Integrand size = 68, antiderivative size = 30 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=\frac {-1+e^{x^2}+2 x}{x \left (25+\frac {3+\log \left (\frac {e}{2}\right )}{x}\right )} \] Output:
(exp(x^2)+2*x-1)/((ln(1/2*exp(1))+3)/x+25)/x
Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=\frac {25 e^{x^2} (8+50 x-\log (4))+2 (4+25 x-\log (2)) (-33+\log (4))}{50 (-4-25 x+\log (2))^2} \] Input:
Integrate[(31 + E^x^2*(-25 + 6*x + 50*x^2) + (2 + 2*E^x^2*x)*Log[E/2])/(9 + 150*x + 625*x^2 + (6 + 50*x)*Log[E/2] + Log[E/2]^2),x]
Output:
(25*E^x^2*(8 + 50*x - Log[4]) + 2*(4 + 25*x - Log[2])*(-33 + Log[4]))/(50* (-4 - 25*x + Log[2])^2)
Time = 0.83 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {7292, 7277, 27, 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.
\(\displaystyle \int \frac {e^{x^2} \left (50 x^2+6 x-25\right )+\left (2 e^{x^2} x+2\right ) \log \left (\frac {e}{2}\right )+31}{625 x^2+150 x+(50 x+6) \log \left (\frac {e}{2}\right )+9+\log ^2\left (\frac {e}{2}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x^2} \left (50 x^2+6 x-25\right )+\left (2 e^{x^2} x+2\right ) \log \left (\frac {e}{2}\right )+31}{625 x^2+50 x (4-\log (2))+(\log (2)-4)^2}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 2500 \int \frac {2 \log \left (\frac {e}{2}\right ) \left (e^{x^2} x+1\right )-e^{x^2} \left (-50 x^2-6 x+25\right )+31}{2500 (25 x-\log (2)+4)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {-e^{x^2} \left (-50 x^2-6 x+25\right )+2 \left (e^{x^2} x+1\right ) \log \left (\frac {e}{2}\right )+31}{(25 x+4-\log (2))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x^2} \left (50 x^2+x (8-\log (4))-25\right )}{(25 x+4-\log (2))^2}+\frac {33-\log (4)}{(25 x+4-\log (2))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{x^2} \left (50 x^2+x (8-\log (4))\right )}{2 x (25 x+4-\log (2))^2}-\frac {33-\log (4)}{25 (25 x+4-\log (2))}\) |
Input:
Int[(31 + E^x^2*(-25 + 6*x + 50*x^2) + (2 + 2*E^x^2*x)*Log[E/2])/(9 + 150* x + 625*x^2 + (6 + 50*x)*Log[E/2] + Log[E/2]^2),x]
Output:
(E^x^2*(50*x^2 + x*(8 - Log[4])))/(2*x*(4 + 25*x - Log[2])^2) - (33 - Log[ 4])/(25*(4 + 25*x - Log[2]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.81 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77
method | result | size |
norman | \(\frac {-{\mathrm e}^{x^{2}}+\frac {33}{25}-\frac {2 \ln \left (2\right )}{25}}{\ln \left (2\right )-25 x -4}\) | \(23\) |
parallelrisch | \(\frac {-31-2 \ln \left (\frac {{\mathrm e}}{2}\right )+25 \,{\mathrm e}^{x^{2}}}{75+25 \ln \left (\frac {{\mathrm e}}{2}\right )+625 x}\) | \(30\) |
parts | \(-\frac {2 \ln \left (\frac {{\mathrm e}}{2}\right )+31}{25 \left (\ln \left (\frac {{\mathrm e}}{2}\right )+25 x +3\right )}-\frac {{\mathrm e}^{x^{2}}}{\ln \left (2\right )-25 x -4}\) | \(40\) |
Input:
int(((2*exp(x^2)*x+2)*ln(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/(ln(1/2* exp(1))^2+(50*x+6)*ln(1/2*exp(1))+625*x^2+150*x+9),x,method=_RETURNVERBOSE )
Output:
(-exp(x^2)+33/25-2/25*ln(2))/(ln(2)-25*x-4)
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=\frac {25 \, e^{\left (x^{2}\right )} + 2 \, \log \left (2\right ) - 33}{25 \, {\left (25 \, x - \log \left (2\right ) + 4\right )}} \] Input:
integrate(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/( log(1/2*exp(1))^2+(50*x+6)*log(1/2*exp(1))+625*x^2+150*x+9),x, algorithm=" fricas")
Output:
1/25*(25*e^(x^2) + 2*log(2) - 33)/(25*x - log(2) + 4)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=- \frac {33 - 2 \log {\left (2 \right )}}{625 x - 25 \log {\left (2 \right )} + 100} + \frac {e^{x^{2}}}{25 x - \log {\left (2 \right )} + 4} \] Input:
integrate(((2*exp(x**2)*x+2)*ln(1/2*exp(1))+(50*x**2+6*x-25)*exp(x**2)+31) /(ln(1/2*exp(1))**2+(50*x+6)*ln(1/2*exp(1))+625*x**2+150*x+9),x)
Output:
-(33 - 2*log(2))/(625*x - 25*log(2) + 100) + exp(x**2)/(25*x - log(2) + 4)
Time = 0.14 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=\frac {e^{\left (x^{2}\right )}}{25 \, x - \log \left (2\right ) + 4} - \frac {2 \, \log \left (\frac {1}{2} \, e\right )}{25 \, {\left (25 \, x + \log \left (\frac {1}{2} \, e\right ) + 3\right )}} - \frac {31}{25 \, {\left (25 \, x + \log \left (\frac {1}{2} \, e\right ) + 3\right )}} \] Input:
integrate(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/( log(1/2*exp(1))^2+(50*x+6)*log(1/2*exp(1))+625*x^2+150*x+9),x, algorithm=" maxima")
Output:
e^(x^2)/(25*x - log(2) + 4) - 2/25*log(1/2*e)/(25*x + log(1/2*e) + 3) - 31 /25/(25*x + log(1/2*e) + 3)
Time = 0.12 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=\frac {25 \, e^{\left (x^{2}\right )} + 2 \, \log \left (2\right ) - 33}{25 \, {\left (25 \, x - \log \left (2\right ) + 4\right )}} \] Input:
integrate(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/( log(1/2*exp(1))^2+(50*x+6)*log(1/2*exp(1))+625*x^2+150*x+9),x, algorithm=" giac")
Output:
1/25*(25*e^(x^2) + 2*log(2) - 33)/(25*x - log(2) + 4)
Time = 3.75 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=-\frac {\frac {\ln \left (\frac {{\mathrm {e}}^2}{4}\right )}{25}-{\mathrm {e}}^{x^2}+\frac {31}{25}}{25\,x+\ln \left (\frac {\mathrm {e}}{2}\right )+3} \] Input:
int((log(exp(1)/2)*(2*x*exp(x^2) + 2) + exp(x^2)*(6*x + 50*x^2 - 25) + 31) /(150*x + log(exp(1)/2)^2 + log(exp(1)/2)*(50*x + 6) + 625*x^2 + 9),x)
Output:
-(log(exp(2)/4)/25 - exp(x^2) + 31/25)/(25*x + log(exp(1)/2) + 3)
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.83 \[ \int \frac {31+e^{x^2} \left (-25+6 x+50 x^2\right )+\left (2+2 e^{x^2} x\right ) \log \left (\frac {e}{2}\right )}{9+150 x+625 x^2+(6+50 x) \log \left (\frac {e}{2}\right )+\log ^2\left (\frac {e}{2}\right )} \, dx=\frac {e^{x^{2}} \mathrm {log}\left (\frac {e}{2}\right )+3 e^{x^{2}}+2 \,\mathrm {log}\left (\frac {e}{2}\right ) x +31 x}{\mathrm {log}\left (\frac {e}{2}\right )^{2}+25 \,\mathrm {log}\left (\frac {e}{2}\right ) x +6 \,\mathrm {log}\left (\frac {e}{2}\right )+75 x +9} \] Input:
int(((2*exp(x^2)*x+2)*log(1/2*exp(1))+(50*x^2+6*x-25)*exp(x^2)+31)/(log(1/ 2*exp(1))^2+(50*x+6)*log(1/2*exp(1))+625*x^2+150*x+9),x)
Output:
(e**(x**2)*log(e/2) + 3*e**(x**2) + 2*log(e/2)*x + 31*x)/(log(e/2)**2 + 25 *log(e/2)*x + 6*log(e/2) + 75*x + 9)