Integrand size = 215, antiderivative size = 24 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^2}+\frac {5}{-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \]
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^2}+\frac {5}{-5+\log \left (e^{e^{3+x}}+\frac {1}{x}\right )} \]
Integrate[(5 + 50*E^x^2*x^2 + E^E^(3 + x)*(-5*E^(3 + x)*x^2 + 50*E^x^2*x^3 ) + (-20*E^x^2*x^2 - 20*E^(E^(3 + x) + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x ] + (2*E^x^2*x^2 + 2*E^(E^(3 + x) + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x]^2 )/(25*x + 25*E^E^(3 + x)*x^2 + (-10*x - 10*E^E^(3 + x)*x^2)*Log[(1 + E^E^( 3 + x)*x)/x] + (x + E^E^(3 + x)*x^2)*Log[(1 + E^E^(3 + x)*x)/x]^2),x]
Time = 2.60 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {50 e^{x^2} x^2+e^{e^{x+3}} \left (50 e^{x^2} x^3-5 e^{x+3} x^2\right )+\left (2 e^{x^2} x^2+2 e^{x^2+e^{x+3}} x^3\right ) \log ^2\left (\frac {e^{e^{x+3}} x+1}{x}\right )+\left (-20 e^{x^2} x^2-20 e^{x^2+e^{x+3}} x^3\right ) \log \left (\frac {e^{e^{x+3}} x+1}{x}\right )+5}{25 e^{e^{x+3}} x^2+\left (e^{e^{x+3}} x^2+x\right ) \log ^2\left (\frac {e^{e^{x+3}} x+1}{x}\right )+\left (-10 e^{e^{x+3}} x^2-10 x\right ) \log \left (\frac {e^{e^{x+3}} x+1}{x}\right )+25 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {50 e^{x^2} x^2+e^{e^{x+3}} \left (50 e^{x^2} x^3-5 e^{x+3} x^2\right )+\left (2 e^{x^2} x^2+2 e^{x^2+e^{x+3}} x^3\right ) \log ^2\left (\frac {e^{e^{x+3}} x+1}{x}\right )+\left (-20 e^{x^2} x^2-20 e^{x^2+e^{x+3}} x^3\right ) \log \left (\frac {e^{e^{x+3}} x+1}{x}\right )+5}{x \left (e^{e^{x+3}} x+1\right ) \left (5-\log \left (e^{e^{x+3}}+\frac {1}{x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 e^{x^2} x-\frac {5 \left (e^{x+e^{x+3}+3} x^2-1\right )}{x \left (e^{e^{x+3}} x+1\right ) \left (\log \left (e^{e^{x+3}}+\frac {1}{x}\right )-5\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^{x^2}-\frac {5}{5-\log \left (e^{e^{x+3}}+\frac {1}{x}\right )}\) |
Int[(5 + 50*E^x^2*x^2 + E^E^(3 + x)*(-5*E^(3 + x)*x^2 + 50*E^x^2*x^3) + (- 20*E^x^2*x^2 - 20*E^(E^(3 + x) + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x] + (2 *E^x^2*x^2 + 2*E^(E^(3 + x) + x^2)*x^3)*Log[(1 + E^E^(3 + x)*x)/x]^2)/(25* x + 25*E^E^(3 + x)*x^2 + (-10*x - 10*E^E^(3 + x)*x^2)*Log[(1 + E^E^(3 + x) *x)/x] + (x + E^E^(3 + x)*x^2)*Log[(1 + E^E^(3 + x)*x)/x]^2),x]
3.7.98.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(63\) vs. \(2(21)=42\).
Time = 127.14 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67
method | result | size |
parallelrisch | \(-\frac {-10 \ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1}{x}\right ) {\mathrm e}^{x^{2}}+50 \,{\mathrm e}^{x^{2}}-10 \ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1}{x}\right )}{10 \left (\ln \left (\frac {x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1}{x}\right )-5\right )}\) | \(64\) |
risch | \({\mathrm e}^{x^{2}}-\frac {10 i}{\pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )\right ) {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )\right ) \operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-\pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )}^{3}+\pi {\operatorname {csgn}\left (\frac {i \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i \ln \left (x \right )-2 i \ln \left (x \,{\mathrm e}^{{\mathrm e}^{3+x}}+1\right )+10 i}\) | \(152\) |
int(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*ln((x*exp(exp(3+x))+1)/ x)^2+(-20*x^3*exp(x^2)*exp(exp(3+x))-20*x^2*exp(x^2))*ln((x*exp(exp(3+x))+ 1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(x^2)+5)/(( x^2*exp(exp(3+x))+x)*ln((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp(3+x))-10 *x)*ln((x*exp(exp(3+x))+1)/x)+25*x^2*exp(exp(3+x))+25*x),x,method=_RETURNV ERBOSE)
-1/10*(-10*ln((x*exp(exp(3+x))+1)/x)*exp(x^2)+50*exp(x^2)-10*ln((x*exp(exp (3+x))+1)/x))/(ln((x*exp(exp(3+x))+1)/x)-5)
Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (21) = 42\).
Time = 0.42 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=\frac {e^{\left (x^{2}\right )} \log \left (\frac {{\left (x e^{\left (x^{2} + e^{\left (x + 3\right )}\right )} + e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{x}\right ) - 5 \, e^{\left (x^{2}\right )} + 5}{\log \left (\frac {{\left (x e^{\left (x^{2} + e^{\left (x + 3\right )}\right )} + e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{x}\right ) - 5} \]
integrate(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*log((x*exp(exp(3+ x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(exp(3+x))-20*x^2*exp(x^2))*log((x*exp(ex p(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(x^ 2)+5)/((x^2*exp(exp(3+x))+x)*log((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp (3+x))-10*x)*log((x*exp(exp(3+x))+1)/x)+25*x^2*exp(exp(3+x))+25*x),x, algo rithm=\
(e^(x^2)*log((x*e^(x^2 + e^(x + 3)) + e^(x^2))*e^(-x^2)/x) - 5*e^(x^2) + 5 )/(log((x*e^(x^2 + e^(x + 3)) + e^(x^2))*e^(-x^2)/x) - 5)
Time = 0.54 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=e^{x^{2}} + \frac {5}{\log {\left (\frac {x e^{e^{x + 3}} + 1}{x} \right )} - 5} \]
integrate(((2*x**3*exp(x**2)*exp(exp(3+x))+2*x**2*exp(x**2))*ln((x*exp(exp (3+x))+1)/x)**2+(-20*x**3*exp(x**2)*exp(exp(3+x))-20*x**2*exp(x**2))*ln((x *exp(exp(3+x))+1)/x)+(50*x**3*exp(x**2)-5*x**2*exp(3+x))*exp(exp(3+x))+50* x**2*exp(x**2)+5)/((x**2*exp(exp(3+x))+x)*ln((x*exp(exp(3+x))+1)/x)**2+(-1 0*x**2*exp(exp(3+x))-10*x)*ln((x*exp(exp(3+x))+1)/x)+25*x**2*exp(exp(3+x)) +25*x),x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=-\frac {{\left (\log \left (x\right ) + 5\right )} e^{\left (x^{2}\right )} - e^{\left (x^{2}\right )} \log \left (x e^{\left (e^{\left (x + 3\right )}\right )} + 1\right ) - 5}{\log \left (x e^{\left (e^{\left (x + 3\right )}\right )} + 1\right ) - \log \left (x\right ) - 5} \]
integrate(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*log((x*exp(exp(3+ x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(exp(3+x))-20*x^2*exp(x^2))*log((x*exp(ex p(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(x^ 2)+5)/((x^2*exp(exp(3+x))+x)*log((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp (3+x))-10*x)*log((x*exp(exp(3+x))+1)/x)+25*x^2*exp(exp(3+x))+25*x),x, algo rithm=\
-((log(x) + 5)*e^(x^2) - e^(x^2)*log(x*e^(e^(x + 3)) + 1) - 5)/(log(x*e^(e ^(x + 3)) + 1) - log(x) - 5)
Leaf count of result is larger than twice the leaf count of optimal. 932 vs. \(2 (21) = 42\).
Time = 0.53 (sec) , antiderivative size = 932, normalized size of antiderivative = 38.83 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx=\text {Too large to display} \]
integrate(((2*x^3*exp(x^2)*exp(exp(3+x))+2*x^2*exp(x^2))*log((x*exp(exp(3+ x))+1)/x)^2+(-20*x^3*exp(x^2)*exp(exp(3+x))-20*x^2*exp(x^2))*log((x*exp(ex p(3+x))+1)/x)+(50*x^3*exp(x^2)-5*x^2*exp(3+x))*exp(exp(3+x))+50*x^2*exp(x^ 2)+5)/((x^2*exp(exp(3+x))+x)*log((x*exp(exp(3+x))+1)/x)^2+(-10*x^2*exp(exp (3+x))-10*x)*log((x*exp(exp(3+x))+1)/x)+25*x^2*exp(exp(3+x))+25*x),x, algo rithm=\
(x^4*e^(x^2 + 2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x + 3)) + 1)/x) - 5*x^4*e^(x^2 + 2*x + 2*e ^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5* x^4*e^(x^2 + 2*x + 2*e^(x + 3) + 6)*log((x*e^(e^(x + 3)) + 1)/x) + 5*x^4*e ^(2*x + 2*e^(x + 3) + 6)*log((x*e^(e^(x + 3)) + 1)/x) + 25*x^4*e^(x^2 + 2* x + 2*e^(x + 3) + 6) - 25*x^4*e^(2*x + 2*e^(x + 3) + 6) - 2*x^2*e^(x^2 + x + e^(x + 3) + 3)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)* log((x*e^(e^(x + 3)) + 1)/x) + 10*x^2*e^(x^2 + x + e^(x + 3) + 3)*log((x*e ^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5*x^2*e^(x + e^(x + 3) + 3)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) + 10*x^2*e^(x^ 2 + x + e^(x + 3) + 3)*log((x*e^(e^(x + 3)) + 1)/x) - 5*x^2*e^(x + e^(x + 3) + 3)*log((x*e^(e^(x + 3)) + 1)/x) - 50*x^2*e^(x^2 + x + e^(x + 3) + 3) + 50*x^2*e^(x + e^(x + 3) + 3) + e^(x^2)*log((x*e^(x + e^(x + 3) + 3) + e^ (x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x + 3)) + 1)/x) - 5*e^(x^2)*log((x*e^( x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x) - 5*e^(x^2)*log((x*e^(e^(x + 3)) + 1)/x) + 25*e^(x^2) + 5*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^ (-x - 3)/x) - 25)/(x^4*e^(2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x)*log((x*e^(e^(x + 3)) + 1)/x) - 5*x^4*e^(2*x + 2*e^(x + 3) + 6)*log((x*e^(x + e^(x + 3) + 3) + e^(x + 3))*e^(-x - 3)/x ) - 5*x^4*e^(2*x + 2*e^(x + 3) + 6)*log((x*e^(e^(x + 3)) + 1)/x) + 25*x...
Time = 12.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {5+50 e^{x^2} x^2+e^{e^{3+x}} \left (-5 e^{3+x} x^2+50 e^{x^2} x^3\right )+\left (-20 e^{x^2} x^2-20 e^{e^{3+x}+x^2} x^3\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (2 e^{x^2} x^2+2 e^{e^{3+x}+x^2} x^3\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )}{25 x+25 e^{e^{3+x}} x^2+\left (-10 x-10 e^{e^{3+x}} x^2\right ) \log \left (\frac {1+e^{e^{3+x}} x}{x}\right )+\left (x+e^{e^{3+x}} x^2\right ) \log ^2\left (\frac {1+e^{e^{3+x}} x}{x}\right )} \, dx={\mathrm {e}}^{x^2}+\frac {5}{\ln \left (\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^3\,{\mathrm {e}}^x}+1}{x}\right )-5} \]
int((log((x*exp(exp(x + 3)) + 1)/x)^2*(2*x^2*exp(x^2) + 2*x^3*exp(x^2)*exp (exp(x + 3))) - exp(exp(x + 3))*(5*x^2*exp(x + 3) - 50*x^3*exp(x^2)) + 50* x^2*exp(x^2) - log((x*exp(exp(x + 3)) + 1)/x)*(20*x^2*exp(x^2) + 20*x^3*ex p(x^2)*exp(exp(x + 3))) + 5)/(25*x + 25*x^2*exp(exp(x + 3)) - log((x*exp(e xp(x + 3)) + 1)/x)*(10*x + 10*x^2*exp(exp(x + 3))) + log((x*exp(exp(x + 3) ) + 1)/x)^2*(x + x^2*exp(exp(x + 3)))),x)