Integrand size = 237, antiderivative size = 27 \[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=e^x \log \left (x \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(1+x)\right )\right )\right ) \] Output:
exp(x)*ln(x*ln(-2/3*exp(x)^4*ln(1+x)^4+2/3*x))
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=e^x \log \left (x \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(1+x)\right )\right )\right ) \] Input:
Integrate[(E^x*(-x - x^2) + 4*E^(5*x)*x*Log[1 + x]^3 + E^(5*x)*(4*x + 4*x^ 2)*Log[1 + x]^4 + (E^x*(-x - x^2) + E^(5*x)*(1 + x)*Log[1 + x]^4)*Log[(2*x - 2*E^(4*x)*Log[1 + x]^4)/3] + (E^x*(-x^2 - x^3) + E^(5*x)*(x + x^2)*Log[ 1 + x]^4)*Log[(2*x - 2*E^(4*x)*Log[1 + x]^4)/3]*Log[x*Log[(2*x - 2*E^(4*x) *Log[1 + x]^4)/3]])/((-x^2 - x^3 + E^(4*x)*(x + x^2)*Log[1 + x]^4)*Log[(2* x - 2*E^(4*x)*Log[1 + x]^4)/3]),x]
Output:
E^x*Log[x*Log[(2*(x - E^(4*x)*Log[1 + x]^4))/3]]
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.
| \(\displaystyle \int \frac {e^x \left (-x^2-x\right )+e^{5 x} \left (4 x^2+4 x\right ) \log ^4(x+1)+\left (e^x \left (-x^2-x\right )+e^{5 x} (x+1) \log ^4(x+1)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right )+\left (e^{5 x} \left (x^2+x\right ) \log ^4(x+1)+e^x \left (-x^3-x^2\right )\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right )\right )+4 e^{5 x} x \log ^3(x+1)}{\left (-x^3-x^2+e^{4 x} \left (x^2+x\right ) \log ^4(x+1)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-e^x \left (-x^2-x\right )-e^{5 x} \left (4 x^2+4 x\right ) \log ^4(x+1)-\left (e^x \left (-x^2-x\right )+e^{5 x} (x+1) \log ^4(x+1)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right )-\left (e^{5 x} \left (x^2+x\right ) \log ^4(x+1)+e^x \left (-x^3-x^2\right )\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(x+1)\right )\right )\right )-4 e^{5 x} x \log ^3(x+1)}{x (x+1) \left (x-e^{4 x} \log ^4(x+1)\right ) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^x \left (4 x^2 \log (x+1)+4 x+3 x \log (x+1)-\log (x+1)\right )}{(x+1) \log (x+1) \left (e^{4 x} \log ^4(x+1)-x\right ) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )}+\frac {e^x \left (x^2 \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right ) \log \left (x \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )\right )+4 x^2 \log (x+1)+4 x+x \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )+x \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right ) \log \left (x \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )\right )+\log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )+4 x \log (x+1)\right )}{x (x+1) \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {e^x \left (4 x^2 \log (x+1)+4 x+3 x \log (x+1)-\log (x+1)\right )}{(x+1) \log (x+1) \left (e^{4 x} \log ^4(x+1)-x\right ) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )}+\frac {e^x \left (x^2 \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right ) \log \left (x \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )\right )+4 x^2 \log (x+1)+4 x+x \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )+x \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right ) \log \left (x \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )\right )+\log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )+4 x \log (x+1)\right )}{x (x+1) \log (x+1) \log \left (\frac {2}{3} \left (x-e^{4 x} \log ^4(x+1)\right )\right )}\right )dx\) |
Input:
Int[(E^x*(-x - x^2) + 4*E^(5*x)*x*Log[1 + x]^3 + E^(5*x)*(4*x + 4*x^2)*Log [1 + x]^4 + (E^x*(-x - x^2) + E^(5*x)*(1 + x)*Log[1 + x]^4)*Log[(2*x - 2*E ^(4*x)*Log[1 + x]^4)/3] + (E^x*(-x^2 - x^3) + E^(5*x)*(x + x^2)*Log[1 + x] ^4)*Log[(2*x - 2*E^(4*x)*Log[1 + x]^4)/3]*Log[x*Log[(2*x - 2*E^(4*x)*Log[1 + x]^4)/3]])/((-x^2 - x^3 + E^(4*x)*(x + x^2)*Log[1 + x]^4)*Log[(2*x - 2* E^(4*x)*Log[1 + x]^4)/3]),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 198, normalized size of antiderivative = 7.33
\[{\mathrm e}^{x} \ln \left (\ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right )+{\mathrm e}^{x} \ln \left (x \right )+\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right ) {\operatorname {csgn}\left (i x \ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right )}^{2} {\mathrm e}^{x}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right ) \operatorname {csgn}\left (i x \ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right ) \operatorname {csgn}\left (i x \right ) {\mathrm e}^{x}}{2}-\frac {i \pi {\operatorname {csgn}\left (i x \ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right )}^{3} {\mathrm e}^{x}}{2}+\frac {i \pi {\operatorname {csgn}\left (i x \ln \left (-\frac {2 \,{\mathrm e}^{4 x} \ln \left (1+x \right )^{4}}{3}+\frac {2 x}{3}\right )\right )}^{2} \operatorname {csgn}\left (i x \right ) {\mathrm e}^{x}}{2}\]
Input:
int((((x^2+x)*exp(x)^5*ln(1+x)^4+(-x^3-x^2)*exp(x))*ln(-2/3*exp(x)^4*ln(1+ x)^4+2/3*x)*ln(x*ln(-2/3*exp(x)^4*ln(1+x)^4+2/3*x))+((1+x)*exp(x)^5*ln(1+x )^4+(-x^2-x)*exp(x))*ln(-2/3*exp(x)^4*ln(1+x)^4+2/3*x)+(4*x^2+4*x)*exp(x)^ 5*ln(1+x)^4+4*x*exp(x)^5*ln(1+x)^3+(-x^2-x)*exp(x))/((x^2+x)*exp(x)^4*ln(1 +x)^4-x^3-x^2)/ln(-2/3*exp(x)^4*ln(1+x)^4+2/3*x),x)
Output:
exp(x)*ln(ln(-2/3*exp(4*x)*ln(1+x)^4+2/3*x))+exp(x)*ln(x)+1/2*I*Pi*csgn(I* ln(-2/3*exp(4*x)*ln(1+x)^4+2/3*x))*csgn(I*x*ln(-2/3*exp(4*x)*ln(1+x)^4+2/3 *x))^2*exp(x)-1/2*I*Pi*csgn(I*ln(-2/3*exp(4*x)*ln(1+x)^4+2/3*x))*csgn(I*x* ln(-2/3*exp(4*x)*ln(1+x)^4+2/3*x))*csgn(I*x)*exp(x)-1/2*I*Pi*csgn(I*x*ln(- 2/3*exp(4*x)*ln(1+x)^4+2/3*x))^3*exp(x)+1/2*I*Pi*csgn(I*x*ln(-2/3*exp(4*x) *ln(1+x)^4+2/3*x))^2*csgn(I*x)*exp(x)
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=e^{x} \log \left (x \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )\right ) \] Input:
integrate((((x^2+x)*exp(x)^5*log(1+x)^4+(-x^3-x^2)*exp(x))*log(-2/3*exp(x) ^4*log(1+x)^4+2/3*x)*log(x*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x))+((1+x)*exp (x)^5*log(1+x)^4+(-x^2-x)*exp(x))*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x)+(4*x ^2+4*x)*exp(x)^5*log(1+x)^4+4*x*exp(x)^5*log(1+x)^3+(-x^2-x)*exp(x))/((x^2 +x)*exp(x)^4*log(1+x)^4-x^3-x^2)/log(-2/3*exp(x)^4*log(1+x)^4+2/3*x),x, al gorithm="fricas")
Output:
e^x*log(x*log(-2/3*e^(4*x)*log(x + 1)^4 + 2/3*x))
Timed out. \[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=\text {Timed out} \] Input:
integrate((((x**2+x)*exp(x)**5*ln(1+x)**4+(-x**3-x**2)*exp(x))*ln(-2/3*exp (x)**4*ln(1+x)**4+2/3*x)*ln(x*ln(-2/3*exp(x)**4*ln(1+x)**4+2/3*x))+((1+x)* exp(x)**5*ln(1+x)**4+(-x**2-x)*exp(x))*ln(-2/3*exp(x)**4*ln(1+x)**4+2/3*x) +(4*x**2+4*x)*exp(x)**5*ln(1+x)**4+4*x*exp(x)**5*ln(1+x)**3+(-x**2-x)*exp( x))/((x**2+x)*exp(x)**4*ln(1+x)**4-x**3-x**2)/ln(-2/3*exp(x)**4*ln(1+x)**4 +2/3*x),x)
Output:
Timed out
\[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=\int { \frac {4 \, {\left (x^{2} + x\right )} e^{\left (5 \, x\right )} \log \left (x + 1\right )^{4} + 4 \, x e^{\left (5 \, x\right )} \log \left (x + 1\right )^{3} + {\left ({\left (x^{2} + x\right )} e^{\left (5 \, x\right )} \log \left (x + 1\right )^{4} - {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right ) \log \left (x \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )\right ) - {\left (x^{2} + x\right )} e^{x} + {\left ({\left (x + 1\right )} e^{\left (5 \, x\right )} \log \left (x + 1\right )^{4} - {\left (x^{2} + x\right )} e^{x}\right )} \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )}{{\left ({\left (x^{2} + x\right )} e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} - x^{3} - x^{2}\right )} \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )} \,d x } \] Input:
integrate((((x^2+x)*exp(x)^5*log(1+x)^4+(-x^3-x^2)*exp(x))*log(-2/3*exp(x) ^4*log(1+x)^4+2/3*x)*log(x*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x))+((1+x)*exp (x)^5*log(1+x)^4+(-x^2-x)*exp(x))*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x)+(4*x ^2+4*x)*exp(x)^5*log(1+x)^4+4*x*exp(x)^5*log(1+x)^3+(-x^2-x)*exp(x))/((x^2 +x)*exp(x)^4*log(1+x)^4-x^3-x^2)/log(-2/3*exp(x)^4*log(1+x)^4+2/3*x),x, al gorithm="maxima")
Output:
e^x*log(x) + e^x*log(-log(3) + log(2) + log(-e^(4*x)*log(x + 1)^4 + x)) + integrate((4*(x + 1)*e^(5*x)*log(x + 1)^4 + 4*e^(5*x)*log(x + 1)^3 - (x + 1)*e^x)/((I*pi + (I*pi - log(3) + log(2))*x - log(3) + log(2))*e^(4*x)*log (x + 1)^4 + (-I*pi + log(3) - log(2))*x^2 + (-I*pi + log(3) - log(2))*x + ((x + 1)*e^(4*x)*log(x + 1)^4 - x^2 - x)*log(e^(4*x)*log(x + 1)^4 - x)), x ) - integrate(-(4*(x + 1)*e^(5*x)*log(x + 1)^4 + 4*e^(5*x)*log(x + 1)^3 - (x + 1)*e^x)/((x*(log(3) - log(2)) + log(3) - log(2))*e^(4*x)*log(x + 1)^4 - x^2*(log(3) - log(2)) - x*(log(3) - log(2)) - ((x + 1)*e^(4*x)*log(x + 1)^4 - x^2 - x)*log(-e^(4*x)*log(x + 1)^4 + x)), x)
\[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=\int { \frac {4 \, {\left (x^{2} + x\right )} e^{\left (5 \, x\right )} \log \left (x + 1\right )^{4} + 4 \, x e^{\left (5 \, x\right )} \log \left (x + 1\right )^{3} + {\left ({\left (x^{2} + x\right )} e^{\left (5 \, x\right )} \log \left (x + 1\right )^{4} - {\left (x^{3} + x^{2}\right )} e^{x}\right )} \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right ) \log \left (x \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )\right ) - {\left (x^{2} + x\right )} e^{x} + {\left ({\left (x + 1\right )} e^{\left (5 \, x\right )} \log \left (x + 1\right )^{4} - {\left (x^{2} + x\right )} e^{x}\right )} \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )}{{\left ({\left (x^{2} + x\right )} e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} - x^{3} - x^{2}\right )} \log \left (-\frac {2}{3} \, e^{\left (4 \, x\right )} \log \left (x + 1\right )^{4} + \frac {2}{3} \, x\right )} \,d x } \] Input:
integrate((((x^2+x)*exp(x)^5*log(1+x)^4+(-x^3-x^2)*exp(x))*log(-2/3*exp(x) ^4*log(1+x)^4+2/3*x)*log(x*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x))+((1+x)*exp (x)^5*log(1+x)^4+(-x^2-x)*exp(x))*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x)+(4*x ^2+4*x)*exp(x)^5*log(1+x)^4+4*x*exp(x)^5*log(1+x)^3+(-x^2-x)*exp(x))/((x^2 +x)*exp(x)^4*log(1+x)^4-x^3-x^2)/log(-2/3*exp(x)^4*log(1+x)^4+2/3*x),x, al gorithm="giac")
Output:
integrate((4*(x^2 + x)*e^(5*x)*log(x + 1)^4 + 4*x*e^(5*x)*log(x + 1)^3 + ( (x^2 + x)*e^(5*x)*log(x + 1)^4 - (x^3 + x^2)*e^x)*log(-2/3*e^(4*x)*log(x + 1)^4 + 2/3*x)*log(x*log(-2/3*e^(4*x)*log(x + 1)^4 + 2/3*x)) - (x^2 + x)*e ^x + ((x + 1)*e^(5*x)*log(x + 1)^4 - (x^2 + x)*e^x)*log(-2/3*e^(4*x)*log(x + 1)^4 + 2/3*x))/(((x^2 + x)*e^(4*x)*log(x + 1)^4 - x^3 - x^2)*log(-2/3*e ^(4*x)*log(x + 1)^4 + 2/3*x)), x)
Time = 3.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=\ln \left (x\,\ln \left (\frac {2\,x}{3}-\frac {2\,{\ln \left (x+1\right )}^4\,{\mathrm {e}}^{4\,x}}{3}\right )\right )\,{\mathrm {e}}^x \] Input:
int((log((2*x)/3 - (2*log(x + 1)^4*exp(4*x))/3)*(exp(x)*(x + x^2) - log(x + 1)^4*exp(5*x)*(x + 1)) + exp(x)*(x + x^2) + log(x*log((2*x)/3 - (2*log(x + 1)^4*exp(4*x))/3))*log((2*x)/3 - (2*log(x + 1)^4*exp(4*x))/3)*(exp(x)*( x^2 + x^3) - log(x + 1)^4*exp(5*x)*(x + x^2)) - log(x + 1)^4*exp(5*x)*(4*x + 4*x^2) - 4*x*log(x + 1)^3*exp(5*x))/(log((2*x)/3 - (2*log(x + 1)^4*exp( 4*x))/3)*(x^2 + x^3 - log(x + 1)^4*exp(4*x)*(x + x^2))),x)
Output:
log(x*log((2*x)/3 - (2*log(x + 1)^4*exp(4*x))/3))*exp(x)
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-x-x^2\right )+4 e^{5 x} x \log ^3(1+x)+e^{5 x} \left (4 x+4 x^2\right ) \log ^4(1+x)+\left (e^x \left (-x-x^2\right )+e^{5 x} (1+x) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )+\left (e^x \left (-x^2-x^3\right )+e^{5 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right ) \log \left (x \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )\right )}{\left (-x^2-x^3+e^{4 x} \left (x+x^2\right ) \log ^4(1+x)\right ) \log \left (\frac {1}{3} \left (2 x-2 e^{4 x} \log ^4(1+x)\right )\right )} \, dx=e^{x} \mathrm {log}\left (\mathrm {log}\left (-\frac {2 e^{4 x} \mathrm {log}\left (x +1\right )^{4}}{3}+\frac {2 x}{3}\right ) x \right ) \] Input:
int((((x^2+x)*exp(x)^5*log(1+x)^4+(-x^3-x^2)*exp(x))*log(-2/3*exp(x)^4*log (1+x)^4+2/3*x)*log(x*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x))+((1+x)*exp(x)^5* log(1+x)^4+(-x^2-x)*exp(x))*log(-2/3*exp(x)^4*log(1+x)^4+2/3*x)+(4*x^2+4*x )*exp(x)^5*log(1+x)^4+4*x*exp(x)^5*log(1+x)^3+(-x^2-x)*exp(x))/((x^2+x)*ex p(x)^4*log(1+x)^4-x^3-x^2)/log(-2/3*exp(x)^4*log(1+x)^4+2/3*x),x)
Output:
e**x*log(log(( - 2*e**(4*x)*log(x + 1)**4 + 2*x)/3)*x)