Integrand size = 102, antiderivative size = 28 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=-4+e^{\frac {-3+e^5+(3+x)^2}{5 (5+x-\log (8))}}+x \]
Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(28)=56\).
Time = 0.39 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.25 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=\frac {1}{5} \left (5\ 2^{-\frac {3 \log (8)}{5 (5+x-\log (8))}} e^{\frac {6+e^5+6 x+x^2+\log ^2(8)}{5 (5+x-\log (8))}}+5 (5+x-\log (8))\right ) \]
Integrate[(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2 + E^((-6 - E^5 - 6*x - x^2)/(-25 - 5*x + 5*Log[8]))*(24 - E^5 + 10*x + x^2 + (-6 - 2*x)*Log[8]))/(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2),x]
((5*E^((6 + E^5 + 6*x + x^2 + Log[8]^2)/(5*(5 + x - Log[8]))))/2^((3*Log[8 ])/(5*(5 + x - Log[8]))) + 5*(5 + x - Log[8]))/5
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.
\(\displaystyle \int \frac {5 x^2+e^{\frac {-x^2-6 x-e^5-6}{-5 x-25+5 \log (8)}} \left (x^2+10 x+(-2 x-6) \log (8)-e^5+24\right )+50 x+(-10 x-50) \log (8)+125+5 \log ^2(8)}{5 x^2+50 x+(-10 x-50) \log (8)+125+5 \log ^2(8)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 x^2+e^{\frac {-x^2-6 x-e^5-6}{-5 x-25+5 \log (8)}} \left (x^2+10 x+(-2 x-6) \log (8)-e^5+24\right )+50 x+(-10 x-50) \log (8)+125 \left (1+\frac {\log ^2(8)}{25}\right )}{5 x^2+10 x (5-\log (8))+5 (5-\log (8))^2}dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 20 \int \frac {5 x^2+50 x+e^{\frac {x^2+6 x+e^5+6}{5 (x-\log (8)+5)}} \left (x^2+10 x-2 (x+3) \log (8)-e^5+24\right )+5 \left (25+\log ^2(8)\right )-10 (x+5) \log (8)}{100 (x-\log (8)+5)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {5 x^2+50 x+e^{\frac {x^2+6 x+e^5+6}{5 (x-\log (8)+5)}} \left (x^2+10 x-2 (x+3) \log (8)-e^5+24\right )+5 \left (25+\log ^2(8)\right )-10 (x+5) \log (8)}{(x-\log (8)+5)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {5 x^2}{(x-\log (8)+5)^2}+\frac {50 x}{(x-\log (8)+5)^2}+\frac {e^{\frac {x^2+6 x+e^5+6}{5 (x-\log (8)+5)}} \left (x^2+2 (5-\log (8)) x-6 \log (8)-e^5+24\right )}{(x-\log (8)+5)^2}+\frac {5 \left (25+\log ^2(8)\right )}{(x-\log (8)+5)^2}-\frac {10 (x+5) \log (8)}{(x-\log (8)+5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-\left (1+e^5+\log ^2(8)-4 \log (8)\right ) \int \frac {e^{\frac {x^2+6 x+e^5+6}{5 (x-\log (8)+5)}}}{(x-\log (8)+5)^2}dx+\int e^{\frac {x^2+6 x+e^5+6}{5 (x-\log (8)+5)}}dx+5 x-\frac {5 \left (25+\log ^2(8)\right )}{x+5-\log (8)}+\frac {10 \log ^2(8)}{x+5-\log (8)}-10 \log (8) \log (x+5-\log (8))-10 (5-\log (8)) \log (x+5-\log (8))+50 \log (x+5-\log (8))-\frac {5 (5-\log (8))^2}{x+5-\log (8)}+\frac {50 (5-\log (8))}{x+5-\log (8)}\right )\) |
Int[(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2 + E^((-6 - E^5 - 6*x - x^2)/(-25 - 5*x + 5*Log[8]))*(24 - E^5 + 10*x + x^2 + (-6 - 2*x)*L og[8]))/(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2),x]
3.21.17.3.1 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 = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x +{\mathrm e}^{-\frac {x^{2}+{\mathrm e}^{5}+6 x +6}{5 \left (3 \ln \left (2\right )-x -5\right )}}\) | \(27\) |
parallelrisch | \(12 \ln \left (2\right )+x +{\mathrm e}^{-\frac {x^{2}+{\mathrm e}^{5}+6 x +6}{5 \left (3 \ln \left (2\right )-x -5\right )}}-20\) | \(32\) |
parts | \(x +\frac {\left (3 \ln \left (2\right )-5\right ) {\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}-x \,{\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}}{3 \ln \left (2\right )-x -5}\) | \(80\) |
norman | \(\frac {\left (3 \ln \left (2\right )-5\right ) {\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}-x^{2}-x \,{\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \left (2\right )-5 x -25}}+9 \ln \left (2\right )^{2}-30 \ln \left (2\right )+25}{3 \ln \left (2\right )-x -5}\) | \(94\) |
int(((3*(-2*x-6)*ln(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/(15*ln( 2)-5*x-25))+45*ln(2)^2+3*(-10*x-50)*ln(2)+5*x^2+50*x+125)/(45*ln(2)^2+3*(- 10*x-50)*ln(2)+5*x^2+50*x+125),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=x + e^{\left (\frac {x^{2} + 6 \, x + e^{5} + 6}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}}\right )} \]
integrate(((3*(-2*x-6)*log(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/ (15*log(2)-5*x-25))+45*log(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125)/(45*lo g(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125),x, algorithm=\
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=x + e^{\frac {- x^{2} - 6 x - e^{5} - 6}{- 5 x - 25 + 15 \log {\left (2 \right )}}} \]
integrate(((3*(-2*x-6)*ln(2)-exp(5)+x**2+10*x+24)*exp((-exp(5)-x**2-6*x-6) /(15*ln(2)-5*x-25))+45*ln(2)**2+3*(-10*x-50)*ln(2)+5*x**2+50*x+125)/(45*ln (2)**2+3*(-10*x-50)*ln(2)+5*x**2+50*x+125),x)
Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (24) = 48\).
Time = 0.41 (sec) , antiderivative size = 200, normalized size of antiderivative = 7.14 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=6 \, {\left (\frac {3 \, \log \left (2\right ) - 5}{x - 3 \, \log \left (2\right ) + 5} - \log \left (x - 3 \, \log \left (2\right ) + 5\right )\right )} \log \left (2\right ) + 2 \, {\left (3 \, \log \left (2\right ) - 5\right )} \log \left (x - 3 \, \log \left (2\right ) + 5\right ) + 2^{\frac {3}{5}} e^{\left (\frac {1}{5} \, x + \frac {9 \, \log \left (2\right )^{2}}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} + \frac {e^{5}}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} - \frac {12 \, \log \left (2\right )}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} + \frac {1}{5 \, {\left (x - 3 \, \log \left (2\right ) + 5\right )}} + \frac {1}{5}\right )} + x - \frac {9 \, \log \left (2\right )^{2}}{x - 3 \, \log \left (2\right ) + 5} - \frac {9 \, \log \left (2\right )^{2} - 30 \, \log \left (2\right ) + 25}{x - 3 \, \log \left (2\right ) + 5} - \frac {10 \, {\left (3 \, \log \left (2\right ) - 5\right )}}{x - 3 \, \log \left (2\right ) + 5} + \frac {30 \, \log \left (2\right )}{x - 3 \, \log \left (2\right ) + 5} - \frac {25}{x - 3 \, \log \left (2\right ) + 5} + 10 \, \log \left (x - 3 \, \log \left (2\right ) + 5\right ) \]
integrate(((3*(-2*x-6)*log(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/ (15*log(2)-5*x-25))+45*log(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125)/(45*lo g(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125),x, algorithm=\
6*((3*log(2) - 5)/(x - 3*log(2) + 5) - log(x - 3*log(2) + 5))*log(2) + 2*( 3*log(2) - 5)*log(x - 3*log(2) + 5) + 2^(3/5)*e^(1/5*x + 9/5*log(2)^2/(x - 3*log(2) + 5) + 1/5*e^5/(x - 3*log(2) + 5) - 12/5*log(2)/(x - 3*log(2) + 5) + 1/5/(x - 3*log(2) + 5) + 1/5) + x - 9*log(2)^2/(x - 3*log(2) + 5) - ( 9*log(2)^2 - 30*log(2) + 25)/(x - 3*log(2) + 5) - 10*(3*log(2) - 5)/(x - 3 *log(2) + 5) + 30*log(2)/(x - 3*log(2) + 5) - 25/(x - 3*log(2) + 5) + 10*l og(x - 3*log(2) + 5)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (24) = 48\).
Time = 0.49 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.39 \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=x + e^{\left (\frac {3 \, x^{2} \log \left (2\right ) - 5 \, x^{2} + x e^{5} + 18 \, x \log \left (2\right ) - 24 \, x}{5 \, {\left (3 \, x \log \left (2\right ) - 9 \, \log \left (2\right )^{2} - 5 \, x + 30 \, \log \left (2\right ) - 25\right )}} - \frac {e^{5} + 6}{5 \, {\left (3 \, \log \left (2\right ) - 5\right )}}\right )} \]
integrate(((3*(-2*x-6)*log(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/ (15*log(2)-5*x-25))+45*log(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125)/(45*lo g(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125),x, algorithm=\
x + e^(1/5*(3*x^2*log(2) - 5*x^2 + x*e^5 + 18*x*log(2) - 24*x)/(3*x*log(2) - 9*log(2)^2 - 5*x + 30*log(2) - 25) - 1/5*(e^5 + 6)/(3*log(2) - 5))
Timed out. \[ \int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx=\int \frac {50\,x-3\,\ln \left (2\right )\,\left (10\,x+50\right )+45\,{\ln \left (2\right )}^2+{\mathrm {e}}^{\frac {x^2+6\,x+{\mathrm {e}}^5+6}{5\,x-15\,\ln \left (2\right )+25}}\,\left (10\,x-{\mathrm {e}}^5-3\,\ln \left (2\right )\,\left (2\,x+6\right )+x^2+24\right )+5\,x^2+125}{50\,x-3\,\ln \left (2\right )\,\left (10\,x+50\right )+45\,{\ln \left (2\right )}^2+5\,x^2+125} \,d x \]
int((50*x - 3*log(2)*(10*x + 50) + 45*log(2)^2 + exp((6*x + exp(5) + x^2 + 6)/(5*x - 15*log(2) + 25))*(10*x - exp(5) - 3*log(2)*(2*x + 6) + x^2 + 24 ) + 5*x^2 + 125)/(50*x - 3*log(2)*(10*x + 50) + 45*log(2)^2 + 5*x^2 + 125) ,x)