Integrand size = 89, antiderivative size = 24 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (e^{25+x}-(-1-e-x+x \log (\log (2)))^2\right ) \] Output:
ln(exp(x+25)-(x*ln(ln(2))-exp(1)-1-x)^2)
Time = 0.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (-e^2+e^{25+x}+2 e (-1+x (-1+\log (\log (2))))-(-1+x (-1+\log (\log (2))))^2\right ) \] Input:
Integrate[(2 + 2*E - E^(25 + x) + 2*x + (-2 - 2*E - 4*x)*Log[Log[2]] + 2*x *Log[Log[2]]^2)/(1 + E^2 - E^(25 + x) + 2*x + x^2 + E*(2 + 2*x) + (-2*x - 2*E*x - 2*x^2)*Log[Log[2]] + x^2*Log[Log[2]]^2),x]
Output:
Log[-E^2 + E^(25 + x) + 2*E*(-1 + x*(-1 + Log[Log[2]])) - (-1 + x*(-1 + Lo g[Log[2]]))^2]
Time = 0.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6, 6, 7235}
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.
\(\displaystyle \int \frac {-e^{x+25}+2 x+2 x \log ^2(\log (2))+(-4 x-2 e-2) \log (\log (2))+2 e+2}{x^2+x^2 \log ^2(\log (2))+\left (-2 x^2-2 e x-2 x\right ) \log (\log (2))+2 x-e^{x+25}+e (2 x+2)+e^2+1} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-e^{x+25}+x \left (2+2 \log ^2(\log (2))\right )+(-4 x-2 e-2) \log (\log (2))+2 e+2}{x^2+x^2 \log ^2(\log (2))+\left (-2 x^2-2 e x-2 x\right ) \log (\log (2))+2 x-e^{x+25}+e (2 x+2)+e^2+1}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-e^{x+25}+x \left (2+2 \log ^2(\log (2))\right )+(-4 x-2 e-2) \log (\log (2))+2 e+2}{x^2 \left (1+\log ^2(\log (2))\right )+\left (-2 x^2-2 e x-2 x\right ) \log (\log (2))+2 x-e^{x+25}+e (2 x+2)+e^2+1}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (x^2 \left (1+\log ^2(\log (2))\right )-2 \left (x^2+e x+x\right ) \log (\log (2))+2 x-e^{x+25}+2 e (x+1)+e^2+1\right )\) |
Input:
Int[(2 + 2*E - E^(25 + x) + 2*x + (-2 - 2*E - 4*x)*Log[Log[2]] + 2*x*Log[L og[2]]^2)/(1 + E^2 - E^(25 + x) + 2*x + x^2 + E*(2 + 2*x) + (-2*x - 2*E*x - 2*x^2)*Log[Log[2]] + x^2*Log[Log[2]]^2),x]
Output:
Log[1 + E^2 - E^(25 + x) + 2*x + 2*E*(1 + x) - 2*(x + E*x + x^2)*Log[Log[2 ]] + x^2*(1 + Log[Log[2]]^2)]
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(24)=48\).
Time = 0.58 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29
method | result | size |
derivativedivides | \(\ln \left (x^{2} \ln \left (\ln \left (2\right )\right )^{2}+\left (-2 x \,{\mathrm e}-2 x^{2}-2 x \right ) \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{x +25}+{\mathrm e}^{2}+\left (2+2 x \right ) {\mathrm e}+x^{2}+2 x +1\right )\) | \(55\) |
default | \(\ln \left (x^{2} \ln \left (\ln \left (2\right )\right )^{2}+\left (-2 x \,{\mathrm e}-2 x^{2}-2 x \right ) \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{x +25}+{\mathrm e}^{2}+\left (2+2 x \right ) {\mathrm e}+x^{2}+2 x +1\right )\) | \(55\) |
norman | \(\ln \left (x^{2} \ln \left (\ln \left (2\right )\right )^{2}-2 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x -2 x^{2} \ln \left (\ln \left (2\right )\right )-2 x \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}-{\mathrm e}^{x +25}+2 \,{\mathrm e}+2 x +1\right )\) | \(60\) |
risch | \(-25+\ln \left (-x^{2} \ln \left (\ln \left (2\right )\right )^{2}+2 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x +2 x^{2} \ln \left (\ln \left (2\right )\right )+2 x \ln \left (\ln \left (2\right )\right )-{\mathrm e}^{2}-2 x \,{\mathrm e}-x^{2}-2 \,{\mathrm e}-2 x +{\mathrm e}^{x +25}-1\right )\) | \(63\) |
parallelrisch | \(\ln \left (\frac {x^{2} \ln \left (\ln \left (2\right )\right )^{2}-2 \ln \left (\ln \left (2\right )\right ) {\mathrm e} x -2 x^{2} \ln \left (\ln \left (2\right )\right )-2 x \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{2}+2 x \,{\mathrm e}+x^{2}-{\mathrm e}^{x +25}+2 \,{\mathrm e}+2 x +1}{\ln \left (\ln \left (2\right )\right )^{2}-2 \ln \left (\ln \left (2\right )\right )+1}\right )\) | \(75\) |
Input:
int((2*x*ln(ln(2))^2+(-2*exp(1)-4*x-2)*ln(ln(2))-exp(x+25)+2*exp(1)+2*x+2) /(x^2*ln(ln(2))^2+(-2*x*exp(1)-2*x^2-2*x)*ln(ln(2))-exp(x+25)+exp(1)^2+(2+ 2*x)*exp(1)+x^2+2*x+1),x,method=_RETURNVERBOSE)
Output:
ln(x^2*ln(ln(2))^2+(-2*x*exp(1)-2*x^2-2*x)*ln(ln(2))-exp(x+25)+exp(1)^2+(2 +2*x)*exp(1)+x^2+2*x+1)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (-x^{2} \log \left (\log \left (2\right )\right )^{2} - x^{2} - 2 \, {\left (x + 1\right )} e + 2 \, {\left (x^{2} + x e + x\right )} \log \left (\log \left (2\right )\right ) - 2 \, x - e^{2} + e^{\left (x + 25\right )} - 1\right ) \] Input:
integrate((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp (1)+2*x+2)/(x^2*log(log(2))^2+(-2*exp(1)*x-2*x^2-2*x)*log(log(2))-exp(x+25 )+exp(1)^2+(2+2*x)*exp(1)+x^2+2*x+1),x, algorithm="fricas")
Output:
log(-x^2*log(log(2))^2 - x^2 - 2*(x + 1)*e + 2*(x^2 + x*e + x)*log(log(2)) - 2*x - e^2 + e^(x + 25) - 1)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (20) = 40\).
Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.92 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log {\left (- x^{2} + 2 x^{2} \log {\left (\log {\left (2 \right )} \right )} - x^{2} \log {\left (\log {\left (2 \right )} \right )}^{2} - 2 e x - 2 x + 2 e x \log {\left (\log {\left (2 \right )} \right )} + 2 x \log {\left (\log {\left (2 \right )} \right )} + e^{x + 25} - e^{2} - 2 e - 1 \right )} \] Input:
integrate((2*x*ln(ln(2))**2+(-2*exp(1)-4*x-2)*ln(ln(2))-exp(x+25)+2*exp(1) +2*x+2)/(x**2*ln(ln(2))**2+(-2*exp(1)*x-2*x**2-2*x)*ln(ln(2))-exp(x+25)+ex p(1)**2+(2+2*x)*exp(1)+x**2+2*x+1),x)
Output:
log(-x**2 + 2*x**2*log(log(2)) - x**2*log(log(2))**2 - 2*E*x - 2*x + 2*E*x *log(log(2)) + 2*x*log(log(2)) + exp(x + 25) - exp(2) - 2*E - 1)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (x^{2} \log \left (\log \left (2\right )\right )^{2} + x^{2} + 2 \, {\left (x + 1\right )} e - 2 \, {\left (x^{2} + x e + x\right )} \log \left (\log \left (2\right )\right ) + 2 \, x + e^{2} - e^{\left (x + 25\right )} + 1\right ) \] Input:
integrate((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp (1)+2*x+2)/(x^2*log(log(2))^2+(-2*exp(1)*x-2*x^2-2*x)*log(log(2))-exp(x+25 )+exp(1)^2+(2+2*x)*exp(1)+x^2+2*x+1),x, algorithm="maxima")
Output:
log(x^2*log(log(2))^2 + x^2 + 2*(x + 1)*e - 2*(x^2 + x*e + x)*log(log(2)) + 2*x + e^2 - e^(x + 25) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (24) = 48\).
Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.50 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\log \left (-x^{2} \log \left (\log \left (2\right )\right )^{2} + 2 \, x^{2} \log \left (\log \left (2\right )\right ) + 2 \, x e \log \left (\log \left (2\right )\right ) - x^{2} - 2 \, x e + 2 \, x \log \left (\log \left (2\right )\right ) - 2 \, x - e^{2} - 2 \, e + e^{\left (x + 25\right )} - 1\right ) \] Input:
integrate((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp (1)+2*x+2)/(x^2*log(log(2))^2+(-2*exp(1)*x-2*x^2-2*x)*log(log(2))-exp(x+25 )+exp(1)^2+(2+2*x)*exp(1)+x^2+2*x+1),x, algorithm="giac")
Output:
log(-x^2*log(log(2))^2 + 2*x^2*log(log(2)) + 2*x*e*log(log(2)) - x^2 - 2*x *e + 2*x*log(log(2)) - 2*x - e^2 - 2*e + e^(x + 25) - 1)
Time = 3.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.21 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\ln \left (2\,x-{\mathrm {e}}^{x+25}+2\,\mathrm {e}+{\mathrm {e}}^2-2\,x^2\,\ln \left (\ln \left (2\right )\right )+2\,x\,\mathrm {e}+x^2\,{\ln \left (\ln \left (2\right )\right )}^2+x^2-2\,x\,\ln \left (\ln \left (2\right )\right )\,\left (\mathrm {e}+1\right )+1\right ) \] Input:
int((2*x - exp(x + 25) + 2*exp(1) + 2*x*log(log(2))^2 - log(log(2))*(4*x + 2*exp(1) + 2) + 2)/(2*x - exp(x + 25) + exp(2) + x^2*log(log(2))^2 + x^2 - log(log(2))*(2*x + 2*x*exp(1) + 2*x^2) + exp(1)*(2*x + 2) + 1),x)
Output:
log(2*x - exp(x + 25) + 2*exp(1) + exp(2) - 2*x^2*log(log(2)) + 2*x*exp(1) + x^2*log(log(2))^2 + x^2 - 2*x*log(log(2))*(exp(1) + 1) + 1)
Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.54 \[ \int \frac {2+2 e-e^{25+x}+2 x+(-2-2 e-4 x) \log (\log (2))+2 x \log ^2(\log (2))}{1+e^2-e^{25+x}+2 x+x^2+e (2+2 x)+\left (-2 x-2 e x-2 x^2\right ) \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\mathrm {log}\left (e^{x} e^{25}-\mathrm {log}\left (\mathrm {log}\left (2\right )\right )^{2} x^{2}+2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) e x +2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x^{2}+2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x -e^{2}-2 e x -2 e -x^{2}-2 x -1\right ) \] Input:
int((2*x*log(log(2))^2+(-2*exp(1)-4*x-2)*log(log(2))-exp(x+25)+2*exp(1)+2* x+2)/(x^2*log(log(2))^2+(-2*exp(1)*x-2*x^2-2*x)*log(log(2))-exp(x+25)+exp( 1)^2+(2+2*x)*exp(1)+x^2+2*x+1),x)
Output:
log(e**x*e**25 - log(log(2))**2*x**2 + 2*log(log(2))*e*x + 2*log(log(2))*x **2 + 2*log(log(2))*x - e**2 - 2*e*x - 2*e - x**2 - 2*x - 1)