3.67.84 $\int \frac{(-6-60x+24x^2+e^x(-12x-12x^2)){\log }^2(x)+((2+20x-8x^2+e^x(4x+4x^2))\log (x)+((60x+12e^{x}x-12x^2)\log (x)+6{\log }^2(x))\log (10x+2e^{x}x-2x^2+\log (x)))\log (\log (10x+2e^{x}x-2x^2+\log (x)))+(-20x-4e^{x}x+4x^2-2\log (x))\log (10x+2e^{x}x-2x^2+\log (x)){\log }^2(\log (10x+2e^{x}x-2x^2+\log (x)))}{((10x^2+2e^x{x}^2-2x^3){\log }^3(x)+x{\log }^4(x))\log (10x+2e^{x}x-2x^2+\log (x))}dx$

Optimal. Leaf size=31 $$1+{(3-\frac{\log (\log ((4+2(3+e^x-x))x+\log (x)))}{\log (x)})}^2$$

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Rubi [A]  time = 6.86, antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 3, integrand size = 233, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.013, Rules used = {6688, 12, 6712} $$\begin{array}{c}\frac{{\log }^2(\log (2(-x+e^x+5)x+\log (x)))}{{\log }^2(x)}-\frac{6\log (\log (2(-x+e^x+5)x+\log (x)))}{\log (x)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[((-6 - 60*x + 24*x^2 + E^x*(-12*x - 12*x^2))*Log[x]^2 + ((2 + 20*x - 8*x^2 + E^x*(4*x + 4*x^2))*Log[x] + (
(60*x + 12*E^x*x - 12*x^2)*Log[x] + 6*Log[x]^2)*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]])*Log[Log[10*x + 2*E^x*x -
2*x^2 + Log[x]]] + (-20*x - 4*E^x*x + 4*x^2 - 2*Log[x])*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]*Log[Log[10*x + 2
*E^x*x - 2*x^2 + Log[x]]]^2)/(((10*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x]^3 + x*Log[x]^4)*Log[10*x + 2*E^x*x - 2*x^2
+ Log[x]]),x]

[Out]

(-6*Log[Log[2*(5 + E^x - x)*x + Log[x]]])/Log[x] + Log[Log[2*(5 + E^x - x)*x + Log[x]]]^2/Log[x]^2

Rule 12

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

Rule 6688

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

Rule 6712

Int[(u_)*(v_)^(r_.)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x
] - q*v*D[w, x])]}, -Dist[c*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ
[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2(3\log (x)-\log (\log (-2x(-5-e^x+x)+\log (x))))(2x(-5-e^x+x)\log (-2x(-5-e^x+x)+\log (x))\log (\log (-2x(-5-e^x+x)+\log (x)))+\log (x)(1+2(5+e^x)x+2(-2+e^x)x^2-\log (-2x(-5-e^x+x)+\log (x))\log (\log (-2x(-5-e^x+x)+\log (x)))))}{x(2x(-5-e^x+x)-\log (x)){\log }^3(x)\log (-2x(-5-e^x+x)+\log (x))}dx\\ & =2\int \frac{(3\log (x)-\log (\log (-2x(-5-e^x+x)+\log (x))))(2x(-5-e^x+x)\log (-2x(-5-e^x+x)+\log (x))\log (\log (-2x(-5-e^x+x)+\log (x)))+\log (x)(1+2(5+e^x)x+2(-2+e^x)x^2-\log (-2x(-5-e^x+x)+\log (x))\log (\log (-2x(-5-e^x+x)+\log (x)))))}{x(2x(-5-e^x+x)-\log (x)){\log }^3(x)\log (-2x(-5-e^x+x)+\log (x))}dx\\ & =-\left(2\mathrm{Subst}(\int (3-x)dx,x,\frac{\log (\log (-2x(-5-e^x+x)+\log (x)))}{\log (x)})\right)\\ & =-\frac{6\log (\log (2(5+e^x-x)x+\log (x)))}{\log (x)}+\frac{{\log }^2(\log (2(5+e^x-x)x+\log (x)))}{{\log }^2(x)}\end{array}\end{array}$$

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Mathematica [A]  time = 0.20, size = 51, normalized size = 1.65 $$\begin{array}{c}2(-\frac{3\log (\log (-2x(-5-e^x+x)+\log (x)))}{\log (x)}+\frac{{\log }^2(\log (-2x(-5-e^x+x)+\log (x)))}{2{\log }^2(x)})\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[((-6 - 60*x + 24*x^2 + E^x*(-12*x - 12*x^2))*Log[x]^2 + ((2 + 20*x - 8*x^2 + E^x*(4*x + 4*x^2))*Log[
x] + ((60*x + 12*E^x*x - 12*x^2)*Log[x] + 6*Log[x]^2)*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]])*Log[Log[10*x + 2*E
^x*x - 2*x^2 + Log[x]]] + (-20*x - 4*E^x*x + 4*x^2 - 2*Log[x])*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]*Log[Log[10
*x + 2*E^x*x - 2*x^2 + Log[x]]]^2)/(((10*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x]^3 + x*Log[x]^4)*Log[10*x + 2*E^x*x -
2*x^2 + Log[x]]),x]

[Out]

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

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fricas [A]  time = 0.60, size = 51, normalized size = 1.65 $$\begin{array}{c}-\frac{6\log (x)\log (\log (-2x^2+2x{e}^x+10x+\log (x)))-\log {(\log (-2x^2+2x{e}^x+10x+\log (x)))}^2}{\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*log(log(log(x)+2*exp(x)*x-2*x^2
+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)
-8*x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*log(x)^2)/
(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="fricas")

[Out]

-(6*log(x)*log(log(-2*x^2 + 2*x*e^x + 10*x + log(x))) - log(log(-2*x^2 + 2*x*e^x + 10*x + log(x)))^2)/log(x)^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Exception raised: TypeError}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*log(log(log(x)+2*exp(x)*x-2*x^2
+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)
-8*x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*log(x)^2)/
(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Sign error %%%{ln(` w`),0%%%}
Simplificat

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maple [A]  time = 0.04, size = 51, normalized size = 1.65

method	result	size
risch	$\frac{\ln {(\ln (\ln (x)+2{\mathrm{e}}^{x}x-2x^2+10x))}^2}{\ln (x{)}^2}-\frac{6\ln (\ln (\ln (x)+2{\mathrm{e}}^{x}x-2x^2+10x))}{\ln (x)}$	$51$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*ln(x)-4*exp(x)*x+4*x^2-20*x)*ln(ln(x)+2*exp(x)*x-2*x^2+10*x)*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))^2+((
6*ln(x)^2+(12*exp(x)*x-12*x^2+60*x)*ln(x))*ln(ln(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8*x^2+20*x+2)*l
n(x))*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*ln(x)^2)/(x*ln(x)^4+(2*exp(x)*
x^2-2*x^3+10*x^2)*ln(x)^3)/ln(ln(x)+2*exp(x)*x-2*x^2+10*x),x,method=_RETURNVERBOSE)

[Out]

1/ln(x)^2*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))^2-6/ln(x)*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))

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maxima [A]  time = 0.48, size = 51, normalized size = 1.65 $$\begin{array}{c}-\frac{6\log (x)\log (\log (-2x^2+2x{e}^x+10x+\log (x)))-\log {(\log (-2x^2+2x{e}^x+10x+\log (x)))}^2}{\log (x{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*log(log(log(x)+2*exp(x)*x-2*x^2
+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)
-8*x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*log(x)^2)/
(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="maxima")

[Out]

-(6*log(x)*log(log(-2*x^2 + 2*x*e^x + 10*x + log(x))) - log(log(-2*x^2 + 2*x*e^x + 10*x + log(x)))^2)/log(x)^2

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mupad [B]  time = 5.34, size = 46, normalized size = 1.48 $$\begin{array}{c}\frac{\ln (\ln (10x+\ln (x)+2x{\mathrm{e}}^x-2x^2))(\ln (\ln (10x+\ln (x)+2x{\mathrm{e}}^x-2x^2))-6\ln (x))}{{\ln (x)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)^2*(60*x + exp(x)*(12*x + 12*x^2) - 24*x^2 + 6) - log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))*(lo
g(10*x + log(x) + 2*x*exp(x) - 2*x^2)*(6*log(x)^2 + log(x)*(60*x + 12*x*exp(x) - 12*x^2)) + log(x)*(20*x + exp
(x)*(4*x + 4*x^2) - 8*x^2 + 2)) + log(10*x + log(x) + 2*x*exp(x) - 2*x^2)*log(log(10*x + log(x) + 2*x*exp(x) -
2*x^2))^2*(20*x + 2*log(x) + 4*x*exp(x) - 4*x^2))/(log(10*x + log(x) + 2*x*exp(x) - 2*x^2)*(x*log(x)^4 + log(
x)^3*(2*x^2*exp(x) + 10*x^2 - 2*x^3))),x)

[Out]

(log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))*(log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2)) - 6*log(x)))/log(x
)^2

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\text{Timed out}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*ln(x)-4*exp(x)*x+4*x**2-20*x)*ln(ln(x)+2*exp(x)*x-2*x**2+10*x)*ln(ln(ln(x)+2*exp(x)*x-2*x**2+10
*x))**2+((6*ln(x)**2+(12*exp(x)*x-12*x**2+60*x)*ln(x))*ln(ln(x)+2*exp(x)*x-2*x**2+10*x)+((4*x**2+4*x)*exp(x)-8
*x**2+20*x+2)*ln(x))*ln(ln(ln(x)+2*exp(x)*x-2*x**2+10*x))+((-12*x**2-12*x)*exp(x)+24*x**2-60*x-6)*ln(x)**2)/(x
*ln(x)**4+(2*exp(x)*x**2-2*x**3+10*x**2)*ln(x)**3)/ln(ln(x)+2*exp(x)*x-2*x**2+10*x),x)

[Out]

Timed out

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