Integrand size = 85, antiderivative size = 23 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=e^{-1+e^{x^2}} \left (4-2 x+\log \left (\left (5+e^x\right )^2\right )\right ) \] Output:
exp(exp(x^2)-1)*(ln((exp(x)+5)^2)-2*x+4)
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=e^{-1+e^{x^2}} \left (4-2 x+\log \left (\left (5+e^x\right )^2\right )\right ) \] Input:
Integrate[(E^(-1 + E^x^2)*(-10 + E^x^2*(40*x - 20*x^2 + E^x*(8*x - 4*x^2)) ) + E^(-1 + E^x^2 + x^2)*(10*x + 2*E^x*x)*Log[25 + 10*E^x + E^(2*x)])/(5 + E^x),x]
Output:
E^(-1 + E^x^2)*(4 - 2*x + Log[(5 + E^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.
\(\displaystyle \int \frac {e^{e^{x^2}-1} \left (e^{x^2} \left (-20 x^2+e^x \left (8 x-4 x^2\right )+40 x\right )-10\right )+e^{x^2+e^{x^2}-1} \left (2 e^x x+10 x\right ) \log \left (10 e^x+e^{2 x}+25\right )}{e^x+5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {10 e^{e^{x^2}-1}}{e^x+5}-2 e^{x^2+e^{x^2}-1} x \left (2 x-\log \left (\left (e^x+5\right )^2\right )-4\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 \int \frac {e^{-1+e^{x^2}}}{5+e^x}dx-2 \int \frac {e^{x+e^{x^2}-1}}{5+e^x}dx-4 \int e^{x^2+e^{x^2}-1} x^2dx+4 e^{e^{x^2}-1}+e^{e^{x^2}-1} \log \left (\left (e^x+5\right )^2\right )\) |
Input:
Int[(E^(-1 + E^x^2)*(-10 + E^x^2*(40*x - 20*x^2 + E^x*(8*x - 4*x^2))) + E^ (-1 + E^x^2 + x^2)*(10*x + 2*E^x*x)*Log[25 + 10*E^x + E^(2*x)])/(5 + E^x), x]
Output:
$Aborted
Time = 0.58 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(-2 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1} x +\ln \left ({\mathrm e}^{2 x}+10 \,{\mathrm e}^{x}+25\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}-1}+4 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1}\) | \(40\) |
risch | \(2 \,{\mathrm e}^{{\mathrm e}^{x^{2}}-1} \ln \left ({\mathrm e}^{x}+5\right )-\frac {\left (i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )-2 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )}^{2}+i \pi {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+5\right )^{2}\right )}^{3}+4 x -8\right ) {\mathrm e}^{{\mathrm e}^{x^{2}}-1}}{2}\) | \(94\) |
Input:
int(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*ln(exp(x)^2+10*exp(x)+25)+ (((-4*x^2+8*x)*exp(x)-20*x^2+40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(exp(x)+5 ),x,method=_RETURNVERBOSE)
Output:
-2*exp(exp(x^2)-1)*x+ln(exp(x)^2+10*exp(x)+25)*exp(exp(x^2)-1)+4*exp(exp(x ^2)-1)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=-{\left (2 \, {\left (x - 2\right )} e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} - e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} \log \left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )\right )} e^{\left (-x^{2}\right )} \] Input:
integrate(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp( x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(e xp(x)+5),x, algorithm="fricas")
Output:
-(2*(x - 2)*e^(x^2 + e^(x^2) - 1) - e^(x^2 + e^(x^2) - 1)*log(e^(2*x) + 10 *e^x + 25))*e^(-x^2)
Timed out. \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=\text {Timed out} \] Input:
integrate(((2*exp(x)*x+10*x)*exp(x**2)*exp(exp(x**2)-1)*ln(exp(x)**2+10*ex p(x)+25)+(((-4*x**2+8*x)*exp(x)-20*x**2+40*x)*exp(x**2)-10)*exp(exp(x**2)- 1))/(exp(x)+5),x)
Output:
Timed out
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=-2 \, {\left (x - \log \left (e^{x} + 5\right ) - 2\right )} e^{\left (e^{\left (x^{2}\right )} - 1\right )} \] Input:
integrate(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp( x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(e xp(x)+5),x, algorithm="maxima")
Output:
-2*(x - log(e^x + 5) - 2)*e^(e^(x^2) - 1)
\[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=\int { \frac {2 \, {\left ({\left (x e^{x} + 5 \, x\right )} e^{\left (x^{2} + e^{\left (x^{2}\right )} - 1\right )} \log \left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right ) - {\left (2 \, {\left (5 \, x^{2} + {\left (x^{2} - 2 \, x\right )} e^{x} - 10 \, x\right )} e^{\left (x^{2}\right )} + 5\right )} e^{\left (e^{\left (x^{2}\right )} - 1\right )}\right )}}{e^{x} + 5} \,d x } \] Input:
integrate(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp( x)+25)+(((-4*x^2+8*x)*exp(x)-20*x^2+40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(e xp(x)+5),x, algorithm="giac")
Output:
integrate(2*((x*e^x + 5*x)*e^(x^2 + e^(x^2) - 1)*log(e^(2*x) + 10*e^x + 25 ) - (2*(5*x^2 + (x^2 - 2*x)*e^x - 10*x)*e^(x^2) + 5)*e^(e^(x^2) - 1))/(e^x + 5), x)
Time = 0.60 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx={\mathrm {e}}^{{\mathrm {e}}^{x^2}-1}\,\left (\ln \left ({\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^x+25\right )-2\,x+4\right ) \] Input:
int((exp(exp(x^2) - 1)*(exp(x^2)*(40*x + exp(x)*(8*x - 4*x^2) - 20*x^2) - 10) + log(exp(2*x) + 10*exp(x) + 25)*exp(exp(x^2) - 1)*exp(x^2)*(10*x + 2* x*exp(x)))/(exp(x) + 5),x)
Output:
exp(exp(x^2) - 1)*(log(exp(2*x) + 10*exp(x) + 25) - 2*x + 4)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-1+e^{x^2}} \left (-10+e^{x^2} \left (40 x-20 x^2+e^x \left (8 x-4 x^2\right )\right )\right )+e^{-1+e^{x^2}+x^2} \left (10 x+2 e^x x\right ) \log \left (25+10 e^x+e^{2 x}\right )}{5+e^x} \, dx=\frac {e^{e^{x^{2}}} \left (\mathrm {log}\left (e^{2 x}+10 e^{x}+25\right )-2 x +4\right )}{e} \] Input:
int(((2*exp(x)*x+10*x)*exp(x^2)*exp(exp(x^2)-1)*log(exp(x)^2+10*exp(x)+25) +(((-4*x^2+8*x)*exp(x)-20*x^2+40*x)*exp(x^2)-10)*exp(exp(x^2)-1))/(exp(x)+ 5),x)
Output:
(e**(e**(x**2))*(log(e**(2*x) + 10*e**x + 25) - 2*x + 4))/e