Integrand size = 92, antiderivative size = 26 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=3 \left (\log (4)+\frac {4-x}{2 \left (4+e^{2 x}+x+\log (x)\right )}\right ) \] Output:
3/2*(4-x)/(exp(x)^2+x+ln(x)+4)+6*ln(2)
Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=\frac {3 (4-x)}{2 \left (4+e^{2 x}+x+\log (x)\right )} \] Input:
Integrate[(-12 - 21*x + E^(2*x)*(-27*x + 6*x^2) - 3*x*Log[x])/(32*x + 2*E^ (4*x)*x + 16*x^2 + 2*x^3 + E^(2*x)*(16*x + 4*x^2) + (16*x + 4*E^(2*x)*x + 4*x^2)*Log[x] + 2*x*Log[x]^2),x]
Output:
(3*(4 - x))/(2*(4 + E^(2*x) + 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.
\(\displaystyle \int \frac {e^{2 x} \left (6 x^2-27 x\right )-21 x-3 x \log (x)-12}{2 x^3+16 x^2+e^{2 x} \left (4 x^2+16 x\right )+\left (4 x^2+4 e^{2 x} x+16 x\right ) \log (x)+2 e^{4 x} x+32 x+2 x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 \left (2 e^{2 x} x^2-\left (9 e^{2 x}+7\right ) x-x \log (x)-4\right )}{2 x \left (x+e^{2 x}+\log (x)+4\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{2} \int -\frac {-2 e^{2 x} x^2+\left (7+9 e^{2 x}\right ) x+\log (x) x+4}{x \left (x+e^{2 x}+\log (x)+4\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {3}{2} \int \frac {-2 e^{2 x} x^2+\left (7+9 e^{2 x}\right ) x+\log (x) x+4}{x \left (x+e^{2 x}+\log (x)+4\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3}{2} \int \left (\frac {(x-4) \left (2 x^2+2 \log (x) x+7 x-1\right )}{x \left (x+e^{2 x}+\log (x)+4\right )^2}-\frac {2 x-9}{x+e^{2 x}+\log (x)+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3}{2} \left (2 \int \frac {x^2}{\left (x+e^{2 x}+\log (x)+4\right )^2}dx-29 \int \frac {1}{\left (x+e^{2 x}+\log (x)+4\right )^2}dx+4 \int \frac {1}{x \left (x+e^{2 x}+\log (x)+4\right )^2}dx-\int \frac {x}{\left (x+e^{2 x}+\log (x)+4\right )^2}dx-8 \int \frac {\log (x)}{\left (x+e^{2 x}+\log (x)+4\right )^2}dx+2 \int \frac {x \log (x)}{\left (x+e^{2 x}+\log (x)+4\right )^2}dx+9 \int \frac {1}{x+e^{2 x}+\log (x)+4}dx-2 \int \frac {x}{x+e^{2 x}+\log (x)+4}dx\right )\) |
Input:
Int[(-12 - 21*x + E^(2*x)*(-27*x + 6*x^2) - 3*x*Log[x])/(32*x + 2*E^(4*x)* x + 16*x^2 + 2*x^3 + E^(2*x)*(16*x + 4*x^2) + (16*x + 4*E^(2*x)*x + 4*x^2) *Log[x] + 2*x*Log[x]^2),x]
Output:
$Aborted
Time = 0.42 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65
method | result | size |
risch | \(-\frac {3 \left (x -4\right )}{2 \left ({\mathrm e}^{2 x}+x +\ln \left (x \right )+4\right )}\) | \(17\) |
parallelrisch | \(\frac {-3 x +12}{2 \,{\mathrm e}^{2 x}+2 x +2 \ln \left (x \right )+8}\) | \(19\) |
Input:
int((-3*x*ln(x)+(6*x^2-27*x)*exp(x)^2-21*x-12)/(2*x*ln(x)^2+(4*x*exp(x)^2+ 4*x^2+16*x)*ln(x)+2*x*exp(x)^4+(4*x^2+16*x)*exp(x)^2+2*x^3+16*x^2+32*x),x, method=_RETURNVERBOSE)
Output:
-3/2*(x-4)/(exp(2*x)+x+ln(x)+4)
Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=-\frac {3 \, {\left (x - 4\right )}}{2 \, {\left (x + e^{\left (2 \, x\right )} + \log \left (x\right ) + 4\right )}} \] Input:
integrate((-3*x*log(x)+(6*x^2-27*x)*exp(x)^2-21*x-12)/(2*x*log(x)^2+(4*x*e xp(x)^2+4*x^2+16*x)*log(x)+2*x*exp(x)^4+(4*x^2+16*x)*exp(x)^2+2*x^3+16*x^2 +32*x),x, algorithm="fricas")
Output:
-3/2*(x - 4)/(x + e^(2*x) + log(x) + 4)
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=\frac {12 - 3 x}{2 x + 2 e^{2 x} + 2 \log {\left (x \right )} + 8} \] Input:
integrate((-3*x*ln(x)+(6*x**2-27*x)*exp(x)**2-21*x-12)/(2*x*ln(x)**2+(4*x* exp(x)**2+4*x**2+16*x)*ln(x)+2*x*exp(x)**4+(4*x**2+16*x)*exp(x)**2+2*x**3+ 16*x**2+32*x),x)
Output:
(12 - 3*x)/(2*x + 2*exp(2*x) + 2*log(x) + 8)
Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=-\frac {3 \, {\left (x - 4\right )}}{2 \, {\left (x + e^{\left (2 \, x\right )} + \log \left (x\right ) + 4\right )}} \] Input:
integrate((-3*x*log(x)+(6*x^2-27*x)*exp(x)^2-21*x-12)/(2*x*log(x)^2+(4*x*e xp(x)^2+4*x^2+16*x)*log(x)+2*x*exp(x)^4+(4*x^2+16*x)*exp(x)^2+2*x^3+16*x^2 +32*x),x, algorithm="maxima")
Output:
-3/2*(x - 4)/(x + e^(2*x) + log(x) + 4)
Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=-\frac {3 \, {\left (x - 4\right )}}{2 \, {\left (x + e^{\left (2 \, x\right )} + \log \left (x\right ) + 4\right )}} \] Input:
integrate((-3*x*log(x)+(6*x^2-27*x)*exp(x)^2-21*x-12)/(2*x*log(x)^2+(4*x*e xp(x)^2+4*x^2+16*x)*log(x)+2*x*exp(x)^4+(4*x^2+16*x)*exp(x)^2+2*x^3+16*x^2 +32*x),x, algorithm="giac")
Output:
-3/2*(x - 4)/(x + e^(2*x) + log(x) + 4)
Time = 4.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=-\frac {3\,\left (x-4\right )}{2\,\left (x+{\mathrm {e}}^{2\,x}+\ln \left (x\right )+4\right )} \] Input:
int(-(21*x + exp(2*x)*(27*x - 6*x^2) + 3*x*log(x) + 12)/(32*x + exp(2*x)*( 16*x + 4*x^2) + 2*x*exp(4*x) + 2*x*log(x)^2 + log(x)*(16*x + 4*x*exp(2*x) + 4*x^2) + 16*x^2 + 2*x^3),x)
Output:
-(3*(x - 4))/(2*(x + exp(2*x) + log(x) + 4))
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {-12-21 x+e^{2 x} \left (-27 x+6 x^2\right )-3 x \log (x)}{32 x+2 e^{4 x} x+16 x^2+2 x^3+e^{2 x} \left (16 x+4 x^2\right )+\left (16 x+4 e^{2 x} x+4 x^2\right ) \log (x)+2 x \log ^2(x)} \, dx=\frac {-3 x +12}{2 e^{2 x}+2 \,\mathrm {log}\left (x \right )+2 x +8} \] Input:
int((-3*x*log(x)+(6*x^2-27*x)*exp(x)^2-21*x-12)/(2*x*log(x)^2+(4*x*exp(x)^ 2+4*x^2+16*x)*log(x)+2*x*exp(x)^4+(4*x^2+16*x)*exp(x)^2+2*x^3+16*x^2+32*x) ,x)
Output:
(3*( - x + 4))/(2*(e**(2*x) + log(x) + x + 4))