\(\int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+(-45-51 x-21 x^2-3 x^3) \log (5+4 x+x^2)+(-15 x^2-12 x^3-3 x^4+(-15 x-12 x^2-3 x^3) \log (5+4 x+x^2)+(5 x^3+4 x^4+x^5+(5 x^2+4 x^3+x^4) \log (5+4 x+x^2)) \log (x+\log (5+4 x+x^2))) \log (\frac {-3+x \log (x+\log (5+4 x+x^2))}{x}) \log (\log (\frac {-3+x \log (x+\log (5+4 x+x^2))}{x}))}{(-15 x^2-12 x^3-3 x^4+(-15 x-12 x^2-3 x^3) \log (5+4 x+x^2)+(5 x^3+4 x^4+x^5+(5 x^2+4 x^3+x^4) \log (5+4 x+x^2)) \log (x+\log (5+4 x+x^2))) \log (\frac {-3+x \log (x+\log (5+4 x+x^2))}{x}) \log ^2(\log (\frac {-3+x \log (x+\log (5+4 x+x^2))}{x}))} \, dx\) [7412]

Optimal result

Integrand size = 325, antiderivative size = 25 \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\frac {3+x}{\log \left (\log \left (-\frac {3}{x}+\log (x+\log (5+x (4+x)))\right )\right )} \]

[Out]

(3+x)/ln(ln(ln(x+ln((4+x)*x+5))-3/x))

Rubi [F]

\[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx \]

[In]

Int[(-45*x - 78*x^2 - 48*x^3 - 12*x^4 - x^5 + (-45 - 51*x - 21*x^2 - 3*x^3)*Log[5 + 4*x + x^2] + (-15*x^2 - 12
*x^3 - 3*x^4 + (-15*x - 12*x^2 - 3*x^3)*Log[5 + 4*x + x^2] + (5*x^3 + 4*x^4 + x^5 + (5*x^2 + 4*x^3 + x^4)*Log[
5 + 4*x + x^2])*Log[x + Log[5 + 4*x + x^2]])*Log[(-3 + x*Log[x + Log[5 + 4*x + x^2]])/x]*Log[Log[(-3 + x*Log[x
+ Log[5 + 4*x + x^2]])/x]])/((-15*x^2 - 12*x^3 - 3*x^4 + (-15*x - 12*x^2 - 3*x^3)*Log[5 + 4*x + x^2] + (5*x^3
+ 4*x^4 + x^5 + (5*x^2 + 4*x^3 + x^4)*Log[5 + 4*x + x^2])*Log[x + Log[5 + 4*x + x^2]])*Log[(-3 + x*Log[x + Lo
g[5 + 4*x + x^2]])/x]*Log[Log[(-3 + x*Log[x + Log[5 + 4*x + x^2]])/x]]^2),x]

[Out]

-11*Defer[Int][1/((x + Log[5 + 4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x + x^2]])*Log[-3/x + Log[x + Log[5 + 4*x
+ x^2]]]*Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]]^2), x] + (10*I)*Defer[Int][1/(((-4 + 2*I) - 2*x)*(x + L
og[5 + 4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x + x^2]])*Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]*Log[Log[-3/x +
Log[x + Log[5 + 4*x + x^2]]]]^2), x] - 8*Defer[Int][x/((x + Log[5 + 4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x +
x^2]])*Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]*Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]]^2), x] - Defer[In
t][x^2/((x + Log[5 + 4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x + x^2]])*Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]*
Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]]^2), x] + (6 + 12*I)*Defer[Int][1/(((4 - 2*I) + 2*x)*(x + Log[5 +
4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x + x^2]])*Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]*Log[Log[-3/x + Log[x
+ Log[5 + 4*x + x^2]]]]^2), x] + (6 - 2*I)*Defer[Int][1/(((4 + 2*I) + 2*x)*(x + Log[5 + 4*x + x^2])*(-3 + x*Lo
g[x + Log[5 + 4*x + x^2]])*Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]*Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]
]^2), x] - 3*Defer[Int][Log[5 + 4*x + x^2]/((x + Log[5 + 4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x + x^2]])*Log[
-3/x + Log[x + Log[5 + 4*x + x^2]]]*Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]]^2), x] - 9*Defer[Int][Log[5 +
4*x + x^2]/(x*(x + Log[5 + 4*x + x^2])*(-3 + x*Log[x + Log[5 + 4*x + x^2]])*Log[-3/x + Log[x + Log[5 + 4*x +
x^2]]]*Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]]^2), x] + Defer[Int][Log[Log[-3/x + Log[x + Log[5 + 4*x + x
^2]]]]^(-1), x]

Rubi steps Aborted

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=-\frac {-3-x}{\log \left (\log \left (-\frac {3}{x}+\log \left (x+\log \left (5+4 x+x^2\right )\right )\right )\right )} \]

[In]

Integrate[(-45*x - 78*x^2 - 48*x^3 - 12*x^4 - x^5 + (-45 - 51*x - 21*x^2 - 3*x^3)*Log[5 + 4*x + x^2] + (-15*x^
2 - 12*x^3 - 3*x^4 + (-15*x - 12*x^2 - 3*x^3)*Log[5 + 4*x + x^2] + (5*x^3 + 4*x^4 + x^5 + (5*x^2 + 4*x^3 + x^4
)*Log[5 + 4*x + x^2])*Log[x + Log[5 + 4*x + x^2]])*Log[(-3 + x*Log[x + Log[5 + 4*x + x^2]])/x]*Log[Log[(-3 + x
*Log[x + Log[5 + 4*x + x^2]])/x]])/((-15*x^2 - 12*x^3 - 3*x^4 + (-15*x - 12*x^2 - 3*x^3)*Log[5 + 4*x + x^2] +
(5*x^3 + 4*x^4 + x^5 + (5*x^2 + 4*x^3 + x^4)*Log[5 + 4*x + x^2])*Log[x + Log[5 + 4*x + x^2]])*Log[(-3 + x*Log[
x + Log[5 + 4*x + x^2]])/x]*Log[Log[(-3 + x*Log[x + Log[5 + 4*x + x^2]])/x]]^2),x]

[Out]

-((-3 - x)/Log[Log[-3/x + Log[x + Log[5 + 4*x + x^2]]]])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.15 (sec) , antiderivative size = 136, normalized size of antiderivative = 5.44

\[\frac {3+x}{\ln \left (-\ln \left (x \right )+\ln \left (x \ln \left (\ln \left (x^{2}+4 x +5\right )+x \right )-3\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (x \ln \left (\ln \left (x^{2}+4 x +5\right )+x \right )-3\right )}{x}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (x \ln \left (\ln \left (x^{2}+4 x +5\right )+x \right )-3\right )}{x}\right )+\operatorname {csgn}\left (\frac {i}{x}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (x \ln \left (\ln \left (x^{2}+4 x +5\right )+x \right )-3\right )}{x}\right )+\operatorname {csgn}\left (i \left (x \ln \left (\ln \left (x^{2}+4 x +5\right )+x \right )-3\right )\right )\right )}{2}\right )}\]

[In]

int(((((x^4+4*x^3+5*x^2)*ln(x^2+4*x+5)+x^5+4*x^4+5*x^3)*ln(ln(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*ln(x^2+4*x+5)
-3*x^4-12*x^3-15*x^2)*ln((x*ln(ln(x^2+4*x+5)+x)-3)/x)*ln(ln((x*ln(ln(x^2+4*x+5)+x)-3)/x))+(-3*x^3-21*x^2-51*x-
45)*ln(x^2+4*x+5)-x^5-12*x^4-48*x^3-78*x^2-45*x)/(((x^4+4*x^3+5*x^2)*ln(x^2+4*x+5)+x^5+4*x^4+5*x^3)*ln(ln(x^2+
4*x+5)+x)+(-3*x^3-12*x^2-15*x)*ln(x^2+4*x+5)-3*x^4-12*x^3-15*x^2)/ln((x*ln(ln(x^2+4*x+5)+x)-3)/x)/ln(ln((x*ln(
ln(x^2+4*x+5)+x)-3)/x))^2,x)

[Out]

(3+x)/ln(-ln(x)+ln(x*ln(ln(x^2+4*x+5)+x)-3)-1/2*I*Pi*csgn(I/x*(x*ln(ln(x^2+4*x+5)+x)-3))*(-csgn(I/x*(x*ln(ln(x
^2+4*x+5)+x)-3))+csgn(I/x))*(-csgn(I/x*(x*ln(ln(x^2+4*x+5)+x)-3))+csgn(I*(x*ln(ln(x^2+4*x+5)+x)-3))))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\frac {x + 3}{\log \left (\log \left (\frac {x \log \left (x + \log \left (x^{2} + 4 \, x + 5\right )\right ) - 3}{x}\right )\right )} \]

[In]

integrate(((((x^4+4*x^3+5*x^2)*log(x^2+4*x+5)+x^5+4*x^4+5*x^3)*log(log(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*log(
x^2+4*x+5)-3*x^4-12*x^3-15*x^2)*log((x*log(log(x^2+4*x+5)+x)-3)/x)*log(log((x*log(log(x^2+4*x+5)+x)-3)/x))+(-3
*x^3-21*x^2-51*x-45)*log(x^2+4*x+5)-x^5-12*x^4-48*x^3-78*x^2-45*x)/(((x^4+4*x^3+5*x^2)*log(x^2+4*x+5)+x^5+4*x^
4+5*x^3)*log(log(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*log(x^2+4*x+5)-3*x^4-12*x^3-15*x^2)/log((x*log(log(x^2+4*x
+5)+x)-3)/x)/log(log((x*log(log(x^2+4*x+5)+x)-3)/x))^2,x, algorithm="fricas")

[Out]

(x + 3)/log(log((x*log(x + log(x^2 + 4*x + 5)) - 3)/x))

Sympy [F(-1)]

Timed out. \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\text {Timed out} \]

[In]

integrate(((((x**4+4*x**3+5*x**2)*ln(x**2+4*x+5)+x**5+4*x**4+5*x**3)*ln(ln(x**2+4*x+5)+x)+(-3*x**3-12*x**2-15*
x)*ln(x**2+4*x+5)-3*x**4-12*x**3-15*x**2)*ln((x*ln(ln(x**2+4*x+5)+x)-3)/x)*ln(ln((x*ln(ln(x**2+4*x+5)+x)-3)/x)
)+(-3*x**3-21*x**2-51*x-45)*ln(x**2+4*x+5)-x**5-12*x**4-48*x**3-78*x**2-45*x)/(((x**4+4*x**3+5*x**2)*ln(x**2+4
*x+5)+x**5+4*x**4+5*x**3)*ln(ln(x**2+4*x+5)+x)+(-3*x**3-12*x**2-15*x)*ln(x**2+4*x+5)-3*x**4-12*x**3-15*x**2)/l
n((x*ln(ln(x**2+4*x+5)+x)-3)/x)/ln(ln((x*ln(ln(x**2+4*x+5)+x)-3)/x))**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\frac {x + 3}{\log \left (\log \left (x \log \left (x + \log \left (x^{2} + 4 \, x + 5\right )\right ) - 3\right ) - \log \left (x\right )\right )} \]

[In]

integrate(((((x^4+4*x^3+5*x^2)*log(x^2+4*x+5)+x^5+4*x^4+5*x^3)*log(log(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*log(
x^2+4*x+5)-3*x^4-12*x^3-15*x^2)*log((x*log(log(x^2+4*x+5)+x)-3)/x)*log(log((x*log(log(x^2+4*x+5)+x)-3)/x))+(-3
*x^3-21*x^2-51*x-45)*log(x^2+4*x+5)-x^5-12*x^4-48*x^3-78*x^2-45*x)/(((x^4+4*x^3+5*x^2)*log(x^2+4*x+5)+x^5+4*x^
4+5*x^3)*log(log(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*log(x^2+4*x+5)-3*x^4-12*x^3-15*x^2)/log((x*log(log(x^2+4*x
+5)+x)-3)/x)/log(log((x*log(log(x^2+4*x+5)+x)-3)/x))^2,x, algorithm="maxima")

[Out]

(x + 3)/log(log(x*log(x + log(x^2 + 4*x + 5)) - 3) - log(x))

Giac [A] (verification not implemented)

none

Time = 5.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\frac {x + 3}{\log \left (\log \left (x \log \left (x + \log \left (x^{2} + 4 \, x + 5\right )\right ) - 3\right ) - \log \left (x\right )\right )} \]

[In]

integrate(((((x^4+4*x^3+5*x^2)*log(x^2+4*x+5)+x^5+4*x^4+5*x^3)*log(log(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*log(
x^2+4*x+5)-3*x^4-12*x^3-15*x^2)*log((x*log(log(x^2+4*x+5)+x)-3)/x)*log(log((x*log(log(x^2+4*x+5)+x)-3)/x))+(-3
*x^3-21*x^2-51*x-45)*log(x^2+4*x+5)-x^5-12*x^4-48*x^3-78*x^2-45*x)/(((x^4+4*x^3+5*x^2)*log(x^2+4*x+5)+x^5+4*x^
4+5*x^3)*log(log(x^2+4*x+5)+x)+(-3*x^3-12*x^2-15*x)*log(x^2+4*x+5)-3*x^4-12*x^3-15*x^2)/log((x*log(log(x^2+4*x
+5)+x)-3)/x)/log(log((x*log(log(x^2+4*x+5)+x)-3)/x))^2,x, algorithm="giac")

[Out]

(x + 3)/log(log(x*log(x + log(x^2 + 4*x + 5)) - 3) - log(x))

Mupad [B] (verification not implemented)

Time = 18.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {-45 x-78 x^2-48 x^3-12 x^4-x^5+\left (-45-51 x-21 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log \left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )}{\left (-15 x^2-12 x^3-3 x^4+\left (-15 x-12 x^2-3 x^3\right ) \log \left (5+4 x+x^2\right )+\left (5 x^3+4 x^4+x^5+\left (5 x^2+4 x^3+x^4\right ) \log \left (5+4 x+x^2\right )\right ) \log \left (x+\log \left (5+4 x+x^2\right )\right )\right ) \log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right ) \log ^2\left (\log \left (\frac {-3+x \log \left (x+\log \left (5+4 x+x^2\right )\right )}{x}\right )\right )} \, dx=\frac {x+3}{\ln \left (\ln \left (\frac {x\,\ln \left (x+\ln \left (x^2+4\,x+5\right )\right )-3}{x}\right )\right )} \]

[In]

int((45*x + log(4*x + x^2 + 5)*(51*x + 21*x^2 + 3*x^3 + 45) + 78*x^2 + 48*x^3 + 12*x^4 + x^5 + log(log((x*log(
x + log(4*x + x^2 + 5)) - 3)/x))*log((x*log(x + log(4*x + x^2 + 5)) - 3)/x)*(log(4*x + x^2 + 5)*(15*x + 12*x^2
+ 3*x^3) - log(x + log(4*x + x^2 + 5))*(log(4*x + x^2 + 5)*(5*x^2 + 4*x^3 + x^4) + 5*x^3 + 4*x^4 + x^5) + 15*
x^2 + 12*x^3 + 3*x^4))/(log(log((x*log(x + log(4*x + x^2 + 5)) - 3)/x))^2*log((x*log(x + log(4*x + x^2 + 5)) -
3)/x)*(log(4*x + x^2 + 5)*(15*x + 12*x^2 + 3*x^3) - log(x + log(4*x + x^2 + 5))*(log(4*x + x^2 + 5)*(5*x^2 +
4*x^3 + x^4) + 5*x^3 + 4*x^4 + x^5) + 15*x^2 + 12*x^3 + 3*x^4)),x)

[Out]

(x + 3)/log(log((x*log(x + log(4*x + x^2 + 5)) - 3)/x))