\(\int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+(e^{10} (5+15 x)+5 e^{10} x \log (\log (3))) \log (\frac {1+3 x+x \log (\log (3))}{x})+(e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))) \log ^2(\frac {1+3 x+x \log (\log (3))}{x})}{x^2+3 x^3+x^3 \log (\log (3))+(-8 x^2-24 x^3-8 x^3 \log (\log (3))) \log (\frac {1+3 x+x \log (\log (3))}{x})+(16 x^2+48 x^3+16 x^3 \log (\log (3))) \log ^2(\frac {1+3 x+x \log (\log (3))}{x})} \, dx\) [6761]

Optimal result

Integrand size = 182, antiderivative size = 28 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10}}{x \left (4-\frac {3}{1-\log \left (3+\frac {1}{x}+\log (\log (3))\right )}\right )} \]

[Out]

exp(5)^2/x/(4-3/(1-ln(3+1/x+ln(ln(3)))))

Rubi [F]

\[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx \]

[In]

Int[(E^10*(-4 - 3*x) - E^10*x*Log[Log[3]] + (E^10*(5 + 15*x) + 5*E^10*x*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[
3]])/x] + (E^10*(-4 - 12*x) - 4*E^10*x*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x]^2)/(x^2 + 3*x^3 + x^3*Log
[Log[3]] + (-8*x^2 - 24*x^3 - 8*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x] + (16*x^2 + 48*x^3 + 16*x^3*
Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x]^2),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+x^3 (3+\log (\log (3)))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx \\ & = \int \frac {e^{10} \left (-4-x (3+\log (\log (3)))+5 (1+x (3+\log (\log (3)))) \log \left (3+\frac {1}{x}+\log (\log (3))\right )-4 (1+x (3+\log (\log (3)))) \log ^2\left (3+\frac {1}{x}+\log (\log (3))\right )\right )}{(1+x (3+\log (\log (3)))) \left (x-4 x \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = e^{10} \int \frac {-4-x (3+\log (\log (3)))+5 (1+x (3+\log (\log (3)))) \log \left (3+\frac {1}{x}+\log (\log (3))\right )-4 (1+x (3+\log (\log (3)))) \log ^2\left (3+\frac {1}{x}+\log (\log (3))\right )}{(1+x (3+\log (\log (3)))) \left (x-4 x \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = e^{10} \int \left (-\frac {1}{4 x^2}+\frac {3}{x^2 (-1-x (3+\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2}+\frac {3}{4 x^2 \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )}\right ) \, dx \\ & = \frac {e^{10}}{4 x}+\frac {1}{4} \left (3 e^{10}\right ) \int \frac {1}{x^2 \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )} \, dx+\left (3 e^{10}\right ) \int \frac {1}{x^2 (-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = \frac {e^{10}}{4 x}-\frac {1}{4} \left (3 e^{10}\right ) \text {Subst}\left (\int \frac {1}{-1+4 \log (3+x+\log (\log (3)))} \, dx,x,\frac {1}{x}\right )+\left (3 e^{10}\right ) \int \left (-\frac {1}{x^2 \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2}+\frac {3+\log (\log (3))}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2}+\frac {(3+\log (\log (3)))^2}{(-1-x (3+\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2}\right ) \, dx \\ & = \frac {e^{10}}{4 x}-\frac {1}{4} \left (3 e^{10}\right ) \text {Subst}\left (\int \frac {1}{-1+4 \log (x)} \, dx,x,3+\frac {1}{x}+\log (\log (3))\right )-\left (3 e^{10}\right ) \int \frac {1}{x^2 \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))\right ) \int \frac {1}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))^2\right ) \int \frac {1}{(-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = \frac {e^{10}}{4 x}-\frac {1}{4} \left (3 e^{10}\right ) \text {Subst}\left (\int \frac {e^x}{-1+4 x} \, dx,x,\log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )+\left (3 e^{10}\right ) \text {Subst}\left (\int \frac {1}{(1-4 \log (3+x+\log (\log (3))))^2} \, dx,x,\frac {1}{x}\right )+\left (3 e^{10} (3+\log (\log (3)))\right ) \int \frac {1}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))^2\right ) \int \frac {1}{(-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = \frac {e^{10}}{4 x}-\frac {3}{16} e^{41/4} \text {Ei}\left (\frac {1}{4} \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )\right )+\left (3 e^{10}\right ) \text {Subst}\left (\int \frac {1}{(1-4 \log (x))^2} \, dx,x,3+\frac {1}{x}+\log (\log (3))\right )+\left (3 e^{10} (3+\log (\log (3)))\right ) \int \frac {1}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))^2\right ) \int \frac {1}{(-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = \frac {e^{10}}{4 x}-\frac {3}{16} e^{41/4} \text {Ei}\left (\frac {1}{4} \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )\right )+\frac {3 e^{10} \left (3+\frac {1}{x}+\log (\log (3))\right )}{4 \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )}-\frac {1}{4} \left (3 e^{10}\right ) \text {Subst}\left (\int \frac {1}{1-4 \log (x)} \, dx,x,3+\frac {1}{x}+\log (\log (3))\right )+\left (3 e^{10} (3+\log (\log (3)))\right ) \int \frac {1}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))^2\right ) \int \frac {1}{(-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = \frac {e^{10}}{4 x}-\frac {3}{16} e^{41/4} \text {Ei}\left (\frac {1}{4} \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )\right )+\frac {3 e^{10} \left (3+\frac {1}{x}+\log (\log (3))\right )}{4 \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )}-\frac {1}{4} \left (3 e^{10}\right ) \text {Subst}\left (\int \frac {e^x}{1-4 x} \, dx,x,\log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )+\left (3 e^{10} (3+\log (\log (3)))\right ) \int \frac {1}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))^2\right ) \int \frac {1}{(-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ & = \frac {e^{10}}{4 x}+\frac {3 e^{10} \left (3+\frac {1}{x}+\log (\log (3))\right )}{4 \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )}+\left (3 e^{10} (3+\log (\log (3)))\right ) \int \frac {1}{x \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx+\left (3 e^{10} (3+\log (\log (3)))^2\right ) \int \frac {1}{(-1+x (-3-\log (\log (3)))) \left (1-4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=-\frac {e^{10} \left (1-\log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )}{x \left (-1+4 \log \left (3+\frac {1}{x}+\log (\log (3))\right )\right )} \]

[In]

Integrate[(E^10*(-4 - 3*x) - E^10*x*Log[Log[3]] + (E^10*(5 + 15*x) + 5*E^10*x*Log[Log[3]])*Log[(1 + 3*x + x*Lo
g[Log[3]])/x] + (E^10*(-4 - 12*x) - 4*E^10*x*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x]^2)/(x^2 + 3*x^3 + x
^3*Log[Log[3]] + (-8*x^2 - 24*x^3 - 8*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x] + (16*x^2 + 48*x^3 + 1
6*x^3*Log[Log[3]])*Log[(1 + 3*x + x*Log[Log[3]])/x]^2),x]

[Out]

-((E^10*(1 - Log[3 + x^(-1) + Log[Log[3]]]))/(x*(-1 + 4*Log[3 + x^(-1) + Log[Log[3]]])))

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {{\mathrm e}^{10}}{4 x}-\frac {3 \,{\mathrm e}^{10}}{4 x \left (-1+4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )\right )}\)	\(37\)
norman	\(\frac {{\mathrm e}^{10} \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )-{\mathrm e}^{10}}{x \left (-1+4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )\right )}\)	\(53\)
derivativedivides	\(\frac {3 \,{\mathrm e}^{10} \ln \left (\ln \left (3\right )\right )}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}+{\mathrm e}^{10} \left (\frac {9}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}-\frac {3 \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )}{16 \left (\ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-\frac {1}{4}\right )}+\frac {\ln \left (\ln \left (3\right )\right )}{4}+\frac {3}{4}+\frac {1}{4 x}\right )\)	\(83\)
default	\(\frac {3 \,{\mathrm e}^{10} \ln \left (\ln \left (3\right )\right )}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}+{\mathrm e}^{10} \left (\frac {9}{4 \left (4 \ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-1\right )}-\frac {3 \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )}{16 \left (\ln \left (3+\frac {1}{x}+\ln \left (\ln \left (3\right )\right )\right )-\frac {1}{4}\right )}+\frac {\ln \left (\ln \left (3\right )\right )}{4}+\frac {3}{4}+\frac {1}{4 x}\right )\)	\(83\)
parallelrisch	\(-\frac {8 \ln \left (\ln \left (3\right )\right ) {\mathrm e}^{10} x \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )-2 x \,{\mathrm e}^{10} \ln \left (\ln \left (3\right )\right )+24 \,{\mathrm e}^{10} x \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )-6 x \,{\mathrm e}^{10}-4 \,{\mathrm e}^{10} \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )+4 \,{\mathrm e}^{10}}{4 x \left (-1+4 \ln \left (\frac {\ln \left (\ln \left (3\right )\right ) x +3 x +1}{x}\right )\right )}\)	\(119\)

[In]

int(((-4*x*exp(5)^2*ln(ln(3))+(-12*x-4)*exp(5)^2)*ln((ln(ln(3))*x+3*x+1)/x)^2+(5*x*exp(5)^2*ln(ln(3))+(15*x+5)
*exp(5)^2)*ln((ln(ln(3))*x+3*x+1)/x)-x*exp(5)^2*ln(ln(3))+(-3*x-4)*exp(5)^2)/((16*x^3*ln(ln(3))+48*x^3+16*x^2)
*ln((ln(ln(3))*x+3*x+1)/x)^2+(-8*x^3*ln(ln(3))-24*x^3-8*x^2)*ln((ln(ln(3))*x+3*x+1)/x)+x^3*ln(ln(3))+3*x^3+x^2
),x,method=_RETURNVERBOSE)

[Out]

1/4*exp(10)/x-3/4/x*exp(10)/(-1+4*ln((ln(ln(3))*x+3*x+1)/x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \log \left (\frac {x \log \left (\log \left (3\right )\right ) + 3 \, x + 1}{x}\right ) - e^{10}}{4 \, x \log \left (\frac {x \log \left (\log \left (3\right )\right ) + 3 \, x + 1}{x}\right ) - x} \]

[In]

integrate(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3))*x+3*x+1)/x)^2+(5*x*exp(5)^2*log(log
(3))+(15*x+5)*exp(5)^2)*log((log(log(3))*x+3*x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(lo
g(3))+48*x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x^3-8*x^2)*log((log(log(3))*x+3*x+1
)/x)+x^3*log(log(3))+3*x^3+x^2),x, algorithm="fricas")

[Out]

(e^10*log((x*log(log(3)) + 3*x + 1)/x) - e^10)/(4*x*log((x*log(log(3)) + 3*x + 1)/x) - x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=- \frac {3 e^{10}}{16 x \log {\left (\frac {x \log {\left (\log {\left (3 \right )} \right )} + 3 x + 1}{x} \right )} - 4 x} + \frac {e^{10}}{4 x} \]

[In]

integrate(((-4*x*exp(5)**2*ln(ln(3))+(-12*x-4)*exp(5)**2)*ln((ln(ln(3))*x+3*x+1)/x)**2+(5*x*exp(5)**2*ln(ln(3)
)+(15*x+5)*exp(5)**2)*ln((ln(ln(3))*x+3*x+1)/x)-x*exp(5)**2*ln(ln(3))+(-3*x-4)*exp(5)**2)/((16*x**3*ln(ln(3))+
48*x**3+16*x**2)*ln((ln(ln(3))*x+3*x+1)/x)**2+(-8*x**3*ln(ln(3))-24*x**3-8*x**2)*ln((ln(ln(3))*x+3*x+1)/x)+x**
3*ln(ln(3))+3*x**3+x**2),x)

[Out]

-3*exp(10)/(16*x*log((x*log(log(3)) + 3*x + 1)/x) - 4*x) + exp(10)/(4*x)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \log \left (x {\left (\log \left (\log \left (3\right )\right ) + 3\right )} + 1\right ) - e^{10} \log \left (x\right ) - e^{10}}{4 \, x \log \left (x {\left (\log \left (\log \left (3\right )\right ) + 3\right )} + 1\right ) - 4 \, x \log \left (x\right ) - x} \]

[In]

integrate(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3))*x+3*x+1)/x)^2+(5*x*exp(5)^2*log(log
(3))+(15*x+5)*exp(5)^2)*log((log(log(3))*x+3*x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(lo
g(3))+48*x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x^3-8*x^2)*log((log(log(3))*x+3*x+1
)/x)+x^3*log(log(3))+3*x^3+x^2),x, algorithm="maxima")

[Out]

(e^10*log(x*(log(log(3)) + 3) + 1) - e^10*log(x) - e^10)/(4*x*log(x*(log(log(3)) + 3) + 1) - 4*x*log(x) - x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).

Time = 0.35 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.82 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {e^{10} \log \left (x \log \left (\log \left (3\right )\right ) + 3 \, x + 1\right ) - e^{10} \log \left (x\right ) - e^{10}}{4 \, x \log \left (x \log \left (\log \left (3\right )\right ) + 3 \, x + 1\right ) - 4 \, x \log \left (x\right ) - x} \]

[In]

integrate(((-4*x*exp(5)^2*log(log(3))+(-12*x-4)*exp(5)^2)*log((log(log(3))*x+3*x+1)/x)^2+(5*x*exp(5)^2*log(log
(3))+(15*x+5)*exp(5)^2)*log((log(log(3))*x+3*x+1)/x)-x*exp(5)^2*log(log(3))+(-3*x-4)*exp(5)^2)/((16*x^3*log(lo
g(3))+48*x^3+16*x^2)*log((log(log(3))*x+3*x+1)/x)^2+(-8*x^3*log(log(3))-24*x^3-8*x^2)*log((log(log(3))*x+3*x+1
)/x)+x^3*log(log(3))+3*x^3+x^2),x, algorithm="giac")

[Out]

(e^10*log(x*log(log(3)) + 3*x + 1) - e^10*log(x) - e^10)/(4*x*log(x*log(log(3)) + 3*x + 1) - 4*x*log(x) - x)

Mupad [B] (verification not implemented)

Time = 12.83 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.71 \[ \int \frac {e^{10} (-4-3 x)-e^{10} x \log (\log (3))+\left (e^{10} (5+15 x)+5 e^{10} x \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (e^{10} (-4-12 x)-4 e^{10} x \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )}{x^2+3 x^3+x^3 \log (\log (3))+\left (-8 x^2-24 x^3-8 x^3 \log (\log (3))\right ) \log \left (\frac {1+3 x+x \log (\log (3))}{x}\right )+\left (16 x^2+48 x^3+16 x^3 \log (\log (3))\right ) \log ^2\left (\frac {1+3 x+x \log (\log (3))}{x}\right )} \, dx=\frac {{\mathrm {e}}^{10}\,\left (36\,x+12\,x\,\ln \left (\ln \left (3\right )\right )+16\right )}{64\,x}-\frac {\frac {{\mathrm {e}}^{10}\,\left (9\,x+3\,x\,\ln \left (\ln \left (3\right )\right )+16\right )}{16}-\frac {{\mathrm {e}}^{10}\,\left (36\,x+12\,x\,\ln \left (\ln \left (3\right )\right )+16\right )}{64}}{x\,\left (4\,\ln \left (\frac {3\,x+x\,\ln \left (\ln \left (3\right )\right )+1}{x}\right )-1\right )} \]

[In]

int(-(log((3*x + x*log(log(3)) + 1)/x)^2*(exp(10)*(12*x + 4) + 4*x*exp(10)*log(log(3))) - log((3*x + x*log(log
(3)) + 1)/x)*(exp(10)*(15*x + 5) + 5*x*exp(10)*log(log(3))) + exp(10)*(3*x + 4) + x*exp(10)*log(log(3)))/(x^3*
log(log(3)) + log((3*x + x*log(log(3)) + 1)/x)^2*(16*x^3*log(log(3)) + 16*x^2 + 48*x^3) - log((3*x + x*log(log
(3)) + 1)/x)*(8*x^3*log(log(3)) + 8*x^2 + 24*x^3) + x^2 + 3*x^3),x)

[Out]

(exp(10)*(36*x + 12*x*log(log(3)) + 16))/(64*x) - ((exp(10)*(9*x + 3*x*log(log(3)) + 16))/16 - (exp(10)*(36*x
+ 12*x*log(log(3)) + 16))/64)/(x*(4*log((3*x + x*log(log(3)) + 1)/x) - 1))