3.72.41 \(\int \frac {-20+6 x^2+(20-6 x^2) \log (x) \log (4 \log (x))+(-10-3 x^2) \log (x) \log (4 \log (x)) \log (\frac {\log (4 \log (x))}{2 x}) \log (\log ^2(\frac {\log (4 \log (x))}{2 x}))}{(-10 x+3 x^3) \log (x) \log (4 \log (x)) \log (\frac {\log (4 \log (x))}{2 x}) \log (\log ^2(\frac {\log (4 \log (x))}{2 x}))} \, dx\) [7141]

Optimal. Leaf size=32 \[ \log \left (\frac {4 \log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )}{-\frac {2}{x}+\frac {3 x}{5}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 2.98, antiderivative size = 30, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 5, integrand size = 122, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {1607, 6857, 457, 78, 6816} \begin {gather*} -\log \left (10-3 x^2\right )+\log \left (\log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )\right )+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 + 6*x^2 + (20 - 6*x^2)*Log[x]*Log[4*Log[x]] + (-10 - 3*x^2)*Log[x]*Log[4*Log[x]]*Log[Log[4*Log[x]]/(2
*x)]*Log[Log[Log[4*Log[x]]/(2*x)]^2])/((-10*x + 3*x^3)*Log[x]*Log[4*Log[x]]*Log[Log[4*Log[x]]/(2*x)]*Log[Log[L
og[4*Log[x]]/(2*x)]^2]),x]

[Out]

Log[x] - Log[10 - 3*x^2] + Log[Log[Log[Log[4*Log[x]]/(2*x)]^2]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-20+6 x^2+\left (20-6 x^2\right ) \log (x) \log (4 \log (x))+\left (-10-3 x^2\right ) \log (x) \log (4 \log (x)) \log \left (\frac {\log (4 \log (x))}{2 x}\right ) \log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )}{x \left (-10+3 x^2\right ) \log (x) \log (4 \log (x)) \log \left (\frac {\log (4 \log (x))}{2 x}\right ) \log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )} \, dx\\ &=\int \left (\frac {-10-3 x^2}{x \left (-10+3 x^2\right )}-\frac {2 (-1+\log (x) \log (4 \log (x)))}{x \log (x) \log (4 \log (x)) \log \left (\frac {\log (4 \log (x))}{2 x}\right ) \log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )}\right ) \, dx\\ &=-\left (2 \int \frac {-1+\log (x) \log (4 \log (x))}{x \log (x) \log (4 \log (x)) \log \left (\frac {\log (4 \log (x))}{2 x}\right ) \log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )} \, dx\right )+\int \frac {-10-3 x^2}{x \left (-10+3 x^2\right )} \, dx\\ &=\log \left (\log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )\right )+\frac {1}{2} \text {Subst}\left (\int \frac {-10-3 x}{x (-10+3 x)} \, dx,x,x^2\right )\\ &=\log \left (\log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )\right )+\frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x}-\frac {6}{-10+3 x}\right ) \, dx,x,x^2\right )\\ &=\log (x)-\log \left (10-3 x^2\right )+\log \left (\log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]
time = 0.11, size = 30, normalized size = 0.94 \begin {gather*} \log (x)-\log \left (10-3 x^2\right )+\log \left (\log \left (\log ^2\left (\frac {\log (4 \log (x))}{2 x}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + 6*x^2 + (20 - 6*x^2)*Log[x]*Log[4*Log[x]] + (-10 - 3*x^2)*Log[x]*Log[4*Log[x]]*Log[Log[4*Log[
x]]/(2*x)]*Log[Log[Log[4*Log[x]]/(2*x)]^2])/((-10*x + 3*x^3)*Log[x]*Log[4*Log[x]]*Log[Log[4*Log[x]]/(2*x)]*Log
[Log[Log[4*Log[x]]/(2*x)]^2]),x]

[Out]

Log[x] - Log[10 - 3*x^2] + Log[Log[Log[Log[4*Log[x]]/(2*x)]^2]]

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 39.48, size = 892, normalized size = 27.88

method	result	size
risch	\(\text {Expression too large to display}\)	\(892\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2-10)*ln(x)*ln(4*ln(x))*ln(1/2*ln(4*ln(x))/x)*ln(ln(1/2*ln(4*ln(x))/x)^2)+(-6*x^2+20)*ln(x)*ln(4*ln
(x))+6*x^2-20)/(3*x^3-10*x)/ln(x)/ln(4*ln(x))/ln(1/2*ln(4*ln(x))/x)/ln(ln(1/2*ln(4*ln(x))/x)^2),x,method=_RETU
RNVERBOSE)

[Out]

ln(x)-ln(3*x^2-10)+ln(ln(Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^2*csg
n(I/x)-Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln(4*ln(x)))^3-2*I*ln(2)-2*I*ln(x)+2*I*ln(ln
(4*ln(x))))-1/4*I*(2*Pi*csgn(I*(-Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln(4*ln(x)
))^2*csgn(I/x)+Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^3+2*I*ln(2)+2*I*ln(x)-2
*I*ln(ln(4*ln(x))))^2)^2+Pi*csgn(I*(-Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln(4*l
n(x)))^2*csgn(I/x)+Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^3+2*I*ln(2)+2*I*ln(
x)-2*I*ln(ln(4*ln(x)))))^2*csgn(I*(-Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln(4*ln
(x)))^2*csgn(I/x)+Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^3+2*I*ln(2)+2*I*ln(x
)-2*I*ln(ln(4*ln(x))))^2)+2*Pi*csgn(I*(-Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln(
4*ln(x)))^2*csgn(I/x)+Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^3+2*I*ln(2)+2*I*
ln(x)-2*I*ln(ln(4*ln(x)))))*csgn(I*(-Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln(4*l
n(x)))^2*csgn(I/x)+Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^3+2*I*ln(2)+2*I*ln(
x)-2*I*ln(ln(4*ln(x))))^2)^2-Pi*csgn(I*(-Pi*csgn(I/x*ln(4*ln(x)))*csgn(I/x)*csgn(I*ln(4*ln(x)))+Pi*csgn(I/x*ln
(4*ln(x)))^2*csgn(I/x)+Pi*csgn(I/x*ln(4*ln(x)))^2*csgn(I*ln(4*ln(x)))-Pi*csgn(I/x*ln(4*ln(x)))^3+2*I*ln(2)+2*I
*ln(x)-2*I*ln(ln(4*ln(x))))^2)^3-2*Pi-4*I*ln(2)))

________________________________________________________________________________________

Maxima [A]
time = 0.56, size = 31, normalized size = 0.97 \begin {gather*} -\log \left (3 \, x^{2} - 10\right ) + \log \left (x\right ) + \log \left (\log \left (\log \left (2\right ) + \log \left (x\right ) - \log \left (2 \, \log \left (2\right ) + \log \left (\log \left (x\right )\right )\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-10)*log(x)*log(4*log(x))*log(1/2*log(4*log(x))/x)*log(log(1/2*log(4*log(x))/x)^2)+(-6*x^2+2
0)*log(x)*log(4*log(x))+6*x^2-20)/(3*x^3-10*x)/log(x)/log(4*log(x))/log(1/2*log(4*log(x))/x)/log(log(1/2*log(4
*log(x))/x)^2),x, algorithm="maxima")

[Out]

-log(3*x^2 - 10) + log(x) + log(log(log(2) + log(x) - log(2*log(2) + log(log(x)))))

________________________________________________________________________________________

Fricas [A]
time = 0.39, size = 28, normalized size = 0.88 \begin {gather*} -\log \left (3 \, x^{2} - 10\right ) + \log \left (x\right ) + \log \left (\log \left (\log \left (\frac {\log \left (4 \, \log \left (x\right )\right )}{2 \, x}\right )^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-10)*log(x)*log(4*log(x))*log(1/2*log(4*log(x))/x)*log(log(1/2*log(4*log(x))/x)^2)+(-6*x^2+2
0)*log(x)*log(4*log(x))+6*x^2-20)/(3*x^3-10*x)/log(x)/log(4*log(x))/log(1/2*log(4*log(x))/x)/log(log(1/2*log(4
*log(x))/x)^2),x, algorithm="fricas")

[Out]

-log(3*x^2 - 10) + log(x) + log(log(log(1/2*log(4*log(x))/x)^2))

________________________________________________________________________________________

Sympy [A]
time = 0.53, size = 27, normalized size = 0.84 \begin {gather*} \log {\left (x \right )} - \log {\left (3 x^{2} - 10 \right )} + \log {\left (\log {\left (\log {\left (\frac {\log {\left (4 \log {\left (x \right )} \right )}}{2 x} \right )}^{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2-10)*ln(x)*ln(4*ln(x))*ln(1/2*ln(4*ln(x))/x)*ln(ln(1/2*ln(4*ln(x))/x)**2)+(-6*x**2+20)*ln(x
)*ln(4*ln(x))+6*x**2-20)/(3*x**3-10*x)/ln(x)/ln(4*ln(x))/ln(1/2*ln(4*ln(x))/x)/ln(ln(1/2*ln(4*ln(x))/x)**2),x)

[Out]

log(x) - log(3*x**2 - 10) + log(log(log(log(4*log(x))/(2*x))**2))

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-10)*log(x)*log(4*log(x))*log(1/2*log(4*log(x))/x)*log(log(1/2*log(4*log(x))/x)^2)+(-6*x^2+2
0)*log(x)*log(4*log(x))+6*x^2-20)/(3*x^3-10*x)/log(x)/log(4*log(x))/log(1/2*log(4*log(x))/x)/log(log(1/2*log(4
*log(x))/x)^2),x, algorithm="giac")

[Out]

Timed out

________________________________________________________________________________________

Mupad [B]
time = 8.25, size = 26, normalized size = 0.81 \begin {gather*} \ln \left (\ln \left ({\ln \left (\frac {\ln \left (4\,\ln \left (x\right )\right )}{2\,x}\right )}^2\right )\right )-\ln \left (x^2-\frac {10}{3}\right )+\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(4*log(x))*log(x)*(6*x^2 - 20) - 6*x^2 + log(log(log(4*log(x))/(2*x))^2)*log(log(4*log(x))/(2*x))*log(
4*log(x))*log(x)*(3*x^2 + 10) + 20)/(log(log(log(4*log(x))/(2*x))^2)*log(log(4*log(x))/(2*x))*log(4*log(x))*lo
g(x)*(10*x - 3*x^3)),x)

[Out]

log(log(log(log(4*log(x))/(2*x))^2)) - log(x^2 - 10/3) + log(x)

________________________________________________________________________________________