Integrand size = 72, antiderivative size = 27 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=2+x (-3+x+x (x+\log (x)))-\left (x+\log \left (e^x+2 x\right )\right )^2 \] Output:
2+x*(x-3+(x+ln(x))*x)-(x+ln(exp(x)+2*x))^2
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=-3 x+x^3+x^2 \log (x)-2 x \log \left (e^x+2 x\right )-\log ^2\left (e^x+2 x\right ) \] Input:
Integrate[(-10*x + 2*x^2 + 6*x^3 + E^x*(-3 - x + 3*x^2) + (2*E^x*x + 4*x^2 )*Log[x] + (-4 - 4*E^x - 4*x)*Log[E^x + 2*x])/(E^x + 2*x),x]
Output:
-3*x + x^3 + x^2*Log[x] - 2*x*Log[E^x + 2*x] - Log[E^x + 2*x]^2
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.
\(\displaystyle \int \frac {6 x^3+2 x^2+e^x \left (3 x^2-x-3\right )+\left (4 x^2+2 e^x x\right ) \log (x)-10 x+\left (-4 x-4 e^x-4\right ) \log \left (2 x+e^x\right )}{2 x+e^x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (3 x^2-x+2 x \log (x)-4 \log \left (2 x+e^x\right )+\frac {4 (x-1) \left (x+\log \left (2 x+e^x\right )\right )}{2 x+e^x}-3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {x^2}{2 x+e^x}dx+4 \int \frac {x}{2 x+e^x}dx+4 \int \int \frac {1}{2 x+e^x}dxdx+8 \int \frac {\int \frac {1}{2 x+e^x}dx}{2 x+e^x}dx-8 \int \frac {x \int \frac {1}{2 x+e^x}dx}{2 x+e^x}dx-4 \int \int \frac {x}{2 x+e^x}dxdx-8 \int \frac {\int \frac {x}{2 x+e^x}dx}{2 x+e^x}dx+8 \int \frac {x \int \frac {x}{2 x+e^x}dx}{2 x+e^x}dx-4 \log \left (2 x+e^x\right ) \int \frac {1}{2 x+e^x}dx+4 \log \left (2 x+e^x\right ) \int \frac {x}{2 x+e^x}dx+x^3+x^2+x^2 \log (x)-3 x-4 x \log \left (2 x+e^x\right )\) |
Input:
Int[(-10*x + 2*x^2 + 6*x^3 + E^x*(-3 - x + 3*x^2) + (2*E^x*x + 4*x^2)*Log[ x] + (-4 - 4*E^x - 4*x)*Log[E^x + 2*x])/(E^x + 2*x),x]
Output:
$Aborted
Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30
method | result | size |
risch | \(x^{3}+x^{2} \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}+2 x \right ) x -\ln \left ({\mathrm e}^{x}+2 x \right )^{2}-3 x\) | \(35\) |
parallelrisch | \(x^{3}+x^{2} \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x}+2 x \right ) x -\ln \left ({\mathrm e}^{x}+2 x \right )^{2}-3 x\) | \(35\) |
Input:
int(((-4*exp(x)-4*x-4)*ln(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*ln(x)+(3*x^2-x-3) *exp(x)+6*x^3+2*x^2-10*x)/(exp(x)+2*x),x,method=_RETURNVERBOSE)
Output:
x^3+x^2*ln(x)-2*ln(exp(x)+2*x)*x-ln(exp(x)+2*x)^2-3*x
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=x^{3} + x^{2} \log \left (x\right ) - 2 \, x \log \left (2 \, x + e^{x}\right ) - \log \left (2 \, x + e^{x}\right )^{2} - 3 \, x \] Input:
integrate(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3* x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(exp(x)+2*x),x, algorithm="fricas")
Output:
x^3 + x^2*log(x) - 2*x*log(2*x + e^x) - log(2*x + e^x)^2 - 3*x
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=x^{3} + x^{2} \log {\left (x \right )} - 2 x \log {\left (2 x + e^{x} \right )} - 3 x - \log {\left (2 x + e^{x} \right )}^{2} \] Input:
integrate(((-4*exp(x)-4*x-4)*ln(exp(x)+2*x)+(2*exp(x)*x+4*x**2)*ln(x)+(3*x **2-x-3)*exp(x)+6*x**3+2*x**2-10*x)/(exp(x)+2*x),x)
Output:
x**3 + x**2*log(x) - 2*x*log(2*x + exp(x)) - 3*x - log(2*x + exp(x))**2
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=x^{3} + x^{2} \log \left (x\right ) - 2 \, x \log \left (2 \, x + e^{x}\right ) - \log \left (2 \, x + e^{x}\right )^{2} - 3 \, x \] Input:
integrate(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3* x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(exp(x)+2*x),x, algorithm="maxima")
Output:
x^3 + x^2*log(x) - 2*x*log(2*x + e^x) - log(2*x + e^x)^2 - 3*x
\[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=\int { \frac {6 \, x^{3} + 2 \, x^{2} + {\left (3 \, x^{2} - x - 3\right )} e^{x} - 4 \, {\left (x + e^{x} + 1\right )} \log \left (2 \, x + e^{x}\right ) + 2 \, {\left (2 \, x^{2} + x e^{x}\right )} \log \left (x\right ) - 10 \, x}{2 \, x + e^{x}} \,d x } \] Input:
integrate(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3* x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(exp(x)+2*x),x, algorithm="giac")
Output:
integrate((6*x^3 + 2*x^2 + (3*x^2 - x - 3)*e^x - 4*(x + e^x + 1)*log(2*x + e^x) + 2*(2*x^2 + x*e^x)*log(x) - 10*x)/(2*x + e^x), x)
Time = 2.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=x^2\,\ln \left (x\right )-3\,x-2\,x\,\ln \left (2\,x+{\mathrm {e}}^x\right )+x^3-{\ln \left (2\,x+{\mathrm {e}}^x\right )}^2 \] Input:
int(-(10*x + exp(x)*(x - 3*x^2 + 3) + log(2*x + exp(x))*(4*x + 4*exp(x) + 4) - log(x)*(2*x*exp(x) + 4*x^2) - 2*x^2 - 6*x^3)/(2*x + exp(x)),x)
Output:
x^2*log(x) - 3*x - 2*x*log(2*x + exp(x)) + x^3 - log(2*x + exp(x))^2
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx=-\mathrm {log}\left (e^{x}+2 x \right )^{2}-2 \,\mathrm {log}\left (e^{x}+2 x \right ) x +\mathrm {log}\left (x \right ) x^{2}+x^{3}-3 x \] Input:
int(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3*x^2-x- 3)*exp(x)+6*x^3+2*x^2-10*x)/(exp(x)+2*x),x)
Output:
- log(e**x + 2*x)**2 - 2*log(e**x + 2*x)*x + log(x)*x**2 + x**3 - 3*x