Integrand size = 86, antiderivative size = 25 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {x}{4 \left (-3 e^{2 x} (4-x)+x-\log (x)\right )} \] Output:
1/4*x/(x-ln(x)-3*exp(2*x)*(4-x))
Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {x}{4 \left (3 e^{2 x} (-4+x)+x-\log (x)\right )} \] Input:
Integrate[(1 + E^(2*x)*(-12 + 24*x - 6*x^2) - Log[x])/(4*x^2 + E^(2*x)*(-9 6*x + 24*x^2) + E^(4*x)*(576 - 288*x + 36*x^2) + (E^(2*x)*(96 - 24*x) - 8* x)*Log[x] + 4*Log[x]^2),x]
Output:
x/(4*(3*E^(2*x)*(-4 + 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.
\(\displaystyle \int \frac {e^{2 x} \left (-6 x^2+24 x-12\right )-\log (x)+1}{4 x^2+e^{2 x} \left (24 x^2-96 x\right )+e^{4 x} \left (36 x^2-288 x+576\right )+4 \log ^2(x)+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-6 e^{2 x} \left (x^2-4 x+2\right )-\log (x)+1}{4 \left (3 e^{2 x} (x-4)+x-\log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {-6 e^{2 x} \left (x^2-4 x+2\right )-\log (x)+1}{\left (3 e^{2 x} (4-x)-x+\log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{4} \int \left (\frac {2 x^3-2 \log (x) x^2-8 x^2+7 \log (x) x+5 x-4}{(x-4) \left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}-\frac {2 \left (x^2-4 x+2\right )}{(x-4) \left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (2 \int \frac {x^2}{\left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}dx+5 \int \frac {1}{\left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}dx+16 \int \frac {1}{(x-4) \left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}dx-4 \int \frac {1}{(x-4) \left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )}dx-2 \int \frac {x}{3 e^{2 x} x+x-12 e^{2 x}-\log (x)}dx-\int \frac {\log (x)}{\left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}dx-4 \int \frac {\log (x)}{(x-4) \left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}dx-2 \int \frac {x \log (x)}{\left (3 e^{2 x} x+x-12 e^{2 x}-\log (x)\right )^2}dx\right )\) |
Input:
Int[(1 + E^(2*x)*(-12 + 24*x - 6*x^2) - Log[x])/(4*x^2 + E^(2*x)*(-96*x + 24*x^2) + E^(4*x)*(576 - 288*x + 36*x^2) + (E^(2*x)*(96 - 24*x) - 8*x)*Log [x] + 4*Log[x]^2),x]
Output:
$Aborted
Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {x}{12 x \,{\mathrm e}^{2 x}+4 x -48 \,{\mathrm e}^{2 x}-4 \ln \left (x \right )}\) | \(25\) |
parallelrisch | \(\frac {x}{12 x \,{\mathrm e}^{2 x}+4 x -48 \,{\mathrm e}^{2 x}-4 \ln \left (x \right )}\) | \(25\) |
Input:
int((-ln(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*ln(x)^2+((-24*x+96)*exp(2*x)-8 *x)*ln(x)+(36*x^2-288*x+576)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4*x^2),x,me thod=_RETURNVERBOSE)
Output:
1/4*x/(3*x*exp(2*x)+x-12*exp(2*x)-ln(x))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {x}{4 \, {\left (3 \, {\left (x - 4\right )} e^{\left (2 \, x\right )} + x - \log \left (x\right )\right )}} \] Input:
integrate((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*ex p(2*x)-8*x)*log(x)+(36*x^2-288*x+576)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4* x^2),x, algorithm="fricas")
Output:
1/4*x/(3*(x - 4)*e^(2*x) + x - log(x))
Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {x}{4 x + \left (12 x - 48\right ) e^{2 x} - 4 \log {\left (x \right )}} \] Input:
integrate((-ln(x)+(-6*x**2+24*x-12)*exp(2*x)+1)/(4*ln(x)**2+((-24*x+96)*ex p(2*x)-8*x)*ln(x)+(36*x**2-288*x+576)*exp(2*x)**2+(24*x**2-96*x)*exp(2*x)+ 4*x**2),x)
Output:
x/(4*x + (12*x - 48)*exp(2*x) - 4*log(x))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {x}{4 \, {\left (3 \, {\left (x - 4\right )} e^{\left (2 \, x\right )} + x - \log \left (x\right )\right )}} \] Input:
integrate((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*ex p(2*x)-8*x)*log(x)+(36*x^2-288*x+576)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4* x^2),x, algorithm="maxima")
Output:
1/4*x/(3*(x - 4)*e^(2*x) + x - log(x))
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {x}{4 \, {\left (3 \, x e^{\left (2 \, x\right )} + x - 12 \, e^{\left (2 \, x\right )} - \log \left (x\right )\right )}} \] Input:
integrate((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*ex p(2*x)-8*x)*log(x)+(36*x^2-288*x+576)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4* x^2),x, algorithm="giac")
Output:
1/4*x/(3*x*e^(2*x) + x - 12*e^(2*x) - log(x))
Timed out. \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\int -\frac {\ln \left (x\right )+{\mathrm {e}}^{2\,x}\,\left (6\,x^2-24\,x+12\right )-1}{{\mathrm {e}}^{4\,x}\,\left (36\,x^2-288\,x+576\right )-{\mathrm {e}}^{2\,x}\,\left (96\,x-24\,x^2\right )+4\,{\ln \left (x\right )}^2-\ln \left (x\right )\,\left (8\,x+{\mathrm {e}}^{2\,x}\,\left (24\,x-96\right )\right )+4\,x^2} \,d x \] Input:
int(-(log(x) + exp(2*x)*(6*x^2 - 24*x + 12) - 1)/(exp(4*x)*(36*x^2 - 288*x + 576) - exp(2*x)*(96*x - 24*x^2) + 4*log(x)^2 - log(x)*(8*x + exp(2*x)*( 24*x - 96)) + 4*x^2),x)
Output:
int(-(log(x) + exp(2*x)*(6*x^2 - 24*x + 12) - 1)/(exp(4*x)*(36*x^2 - 288*x + 576) - exp(2*x)*(96*x - 24*x^2) + 4*log(x)^2 - log(x)*(8*x + exp(2*x)*( 24*x - 96)) + 4*x^2), x)
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx=\frac {-3 e^{2 x} x +12 e^{2 x}+\mathrm {log}\left (x \right )}{12 e^{2 x} x -48 e^{2 x}-4 \,\mathrm {log}\left (x \right )+4 x} \] Input:
int((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*exp(2*x) -8*x)*log(x)+(36*x^2-288*x+576)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4*x^2),x )
Output:
( - 3*e**(2*x)*x + 12*e**(2*x) + log(x))/(4*(3*e**(2*x)*x - 12*e**(2*x) - log(x) + x))