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

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

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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 {gather*} \frac {\log \left (x \log \left (\frac {9 x}{4}+e^{x-2}-\log (x)\right )\right )}{x} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {e^2 \left (4-9 x+9 x^2-4 x \log (x)\right )}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}+\frac {x+\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )-\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right ) \log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x^2 \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}\right ) \, dx\\ &=-\left (e^2 \int \frac {4-9 x+9 x^2-4 x \log (x)}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\right )+\int \frac {x+\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )-\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right ) \log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x^2 \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\\ &=-\left (e^2 \int \left (\frac {9}{\left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}+\frac {4}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}-\frac {9}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}-\frac {4 \log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}\right ) \, dx\right )+\int \frac {1+\frac {x}{\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}-\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x^2} \, dx\\ &=-\left (\left (4 e^2\right ) \int \frac {1}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\right )+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (9 e^2\right ) \int \frac {1}{\left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (9 e^2\right ) \int \frac {1}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\int \left (\frac {x+\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}{x^2 \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}-\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x^2}\right ) \, dx\\ &=-\left (\left (4 e^2\right ) \int \frac {1}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\right )+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (9 e^2\right ) \int \frac {1}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (36 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (4 e^{4 x}+36 e^2 x-4 e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )+\int \frac {x+\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}{x^2 \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\int \frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x^2} \, dx\\ &=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}-\left (4 e^2\right ) \int \frac {1}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (9 e^2\right ) \int \frac {1}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (36 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )+\int \frac {1+\frac {x}{\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}}{x^2} \, dx-\int \frac {\frac {1}{x}+\frac {\frac {9}{4}+e^{-2+x}-\frac {1}{x}}{\left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}}{x} \, dx\\ &=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}-\left (4 e^2\right ) \int \frac {1}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (9 e^2\right ) \int \frac {1}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )+\int \left (\frac {1}{x^2}+\frac {1}{x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}\right ) \, dx-\int \left (-\frac {e^2 \left (4-9 x+9 x^2-4 x \log (x)\right )}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}+\frac {x+\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}{x^2 \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}\right ) \, dx\\ &=-\frac {1}{x}+\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}+e^2 \int \frac {4-9 x+9 x^2-4 x \log (x)}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (4 e^2\right ) \int \frac {1}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (9 e^2\right ) \int \frac {1}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )+\int \frac {1}{x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\int \frac {x+\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}{x^2 \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\\ &=-\frac {1}{x}+\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}+e^2 \int \left (\frac {9}{\left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}+\frac {4}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}-\frac {9}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}-\frac {4 \log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}\right ) \, dx-\left (4 e^2\right ) \int \frac {1}{x^2 \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (4 e^2\right ) \int \frac {\log (x)}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx+\left (9 e^2\right ) \int \frac {1}{x \left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )-\int \frac {1+\frac {x}{\log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}}{x^2} \, dx+\int \frac {1}{x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\\ &=-\frac {1}{x}+\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}+\left (9 e^2\right ) \int \frac {1}{\left (4 e^x+9 e^2 x-4 e^2 \log (x)\right ) \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx-\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )-\int \left (\frac {1}{x^2}+\frac {1}{x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )}\right ) \, dx+\int \frac {1}{x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )} \, dx\\ &=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}-\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )+\left (36 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (4 e^{4 x}+36 e^2 x-4 e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )\\ &=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}-\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )+\left (36 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (e^{4 x}+9 e^2 x-e^2 \log (4 x)\right ) \log \left (e^{-2+4 x}+9 x-\log (4 x)\right )} \, dx,x,\frac {x}{4}\right )\\ &=\frac {\log \left (x \log \left (e^{-2+x}+\frac {9 x}{4}-\log (x)\right )\right )}{x}\\ \end {aligned} \end {gather*}

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

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 {gather*} \frac {\log \left (x \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \relax (x)\right )\right )}{x} \end {gather*}

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 {gather*} \int -\frac {{\left (9 \, x + 4 \, e^{\left (x - 2\right )} - 4 \, \log \relax (x)\right )} \log \left (x \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \relax (x)\right )\right ) \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \relax (x)\right ) - 4 \, x e^{\left (x - 2\right )} - {\left (9 \, x + 4 \, e^{\left (x - 2\right )} - 4 \, \log \relax (x)\right )} \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \relax (x)\right ) - 9 \, x + 4}{{\left (9 \, x^{3} + 4 \, x^{2} e^{\left (x - 2\right )} - 4 \, x^{2} \log \relax (x)\right )} \log \left (\frac {9}{4} \, x + e^{\left (x - 2\right )} - \log \relax (x)\right )}\,{d x} \end {gather*}

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	\(\frac {\ln \left (\ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right )}{x}+\frac {-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right ) \mathrm {csgn}\left (i x \ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right )+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right ) \mathrm {csgn}\left (i x \ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right )^{2}-i \pi \mathrm {csgn}\left (i x \ln \left (-\ln \relax (x )+{\mathrm e}^{x -2}+\frac {9 x}{4}\right )\right )^{3}+2 \ln \relax (x )}{2 x}\)	\(168\)

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 {gather*} \frac {\log \relax (x) + \log \left (-2 \, \log \relax (2) + \log \left (9 \, x e^{2} - 4 \, e^{2} \log \relax (x) + 4 \, e^{x}\right ) - 2\right )}{x} \end {gather*}

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 {gather*} \frac {\ln \left (x\,\ln \left (\frac {9\,x}{4}-\ln \relax (x)+{\mathrm {e}}^{-2}\,{\mathrm {e}}^x\right )\right )}{x} \end {gather*}

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 {gather*} \text {Timed out} \end {gather*}

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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