3.87.21 \(\int \frac {2 e^6 (17+8 x-\log (x))}{(18 x+4 x^2-x \log (x)) (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 (-18 x-4 x^2+x \log (x))}{18 x+4 x^2-x \log (x)})} \, dx\)

Optimal. Leaf size=24 \[ \log \left (3-\frac {e^6}{x \left (9+2 x-\frac {\log (x)}{2}\right )}\right ) \]

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Rubi [F]  time = 1.03, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {2 e^6 (17+8 x-\log (x))}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*E^6*(17 + 8*x - Log[x]))/((18*x + 4*x^2 - x*Log[x])*(54*x + 12*x^2 - 3*x*Log[x] + (2*E^6*(-18*x - 4*x^2
+ x*Log[x]))/(18*x + 4*x^2 - x*Log[x]))),x]

[Out]

-Log[18 + 4*x - Log[x]] + 2*E^6*Defer[Int][1/(x*(-2*E^6 + 54*x + 12*x^2 - 3*x*Log[x])), x] + 12*Defer[Int][x/(
-2*E^6 + 54*x + 12*x^2 - 3*x*Log[x]), x] + 3*Defer[Int][(2*E^6 - 54*x - 12*x^2 + 3*x*Log[x])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (2 e^6\right ) \int \frac {17+8 x-\log (x)}{\left (18 x+4 x^2-x \log (x)\right ) \left (54 x+12 x^2-3 x \log (x)+\frac {2 e^6 \left (-18 x-4 x^2+x \log (x)\right )}{18 x+4 x^2-x \log (x)}\right )} \, dx\\ &=\left (2 e^6\right ) \int \frac {-17-8 x+\log (x)}{x (18+4 x-\log (x)) \left (2 e^6-6 x (9+2 x)+3 x \log (x)\right )} \, dx\\ &=\left (2 e^6\right ) \int \left (\frac {1-4 x}{2 e^6 x (18+4 x-\log (x))}+\frac {-2 e^6+3 x-12 x^2}{2 e^6 x \left (2 e^6-54 x-12 x^2+3 x \log (x)\right )}\right ) \, dx\\ &=\int \frac {1-4 x}{x (18+4 x-\log (x))} \, dx+\int \frac {-2 e^6+3 x-12 x^2}{x \left (2 e^6-54 x-12 x^2+3 x \log (x)\right )} \, dx\\ &=-\log (18+4 x-\log (x))+\int \left (\frac {2 e^6}{x \left (-2 e^6+54 x+12 x^2-3 x \log (x)\right )}+\frac {12 x}{-2 e^6+54 x+12 x^2-3 x \log (x)}+\frac {3}{2 e^6-54 x-12 x^2+3 x \log (x)}\right ) \, dx\\ &=-\log (18+4 x-\log (x))+3 \int \frac {1}{2 e^6-54 x-12 x^2+3 x \log (x)} \, dx+12 \int \frac {x}{-2 e^6+54 x+12 x^2-3 x \log (x)} \, dx+\left (2 e^6\right ) \int \frac {1}{x \left (-2 e^6+54 x+12 x^2-3 x \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.25, size = 59, normalized size = 2.46 \begin {gather*} 2 e^6 \left (-\frac {\log (x)}{2 e^6}-\frac {\log (18+4 x-\log (x))}{2 e^6}+\frac {\log \left (2 e^6-54 x-12 x^2+3 x \log (x)\right )}{2 e^6}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^6*(17 + 8*x - Log[x]))/((18*x + 4*x^2 - x*Log[x])*(54*x + 12*x^2 - 3*x*Log[x] + (2*E^6*(-18*x -
4*x^2 + x*Log[x]))/(18*x + 4*x^2 - x*Log[x]))),x]

[Out]

2*E^6*(-1/2*Log[x]/E^6 - Log[18 + 4*x - Log[x]]/(2*E^6) + Log[2*E^6 - 54*x - 12*x^2 + 3*x*Log[x]]/(2*E^6))

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fricas [A]  time = 0.46, size = 35, normalized size = 1.46 \begin {gather*} -\log \left (-4 \, x + \log \relax (x) - 18\right ) + \log \left (-\frac {12 \, x^{2} - 3 \, x \log \relax (x) + 54 \, x - 2 \, e^{6}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)+8*x+17)*exp(-log(-1/2*x*log(x)+2*x^2+9*x)+6)/((x*log(x)-4*x^2-18*x)*exp(-log(-1/2*x*log(x)+
2*x^2+9*x)+6)-3*x*log(x)+12*x^2+54*x),x, algorithm="fricas")

[Out]

-log(-4*x + log(x) - 18) + log(-(12*x^2 - 3*x*log(x) + 54*x - 2*e^6)/x)

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giac [A]  time = 0.14, size = 36, normalized size = 1.50 \begin {gather*} \log \left (-12 \, x^{2} + 3 \, x \log \relax (x) - 54 \, x + 2 \, e^{6}\right ) - \log \left (4 \, x - \log \relax (x) + 18\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)+8*x+17)*exp(-log(-1/2*x*log(x)+2*x^2+9*x)+6)/((x*log(x)-4*x^2-18*x)*exp(-log(-1/2*x*log(x)+
2*x^2+9*x)+6)-3*x*log(x)+12*x^2+54*x),x, algorithm="giac")

[Out]

log(-12*x^2 + 3*x*log(x) - 54*x + 2*e^6) - log(4*x - log(x) + 18) - log(x)

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maple [C]  time = 0.14, size = 230, normalized size = 9.58

method	result	size
risch	\(-\ln \left (\ln \relax (x )-4 x -18\right )-\ln \relax (2)+\ln \left (-\frac {\ln \relax (x )}{4}+x +\frac {9}{2}\right )+\frac {i \pi \,\mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right ) \left (\mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (\mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )-\mathrm {csgn}\left (i \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )\right )}{2}-6+\ln \left (\frac {2 \,{\mathrm e}^{6} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )^{2} \mathrm {csgn}\left (i x \right )}{2}} {\mathrm e}^{\frac {i \pi \mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )^{2} \mathrm {csgn}\left (i \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )}{2}} {\mathrm e}^{\frac {i \pi \,\mathrm {csgn}\left (i x \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (\frac {\ln \relax (x )}{4}-x -\frac {9}{2}\right )\right )}{2}}}{x \left (-\frac {\ln \relax (x )}{4}+x +\frac {9}{2}\right )}-3\right )\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)+8*x+17)*exp(-ln(-1/2*x*ln(x)+2*x^2+9*x)+6)/((x*ln(x)-4*x^2-18*x)*exp(-ln(-1/2*x*ln(x)+2*x^2+9*x)+6
)-3*x*ln(x)+12*x^2+54*x),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x)-4*x-18)-ln(2)+ln(-1/4*ln(x)+x+9/2)+1/2*I*Pi*csgn(I*x*(1/4*ln(x)-x-9/2))*(csgn(I*x*(1/4*ln(x)-x-9/2))
+csgn(I*x))*(csgn(I*x*(1/4*ln(x)-x-9/2))-csgn(I*(1/4*ln(x)-x-9/2)))-6+ln(2/x/(-1/4*ln(x)+x+9/2)*exp(6)*exp(-1/
2*I*Pi*csgn(I*x*(1/4*ln(x)-x-9/2))^3)*exp(-1/2*I*Pi*csgn(I*x*(1/4*ln(x)-x-9/2))^2*csgn(I*x))*exp(1/2*I*Pi*csgn
(I*x*(1/4*ln(x)-x-9/2))^2*csgn(I*(1/4*ln(x)-x-9/2)))*exp(1/2*I*Pi*csgn(I*x*(1/4*ln(x)-x-9/2))*csgn(I*x)*csgn(I
*(1/4*ln(x)-x-9/2)))-3)

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maxima [A]  time = 0.42, size = 44, normalized size = 1.83 \begin {gather*} -{\left (e^{\left (-6\right )} \log \left (-4 \, x + \log \relax (x) - 18\right ) - e^{\left (-6\right )} \log \left (-\frac {12 \, x^{2} - 3 \, x \log \relax (x) + 54 \, x - 2 \, e^{6}}{3 \, x}\right )\right )} e^{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)+8*x+17)*exp(-log(-1/2*x*log(x)+2*x^2+9*x)+6)/((x*log(x)-4*x^2-18*x)*exp(-log(-1/2*x*log(x)+
2*x^2+9*x)+6)-3*x*log(x)+12*x^2+54*x),x, algorithm="maxima")

[Out]

-(e^(-6)*log(-4*x + log(x) - 18) - e^(-6)*log(-1/3*(12*x^2 - 3*x*log(x) + 54*x - 2*e^6)/x))*e^6

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mupad [B]  time = 5.70, size = 36, normalized size = 1.50 \begin {gather*} \ln \left (54\,x-2\,{\mathrm {e}}^6-3\,x\,\ln \relax (x)+12\,x^2\right )-\ln \relax (x)-\ln \left (4\,x-\ln \relax (x)+18\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(6 - log(9*x - (x*log(x))/2 + 2*x^2))*(8*x - log(x) + 17))/(54*x - exp(6 - log(9*x - (x*log(x))/2 + 2*
x^2))*(18*x - x*log(x) + 4*x^2) - 3*x*log(x) + 12*x^2),x)

[Out]

log(54*x - 2*exp(6) - 3*x*log(x) + 12*x^2) - log(x) - log(4*x - log(x) + 18)

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sympy [B]  time = 0.43, size = 39, normalized size = 1.62 \begin {gather*} - \log {\left (\log {\relax (x )} + \frac {- 12 x^{2} - 54 x}{3 x} \right )} + \log {\left (\log {\relax (x )} + \frac {- 12 x^{2} - 54 x + 2 e^{6}}{3 x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)+8*x+17)*exp(-ln(-1/2*x*ln(x)+2*x**2+9*x)+6)/((x*ln(x)-4*x**2-18*x)*exp(-ln(-1/2*x*ln(x)+2*x*
*2+9*x)+6)-3*x*ln(x)+12*x**2+54*x),x)

[Out]

-log(log(x) + (-12*x**2 - 54*x)/(3*x)) + log(log(x) + (-12*x**2 - 54*x + 2*exp(6))/(3*x))

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