (4\,x^5-24\,x^4+11\,x^3+20\,x^2+2\,x-6\right )}{\left (x+{\mathrm {e}}^{2\,x}\right )\,\left (2\,x-1\right )} \] Input:
int((exp(4*x)*(8*x - 2*x^2 + 6) - log(x)*(exp(4*x)*(8*x - 4*x^2) - exp(2*x )*(6*x - 12*x^2 - 14*x^3 + 4*x^4)) + exp(2*x)*(6*x + 8*x^2 - 2*x^3) - log( x)^2*(12*x + exp(2*x)*(10*x + 36*x^2 - 8*x^3) + 4*x^3))/(log(x)^2*(x*exp(4 *x) + 2*x^2*exp(2*x) + x^3)),x)
Output:
4*x^2 - ((2*exp(2*x)*(4*x - x^2 + 3))/(x + exp(2*x)) - (2*x*exp(2*x)*log(x )*(6*x + 4*exp(2*x) - 2*x*exp(2*x) + 7*x^2 - 2*x^3 - 3))/(x + exp(2*x))^2) /log(x) - 8*x - (2*(8*x^3 - 3*x^2 + 5*x^4 - 20*x^5 + 4*x^6))/((2*x - 1)*(e xp(4*x) + 2*x*exp(2*x) + x^2)) + (2*(2*x + 20*x^2 + 11*x^3 - 24*x^4 + 4*x^ 5 - 6))/((x + exp(2*x))*(2*x - 1))
Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.71 \[ \int \frac {e^{4 x} \left (6+8 x-2 x^2\right )+e^{2 x} \left (6 x+8 x^2-2 x^3\right )+\left (e^{4 x} \left (-8 x+4 x^2\right )+e^{2 x} \left (6 x-12 x^2-14 x^3+4 x^4\right )\right ) \log (x)+\left (-12 x-4 x^3+e^{2 x} \left (-10 x-36 x^2+8 x^3\right )\right ) \log ^2(x)}{\left (e^{4 x} x+2 e^{2 x} x^2+x^3\right ) \log ^2(x)} \, dx=\frac {-14 e^{2 x} \mathrm {log}\left (x \right )+2 e^{2 x} x^{2}-8 e^{2 x} x -6 e^{2 x}-4 \,\mathrm {log}\left (x \right ) x^{2}+12 \,\mathrm {log}\left (x \right )}{\mathrm {log}\left (x \right ) \left (e^{2 x}+x \right )} \] Input:
int((((8*x^3-36*x^2-10*x)*exp(x)^2-4*x^3-12*x)*log(x)^2+((4*x^2-8*x)*exp(x )^4+(4*x^4-14*x^3-12*x^2+6*x)*exp(x)^2)*log(x)+(-2*x^2+8*x+6)*exp(x)^4+(-2 *x^3+8*x^2+6*x)*exp(x)^2)/(x*exp(x)^4+2*exp(x)^2*x^2+x^3)/log(x)^2,x)
Output:
(2*( - 7*e**(2*x)*log(x) + e**(2*x)*x**2 - 4*e**(2*x)*x - 3*e**(2*x) - 2*l og(x)*x**2 + 6*log(x)))/(log(x)*(e**(2*x) + x))