\(\int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} (10 e^5 x-5 x^2-50 x^4+(e^5-20 x^3) \log (\log (4))-2 x^2 \log ^2(\log (4)))}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx\) [4081]

Optimal result

Integrand size = 129, antiderivative size = 34 \[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=1+e^{-x+\frac {-e^5+x}{2 x (5 x+\log (\log (4)))}}-x \]

[Out]

exp(1/2*(-exp(5)+x)/(ln(2*ln(2))+5*x)/x-x)+1-x

Rubi [F]

\[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=\int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx \]

[In]

Int[(-50*x^4 - 20*x^3*Log[Log[4]] - 2*x^2*Log[Log[4]]^2 + E^((-E^5 + x - 10*x^3 - 2*x^2*Log[Log[4]])/(10*x^2 +
2*x*Log[Log[4]]))*(10*E^5*x - 5*x^2 - 50*x^4 + (E^5 - 20*x^3)*Log[Log[4]] - 2*x^2*Log[Log[4]]^2))/(50*x^4 + 2
0*x^3*Log[Log[4]] + 2*x^2*Log[Log[4]]^2),x]

[Out]

-x - Defer[Int][E^((-E^5 + x - 10*x^3)/(2*x*(5*x + Log[Log[4]])))/Log[4]^(x/(5*x + Log[Log[4]])), x] + Defer[I
nt][E^(5 + (-E^5 + x - 10*x^3)/(2*x*(5*x + Log[Log[4]])))/(x^2*Log[4]^(x/(5*x + Log[Log[4]]))), x]/(2*Log[Log[
4]]) - (5*(1 + (5*E^5)/Log[Log[4]])*Defer[Int][E^((-E^5 + x - 10*x^3)/(2*x*(5*x + Log[Log[4]])))/(Log[4]^(x/(5
*x + Log[Log[4]]))*(5*x + Log[Log[4]])^2), x])/2

Rubi steps \begin{align*} \text {integral}& = \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{x^2 \left (50 x^2+20 x \log (\log (4))+2 \log ^2(\log (4))\right )} \, dx \\ & = \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{2 x^2 (5 x+\log (\log (4)))^2} \, dx \\ & = \frac {1}{2} \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+\exp \left (\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}\right ) \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{x^2 (5 x+\log (\log (4)))^2} \, dx \\ & = \frac {1}{2} \int \left (-2+\frac {e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \left (10 e^5 x-50 x^4+e^5 \log (\log (4))-20 x^3 \log (\log (4))-x^2 \left (5+2 \log ^2(\log (4))\right )\right )}{x^2 (5 x+\log (\log (4)))^2}\right ) \, dx \\ & = -x+\frac {1}{2} \int \frac {e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \left (10 e^5 x-50 x^4+e^5 \log (\log (4))-20 x^3 \log (\log (4))-x^2 \left (5+2 \log ^2(\log (4))\right )\right )}{x^2 (5 x+\log (\log (4)))^2} \, dx \\ & = -x+\frac {1}{2} \int \left (-2 e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)+\frac {\exp \left (5+\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}\right ) \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)}{x^2 \log (\log (4))}-\frac {5 e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \left (5 e^5+\log (\log (4))\right )}{\log (\log (4)) (5 x+\log (\log (4)))^2}\right ) \, dx \\ & = -x-\frac {1}{2} \left (5 \left (1+\frac {5 e^5}{\log (\log (4))}\right )\right ) \int \frac {e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)}{(5 x+\log (\log (4)))^2} \, dx+\frac {\int \frac {\exp \left (5+\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}\right ) \log ^{-\frac {x}{5 x+\log (\log (4))}}(4)}{x^2} \, dx}{2 \log (\log (4))}-\int e^{\frac {-e^5+x-10 x^3}{2 x (5 x+\log (\log (4)))}} \log ^{-\frac {x}{5 x+\log (\log (4))}}(4) \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.38 \[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=\frac {1}{2} \left (2 e^{-\frac {e^5+x \left (-1+10 x^2+2 x \log (\log (4))\right )}{2 x (5 x+\log (\log (4)))}}-2 x\right ) \]

[In]

Integrate[(-50*x^4 - 20*x^3*Log[Log[4]] - 2*x^2*Log[Log[4]]^2 + E^((-E^5 + x - 10*x^3 - 2*x^2*Log[Log[4]])/(10
*x^2 + 2*x*Log[Log[4]]))*(10*E^5*x - 5*x^2 - 50*x^4 + (E^5 - 20*x^3)*Log[Log[4]] - 2*x^2*Log[Log[4]]^2))/(50*x
^4 + 20*x^3*Log[Log[4]] + 2*x^2*Log[Log[4]]^2),x]

[Out]

(2/E^((E^5 + x*(-1 + 10*x^2 + 2*x*Log[Log[4]]))/(2*x*(5*x + Log[Log[4]]))) - 2*x)/2

Maple [A] (verified)

Time = 4.00 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41

method	result	size
risch	\(-x +{\mathrm e}^{-\frac {2 x^{2} \ln \left (2\right )+2 x^{2} \ln \left (\ln \left (2\right )\right )+10 x^{3}+{\mathrm e}^{5}-x}{2 x \left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+5 x \right )}}\)	\(48\)
parallelrisch	\(\frac {4 \ln \left (2 \ln \left (2\right )\right )}{5}-x +{\mathrm e}^{-\frac {2 x^{2} \ln \left (2 \ln \left (2\right )\right )+10 x^{3}+{\mathrm e}^{5}-x}{2 x \left (\ln \left (2 \ln \left (2\right )\right )+5 x \right )}}\)	\(50\)
parts	\(-x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x \,{\mathrm e}^{\frac {-2 x^{2} \ln \left (2 \ln \left (2\right )\right )-{\mathrm e}^{5}-10 x^{3}+x}{2 x \ln \left (2 \ln \left (2\right )\right )+10 x^{2}}}+5 x^{2} {\mathrm e}^{\frac {-2 x^{2} \ln \left (2 \ln \left (2\right )\right )-{\mathrm e}^{5}-10 x^{3}+x}{2 x \ln \left (2 \ln \left (2\right )\right )+10 x^{2}}}}{x \left (\ln \left (2 \ln \left (2\right )\right )+5 x \right )}\)	\(112\)
norman	\(\frac {\left (\frac {\ln \left (2\right )^{2}}{5}+\frac {2 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )}{5}+\frac {\ln \left (\ln \left (2\right )\right )^{2}}{5}\right ) x +\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x \,{\mathrm e}^{\frac {-2 x^{2} \ln \left (2 \ln \left (2\right )\right )-{\mathrm e}^{5}-10 x^{3}+x}{2 x \ln \left (2 \ln \left (2\right )\right )+10 x^{2}}}-5 x^{3}+5 x^{2} {\mathrm e}^{\frac {-2 x^{2} \ln \left (2 \ln \left (2\right )\right )-{\mathrm e}^{5}-10 x^{3}+x}{2 x \ln \left (2 \ln \left (2\right )\right )+10 x^{2}}}}{x \left (\ln \left (2 \ln \left (2\right )\right )+5 x \right )}\)	\(136\)

[In]

int(((-2*x^2*ln(2*ln(2))^2+(exp(5)-20*x^3)*ln(2*ln(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*ln(2*ln(2))-exp(5
)-10*x^3+x)/(2*x*ln(2*ln(2))+10*x^2))-2*x^2*ln(2*ln(2))^2-20*x^3*ln(2*ln(2))-50*x^4)/(2*x^2*ln(2*ln(2))^2+20*x
^3*ln(2*ln(2))+50*x^4),x,method=_RETURNVERBOSE)

[Out]

-x+exp(-1/2*(2*x^2*ln(2)+2*x^2*ln(ln(2))+10*x^3+exp(5)-x)/x/(ln(2)+ln(ln(2))+5*x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=-x + e^{\left (-\frac {10 \, x^{3} + 2 \, x^{2} \log \left (2 \, \log \left (2\right )\right ) - x + e^{5}}{2 \, {\left (5 \, x^{2} + x \log \left (2 \, \log \left (2\right )\right )\right )}}\right )} \]

[In]

integrate(((-2*x^2*log(2*log(2))^2+(exp(5)-20*x^3)*log(2*log(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*log(2*l
og(2))-exp(5)-10*x^3+x)/(2*x*log(2*log(2))+10*x^2))-2*x^2*log(2*log(2))^2-20*x^3*log(2*log(2))-50*x^4)/(2*x^2*
log(2*log(2))^2+20*x^3*log(2*log(2))+50*x^4),x, algorithm="fricas")

[Out]

-x + e^(-1/2*(10*x^3 + 2*x^2*log(2*log(2)) - x + e^5)/(5*x^2 + x*log(2*log(2))))

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=- x + e^{\frac {- 10 x^{3} - 2 x^{2} \log {\left (2 \log {\left (2 \right )} \right )} + x - e^{5}}{10 x^{2} + 2 x \log {\left (2 \log {\left (2 \right )} \right )}}} \]

[In]

integrate(((-2*x**2*ln(2*ln(2))**2+(exp(5)-20*x**3)*ln(2*ln(2))+10*x*exp(5)-50*x**4-5*x**2)*exp((-2*x**2*ln(2*
ln(2))-exp(5)-10*x**3+x)/(2*x*ln(2*ln(2))+10*x**2))-2*x**2*ln(2*ln(2))**2-20*x**3*ln(2*ln(2))-50*x**4)/(2*x**2
*ln(2*ln(2))**2+20*x**3*ln(2*ln(2))+50*x**4),x)

[Out]

-x + exp((-10*x**3 - 2*x**2*log(2*log(2)) + x - exp(5))/(10*x**2 + 2*x*log(2*log(2))))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (32) = 64\).

Time = 0.47 (sec) , antiderivative size = 141, normalized size of antiderivative = 4.15 \[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=-\frac {2}{5} \, {\left (\frac {\log \left (2 \, \log \left (2\right )\right )}{5 \, x + \log \left (2 \, \log \left (2\right )\right )} + \log \left (5 \, x + \log \left (2 \, \log \left (2\right )\right )\right )\right )} \log \left (2 \, \log \left (2\right )\right ) + \frac {2}{5} \, \log \left (5 \, x + \log \left (2 \, \log \left (2\right )\right )\right ) \log \left (2 \, \log \left (2\right )\right ) - x + \frac {2 \, \log \left (2 \, \log \left (2\right )\right )^{2}}{5 \, {\left (5 \, x + \log \left (2 \, \log \left (2\right )\right )\right )}} + e^{\left (-x + \frac {5 \, e^{5}}{2 \, {\left (5 \, x {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} + \log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )}} + \frac {1}{2 \, {\left (5 \, x + \log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )}} - \frac {e^{5}}{2 \, x {\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )}}\right )} \]

[In]

integrate(((-2*x^2*log(2*log(2))^2+(exp(5)-20*x^3)*log(2*log(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*log(2*l
og(2))-exp(5)-10*x^3+x)/(2*x*log(2*log(2))+10*x^2))-2*x^2*log(2*log(2))^2-20*x^3*log(2*log(2))-50*x^4)/(2*x^2*
log(2*log(2))^2+20*x^3*log(2*log(2))+50*x^4),x, algorithm="maxima")

[Out]

-2/5*(log(2*log(2))/(5*x + log(2*log(2))) + log(5*x + log(2*log(2))))*log(2*log(2)) + 2/5*log(5*x + log(2*log(
2)))*log(2*log(2)) - x + 2/5*log(2*log(2))^2/(5*x + log(2*log(2))) + e^(-x + 5/2*e^5/(5*x*(log(2) + log(log(2)
)) + log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2) + 1/2/(5*x + log(2) + log(log(2))) - 1/2*e^5/(x*(log(2)
+ log(log(2)))))

Giac [F]

\[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=\int { -\frac {50 \, x^{4} + 20 \, x^{3} \log \left (2 \, \log \left (2\right )\right ) + 2 \, x^{2} \log \left (2 \, \log \left (2\right )\right )^{2} + {\left (50 \, x^{4} + 2 \, x^{2} \log \left (2 \, \log \left (2\right )\right )^{2} + 5 \, x^{2} - 10 \, x e^{5} + {\left (20 \, x^{3} - e^{5}\right )} \log \left (2 \, \log \left (2\right )\right )\right )} e^{\left (-\frac {10 \, x^{3} + 2 \, x^{2} \log \left (2 \, \log \left (2\right )\right ) - x + e^{5}}{2 \, {\left (5 \, x^{2} + x \log \left (2 \, \log \left (2\right )\right )\right )}}\right )}}{2 \, {\left (25 \, x^{4} + 10 \, x^{3} \log \left (2 \, \log \left (2\right )\right ) + x^{2} \log \left (2 \, \log \left (2\right )\right )^{2}\right )}} \,d x } \]

[In]

integrate(((-2*x^2*log(2*log(2))^2+(exp(5)-20*x^3)*log(2*log(2))+10*x*exp(5)-50*x^4-5*x^2)*exp((-2*x^2*log(2*l
og(2))-exp(5)-10*x^3+x)/(2*x*log(2*log(2))+10*x^2))-2*x^2*log(2*log(2))^2-20*x^3*log(2*log(2))-50*x^4)/(2*x^2*
log(2*log(2))^2+20*x^3*log(2*log(2))+50*x^4),x, algorithm="giac")

[Out]

undef

Mupad [F(-1)]

Timed out. \[ \int \frac {-50 x^4-20 x^3 \log (\log (4))-2 x^2 \log ^2(\log (4))+e^{\frac {-e^5+x-10 x^3-2 x^2 \log (\log (4))}{10 x^2+2 x \log (\log (4))}} \left (10 e^5 x-5 x^2-50 x^4+\left (e^5-20 x^3\right ) \log (\log (4))-2 x^2 \log ^2(\log (4))\right )}{50 x^4+20 x^3 \log (\log (4))+2 x^2 \log ^2(\log (4))} \, dx=\int -\frac {20\,x^3\,\ln \left (2\,\ln \left (2\right )\right )+2\,x^2\,{\ln \left (2\,\ln \left (2\right )\right )}^2+50\,x^4+{\mathrm {e}}^{-\frac {10\,x^3+2\,\ln \left (2\,\ln \left (2\right )\right )\,x^2-x+{\mathrm {e}}^5}{10\,x^2+2\,\ln \left (2\,\ln \left (2\right )\right )\,x}}\,\left (2\,x^2\,{\ln \left (2\,\ln \left (2\right )\right )}^2-\ln \left (2\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^5-20\,x^3\right )-10\,x\,{\mathrm {e}}^5+5\,x^2+50\,x^4\right )}{50\,x^4+20\,\ln \left (2\,\ln \left (2\right )\right )\,x^3+2\,{\ln \left (2\,\ln \left (2\right )\right )}^2\,x^2} \,d x \]

[In]

int(-(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4 + exp(-(exp(5) - x + 2*x^2*log(2*log(2)) + 10*x^3)
/(2*x*log(2*log(2)) + 10*x^2))*(2*x^2*log(2*log(2))^2 - log(2*log(2))*(exp(5) - 20*x^3) - 10*x*exp(5) + 5*x^2
+ 50*x^4))/(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4),x)

[Out]

int(-(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4 + exp(-(exp(5) - x + 2*x^2*log(2*log(2)) + 10*x^3)
/(2*x*log(2*log(2)) + 10*x^2))*(2*x^2*log(2*log(2))^2 - log(2*log(2))*(exp(5) - 20*x^3) - 10*x*exp(5) + 5*x^2
+ 50*x^4))/(20*x^3*log(2*log(2)) + 2*x^2*log(2*log(2))^2 + 50*x^4), x)