Integrand size = 55, antiderivative size = 32 \[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=(3-x) \left (\frac {10 e^x}{5-x}+\frac {1}{2} e^{-\log ^2(3)} x\right ) \]
Time = 0.18 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=10 e^x \left (1+\frac {2}{-5+x}\right )+\frac {3}{2} e^{-\log ^2(3)} x-\frac {1}{2} e^{-\log ^2(3)} x^2 \]
Integrate[(75 - 80*x + 23*x^2 - 2*x^3 + E^(x + Log[3]^2)*(260 - 160*x + 20 *x^2))/(E^Log[3]^2*(50 - 20*x + 2*x^2)),x]
Time = 0.64 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.50, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {27, 27, 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.
\(\displaystyle \int \frac {e^{-\log ^2(3)} \left (-2 x^3+23 x^2+\left (20 x^2-160 x+260\right ) e^{x+\log ^2(3)}-80 x+75\right )}{2 x^2-20 x+50} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{-\log ^2(3)} \int \frac {-2 x^3+23 x^2-80 x+20 e^{x+\log ^2(3)} \left (x^2-8 x+13\right )+75}{2 \left (x^2-10 x+25\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} e^{-\log ^2(3)} \int \frac {-2 x^3+23 x^2-80 x+20 e^{x+\log ^2(3)} \left (x^2-8 x+13\right )+75}{x^2-10 x+25}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 2 e^{-\log ^2(3)} \int \frac {-2 x^3+23 x^2-80 x+20 e^{x+\log ^2(3)} \left (x^2-8 x+13\right )+75}{4 (5-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} e^{-\log ^2(3)} \int \frac {-2 x^3+23 x^2-80 x+20 e^{x+\log ^2(3)} \left (x^2-8 x+13\right )+75}{(5-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} e^{-\log ^2(3)} \int \left (-\frac {2 x^3}{(x-5)^2}+\frac {23 x^2}{(x-5)^2}-\frac {80 x}{(x-5)^2}+\frac {20 e^{x+\log ^2(3)} \left (x^2-8 x+13\right )}{(x-5)^2}+\frac {75}{(x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} e^{-\log ^2(3)} \left (-x^2+3 x+20 e^{x+\log ^2(3)}-\frac {40 e^{x+\log ^2(3)}}{5-x}\right )\) |
Int[(75 - 80*x + 23*x^2 - 2*x^3 + E^(x + Log[3]^2)*(260 - 160*x + 20*x^2)) /(E^Log[3]^2*(50 - 20*x + 2*x^2)),x]
3.2.47.3.1 Defintions of rubi rules used
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 = 0.76 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97
method | result | size |
parts | \(-\frac {{\mathrm e}^{-\ln \left (3\right )^{2}} \left (x^{2}-3 x \right )}{2}+10 \,{\mathrm e}^{x}+\frac {20 \,{\mathrm e}^{x}}{-5+x}\) | \(31\) |
risch | \(-\frac {{\mathrm e}^{-\ln \left (3\right )^{2}} x^{2}}{2}+\frac {3 x \,{\mathrm e}^{-\ln \left (3\right )^{2}}}{2}+\frac {10 \left (-3+x \right ) {\mathrm e}^{x}}{-5+x}\) | \(36\) |
parallelrisch | \(\frac {{\mathrm e}^{-\ln \left (3\right )^{2}} \left (-75-x^{3}+8 x^{2}+20 \,{\mathrm e}^{x} {\mathrm e}^{\ln \left (3\right )^{2}} x -60 \,{\mathrm e}^{x} {\mathrm e}^{\ln \left (3\right )^{2}}\right )}{2 x -10}\) | \(46\) |
norman | \(\frac {10 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{-\ln \left (3\right )^{2}} x^{2}-\frac {{\mathrm e}^{-\ln \left (3\right )^{2}} x^{3}}{2}-30 \,{\mathrm e}^{x}-\frac {75 \,{\mathrm e}^{-\ln \left (3\right )^{2}}}{2}}{-5+x}\) | \(50\) |
default | \(\frac {{\mathrm e}^{-\ln \left (3\right )^{2}} \left (-x^{2}+3 x +260 \,{\mathrm e}^{\ln \left (3\right )^{2}} \left (-\frac {{\mathrm e}^{x}}{-5+x}-{\mathrm e}^{5} \operatorname {Ei}_{1}\left (5-x \right )\right )-160 \,{\mathrm e}^{\ln \left (3\right )^{2}} \left (-\frac {5 \,{\mathrm e}^{x}}{-5+x}-6 \,{\mathrm e}^{5} \operatorname {Ei}_{1}\left (5-x \right )\right )+20 \,{\mathrm e}^{\ln \left (3\right )^{2}} \left ({\mathrm e}^{x}-\frac {25 \,{\mathrm e}^{x}}{-5+x}-35 \,{\mathrm e}^{5} \operatorname {Ei}_{1}\left (5-x \right )\right )\right )}{2}\) | \(105\) |
int(((20*x^2-160*x+260)*exp(x)*exp(ln(3)^2)-2*x^3+23*x^2-80*x+75)/(2*x^2-2 0*x+50)/exp(ln(3)^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=-\frac {{\left (x^{3} - 8 \, x^{2} - 20 \, {\left (x - 3\right )} e^{\left (\log \left (3\right )^{2} + x\right )} + 15 \, x\right )} e^{\left (-\log \left (3\right )^{2}\right )}}{2 \, {\left (x - 5\right )}} \]
integrate(((20*x^2-160*x+260)*exp(x)*exp(log(3)^2)-2*x^3+23*x^2-80*x+75)/( 2*x^2-20*x+50)/exp(log(3)^2),x, algorithm=\
Time = 0.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=- \frac {x^{2}}{2 e^{\log {\left (3 \right )}^{2}}} + \frac {3 x}{2 e^{\log {\left (3 \right )}^{2}}} + \frac {\left (10 x - 30\right ) e^{x}}{x - 5} \]
integrate(((20*x**2-160*x+260)*exp(x)*exp(ln(3)**2)-2*x**3+23*x**2-80*x+75 )/(2*x**2-20*x+50)/exp(ln(3)**2),x)
\[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=\int { -\frac {{\left (2 \, x^{3} - 23 \, x^{2} - 20 \, {\left (x^{2} - 8 \, x + 13\right )} e^{\left (\log \left (3\right )^{2} + x\right )} + 80 \, x - 75\right )} e^{\left (-\log \left (3\right )^{2}\right )}}{2 \, {\left (x^{2} - 10 \, x + 25\right )}} \,d x } \]
integrate(((20*x^2-160*x+260)*exp(x)*exp(log(3)^2)-2*x^3+23*x^2-80*x+75)/( 2*x^2-20*x+50)/exp(log(3)^2),x, algorithm=\
-1/2*(x^2 - 3*x - 20*(x^2*e^(log(3)^2) - 8*x*e^(log(3)^2))*e^x/(x^2 - 10*x + 25) + 260*e^(log(3)^2 + 5)*exp_integral_e(2, -x + 5)/(x - 5) - integrat e(40*(x*e^(log(3)^2) - 20*e^(log(3)^2))*e^x/(x^3 - 15*x^2 + 75*x - 125), x ))*e^(-log(3)^2)
Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=-\frac {{\left (x^{3} - 8 \, x^{2} - 20 \, x e^{\left (\log \left (3\right )^{2} + x\right )} + 15 \, x + 60 \, e^{\left (\log \left (3\right )^{2} + x\right )}\right )} e^{\left (-\log \left (3\right )^{2}\right )}}{2 \, {\left (x - 5\right )}} \]
integrate(((20*x^2-160*x+260)*exp(x)*exp(log(3)^2)-2*x^3+23*x^2-80*x+75)/( 2*x^2-20*x+50)/exp(log(3)^2),x, algorithm=\
-1/2*(x^3 - 8*x^2 - 20*x*e^(log(3)^2 + x) + 15*x + 60*e^(log(3)^2 + x))*e^ (-log(3)^2)/(x - 5)
Time = 8.55 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\log ^2(3)} \left (75-80 x+23 x^2-2 x^3+e^{x+\log ^2(3)} \left (260-160 x+20 x^2\right )\right )}{50-20 x+2 x^2} \, dx=\frac {{\mathrm {e}}^{-{\ln \left (3\right )}^2}\,\left (x-3\right )\,\left (5\,x+20\,{\mathrm {e}}^{x+{\ln \left (3\right )}^2}-x^2\right )}{2\,\left (x-5\right )} \]