3.69.74 $\int \frac{4-9x-4e^{-2+x}x+(-4e^{-2+x}-9x+4\log (x))\log (\frac{1}{4}(4e^{-2+x}+9x-4\log (x)))+(4e^{-2+x}+9x-4\log (x))\log (\frac{1}{4}(4e^{-2+x}+9x-4\log (x)))\log (x\log (\frac{1}{4}(4e^{-2+x}+9x-4\log (x))))}{(-4e^{-2+x}{x}^2-9x^3+4x^2\log (x))\log (\frac{1}{4}(4e^{-2+x}+9x-4\log (x)))}dx$

Optimal. Leaf size=23 $$\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}$$

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Rubi [A]  time = 8.43, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 7, integrand size = 156, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.045, Rules used = {6742, 6741, 12, 6688, 14, 30, 2555} $$\begin{array}{c}\frac{\log (x\log (\frac{9x}{4}+e^{x-2}-\log (x)))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(4 - 9*x - 4*E^(-2 + x)*x + (-4*E^(-2 + x) - 9*x + 4*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4] + (4*E
^(-2 + x) + 9*x - 4*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4]*Log[x*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4
]])/((-4*E^(-2 + x)*x^2 - 9*x^3 + 4*x^2*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4]),x]

[Out]

Log[x*Log[E^(-2 + x) + (9*x)/4 - Log[x]]]/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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

Rule 6688

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

Rule 6741

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (-\frac{e^2(4-9x+9x^2-4x\log (x))}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}+\frac{x+\log (e^{-2+x}+\frac{9x}{4}-\log (x))-\log (e^{-2+x}+\frac{9x}{4}-\log (x))\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x^2\log (e^{-2+x}+\frac{9x}{4}-\log (x))})dx\\ & =-(e^2\int \frac{4-9x+9x^2-4x\log (x)}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx)+\int \frac{x+\log (e^{-2+x}+\frac{9x}{4}-\log (x))-\log (e^{-2+x}+\frac{9x}{4}-\log (x))\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x^2\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx\\ & =-(e^2\int (\frac{9}{(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}+\frac{4}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}-\frac{9}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}-\frac{4\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))})dx)+\int \frac{1+\frac{x}{\log (e^{-2+x}+\frac{9x}{4}-\log (x))}-\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x^2}dx\\ & =-(\left(4e^2\right)\int \frac{1}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx)+\left(4e^2\right)\int \frac{\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(9e^2\right)\int \frac{1}{(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(9e^2\right)\int \frac{1}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\int (\frac{x+\log (e^{-2+x}+\frac{9x}{4}-\log (x))}{x^2\log (e^{-2+x}+\frac{9x}{4}-\log (x))}-\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x^2})dx\\ & =-(\left(4e^2\right)\int \frac{1}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx)+\left(4e^2\right)\int \frac{\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(9e^2\right)\int \frac{1}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(36e^2\right)\mathrm{Subst}(\int \frac{1}{(4e^{4x}+36e^{2}x-4e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})+\int \frac{x+\log (e^{-2+x}+\frac{9x}{4}-\log (x))}{x^2\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\int \frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x^2}dx\\ & =\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}-\left(4e^2\right)\int \frac{1}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(4e^2\right)\int \frac{\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(9e^2\right)\int \frac{1}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(36e^2\right)\mathrm{Subst}(\int \frac{1}{4(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})+\int \frac{1+\frac{x}{\log (e^{-2+x}+\frac{9x}{4}-\log (x))}}{x^2}dx-\int \frac{\frac{1}{x}+\frac{\frac{9}{4}+e^{-2+x}-\frac{1}{x}}{(e^{-2+x}+\frac{9x}{4}-\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}}{x}dx\\ & =\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}-\left(4e^2\right)\int \frac{1}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(4e^2\right)\int \frac{\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(9e^2\right)\int \frac{1}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(9e^2\right)\mathrm{Subst}(\int \frac{1}{(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})+\int (\frac{1}{x^2}+\frac{1}{x\log (e^{-2+x}+\frac{9x}{4}-\log (x))})dx-\int (-\frac{e^2(4-9x+9x^2-4x\log (x))}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}+\frac{x+\log (e^{-2+x}+\frac{9x}{4}-\log (x))}{x^2\log (e^{-2+x}+\frac{9x}{4}-\log (x))})dx\\ & =-\frac{1}{x}+\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}+e^2\int \frac{4-9x+9x^2-4x\log (x)}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(4e^2\right)\int \frac{1}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(4e^2\right)\int \frac{\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(9e^2\right)\int \frac{1}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(9e^2\right)\mathrm{Subst}(\int \frac{1}{(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})+\int \frac{1}{x\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\int \frac{x+\log (e^{-2+x}+\frac{9x}{4}-\log (x))}{x^2\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx\\ & =-\frac{1}{x}+\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}+e^2\int (\frac{9}{(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}+\frac{4}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}-\frac{9}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}-\frac{4\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))})dx-\left(4e^2\right)\int \frac{1}{x^2(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(4e^2\right)\int \frac{\log (x)}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx+\left(9e^2\right)\int \frac{1}{x(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(9e^2\right)\mathrm{Subst}(\int \frac{1}{(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})-\int \frac{1+\frac{x}{\log (e^{-2+x}+\frac{9x}{4}-\log (x))}}{x^2}dx+\int \frac{1}{x\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx\\ & =-\frac{1}{x}+\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}+\left(9e^2\right)\int \frac{1}{(4e^x+9e^{2}x-4e^2\log (x))\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx-\left(9e^2\right)\mathrm{Subst}(\int \frac{1}{(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})-\int (\frac{1}{x^2}+\frac{1}{x\log (e^{-2+x}+\frac{9x}{4}-\log (x))})dx+\int \frac{1}{x\log (e^{-2+x}+\frac{9x}{4}-\log (x))}dx\\ & =\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}-\left(9e^2\right)\mathrm{Subst}(\int \frac{1}{(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})+\left(36e^2\right)\mathrm{Subst}(\int \frac{1}{(4e^{4x}+36e^{2}x-4e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})\\ & =\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}-\left(9e^2\right)\mathrm{Subst}(\int \frac{1}{(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})+\left(36e^2\right)\mathrm{Subst}(\int \frac{1}{4(e^{4x}+9e^{2}x-e^2\log (4x))\log (e^{-2+4x}+9x-\log (4x))}dx,x,\frac{x}{4})\\ & =\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.17, size = 23, normalized size = 1.00 $$\begin{array}{c}\frac{\log (x\log (e^{-2+x}+\frac{9x}{4}-\log (x)))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(4 - 9*x - 4*E^(-2 + x)*x + (-4*E^(-2 + x) - 9*x + 4*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4]
+ (4*E^(-2 + x) + 9*x - 4*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4]*Log[x*Log[(4*E^(-2 + x) + 9*x - 4*Log
[x])/4]])/((-4*E^(-2 + x)*x^2 - 9*x^3 + 4*x^2*Log[x])*Log[(4*E^(-2 + x) + 9*x - 4*Log[x])/4]),x]

[Out]

Log[x*Log[E^(-2 + x) + (9*x)/4 - Log[x]]]/x

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fricas [A]  time = 0.58, size = 20, normalized size = 0.87 $$\begin{array}{c}\frac{\log (x\log (\frac{9}{4}x+e^{(x-2)}-\log (x)))}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(x)+4*exp(x-2)+9*x)*log(-log(x)+exp(x-2)+9/4*x)*log(x*log(-log(x)+exp(x-2)+9/4*x))+(4*log(x)
-4*exp(x-2)-9*x)*log(-log(x)+exp(x-2)+9/4*x)-4*x*exp(x-2)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(x-2)-9*x^3)/log(-log(
x)+exp(x-2)+9/4*x),x, algorithm="fricas")

[Out]

log(x*log(9/4*x + e^(x - 2) - log(x)))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int -\frac{(9x+4e^{(x-2)}-4\log (x))\log (x\log (\frac{9}{4}x+e^{(x-2)}-\log (x)))\log (\frac{9}{4}x+e^{(x-2)}-\log (x))-4x{e}^{(x-2)}-(9x+4e^{(x-2)}-4\log (x))\log (\frac{9}{4}x+e^{(x-2)}-\log (x))-9x+4}{(9x^3+4x^2{e}^{(x-2)}-4x^2\log (x))\log (\frac{9}{4}x+e^{(x-2)}-\log (x))}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(x)+4*exp(x-2)+9*x)*log(-log(x)+exp(x-2)+9/4*x)*log(x*log(-log(x)+exp(x-2)+9/4*x))+(4*log(x)
-4*exp(x-2)-9*x)*log(-log(x)+exp(x-2)+9/4*x)-4*x*exp(x-2)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(x-2)-9*x^3)/log(-log(
x)+exp(x-2)+9/4*x),x, algorithm="giac")

[Out]

integrate(-((9*x + 4*e^(x - 2) - 4*log(x))*log(x*log(9/4*x + e^(x - 2) - log(x)))*log(9/4*x + e^(x - 2) - log(
x)) - 4*x*e^(x - 2) - (9*x + 4*e^(x - 2) - 4*log(x))*log(9/4*x + e^(x - 2) - log(x)) - 9*x + 4)/((9*x^3 + 4*x^
2*e^(x - 2) - 4*x^2*log(x))*log(9/4*x + e^(x - 2) - log(x))), x)

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maple [C]  time = 0.12, size = 168, normalized size = 7.30

method	result	size
risch	$\frac{\ln (\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))}{x}+\frac{-i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}(i\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))\mathrm{csgn}(ix\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))+i\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{(ix\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))}^2+i\pi \mathrm{csgn}(i\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))\mathrm{csgn}{(ix\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))}^2-i\pi \mathrm{csgn}{(ix\ln (-\ln (x)+{\mathrm{e}}^{x-2}+\frac{9x}{4}))}^3+2\ln (x)}{2x}$	$168$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*ln(x)+4*exp(x-2)+9*x)*ln(-ln(x)+exp(x-2)+9/4*x)*ln(x*ln(-ln(x)+exp(x-2)+9/4*x))+(4*ln(x)-4*exp(x-2)-9
*x)*ln(-ln(x)+exp(x-2)+9/4*x)-4*x*exp(x-2)-9*x+4)/(4*x^2*ln(x)-4*x^2*exp(x-2)-9*x^3)/ln(-ln(x)+exp(x-2)+9/4*x)
,x,method=_RETURNVERBOSE)

[Out]

1/x*ln(ln(-ln(x)+exp(x-2)+9/4*x))+1/2*(-I*Pi*csgn(I*x)*csgn(I*ln(-ln(x)+exp(x-2)+9/4*x))*csgn(I*x*ln(-ln(x)+ex
p(x-2)+9/4*x))+I*Pi*csgn(I*x)*csgn(I*x*ln(-ln(x)+exp(x-2)+9/4*x))^2+I*Pi*csgn(I*ln(-ln(x)+exp(x-2)+9/4*x))*csg
n(I*x*ln(-ln(x)+exp(x-2)+9/4*x))^2-I*Pi*csgn(I*x*ln(-ln(x)+exp(x-2)+9/4*x))^3+2*ln(x))/x

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maxima [A]  time = 0.60, size = 31, normalized size = 1.35 $$\begin{array}{c}\frac{\log (x)+\log (-2\log (2)+\log (9x{e}^2-4e^2\log (x)+4e^x)-2)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*log(x)+4*exp(x-2)+9*x)*log(-log(x)+exp(x-2)+9/4*x)*log(x*log(-log(x)+exp(x-2)+9/4*x))+(4*log(x)
-4*exp(x-2)-9*x)*log(-log(x)+exp(x-2)+9/4*x)-4*x*exp(x-2)-9*x+4)/(4*x^2*log(x)-4*x^2*exp(x-2)-9*x^3)/log(-log(
x)+exp(x-2)+9/4*x),x, algorithm="maxima")

[Out]

(log(x) + log(-2*log(2) + log(9*x*e^2 - 4*e^2*log(x) + 4*e^x) - 2))/x

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mupad [B]  time = 4.99, size = 21, normalized size = 0.91 $$\begin{array}{c}\frac{\ln (x\ln (\frac{9x}{4}-\ln (x)+{\mathrm{e}}^{-2}{\mathrm{e}}^x))}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x + 4*x*exp(x - 2) + log((9*x)/4 + exp(x - 2) - log(x))*(9*x + 4*exp(x - 2) - 4*log(x)) - log((9*x)/4 +
exp(x - 2) - log(x))*log(x*log((9*x)/4 + exp(x - 2) - log(x)))*(9*x + 4*exp(x - 2) - 4*log(x)) - 4)/(log((9*x
)/4 + exp(x - 2) - log(x))*(4*x^2*exp(x - 2) - 4*x^2*log(x) + 9*x^3)),x)

[Out]

log(x*log((9*x)/4 - log(x) + exp(-2)*exp(x)))/x

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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(((-4*ln(x)+4*exp(x-2)+9*x)*ln(-ln(x)+exp(x-2)+9/4*x)*ln(x*ln(-ln(x)+exp(x-2)+9/4*x))+(4*ln(x)-4*exp(
x-2)-9*x)*ln(-ln(x)+exp(x-2)+9/4*x)-4*x*exp(x-2)-9*x+4)/(4*x**2*ln(x)-4*x**2*exp(x-2)-9*x**3)/ln(-ln(x)+exp(x-
2)+9/4*x),x)

[Out]

Timed out

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