Integrand size = 115, antiderivative size = 31 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\log \left (-e^{e^2}+e^x-e^{\left (x+x \log ^2\left (\frac {4}{\log (3)}\right )\right )^2}\right ) \] Output:
ln(exp(x)-exp(exp(1)^2)-exp((x+x*ln(4/ln(3))^2)^2))
Time = 2.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\log \left (e^{e^2}-e^x+e^{x^2 \left (1+\log ^2\left (\frac {4}{\log (3)}\right )\right )^2}\right ) \] Input:
Integrate[(-E^x + E^(x^2 + 2*x^2*Log[4/Log[3]]^2 + x^2*Log[4/Log[3]]^4)*(2 *x + 4*x*Log[4/Log[3]]^2 + 2*x*Log[4/Log[3]]^4))/(E^E^2 - E^x + E^(x^2 + 2 *x^2*Log[4/Log[3]]^2 + x^2*Log[4/Log[3]]^4)),x]
Output:
Log[E^E^2 - E^x + E^(x^2*(1 + Log[4/Log[3]]^2)^2)]
Time = 0.45 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {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.
\(\displaystyle \int \frac {\left (2 x+2 x \log ^4\left (\frac {4}{\log (3)}\right )+4 x \log ^2\left (\frac {4}{\log (3)}\right )\right ) \exp \left (x^2+x^2 \log ^4\left (\frac {4}{\log (3)}\right )+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )\right )-e^x}{\exp \left (x^2+x^2 \log ^4\left (\frac {4}{\log (3)}\right )+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )\right )-e^x+e^{e^2}} \, dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (\exp \left (x^2+x^2 \log ^4\left (\frac {4}{\log (3)}\right )+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )\right )-e^x+e^{e^2}\right )\) |
Input:
Int[(-E^x + E^(x^2 + 2*x^2*Log[4/Log[3]]^2 + x^2*Log[4/Log[3]]^4)*(2*x + 4 *x*Log[4/Log[3]]^2 + 2*x*Log[4/Log[3]]^4))/(E^E^2 - E^x + E^(x^2 + 2*x^2*L og[4/Log[3]]^2 + x^2*Log[4/Log[3]]^4)),x]
Output:
Log[E^E^2 - E^x + E^(x^2 + 2*x^2*Log[4/Log[3]]^2 + x^2*Log[4/Log[3]]^4)]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 1.65 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19
method | result | size |
risch | \(\ln \left ({\mathrm e}^{x^{2} \left (4 \ln \left (2\right )^{2}-4 \ln \left (2\right ) \ln \left (\ln \left (3\right )\right )+\ln \left (\ln \left (3\right )\right )^{2}+1\right )^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\) | \(37\) |
parallelrisch | \(\ln \left (-{\mathrm e}^{x^{2} \left (\ln \left (\frac {4}{\ln \left (3\right )}\right )^{4}+2 \ln \left (\frac {4}{\ln \left (3\right )}\right )^{2}+1\right )}-{\mathrm e}^{{\mathrm e}^{2}}+{\mathrm e}^{x}\right )\) | \(41\) |
derivativedivides | \(\ln \left ({\mathrm e}^{x^{2} \ln \left (\frac {4}{\ln \left (3\right )}\right )^{4}+2 x^{2} \ln \left (\frac {4}{\ln \left (3\right )}\right )^{2}+x^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\) | \(44\) |
default | \(\ln \left ({\mathrm e}^{x^{2} \ln \left (\frac {4}{\ln \left (3\right )}\right )^{4}+2 x^{2} \ln \left (\frac {4}{\ln \left (3\right )}\right )^{2}+x^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\) | \(44\) |
norman | \(\ln \left ({\mathrm e}^{x^{2} \ln \left (\frac {4}{\ln \left (3\right )}\right )^{4}+2 x^{2} \ln \left (\frac {4}{\ln \left (3\right )}\right )^{2}+x^{2}}+{\mathrm e}^{{\mathrm e}^{2}}-{\mathrm e}^{x}\right )\) | \(44\) |
Input:
int(((2*x*ln(4/ln(3))^4+4*x*ln(4/ln(3))^2+2*x)*exp(x^2*ln(4/ln(3))^4+2*x^2 *ln(4/ln(3))^2+x^2)-exp(x))/(exp(x^2*ln(4/ln(3))^4+2*x^2*ln(4/ln(3))^2+x^2 )+exp(exp(1)^2)-exp(x)),x,method=_RETURNVERBOSE)
Output:
ln(exp(x^2*(4*ln(2)^2-4*ln(2)*ln(ln(3))+ln(ln(3))^2+1)^2)+exp(exp(2))-exp( x))
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\log \left (e^{\left (x^{2} \log \left (\frac {4}{\log \left (3\right )}\right )^{4} + 2 \, x^{2} \log \left (\frac {4}{\log \left (3\right )}\right )^{2} + x^{2}\right )} - e^{x} + e^{\left (e^{2}\right )}\right ) \] Input:
integrate(((2*x*log(4/log(3))^4+4*x*log(4/log(3))^2+2*x)*exp(x^2*log(4/log (3))^4+2*x^2*log(4/log(3))^2+x^2)-exp(x))/(exp(x^2*log(4/log(3))^4+2*x^2*l og(4/log(3))^2+x^2)+exp(exp(1)^2)-exp(x)),x, algorithm="fricas")
Output:
log(e^(x^2*log(4/log(3))^4 + 2*x^2*log(4/log(3))^2 + x^2) - e^x + e^(e^2))
Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\log {\left (- e^{x} + e^{x^{2} + x^{2} \log {\left (\frac {4}{\log {\left (3 \right )}} \right )}^{4} + 2 x^{2} \log {\left (\frac {4}{\log {\left (3 \right )}} \right )}^{2}} + e^{e^{2}} \right )} \] Input:
integrate(((2*x*ln(4/ln(3))**4+4*x*ln(4/ln(3))**2+2*x)*exp(x**2*ln(4/ln(3) )**4+2*x**2*ln(4/ln(3))**2+x**2)-exp(x))/(exp(x**2*ln(4/ln(3))**4+2*x**2*l n(4/ln(3))**2+x**2)+exp(exp(1)**2)-exp(x)),x)
Output:
log(-exp(x) + exp(x**2 + x**2*log(4/log(3))**4 + 2*x**2*log(4/log(3))**2) + exp(exp(2)))
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (27) = 54\).
Time = 0.21 (sec) , antiderivative size = 205, normalized size of antiderivative = 6.61 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx={\left (16 \, \log \left (2\right )^{4} - 32 \, \log \left (2\right )^{3} \log \left (\log \left (3\right )\right ) + \log \left (\log \left (3\right )\right )^{4} - 8 \, {\left (\log \left (\log \left (3\right )\right )^{3} + \log \left (\log \left (3\right )\right )\right )} \log \left (2\right ) + 8 \, \log \left (2\right )^{2} + 2 \, \log \left (\log \left (3\right )\right )^{2} + 1\right )} x^{2} + \log \left (-{\left ({\left (e^{x} - e^{\left (e^{2}\right )}\right )} e^{\left (32 \, x^{2} \log \left (2\right )^{3} \log \left (\log \left (3\right )\right ) + 8 \, x^{2} \log \left (2\right ) \log \left (\log \left (3\right )\right )^{3} + 8 \, x^{2} \log \left (2\right ) \log \left (\log \left (3\right )\right )\right )} - e^{\left (16 \, x^{2} \log \left (2\right )^{4} + 24 \, x^{2} \log \left (2\right )^{2} \log \left (\log \left (3\right )\right )^{2} + x^{2} \log \left (\log \left (3\right )\right )^{4} + 8 \, x^{2} \log \left (2\right )^{2} + 2 \, x^{2} \log \left (\log \left (3\right )\right )^{2} + x^{2}\right )}\right )} e^{\left (-16 \, x^{2} \log \left (2\right )^{4} - x^{2} \log \left (\log \left (3\right )\right )^{4} - 8 \, x^{2} \log \left (2\right )^{2} - 2 \, x^{2} \log \left (\log \left (3\right )\right )^{2} - x^{2}\right )}\right ) \] Input:
integrate(((2*x*log(4/log(3))^4+4*x*log(4/log(3))^2+2*x)*exp(x^2*log(4/log (3))^4+2*x^2*log(4/log(3))^2+x^2)-exp(x))/(exp(x^2*log(4/log(3))^4+2*x^2*l og(4/log(3))^2+x^2)+exp(exp(1)^2)-exp(x)),x, algorithm="maxima")
Output:
(16*log(2)^4 - 32*log(2)^3*log(log(3)) + log(log(3))^4 - 8*(log(log(3))^3 + log(log(3)))*log(2) + 8*log(2)^2 + 2*log(log(3))^2 + 1)*x^2 + log(-((e^x - e^(e^2))*e^(32*x^2*log(2)^3*log(log(3)) + 8*x^2*log(2)*log(log(3))^3 + 8*x^2*log(2)*log(log(3))) - e^(16*x^2*log(2)^4 + 24*x^2*log(2)^2*log(log(3 ))^2 + x^2*log(log(3))^4 + 8*x^2*log(2)^2 + 2*x^2*log(log(3))^2 + x^2))*e^ (-16*x^2*log(2)^4 - x^2*log(log(3))^4 - 8*x^2*log(2)^2 - 2*x^2*log(log(3)) ^2 - x^2))
\[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\int { \frac {2 \, {\left (x \log \left (\frac {4}{\log \left (3\right )}\right )^{4} + 2 \, x \log \left (\frac {4}{\log \left (3\right )}\right )^{2} + x\right )} e^{\left (x^{2} \log \left (\frac {4}{\log \left (3\right )}\right )^{4} + 2 \, x^{2} \log \left (\frac {4}{\log \left (3\right )}\right )^{2} + x^{2}\right )} - e^{x}}{e^{\left (x^{2} \log \left (\frac {4}{\log \left (3\right )}\right )^{4} + 2 \, x^{2} \log \left (\frac {4}{\log \left (3\right )}\right )^{2} + x^{2}\right )} - e^{x} + e^{\left (e^{2}\right )}} \,d x } \] Input:
integrate(((2*x*log(4/log(3))^4+4*x*log(4/log(3))^2+2*x)*exp(x^2*log(4/log (3))^4+2*x^2*log(4/log(3))^2+x^2)-exp(x))/(exp(x^2*log(4/log(3))^4+2*x^2*l og(4/log(3))^2+x^2)+exp(exp(1)^2)-exp(x)),x, algorithm="giac")
Output:
undef
Time = 2.95 (sec) , antiderivative size = 110, normalized size of antiderivative = 3.55 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\ln \left ({\mathrm {e}}^{{\mathrm {e}}^2}-{\mathrm {e}}^x+\frac {{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{8\,x^2\,{\ln \left (2\right )}^2}\,{\mathrm {e}}^{16\,x^2\,{\ln \left (2\right )}^4}\,{\mathrm {e}}^{24\,x^2\,{\ln \left (2\right )}^2\,{\ln \left (\ln \left (3\right )\right )}^2}\,{\mathrm {e}}^{x^2\,{\ln \left (\ln \left (3\right )\right )}^4}\,{\mathrm {e}}^{2\,x^2\,{\ln \left (\ln \left (3\right )\right )}^2}}{2^{8\,x^2\,\ln \left (\ln \left (3\right )\right )}\,2^{8\,x^2\,{\ln \left (\ln \left (3\right )\right )}^3}\,{\ln \left (3\right )}^{32\,x^2\,{\ln \left (2\right )}^3}}\right ) \] Input:
int(-(exp(x) - exp(2*x^2*log(4/log(3))^2 + x^2*log(4/log(3))^4 + x^2)*(2*x + 4*x*log(4/log(3))^2 + 2*x*log(4/log(3))^4))/(exp(exp(2)) - exp(x) + exp (2*x^2*log(4/log(3))^2 + x^2*log(4/log(3))^4 + x^2)),x)
Output:
log(exp(exp(2)) - exp(x) + (exp(x^2)*exp(8*x^2*log(2)^2)*exp(16*x^2*log(2) ^4)*exp(24*x^2*log(2)^2*log(log(3))^2)*exp(x^2*log(log(3))^4)*exp(2*x^2*lo g(log(3))^2))/(2^(8*x^2*log(log(3)))*2^(8*x^2*log(log(3))^3)*log(3)^(32*x^ 2*log(2)^3)))
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )} \left (2 x+4 x \log ^2\left (\frac {4}{\log (3)}\right )+2 x \log ^4\left (\frac {4}{\log (3)}\right )\right )}{e^{e^2}-e^x+e^{x^2+2 x^2 \log ^2\left (\frac {4}{\log (3)}\right )+x^2 \log ^4\left (\frac {4}{\log (3)}\right )}} \, dx=\mathrm {log}\left (e^{e^{2}}+e^{\mathrm {log}\left (\frac {4}{\mathrm {log}\left (3\right )}\right )^{4} x^{2}+2 \mathrm {log}\left (\frac {4}{\mathrm {log}\left (3\right )}\right )^{2} x^{2}+x^{2}}-e^{x}\right ) \] Input:
int(((2*x*log(4/log(3))^4+4*x*log(4/log(3))^2+2*x)*exp(x^2*log(4/log(3))^4 +2*x^2*log(4/log(3))^2+x^2)-exp(x))/(exp(x^2*log(4/log(3))^4+2*x^2*log(4/l og(3))^2+x^2)+exp(exp(1)^2)-exp(x)),x)
Output:
log(e**(e**2) + e**(log(4/log(3))**4*x**2 + 2*log(4/log(3))**2*x**2 + x**2 ) - e**x)