Integrand size = 104, antiderivative size = 21 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^{2 x} x}{\frac {8}{e^5+x}+\log (x)} \] Output:
exp(x)^2/(4/(1/2*exp(5)+1/2*x)+ln(x))*x
Time = 2.78 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^{2 x} x \left (e^5+x\right )}{8+e^5 \log (x)+x \log (x)} \] Input:
Integrate[(E^(2*x)*(-E^10 + 16*x + 15*x^2 + E^5*(8 + 14*x)) + E^(2*x)*(x^2 + 2*x^3 + E^10*(1 + 2*x) + E^5*(2*x + 4*x^2))*Log[x])/(64 + (16*E^5 + 16* x)*Log[x] + (E^10 + 2*E^5*x + x^2)*Log[x]^2),x]
Output:
(E^(2*x)*x*(E^5 + x))/(8 + E^5*Log[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.
\(\displaystyle \int \frac {e^{2 x} \left (15 x^2+16 x+e^5 (14 x+8)-e^{10}\right )+e^{2 x} \left (2 x^3+x^2+e^5 \left (4 x^2+2 x\right )+e^{10} (2 x+1)\right ) \log (x)}{\left (x^2+2 e^5 x+e^{10}\right ) \log ^2(x)+\left (16 x+16 e^5\right ) \log (x)+64} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{2 x} \left (2 e^5 (7 x+4)+x (15 x+16)+(2 x+1) \left (x+e^5\right )^2 \log (x)-e^{10}\right )}{\left (\left (x+e^5\right ) \log (x)+8\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{2 x} \left (-x^2+2 \left (4-e^5\right ) x-e^{10}\right )}{\left (x \log (x)+e^5 \log (x)+8\right )^2}+\frac {e^{2 x} \left (x+e^5\right ) (2 x+1)}{x \log (x)+e^5 \log (x)+8}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{2 x} x^2}{\left (x \log (x)+e^5 \log (x)+8\right )^2}dx+2 \int \frac {e^{2 x} x^2}{x \log (x)+e^5 \log (x)+8}dx-\int \frac {e^{2 x+10}}{\left (x \log (x)+e^5 \log (x)+8\right )^2}dx+2 \left (4-e^5\right ) \int \frac {e^{2 x} x}{\left (x \log (x)+e^5 \log (x)+8\right )^2}dx+\int \frac {e^{2 x+5}}{x \log (x)+e^5 \log (x)+8}dx+\left (1+2 e^5\right ) \int \frac {e^{2 x} x}{x \log (x)+e^5 \log (x)+8}dx\) |
Input:
Int[(E^(2*x)*(-E^10 + 16*x + 15*x^2 + E^5*(8 + 14*x)) + E^(2*x)*(x^2 + 2*x ^3 + E^10*(1 + 2*x) + E^5*(2*x + 4*x^2))*Log[x])/(64 + (16*E^5 + 16*x)*Log [x] + (E^10 + 2*E^5*x + x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 1.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{2 x} \left ({\mathrm e}^{5}+x \right )}{{\mathrm e}^{5} \ln \left (x \right )+x \ln \left (x \right )+8}\) | \(24\) |
parallelrisch | \(\frac {{\mathrm e}^{2 x} {\mathrm e}^{5} x +{\mathrm e}^{2 x} x^{2}}{{\mathrm e}^{5} \ln \left (x \right )+x \ln \left (x \right )+8}\) | \(32\) |
Input:
int((((1+2*x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*ln(x)+(-exp( 5)^2+(14*x+8)*exp(5)+15*x^2+16*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x^2)*ln( x)^2+(16*exp(5)+16*x)*ln(x)+64),x,method=_RETURNVERBOSE)
Output:
x*exp(2*x)*(exp(5)+x)/(exp(5)*ln(x)+x*ln(x)+8)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {{\left (x^{2} + x e^{5}\right )} e^{\left (2 \, x\right )}}{{\left (x + e^{5}\right )} \log \left (x\right ) + 8} \] Input:
integrate((((1+2*x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x) +(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+16*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x ^2)*log(x)^2+(16*exp(5)+16*x)*log(x)+64),x, algorithm="fricas")
Output:
(x^2 + x*e^5)*e^(2*x)/((x + e^5)*log(x) + 8)
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {\left (x^{2} + x e^{5}\right ) e^{2 x}}{x \log {\left (x \right )} + e^{5} \log {\left (x \right )} + 8} \] Input:
integrate((((1+2*x)*exp(5)**2+(4*x**2+2*x)*exp(5)+2*x**3+x**2)*exp(x)**2*l n(x)+(-exp(5)**2+(14*x+8)*exp(5)+15*x**2+16*x)*exp(x)**2)/((exp(5)**2+2*x* exp(5)+x**2)*ln(x)**2+(16*exp(5)+16*x)*ln(x)+64),x)
Output:
(x**2 + x*exp(5))*exp(2*x)/(x*log(x) + exp(5)*log(x) + 8)
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {{\left (x^{2} + x e^{5}\right )} e^{\left (2 \, x\right )}}{{\left (x + e^{5}\right )} \log \left (x\right ) + 8} \] Input:
integrate((((1+2*x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x) +(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+16*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x ^2)*log(x)^2+(16*exp(5)+16*x)*log(x)+64),x, algorithm="maxima")
Output:
(x^2 + x*e^5)*e^(2*x)/((x + e^5)*log(x) + 8)
Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {x^{2} e^{\left (2 \, x\right )} + x e^{\left (2 \, x + 5\right )}}{x \log \left (x\right ) + e^{5} \log \left (x\right ) + 8} \] Input:
integrate((((1+2*x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x) +(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+16*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x ^2)*log(x)^2+(16*exp(5)+16*x)*log(x)+64),x, algorithm="giac")
Output:
(x^2*e^(2*x) + x*e^(2*x + 5))/(x*log(x) + e^5*log(x) + 8)
Time = 4.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {x\,{\mathrm {e}}^{2\,x}\,\left (x+{\mathrm {e}}^5\right )}{{\mathrm {e}}^5\,\ln \left (x\right )+x\,\ln \left (x\right )+8} \] Input:
int((exp(2*x)*(16*x - exp(10) + 15*x^2 + exp(5)*(14*x + 8)) + exp(2*x)*log (x)*(exp(5)*(2*x + 4*x^2) + x^2 + 2*x^3 + exp(10)*(2*x + 1)))/(log(x)*(16* x + 16*exp(5)) + log(x)^2*(exp(10) + 2*x*exp(5) + x^2) + 64),x)
Output:
(x*exp(2*x)*(x + exp(5)))/(exp(5)*log(x) + x*log(x) + 8)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {e^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^{2 x} x \left (e^{5}+x \right )}{\mathrm {log}\left (x \right ) e^{5}+\mathrm {log}\left (x \right ) x +8} \] Input:
int((((1+2*x)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x)+(-exp (5)^2+(14*x+8)*exp(5)+15*x^2+16*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x^2)*lo g(x)^2+(16*exp(5)+16*x)*log(x)+64),x)
Output:
(e**(2*x)*x*(e**5 + x))/(log(x)*e**5 + log(x)*x + 8)