Integrand size = 149, antiderivative size = 27 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=x \left (2+x^2+\frac {5 (5+x)}{\log \left (4+e^x-x+\log ^2(5)\right )}\right ) \] Output:
(2+5*(5+x)/ln(exp(x)+ln(5)^2-x+4)+x^2)*x
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=2 x+x^3+\frac {5 x (5+x)}{\log \left (4+e^x-x+\log ^2(5)\right )} \] Input:
Integrate[(25*x + 5*x^2 + E^x*(-25*x - 5*x^2) + (100 + 15*x - 10*x^2 + E^x *(25 + 10*x) + (25 + 10*x)*Log[5]^2)*Log[4 + E^x - x + Log[5]^2] + (8 - 2* x + 12*x^2 - 3*x^3 + E^x*(2 + 3*x^2) + (2 + 3*x^2)*Log[5]^2)*Log[4 + E^x - x + Log[5]^2]^2)/((4 + E^x - x + Log[5]^2)*Log[4 + E^x - x + Log[5]^2]^2) ,x]
Output:
2*x + x^3 + (5*x*(5 + x))/Log[4 + E^x - x + Log[5]^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.
| \(\displaystyle \int \frac {5 x^2+e^x \left (-5 x^2-25 x\right )+\left (-10 x^2+15 x+e^x (10 x+25)+(10 x+25) \log ^2(5)+100\right ) \log \left (-x+e^x+4+\log ^2(5)\right )+\left (-3 x^3+12 x^2+e^x \left (3 x^2+2\right )+\left (3 x^2+2\right ) \log ^2(5)-2 x+8\right ) \log ^2\left (-x+e^x+4+\log ^2(5)\right )+25 x}{\left (-x+e^x+4+\log ^2(5)\right ) \log ^2\left (-x+e^x+4+\log ^2(5)\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {5 x^2+e^x \left (-5 x^2-25 x\right )+\left (-10 x^2+15 x+e^x (10 x+25)+(10 x+25) \log ^2(5)+100\right ) \log \left (-x+e^x+4+\log ^2(5)\right )+\left (-3 x^3+12 x^2+e^x \left (3 x^2+2\right )+\left (3 x^2+2\right ) \log ^2(5)-2 x+8\right ) \log ^2\left (-x+e^x+4+\log ^2(5)\right )+25 x}{\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {-5 x^2+3 x^2 \log ^2\left (-x+e^x+4+\log ^2(5)\right )-25 x+10 x \log \left (-x+e^x+4+\log ^2(5)\right )+2 \log ^2\left (-x+e^x+4+\log ^2(5)\right )+25 \log \left (-x+e^x+4+\log ^2(5)\right )}{\log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {5 x (x+5) \left (-x+5+\log ^2(5)\right )}{\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 5 \int \frac {x^3}{\left (x-e^x-4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx-5 \int \frac {x^2}{\log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx+5 \log ^2(5) \int \frac {x^2}{\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx-25 \int \frac {x}{\log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx+25 \left (5+\log ^2(5)\right ) \int \frac {x}{\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx+25 \int \frac {1}{\log \left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx+10 \int \frac {x}{\log \left (-x+e^x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}dx+x^3+2 x\) |
Input:
Int[(25*x + 5*x^2 + E^x*(-25*x - 5*x^2) + (100 + 15*x - 10*x^2 + E^x*(25 + 10*x) + (25 + 10*x)*Log[5]^2)*Log[4 + E^x - x + Log[5]^2] + (8 - 2*x + 12 *x^2 - 3*x^3 + E^x*(2 + 3*x^2) + (2 + 3*x^2)*Log[5]^2)*Log[4 + E^x - x + L og[5]^2]^2)/((4 + E^x - x + Log[5]^2)*Log[4 + E^x - x + Log[5]^2]^2),x]
Output:
$Aborted
Time = 0.53 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
| method | result | size |
| risch | \(x^{3}+2 x +\frac {5 \left (5+x \right ) x}{\ln \left ({\mathrm e}^{x}+\ln \left (5\right )^{2}-x +4\right )}\) | \(28\) |
| parallelrisch | \(\frac {\ln \left ({\mathrm e}^{x}+\ln \left (5\right )^{2}-x +4\right ) x^{3}+5 x^{2}+2 \ln \left ({\mathrm e}^{x}+\ln \left (5\right )^{2}-x +4\right ) x +25 x}{\ln \left ({\mathrm e}^{x}+\ln \left (5\right )^{2}-x +4\right )}\) | \(56\) |
Input:
int((((3*x^2+2)*exp(x)+(3*x^2+2)*ln(5)^2-3*x^3+12*x^2-2*x+8)*ln(exp(x)+ln( 5)^2-x+4)^2+((10*x+25)*exp(x)+(10*x+25)*ln(5)^2-10*x^2+15*x+100)*ln(exp(x) +ln(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+ln(5)^2-x+4)/ln(exp (x)+ln(5)^2-x+4)^2,x,method=_RETURNVERBOSE)
Output:
x^3+2*x+5*(5+x)*x/ln(exp(x)+ln(5)^2-x+4)
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=\frac {5 \, x^{2} + {\left (x^{3} + 2 \, x\right )} \log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right ) + 25 \, x}{\log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right )} \] Input:
integrate((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(ex p(x)+log(5)^2-x+4)^2+((10*x+25)*exp(x)+(10*x+25)*log(5)^2-10*x^2+15*x+100) *log(exp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5)^ 2-x+4)/log(exp(x)+log(5)^2-x+4)^2,x, algorithm="fricas")
Output:
(5*x^2 + (x^3 + 2*x)*log(log(5)^2 - x + e^x + 4) + 25*x)/log(log(5)^2 - x + e^x + 4)
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=x^{3} + 2 x + \frac {5 x^{2} + 25 x}{\log {\left (- x + e^{x} + \log {\left (5 \right )}^{2} + 4 \right )}} \] Input:
integrate((((3*x**2+2)*exp(x)+(3*x**2+2)*ln(5)**2-3*x**3+12*x**2-2*x+8)*ln (exp(x)+ln(5)**2-x+4)**2+((10*x+25)*exp(x)+(10*x+25)*ln(5)**2-10*x**2+15*x +100)*ln(exp(x)+ln(5)**2-x+4)+(-5*x**2-25*x)*exp(x)+5*x**2+25*x)/(exp(x)+l n(5)**2-x+4)/ln(exp(x)+ln(5)**2-x+4)**2,x)
Output:
x**3 + 2*x + (5*x**2 + 25*x)/log(-x + exp(x) + log(5)**2 + 4)
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=\frac {5 \, x^{2} + {\left (x^{3} + 2 \, x\right )} \log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right ) + 25 \, x}{\log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right )} \] Input:
integrate((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(ex p(x)+log(5)^2-x+4)^2+((10*x+25)*exp(x)+(10*x+25)*log(5)^2-10*x^2+15*x+100) *log(exp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5)^ 2-x+4)/log(exp(x)+log(5)^2-x+4)^2,x, algorithm="maxima")
Output:
(5*x^2 + (x^3 + 2*x)*log(log(5)^2 - x + e^x + 4) + 25*x)/log(log(5)^2 - x + e^x + 4)
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (26) = 52\).
Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.04 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=\frac {x^{3} \log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right ) + 5 \, x^{2} + 2 \, x \log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right ) + 25 \, x}{\log \left (\log \left (5\right )^{2} - x + e^{x} + 4\right )} \] Input:
integrate((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(ex p(x)+log(5)^2-x+4)^2+((10*x+25)*exp(x)+(10*x+25)*log(5)^2-10*x^2+15*x+100) *log(exp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5)^ 2-x+4)/log(exp(x)+log(5)^2-x+4)^2,x, algorithm="giac")
Output:
(x^3*log(log(5)^2 - x + e^x + 4) + 5*x^2 + 2*x*log(log(5)^2 - x + e^x + 4) + 25*x)/log(log(5)^2 - x + e^x + 4)
Time = 1.60 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.52 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=12\,x+\frac {25\,x+10\,x\,{\ln \left (5\right )}^2+25\,{\ln \left (5\right )}^2-10\,x^2+125}{{\mathrm {e}}^x-1}+\frac {5\,x\,\left (x+5\right )-\frac {5\,\ln \left ({\mathrm {e}}^x-x+{\ln \left (5\right )}^2+4\right )\,\left (2\,x+5\right )\,\left ({\mathrm {e}}^x-x+{\ln \left (5\right )}^2+4\right )}{{\mathrm {e}}^x-1}}{\ln \left ({\mathrm {e}}^x-x+{\ln \left (5\right )}^2+4\right )}+x^3 \] Input:
int((25*x + log(exp(x) - x + log(5)^2 + 4)^2*(log(5)^2*(3*x^2 + 2) - 2*x + exp(x)*(3*x^2 + 2) + 12*x^2 - 3*x^3 + 8) - exp(x)*(25*x + 5*x^2) + 5*x^2 + log(exp(x) - x + log(5)^2 + 4)*(15*x + log(5)^2*(10*x + 25) + exp(x)*(10 *x + 25) - 10*x^2 + 100))/(log(exp(x) - x + log(5)^2 + 4)^2*(exp(x) - x + log(5)^2 + 4)),x)
Output:
12*x + (25*x + 10*x*log(5)^2 + 25*log(5)^2 - 10*x^2 + 125)/(exp(x) - 1) + (5*x*(x + 5) - (5*log(exp(x) - x + log(5)^2 + 4)*(2*x + 5)*(exp(x) - x + l og(5)^2 + 4))/(exp(x) - 1))/log(exp(x) - x + log(5)^2 + 4) + x^3
Time = 0.22 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \[ \int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx=\frac {x \left (\mathrm {log}\left (e^{x}+\mathrm {log}\left (5\right )^{2}-x +4\right ) x^{2}+2 \,\mathrm {log}\left (e^{x}+\mathrm {log}\left (5\right )^{2}-x +4\right )+5 x +25\right )}{\mathrm {log}\left (e^{x}+\mathrm {log}\left (5\right )^{2}-x +4\right )} \] Input:
int((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(exp(x)+l og(5)^2-x+4)^2+((10*x+25)*exp(x)+(10*x+25)*log(5)^2-10*x^2+15*x+100)*log(e xp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5)^2-x+4) /log(exp(x)+log(5)^2-x+4)^2,x)
Output:
(x*(log(e**x + log(5)**2 - x + 4)*x**2 + 2*log(e**x + log(5)**2 - x + 4) + 5*x + 25))/log(e**x + log(5)**2 - x + 4)