Integrand size = 118, antiderivative size = 25 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=3+\frac {16 x \left (-2-x+\log \left (e^2-x\right )\right )}{e^x+x} \]
Time = 0.67 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=-\frac {16 x \left (2+x-\log \left (e^2-x\right )\right )}{e^x+x} \]
Integrate[(-16*x^2 - 16*E^2*x^2 + 16*x^3 + E^x*(16*x - 16*x^3 + E^2*(-32 + 16*x^2)) + E^x*(E^2*(16 - 16*x) - 16*x + 16*x^2)*Log[E^2 - x])/(E^(2*x)*( E^2 - x) + E^2*x^2 - x^3 + E^x*(2*E^2*x - 2*x^2)),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.
\(\displaystyle \int \frac {16 x^3-16 e^2 x^2-16 x^2+e^x \left (16 x^2-16 x+e^2 (16-16 x)\right ) \log \left (e^2-x\right )+e^x \left (-16 x^3+e^2 \left (16 x^2-32\right )+16 x\right )}{-x^3+e^2 x^2+e^x \left (2 e^2 x-2 x^2\right )+e^{2 x} \left (e^2-x\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {16 x^3+\left (-16-16 e^2\right ) x^2+e^x \left (16 x^2-16 x+e^2 (16-16 x)\right ) \log \left (e^2-x\right )+e^x \left (-16 x^3+e^2 \left (16 x^2-32\right )+16 x\right )}{-x^3+e^2 x^2+e^x \left (2 e^2 x-2 x^2\right )+e^{2 x} \left (e^2-x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {16 x^3+\left (-16-16 e^2\right ) x^2+e^x \left (16 x^2-16 x+e^2 (16-16 x)\right ) \log \left (e^2-x\right )+e^x \left (-16 x^3+e^2 \left (16 x^2-32\right )+16 x\right )}{\left (e^2-x\right ) \left (x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {16 \left (-x^3+e^2 x^2+x^2 \log \left (e^2-x\right )+x-\left (1+e^2\right ) x \log \left (e^2-x\right )+e^2 \log \left (e^2-x\right )-2 e^2\right )}{\left (e^2-x\right ) \left (x+e^x\right )}-\frac {16 (x-1) x \left (x-\log \left (e^2-x\right )+2\right )}{\left (x+e^x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -16 \int \frac {x^3}{\left (x+e^x\right )^2}dx-16 \int \frac {x^2}{\left (x+e^x\right )^2}dx+16 \int \frac {x^2}{x+e^x}dx+16 \int \frac {\int \frac {x^2}{\left (x+e^x\right )^2}dx}{e^2-x}dx+16 \log \left (e^2-x\right ) \int \frac {x^2}{\left (x+e^x\right )^2}dx+32 \int \frac {1}{\left (x+e^x\right )^2}dx+16 \int \frac {1}{x+e^x}dx-16 e^2 \int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx-16 \int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx-16 \int \frac {\int \frac {1}{\left (x+e^x\right )^2}dx}{e^2-x}dx+16 \left (1+e^2\right ) \int \frac {\int \frac {1}{x+e^x}dx}{e^2-x}dx-16 e^2 \int \frac {\int \frac {1}{x+e^x}dx}{e^2-x}dx-16 \int \frac {\int \frac {1}{x+e^x}dx}{e^2-x}dx-16 e^2 \left (1+e^2\right ) \int \frac {\int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx}{e^2-x}dx+16 e^4 \int \frac {\int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx}{e^2-x}dx+16 e^2 \int \frac {\int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx}{e^2-x}dx-16 \int \frac {\int \frac {x}{x+e^x}dx}{e^2-x}dx-16 \log \left (e^2-x\right ) \int \frac {1}{\left (x+e^x\right )^2}dx+16 \left (1+e^2\right ) \log \left (e^2-x\right ) \int \frac {1}{x+e^x}dx-16 e^2 \log \left (e^2-x\right ) \int \frac {1}{x+e^x}dx-16 \log \left (e^2-x\right ) \int \frac {1}{x+e^x}dx-16 e^2 \left (1+e^2\right ) \log \left (e^2-x\right ) \int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx+16 e^4 \log \left (e^2-x\right ) \int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx+16 e^2 \log \left (e^2-x\right ) \int \frac {1}{\left (e^2-x\right ) \left (x+e^x\right )}dx-16 \log \left (e^2-x\right ) \int \frac {x}{x+e^x}dx+\frac {32}{x+e^x}-\frac {16 \log \left (e^2-x\right )}{x+e^x}\) |
Int[(-16*x^2 - 16*E^2*x^2 + 16*x^3 + E^x*(16*x - 16*x^3 + E^2*(-32 + 16*x^ 2)) + E^x*(E^2*(16 - 16*x) - 16*x + 16*x^2)*Log[E^2 - x])/(E^(2*x)*(E^2 - x) + E^2*x^2 - x^3 + E^x*(2*E^2*x - 2*x^2)),x]
3.18.39.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
method | result | size |
parallelrisch | \(\frac {-16 x^{2}+16 x \ln \left ({\mathrm e}^{2}-x \right )-32 x}{{\mathrm e}^{x}+x}\) | \(27\) |
risch | \(\frac {16 x \ln \left ({\mathrm e}^{2}-x \right )}{{\mathrm e}^{x}+x}-\frac {16 x \left (2+x \right )}{{\mathrm e}^{x}+x}\) | \(30\) |
int((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*ln(exp(2)-x)+((16*x^2-32)*exp( 2)-16*x^3+16*x)*exp(x)-16*x^2*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp(x)^2+( 2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=-\frac {16 \, {\left (x^{2} - x \log \left (-x + e^{2}\right ) + 2 \, x\right )}}{x + e^{x}} \]
integrate((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*log(exp(2)-x)+((16*x^2-3 2)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp (x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=\frac {- 16 x^{2} + 16 x \log {\left (- x + e^{2} \right )} - 32 x}{x + e^{x}} \]
integrate((((-16*x+16)*exp(2)+16*x**2-16*x)*exp(x)*ln(exp(2)-x)+((16*x**2- 32)*exp(2)-16*x**3+16*x)*exp(x)-16*x**2*exp(2)+16*x**3-16*x**2)/((exp(2)-x )*exp(x)**2+(2*exp(2)*x-2*x**2)*exp(x)+x**2*exp(2)-x**3),x)
Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=-\frac {16 \, {\left (x^{2} - x \log \left (-x + e^{2}\right ) + 2 \, x\right )}}{x + e^{x}} \]
integrate((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*log(exp(2)-x)+((16*x^2-3 2)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp (x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=-\frac {16 \, {\left (x^{2} - x \log \left (-x + e^{2}\right ) + 2 \, x\right )}}{x + e^{x}} \]
integrate((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*log(exp(2)-x)+((16*x^2-3 2)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp (x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x, algorithm=\
Time = 11.61 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx=-\frac {16\,x\,\left (x-\ln \left ({\mathrm {e}}^2-x\right )+2\right )}{x+{\mathrm {e}}^x} \]