3.46.51 \(\int \frac {-5-2 x+(-20-4 x+(40+8 x) \log (x)+(-5-x+(10+2 x) \log (x)) \log (5 x+x^2)) \log (4+\log (5 x+x^2))}{(20 x+4 x^2+(5 x+x^2) \log (5 x+x^2)) \log (4+\log (5 x+x^2))} \, dx\)

Optimal. Leaf size=22 \[ \log ^2(x)-\log \left (\frac {1}{9} x \log (4+\log (x (5+x)))\right ) \]

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Rubi [A]  time = 1.02, antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 6742, 2301, 6684} \begin {gather*} \frac {1}{4} (1-2 \log (x))^2-\log (\log (\log (x (x+5))+4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - 2*x + (-20 - 4*x + (40 + 8*x)*Log[x] + (-5 - x + (10 + 2*x)*Log[x])*Log[5*x + x^2])*Log[4 + Log[5*x
+ x^2]])/((20*x + 4*x^2 + (5*x + x^2)*Log[5*x + x^2])*Log[4 + Log[5*x + x^2]]),x]

[Out]

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 6684

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5-2 x+(5+x) (-1+2 \log (x)) (4+\log (x (5+x))) \log (4+\log (x (5+x)))}{x (5+x) (4+\log (x (5+x))) \log (4+\log (x (5+x)))} \, dx\\ &=\int \left (\frac {-1+2 \log (x)}{x}+\frac {-5-2 x}{x (5+x) (4+\log (x (5+x))) \log (4+\log (x (5+x)))}\right ) \, dx\\ &=\int \frac {-1+2 \log (x)}{x} \, dx+\int \frac {-5-2 x}{x (5+x) (4+\log (x (5+x))) \log (4+\log (x (5+x)))} \, dx\\ &=\frac {1}{4} (1-2 \log (x))^2-\log (\log (4+\log (x (5+x))))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 21, normalized size = 0.95 \begin {gather*} -\log (x)+\log ^2(x)-\log (\log (4+\log (x (5+x)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 2*x + (-20 - 4*x + (40 + 8*x)*Log[x] + (-5 - x + (10 + 2*x)*Log[x])*Log[5*x + x^2])*Log[4 + Lo
g[5*x + x^2]])/((20*x + 4*x^2 + (5*x + x^2)*Log[5*x + x^2])*Log[4 + Log[5*x + x^2]]),x]

[Out]

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

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fricas [A]  time = 0.47, size = 23, normalized size = 1.05 \begin {gather*} \log \relax (x)^{2} - \log \relax (x) - \log \left (\log \left (\log \left (x^{2} + 5 \, x\right ) + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+10)*log(x)-x-5)*log(x^2+5*x)+(8*x+40)*log(x)-4*x-20)*log(log(x^2+5*x)+4)-2*x-5)/((x^2+5*x)*l
og(x^2+5*x)+4*x^2+20*x)/log(log(x^2+5*x)+4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.23, size = 21, normalized size = 0.95 \begin {gather*} \log \relax (x)^{2} - \log \relax (x) - \log \left (\log \left (\log \left (x + 5\right ) + \log \relax (x) + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+10)*log(x)-x-5)*log(x^2+5*x)+(8*x+40)*log(x)-4*x-20)*log(log(x^2+5*x)+4)-2*x-5)/((x^2+5*x)*l
og(x^2+5*x)+4*x^2+20*x)/log(log(x^2+5*x)+4),x, algorithm="giac")

[Out]

log(x)^2 - log(x) - log(log(log(x + 5) + log(x) + 4))

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maple [A]  time = 0.13, size = 24, normalized size = 1.09

method	result	size
default	\(\ln \relax (x )^{2}-\ln \relax (x )-\ln \left (\ln \left (\ln \left (x^{2}+5 x \right )+4\right )\right )\)	\(24\)
risch	\(\ln \relax (x )^{2}-\ln \relax (x )-\ln \left (\ln \left (\ln \relax (x )+\ln \left (5+x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x \left (5+x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (5+x \right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (5+x \right )\right )+\mathrm {csgn}\left (i \left (5+x \right )\right )\right )}{2}+4\right )\right )\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((2*x+10)*ln(x)-x-5)*ln(x^2+5*x)+(8*x+40)*ln(x)-4*x-20)*ln(ln(x^2+5*x)+4)-2*x-5)/((x^2+5*x)*ln(x^2+5*x)+
4*x^2+20*x)/ln(ln(x^2+5*x)+4),x,method=_RETURNVERBOSE)

[Out]

ln(x)^2-ln(x)-ln(ln(ln(x^2+5*x)+4))

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maxima [A]  time = 0.60, size = 21, normalized size = 0.95 \begin {gather*} \log \relax (x)^{2} - \log \relax (x) - \log \left (\log \left (\log \left (x + 5\right ) + \log \relax (x) + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+10)*log(x)-x-5)*log(x^2+5*x)+(8*x+40)*log(x)-4*x-20)*log(log(x^2+5*x)+4)-2*x-5)/((x^2+5*x)*l
og(x^2+5*x)+4*x^2+20*x)/log(log(x^2+5*x)+4),x, algorithm="maxima")

[Out]

log(x)^2 - log(x) - log(log(log(x + 5) + log(x) + 4))

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mupad [B]  time = 3.73, size = 23, normalized size = 1.05 \begin {gather*} {\ln \relax (x)}^2-\ln \relax (x)-\ln \left (\ln \left (\ln \left (x^2+5\,x\right )+4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + log(log(5*x + x^2) + 4)*(4*x + log(5*x + x^2)*(x - log(x)*(2*x + 10) + 5) - log(x)*(8*x + 40) + 20
) + 5)/(log(log(5*x + x^2) + 4)*(20*x + 4*x^2 + log(5*x + x^2)*(5*x + x^2))),x)

[Out]

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

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sympy [A]  time = 0.80, size = 20, normalized size = 0.91 \begin {gather*} \log {\relax (x )}^{2} - \log {\relax (x )} - \log {\left (\log {\left (\log {\left (x^{2} + 5 x \right )} + 4 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((2*x+10)*ln(x)-x-5)*ln(x**2+5*x)+(8*x+40)*ln(x)-4*x-20)*ln(ln(x**2+5*x)+4)-2*x-5)/((x**2+5*x)*ln(
x**2+5*x)+4*x**2+20*x)/ln(ln(x**2+5*x)+4),x)

[Out]

log(x)**2 - log(x) - log(log(log(x**2 + 5*x) + 4))

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