3.41.8 $\int \frac{e^{-6+3e^3}(-1-\log (4))+e^{-6+3e^3}\log (x)}{-x\log (4)+x\log (x)}dx$

Optimal. Leaf size=23 $$e^{3(-2+e^3)}(\log (x)-\log (\log (4)-\log (x)))$$

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Rubi [A]  time = 0.05, antiderivative size = 32, normalized size of antiderivative = 1.39, number of steps used = 4, number of rules used = 2, integrand size = 42, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.048, Rules used = {12, 43} $$\begin{array}{c}{e}^{3e^3-6}\log (x)-e^{3e^3-6}\log (\log (4)-\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(E^(-6 + 3*E^3)*(-1 - Log[4]) + E^(-6 + 3*E^3)*Log[x])/(-(x*Log[4]) + x*Log[x]),x]

[Out]

E^(-6 + 3*E^3)*Log[x] - E^(-6 + 3*E^3)*Log[Log[4] - 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\mathrm{Subst}(\int \frac{e^{-6+3e^3}(1-x+\log (4))}{-x+\log (4)}dx,x,\log (x))\\ & =e^{-6+3e^3}\mathrm{Subst}(\int \frac{1-x+\log (4)}{-x+\log (4)}dx,x,\log (x))\\ & =e^{-6+3e^3}\mathrm{Subst}(\int (1+\frac{1}{-x+\log (4)})dx,x,\log (x))\\ & =e^{-6+3e^3}\log (x)-e^{-6+3e^3}\log (\log (4)-\log (x))\end{array}\end{array}$$

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Mathematica [A]  time = 0.04, size = 26, normalized size = 1.13 $$\begin{array}{c}{e}^{-6+3e^3}(\log \left(\frac{x}{4}\right)-\log (\log \left(\frac{x}{4}\right)))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^(-6 + 3*E^3)*(-1 - Log[4]) + E^(-6 + 3*E^3)*Log[x])/(-(x*Log[4]) + x*Log[x]),x]

[Out]

E^(-6 + 3*E^3)*(Log[x/4] - Log[Log[x/4]])

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fricas [A]  time = 0.61, size = 28, normalized size = 1.22 $$\begin{array}{c}{e}^{(3e^3-6)}\log (x)-e^{(3e^3-6)}\log (-2\log (2)+\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3*exp(3)-6)*log(x)+(-1-2*log(2))*exp(3*exp(3)-6))/(x*log(x)-2*x*log(2)),x, algorithm="fricas")

[Out]

e^(3*e^3 - 6)*log(x) - e^(3*e^3 - 6)*log(-2*log(2) + log(x))

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giac [A]  time = 0.15, size = 45, normalized size = 1.96 $$\begin{array}{c}-\frac{1}{2}{e}^{(3e^3-6)}\log (\frac{1}{4}{\pi }^2{(\mathrm{sgn}(x)-1)}^2+{(2\log (2)-\log \left(\left|x\right|\right))}^2)+e^{(3e^3-6)}\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3*exp(3)-6)*log(x)+(-1-2*log(2))*exp(3*exp(3)-6))/(x*log(x)-2*x*log(2)),x, algorithm="giac")

[Out]

-1/2*e^(3*e^3 - 6)*log(1/4*pi^2*(sgn(x) - 1)^2 + (2*log(2) - log(abs(x)))^2) + e^(3*e^3 - 6)*log(x)

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maple [A]  time = 0.04, size = 22, normalized size = 0.96

method	result	size
default	${\mathrm{e}}^{3{\mathrm{e}}^3}{\mathrm{e}}^{-6}(\ln (x)-\ln (\ln (x)-2\ln (2)))$	$22$
risch	${\mathrm{e}}^{3{\mathrm{e}}^3-6}\ln (x)-\ln (\ln (x)-2\ln (2)){\mathrm{e}}^{3{\mathrm{e}}^3-6}$	$29$
norman	${\mathrm{e}}^{3{\mathrm{e}}^3}{\mathrm{e}}^{-6}\ln (x)-{\mathrm{e}}^{3{\mathrm{e}}^3}{\mathrm{e}}^{-6}\ln (2\ln (2)-\ln (x))$	$31$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*exp(3)-6)*ln(x)+(-1-2*ln(2))*exp(3*exp(3)-6))/(x*ln(x)-2*x*ln(2)),x,method=_RETURNVERBOSE)

[Out]

exp(exp(3))^3*exp(-6)*(ln(x)-ln(ln(x)-2*ln(2)))

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maxima [B]  time = 0.43, size = 88, normalized size = 3.83 $$\begin{array}{c}-2e^{(3e^3-6)}\log (2)\log (-2\log (2)+\log (x))+e^{(3e^3-6)}\log (x)\log (-2\log (2)+\log (x))+((2\log (2)-\log (x))\log (-2\log (2)+\log (x))-2\log (2)+\log (x))e^{(3e^3-6)}-e^{(3e^3-6)}\log (-2\log (2)+\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3*exp(3)-6)*log(x)+(-1-2*log(2))*exp(3*exp(3)-6))/(x*log(x)-2*x*log(2)),x, algorithm="maxima")

[Out]

-2*e^(3*e^3 - 6)*log(2)*log(-2*log(2) + log(x)) + e^(3*e^3 - 6)*log(x)*log(-2*log(2) + log(x)) + ((2*log(2) -
log(x))*log(-2*log(2) + log(x)) - 2*log(2) + log(x))*e^(3*e^3 - 6) - e^(3*e^3 - 6)*log(-2*log(2) + log(x))

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mupad [B]  time = 2.59, size = 19, normalized size = 0.83 $$\begin{array}{c}-{\mathrm{e}}^{3{\mathrm{e}}^3-6}(\ln (\ln \left(\frac{x}{4}\right))-\ln (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*exp(3) - 6)*(2*log(2) + 1) - exp(3*exp(3) - 6)*log(x))/(2*x*log(2) - x*log(x)),x)

[Out]

-exp(3*exp(3) - 6)*(log(log(x/4)) - log(x))

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sympy [A]  time = 0.14, size = 32, normalized size = 1.39 $$\begin{array}{c}\frac{e^{3e^3}\log (x)}{e^6}-\frac{e^{3e^3}\log (\log (x)-2\log (2))}{e^6}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(3*exp(3)-6)*ln(x)+(-1-2*ln(2))*exp(3*exp(3)-6))/(x*ln(x)-2*x*ln(2)),x)

[Out]

exp(-6)*exp(3*exp(3))*log(x) - exp(-6)*exp(3*exp(3))*log(log(x) - 2*log(2))

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