Integrand size = 117, antiderivative size = 25 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-4+e^x-\frac {x}{2}-\frac {\log (x)}{(3+x) \log (2+x)} \] Output:
exp(x)-4-1/2*x-1/ln(2+x)*ln(x)/(3+x)
Time = 0.48 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=\frac {1}{2} \left (2 e^x-x-\frac {2 \log (x)}{(3+x) \log (2+x)}\right ) \] Input:
Integrate[((6*x + 2*x^2)*Log[x] + (-12 - 10*x - 2*x^2 + (4*x + 2*x^2)*Log[ x])*Log[2 + x] + (-18*x - 21*x^2 - 8*x^3 - x^4 + E^x*(36*x + 42*x^2 + 16*x ^3 + 2*x^4))*Log[2 + x]^2)/((36*x + 42*x^2 + 16*x^3 + 2*x^4)*Log[2 + x]^2) ,x]
Output:
(2*E^x - x - (2*Log[x])/((3 + x)*Log[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.
\(\displaystyle \int \frac {\left (-2 x^2+\left (2 x^2+4 x\right ) \log (x)-10 x-12\right ) \log (x+2)+\left (2 x^2+6 x\right ) \log (x)+\left (-x^4-8 x^3-21 x^2+e^x \left (2 x^4+16 x^3+42 x^2+36 x\right )-18 x\right ) \log ^2(x+2)}{\left (2 x^4+16 x^3+42 x^2+36 x\right ) \log ^2(x+2)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-2 x^2+\left (2 x^2+4 x\right ) \log (x)-10 x-12\right ) \log (x+2)+\left (2 x^2+6 x\right ) \log (x)+\left (-x^4-8 x^3-21 x^2+e^x \left (2 x^4+16 x^3+42 x^2+36 x\right )-18 x\right ) \log ^2(x+2)}{x \left (2 x^3+16 x^2+42 x+36\right ) \log ^2(x+2)}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-2 x^2+\left (2 x^2+4 x\right ) \log (x)-10 x-12\right ) \log (x+2)+\left (2 x^2+6 x\right ) \log (x)+\left (-x^4-8 x^3-21 x^2+e^x \left (2 x^4+16 x^3+42 x^2+36 x\right )-18 x\right ) \log ^2(x+2)}{2 x (x+2) \log ^2(x+2)}-\frac {\left (-2 x^2+\left (2 x^2+4 x\right ) \log (x)-10 x-12\right ) \log (x+2)+\left (2 x^2+6 x\right ) \log (x)+\left (-x^4-8 x^3-21 x^2+e^x \left (2 x^4+16 x^3+42 x^2+36 x\right )-18 x\right ) \log ^2(x+2)}{2 x (x+3) \log ^2(x+2)}-\frac {\left (-2 x^2+\left (2 x^2+4 x\right ) \log (x)-10 x-12\right ) \log (x+2)+\left (2 x^2+6 x\right ) \log (x)+\left (-x^4-8 x^3-21 x^2+e^x \left (2 x^4+16 x^3+42 x^2+36 x\right )-18 x\right ) \log ^2(x+2)}{2 x (x+3)^2 \log ^2(x+2)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \text {Subst}\left (\int \frac {\log (x-2)}{x \log ^2(x)}dx,x,x+2\right )-\int \frac {\log (x)}{(x+3) \log ^2(x+2)}dx-\frac {1}{3} \int \frac {1}{x \log (x+2)}dx+\frac {1}{3} \int \frac {1}{(x+3) \log (x+2)}dx+\int \frac {\log (x)}{(x+3)^2 \log (x+2)}dx+\frac {x^3}{6}-e^x x^2+\frac {3 x^2}{2}-3 e^x x+4 x-\frac {1}{6} (x+3)^3+e^x (x+3)^2-e^x (x+2)-2 e^x (x+3)\) |
Input:
Int[((6*x + 2*x^2)*Log[x] + (-12 - 10*x - 2*x^2 + (4*x + 2*x^2)*Log[x])*Lo g[2 + x] + (-18*x - 21*x^2 - 8*x^3 - x^4 + E^x*(36*x + 42*x^2 + 16*x^3 + 2 *x^4))*Log[2 + x]^2)/((36*x + 42*x^2 + 16*x^3 + 2*x^4)*Log[2 + x]^2),x]
Output:
$Aborted
Time = 7.40 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
\[-\frac {x}{2}+{\mathrm e}^{x}-\frac {\ln \left (x \right )}{\ln \left (2+x \right ) \left (3+x \right )}\]
Input:
int((((2*x^4+16*x^3+42*x^2+36*x)*exp(x)-x^4-8*x^3-21*x^2-18*x)*ln(2+x)^2+( (2*x^2+4*x)*ln(x)-2*x^2-10*x-12)*ln(2+x)+(2*x^2+6*x)*ln(x))/(2*x^4+16*x^3+ 42*x^2+36*x)/ln(2+x)^2,x)
Output:
-1/2*x+exp(x)-1/ln(2+x)*ln(x)/(3+x)
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-\frac {{\left (x^{2} - 2 \, {\left (x + 3\right )} e^{x} + 3 \, x\right )} \log \left (x + 2\right ) + 2 \, \log \left (x\right )}{2 \, {\left (x + 3\right )} \log \left (x + 2\right )} \] Input:
integrate((((2*x^4+16*x^3+42*x^2+36*x)*exp(x)-x^4-8*x^3-21*x^2-18*x)*log(2 +x)^2+((2*x^2+4*x)*log(x)-2*x^2-10*x-12)*log(2+x)+(2*x^2+6*x)*log(x))/(2*x ^4+16*x^3+42*x^2+36*x)/log(2+x)^2,x, algorithm="fricas")
Output:
-1/2*((x^2 - 2*(x + 3)*e^x + 3*x)*log(x + 2) + 2*log(x))/((x + 3)*log(x + 2))
Time = 0.17 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=- \frac {x}{2} + e^{x} - \frac {\log {\left (x \right )}}{\left (x + 3\right ) \log {\left (x + 2 \right )}} \] Input:
integrate((((2*x**4+16*x**3+42*x**2+36*x)*exp(x)-x**4-8*x**3-21*x**2-18*x) *ln(2+x)**2+((2*x**2+4*x)*ln(x)-2*x**2-10*x-12)*ln(2+x)+(2*x**2+6*x)*ln(x) )/(2*x**4+16*x**3+42*x**2+36*x)/ln(2+x)**2,x)
Output:
-x/2 + exp(x) - log(x)/((x + 3)*log(x + 2))
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-\frac {{\left (x^{2} - 2 \, {\left (x + 3\right )} e^{x} + 3 \, x\right )} \log \left (x + 2\right ) + 2 \, \log \left (x\right )}{2 \, {\left (x + 3\right )} \log \left (x + 2\right )} \] Input:
integrate((((2*x^4+16*x^3+42*x^2+36*x)*exp(x)-x^4-8*x^3-21*x^2-18*x)*log(2 +x)^2+((2*x^2+4*x)*log(x)-2*x^2-10*x-12)*log(2+x)+(2*x^2+6*x)*log(x))/(2*x ^4+16*x^3+42*x^2+36*x)/log(2+x)^2,x, algorithm="maxima")
Output:
-1/2*((x^2 - 2*(x + 3)*e^x + 3*x)*log(x + 2) + 2*log(x))/((x + 3)*log(x + 2))
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.16 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=-\frac {x^{2} \log \left (x + 2\right ) - 2 \, x e^{x} \log \left (x + 2\right ) + 3 \, x \log \left (x + 2\right ) - 6 \, e^{x} \log \left (x + 2\right ) + 2 \, \log \left (x\right )}{2 \, {\left (x \log \left (x + 2\right ) + 3 \, \log \left (x + 2\right )\right )}} \] Input:
integrate((((2*x^4+16*x^3+42*x^2+36*x)*exp(x)-x^4-8*x^3-21*x^2-18*x)*log(2 +x)^2+((2*x^2+4*x)*log(x)-2*x^2-10*x-12)*log(2+x)+(2*x^2+6*x)*log(x))/(2*x ^4+16*x^3+42*x^2+36*x)/log(2+x)^2,x, algorithm="giac")
Output:
-1/2*(x^2*log(x + 2) - 2*x*e^x*log(x + 2) + 3*x*log(x + 2) - 6*e^x*log(x + 2) + 2*log(x))/(x*log(x + 2) + 3*log(x + 2))
Time = 3.57 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.12 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx={\mathrm {e}}^x-\frac {x}{2}-\frac {x+2}{x^2+3\,x}-\frac {\frac {\ln \left (x\right )}{x+3}-\frac {\ln \left (x+2\right )\,\left (x+2\right )\,\left (x-x\,\ln \left (x\right )+3\right )}{x\,{\left (x+3\right )}^2}}{\ln \left (x+2\right )}+\frac {\ln \left (x\right )\,\left (x+2\right )}{x^2+6\,x+9} \] Input:
int(-(log(x + 2)^2*(18*x - exp(x)*(36*x + 42*x^2 + 16*x^3 + 2*x^4) + 21*x^ 2 + 8*x^3 + x^4) - log(x)*(6*x + 2*x^2) + log(x + 2)*(10*x - log(x)*(4*x + 2*x^2) + 2*x^2 + 12))/(log(x + 2)^2*(36*x + 42*x^2 + 16*x^3 + 2*x^4)),x)
Output:
exp(x) - x/2 - (x + 2)/(3*x + x^2) - (log(x)/(x + 3) - (log(x + 2)*(x + 2) *(x - x*log(x) + 3))/(x*(x + 3)^2))/log(x + 2) + (log(x)*(x + 2))/(6*x + x ^2 + 9)
Time = 0.17 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {\left (6 x+2 x^2\right ) \log (x)+\left (-12-10 x-2 x^2+\left (4 x+2 x^2\right ) \log (x)\right ) \log (2+x)+\left (-18 x-21 x^2-8 x^3-x^4+e^x \left (36 x+42 x^2+16 x^3+2 x^4\right )\right ) \log ^2(2+x)}{\left (36 x+42 x^2+16 x^3+2 x^4\right ) \log ^2(2+x)} \, dx=\frac {2 e^{x} \mathrm {log}\left (x +2\right ) x +6 e^{x} \mathrm {log}\left (x +2\right )-\mathrm {log}\left (x +2\right ) x^{2}-3 \,\mathrm {log}\left (x +2\right ) x -2 \,\mathrm {log}\left (x \right )}{2 \,\mathrm {log}\left (x +2\right ) \left (x +3\right )} \] Input:
int((((2*x^4+16*x^3+42*x^2+36*x)*exp(x)-x^4-8*x^3-21*x^2-18*x)*log(2+x)^2+ ((2*x^2+4*x)*log(x)-2*x^2-10*x-12)*log(2+x)+(2*x^2+6*x)*log(x))/(2*x^4+16* x^3+42*x^2+36*x)/log(2+x)^2,x)
Output:
(2*e**x*log(x + 2)*x + 6*e**x*log(x + 2) - log(x + 2)*x**2 - 3*log(x + 2)* x - 2*log(x))/(2*log(x + 2)*(x + 3))