3.5.8 \(\int \frac {-2-20 x-20 x \log (x)}{6480 x+3600 x^2+500 x^3+(6480 x+3600 x^2+500 x^3) \log (x)+(360 x+100 x^2+(360 x+100 x^2) \log (x)) \log (1+\log (x))+(5 x+5 x \log (x)) \log ^2(1+\log (x))} \, dx\)

Optimal. Leaf size=18 \[ \frac {2}{5 (-4+10 (4+x)+\log (1+\log (x)))} \]

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Rubi [A]  time = 0.27, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 12, 6686} \begin {gather*} \frac {2}{5 (10 x+\log (\log (x)+1)+36)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 20*x - 20*x*Log[x])/(6480*x + 3600*x^2 + 500*x^3 + (6480*x + 3600*x^2 + 500*x^3)*Log[x] + (360*x + 1
00*x^2 + (360*x + 100*x^2)*Log[x])*Log[1 + Log[x]] + (5*x + 5*x*Log[x])*Log[1 + Log[x]]^2),x]

[Out]

2/(5*(36 + 10*x + Log[1 + Log[x]]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 (-1-10 x-10 x \log (x))}{5 x (1+\log (x)) (36+10 x+\log (1+\log (x)))^2} \, dx\\ &=\frac {2}{5} \int \frac {-1-10 x-10 x \log (x)}{x (1+\log (x)) (36+10 x+\log (1+\log (x)))^2} \, dx\\ &=\frac {2}{5 (36+10 x+\log (1+\log (x)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.89 \begin {gather*} \frac {2}{5 (36+10 x+\log (1+\log (x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 20*x - 20*x*Log[x])/(6480*x + 3600*x^2 + 500*x^3 + (6480*x + 3600*x^2 + 500*x^3)*Log[x] + (360
*x + 100*x^2 + (360*x + 100*x^2)*Log[x])*Log[1 + Log[x]] + (5*x + 5*x*Log[x])*Log[1 + Log[x]]^2),x]

[Out]

2/(5*(36 + 10*x + Log[1 + Log[x]]))

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fricas [A]  time = 0.70, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{5 \, {\left (10 \, x + \log \left (\log \relax (x) + 1\right ) + 36\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(x)-20*x-2)/((5*x*log(x)+5*x)*log(log(x)+1)^2+((100*x^2+360*x)*log(x)+100*x^2+360*x)*log(l
og(x)+1)+(500*x^3+3600*x^2+6480*x)*log(x)+500*x^3+3600*x^2+6480*x),x, algorithm="fricas")

[Out]

2/5/(10*x + log(log(x) + 1) + 36)

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giac [A]  time = 0.38, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{5 \, {\left (10 \, x + \log \left (\log \relax (x) + 1\right ) + 36\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(x)-20*x-2)/((5*x*log(x)+5*x)*log(log(x)+1)^2+((100*x^2+360*x)*log(x)+100*x^2+360*x)*log(l
og(x)+1)+(500*x^3+3600*x^2+6480*x)*log(x)+500*x^3+3600*x^2+6480*x),x, algorithm="giac")

[Out]

2/5/(10*x + log(log(x) + 1) + 36)

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maple [A]  time = 0.02, size = 15, normalized size = 0.83

method	result	size
risch	\(\frac {2}{5 \left (\ln \left (\ln \relax (x )+1\right )+10 x +36\right )}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*x*ln(x)-20*x-2)/((5*x*ln(x)+5*x)*ln(ln(x)+1)^2+((100*x^2+360*x)*ln(x)+100*x^2+360*x)*ln(ln(x)+1)+(500
*x^3+3600*x^2+6480*x)*ln(x)+500*x^3+3600*x^2+6480*x),x,method=_RETURNVERBOSE)

[Out]

2/5/(ln(ln(x)+1)+10*x+36)

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maxima [A]  time = 0.79, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{5 \, {\left (10 \, x + \log \left (\log \relax (x) + 1\right ) + 36\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(x)-20*x-2)/((5*x*log(x)+5*x)*log(log(x)+1)^2+((100*x^2+360*x)*log(x)+100*x^2+360*x)*log(l
og(x)+1)+(500*x^3+3600*x^2+6480*x)*log(x)+500*x^3+3600*x^2+6480*x),x, algorithm="maxima")

[Out]

2/5/(10*x + log(log(x) + 1) + 36)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int -\frac {20\,x+20\,x\,\ln \relax (x)+2}{6480\,x+\ln \left (\ln \relax (x)+1\right )\,\left (360\,x+\ln \relax (x)\,\left (100\,x^2+360\,x\right )+100\,x^2\right )+{\ln \left (\ln \relax (x)+1\right )}^2\,\left (5\,x+5\,x\,\ln \relax (x)\right )+3600\,x^2+500\,x^3+\ln \relax (x)\,\left (500\,x^3+3600\,x^2+6480\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + 20*x*log(x) + 2)/(6480*x + log(log(x) + 1)*(360*x + log(x)*(360*x + 100*x^2) + 100*x^2) + log(log
(x) + 1)^2*(5*x + 5*x*log(x)) + 3600*x^2 + 500*x^3 + log(x)*(6480*x + 3600*x^2 + 500*x^3)),x)

[Out]

int(-(20*x + 20*x*log(x) + 2)/(6480*x + log(log(x) + 1)*(360*x + log(x)*(360*x + 100*x^2) + 100*x^2) + log(log
(x) + 1)^2*(5*x + 5*x*log(x)) + 3600*x^2 + 500*x^3 + log(x)*(6480*x + 3600*x^2 + 500*x^3)), x)

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sympy [A]  time = 0.50, size = 14, normalized size = 0.78 \begin {gather*} \frac {2}{50 x + 5 \log {\left (\log {\relax (x )} + 1 \right )} + 180} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*ln(x)-20*x-2)/((5*x*ln(x)+5*x)*ln(ln(x)+1)**2+((100*x**2+360*x)*ln(x)+100*x**2+360*x)*ln(ln(x
)+1)+(500*x**3+3600*x**2+6480*x)*ln(x)+500*x**3+3600*x**2+6480*x),x)

[Out]

2/(50*x + 5*log(log(x) + 1) + 180)

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