Integrand size = 101, antiderivative size = 20 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \log ^2\left (3 \left (4+e^x+x\right )^2\right )}{4 x} \] Output:
3/4*ln(3*(4+x+exp(x))^2)^2/x
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \log ^2\left (3 \left (4+e^x+x\right )^2\right )}{4 x} \] Input:
Integrate[((12*x + 12*E^x*x)*Log[48 + 3*E^(2*x) + 24*x + 3*x^2 + E^x*(24 + 6*x)] + (-12 - 3*E^x - 3*x)*Log[48 + 3*E^(2*x) + 24*x + 3*x^2 + E^x*(24 + 6*x)]^2)/(16*x^2 + 4*E^x*x^2 + 4*x^3),x]
Output:
(3*Log[3*(4 + E^x + x)^2]^2)/(4*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.
| \(\displaystyle \int \frac {\left (-3 x-3 e^x-12\right ) \log ^2\left (3 x^2+24 x+3 e^{2 x}+e^x (6 x+24)+48\right )+\left (12 e^x x+12 x\right ) \log \left (3 x^2+24 x+3 e^{2 x}+e^x (6 x+24)+48\right )}{4 x^3+4 e^x x^2+16 x^2} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {3 \left (\frac {4 \left (e^x+1\right ) x}{x+e^x+4}-\log \left (3 \left (x+e^x+4\right )^2\right )\right ) \log \left (3 \left (x+e^x+4\right )^2\right )}{4 x^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3}{4} \int \frac {\left (\frac {4 \left (1+e^x\right ) x}{x+e^x+4}-\log \left (3 \left (x+e^x+4\right )^2\right )\right ) \log \left (3 \left (x+e^x+4\right )^2\right )}{x^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {3}{4} \int \left (\frac {\left (4 x-\log \left (3 \left (x+e^x+4\right )^2\right )\right ) \log \left (3 \left (x+e^x+4\right )^2\right )}{x^2}-\frac {4 (x+3) \log \left (3 \left (x+e^x+4\right )^2\right )}{x \left (x+e^x+4\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3}{4} \left (-\int \frac {\log ^2\left (3 \left (x+e^x+4\right )^2\right )}{x^2}dx+8 \int \int \frac {1}{x+e^x+4}dxdx-24 \int \frac {\int \frac {1}{x+e^x+4}dx}{x+e^x+4}dx-8 \int \frac {x \int \frac {1}{x+e^x+4}dx}{x+e^x+4}dx+24 \int \int \frac {1}{x \left (x+e^x+4\right )}dxdx-72 \int \frac {\int \frac {1}{x \left (x+e^x+4\right )}dx}{x+e^x+4}dx-24 \int \frac {x \int \frac {1}{x \left (x+e^x+4\right )}dx}{x+e^x+4}dx-4 \log \left (3 \left (x+e^x+4\right )^2\right ) \int \frac {1}{x+e^x+4}dx-12 \log \left (3 \left (x+e^x+4\right )^2\right ) \int \frac {1}{x \left (x+e^x+4\right )}dx+4 \int \frac {\log \left (3 \left (x+e^x+4\right )^2\right )}{x}dx\right )\) |
Input:
Int[((12*x + 12*E^x*x)*Log[48 + 3*E^(2*x) + 24*x + 3*x^2 + E^x*(24 + 6*x)] + (-12 - 3*E^x - 3*x)*Log[48 + 3*E^(2*x) + 24*x + 3*x^2 + E^x*(24 + 6*x)] ^2)/(16*x^2 + 4*E^x*x^2 + 4*x^3),x]
Output:
$Aborted
Time = 0.55 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65
| method | result | size |
| parallelrisch | \(\frac {3 \ln \left (3 \,{\mathrm e}^{2 x}+\left (24+6 x \right ) {\mathrm e}^{x}+3 x^{2}+24 x +48\right )^{2}}{4 x}\) | \(33\) |
| risch | \(\frac {3 \ln \left (4+x +{\mathrm e}^{x}\right )^{2}}{x}+\frac {3 \left (-i \pi {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right )}^{2} \operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{2}-i \pi {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{3}+2 \ln \left (3\right )\right ) \ln \left (4+x +{\mathrm e}^{x}\right )}{2 x}+\frac {\frac {3 \ln \left (3\right )^{2}}{4}-\frac {3 \pi ^{2} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{6}}{16}-\frac {3 i \ln \left (3\right ) \pi {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{3}}{4}-\frac {3 \pi ^{2} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right )}^{4} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{2}}{16}+\frac {3 \pi ^{2} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right )}^{3} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{3}}{4}-\frac {9 \pi ^{2} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right )}^{2} {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{4}}{8}+\frac {3 \pi ^{2} \operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{5}}{4}-\frac {3 i \ln \left (3\right ) \pi {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right )}^{2} \operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}{4}+\frac {3 i \ln \left (3\right ) \pi \,\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (i \left (4+x +{\mathrm e}^{x}\right )^{2}\right )}^{2}}{2}}{x}\) | \(319\) |
Input:
int(((-3*exp(x)-3*x-12)*ln(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48)^2+(12 *exp(x)*x+12*x)*ln(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48))/(4*exp(x)*x^ 2+4*x^3+16*x^2),x,method=_RETURNVERBOSE)
Output:
3/4*ln(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48)^2/x
Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \, \log \left (3 \, x^{2} + 6 \, {\left (x + 4\right )} e^{x} + 24 \, x + 3 \, e^{\left (2 \, x\right )} + 48\right )^{2}}{4 \, x} \] Input:
integrate(((-3*exp(x)-3*x-12)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48 )^2+(12*exp(x)*x+12*x)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48))/(4*e xp(x)*x^2+4*x^3+16*x^2),x, algorithm="fricas")
Output:
3/4*log(3*x^2 + 6*(x + 4)*e^x + 24*x + 3*e^(2*x) + 48)^2/x
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \log {\left (3 x^{2} + 24 x + \left (6 x + 24\right ) e^{x} + 3 e^{2 x} + 48 \right )}^{2}}{4 x} \] Input:
integrate(((-3*exp(x)-3*x-12)*ln(3*exp(x)**2+(24+6*x)*exp(x)+3*x**2+24*x+4 8)**2+(12*exp(x)*x+12*x)*ln(3*exp(x)**2+(24+6*x)*exp(x)+3*x**2+24*x+48))/( 4*exp(x)*x**2+4*x**3+16*x**2),x)
Output:
3*log(3*x**2 + 24*x + (6*x + 24)*exp(x) + 3*exp(2*x) + 48)**2/(4*x)
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \, {\left (\log \left (3\right )^{2} + 4 \, \log \left (3\right ) \log \left (x + e^{x} + 4\right ) + 4 \, \log \left (x + e^{x} + 4\right )^{2}\right )}}{4 \, x} \] Input:
integrate(((-3*exp(x)-3*x-12)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48 )^2+(12*exp(x)*x+12*x)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48))/(4*e xp(x)*x^2+4*x^3+16*x^2),x, algorithm="maxima")
Output:
3/4*(log(3)^2 + 4*log(3)*log(x + e^x + 4) + 4*log(x + e^x + 4)^2)/x
Time = 0.42 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.65 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \, \log \left (3 \, x^{2} + 6 \, x e^{x} + 24 \, x + 3 \, e^{\left (2 \, x\right )} + 24 \, e^{x} + 48\right )^{2}}{4 \, x} \] Input:
integrate(((-3*exp(x)-3*x-12)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48 )^2+(12*exp(x)*x+12*x)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48))/(4*e xp(x)*x^2+4*x^3+16*x^2),x, algorithm="giac")
Output:
3/4*log(3*x^2 + 6*x*e^x + 24*x + 3*e^(2*x) + 24*e^x + 48)^2/x
Time = 0.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3\,{\ln \left (24\,x+3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (6\,x+24\right )+3\,x^2+48\right )}^2}{4\,x} \] Input:
int(-(log(24*x + 3*exp(2*x) + exp(x)*(6*x + 24) + 3*x^2 + 48)^2*(3*x + 3*e xp(x) + 12) - log(24*x + 3*exp(2*x) + exp(x)*(6*x + 24) + 3*x^2 + 48)*(12* x + 12*x*exp(x)))/(4*x^2*exp(x) + 16*x^2 + 4*x^3),x)
Output:
(3*log(24*x + 3*exp(2*x) + exp(x)*(6*x + 24) + 3*x^2 + 48)^2)/(4*x)
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.80 \[ \int \frac {\left (12 x+12 e^x x\right ) \log \left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )+\left (-12-3 e^x-3 x\right ) \log ^2\left (48+3 e^{2 x}+24 x+3 x^2+e^x (24+6 x)\right )}{16 x^2+4 e^x x^2+4 x^3} \, dx=\frac {3 \mathrm {log}\left (3 e^{2 x}+6 e^{x} x +24 e^{x}+3 x^{2}+24 x +48\right )^{2}}{4 x} \] Input:
int(((-3*exp(x)-3*x-12)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48)^2+(1 2*exp(x)*x+12*x)*log(3*exp(x)^2+(24+6*x)*exp(x)+3*x^2+24*x+48))/(4*exp(x)* x^2+4*x^3+16*x^2),x)
Output:
(3*log(3*e**(2*x) + 6*e**x*x + 24*e**x + 3*x**2 + 24*x + 48)**2)/(4*x)