3.4.7 \(\int \frac {e^3 (-3+5 x-x^2)+e^6 (-12+20 x-4 x^2) \log (x)+(e^6 (-12 x+4 x^2)+e^6 (12-4 x) \log (3 x-x^2)) \log (-x+\log (3 x-x^2))}{30 x^2-10 x^3+(-30 x+10 x^2) \log (3 x-x^2)+\log (x) (e^3 (240 x^2-80 x^3)+e^3 (-240 x+80 x^2) \log (3 x-x^2))+\log ^2(x) (e^6 (480 x^2-160 x^3)+e^6 (-480 x+160 x^2) \log (3 x-x^2))} \, dx\)

Optimal. Leaf size=29 \[ \frac {\log (-x+\log (x+(2-x) x))}{10 \left (\frac {1}{e^3}+4 \log (x)\right )} \]

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Rubi [F]  time = 12.05, 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 {e^3 \left (-3+5 x-x^2\right )+e^6 \left (-12+20 x-4 x^2\right ) \log (x)+\left (e^6 \left (-12 x+4 x^2\right )+e^6 (12-4 x) \log \left (3 x-x^2\right )\right ) \log \left (-x+\log \left (3 x-x^2\right )\right )}{30 x^2-10 x^3+\left (-30 x+10 x^2\right ) \log \left (3 x-x^2\right )+\log (x) \left (e^3 \left (240 x^2-80 x^3\right )+e^3 \left (-240 x+80 x^2\right ) \log \left (3 x-x^2\right )\right )+\log ^2(x) \left (e^6 \left (480 x^2-160 x^3\right )+e^6 \left (-480 x+160 x^2\right ) \log \left (3 x-x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^3*(-3 + 5*x - x^2) + E^6*(-12 + 20*x - 4*x^2)*Log[x] + (E^6*(-12*x + 4*x^2) + E^6*(12 - 4*x)*Log[3*x -
x^2])*Log[-x + Log[3*x - x^2]])/(30*x^2 - 10*x^3 + (-30*x + 10*x^2)*Log[3*x - x^2] + Log[x]*(E^3*(240*x^2 - 80
*x^3) + E^3*(-240*x + 80*x^2)*Log[3*x - x^2]) + Log[x]^2*(E^6*(480*x^2 - 160*x^3) + E^6*(-480*x + 160*x^2)*Log
[3*x - x^2])),x]

[Out]

(E^3*Defer[Int][1/((1 + 4*E^3*Log[x])^2*(x - Log[-((-3 + x)*x)])), x])/10 - (E^3*Defer[Int][1/((-3 + x)*(1 + 4
*E^3*Log[x])^2*(x - Log[-((-3 + x)*x)])), x])/10 - (E^3*Defer[Int][1/(x*(1 + 4*E^3*Log[x])^2*(x - Log[-((-3 +
x)*x)])), x])/10 + (2*E^6*Defer[Int][Log[x]/((1 + 4*E^3*Log[x])^2*(x - Log[-((-3 + x)*x)])), x])/5 - (2*E^6*De
fer[Int][Log[x]/((-3 + x)*(1 + 4*E^3*Log[x])^2*(x - Log[-((-3 + x)*x)])), x])/5 - (2*E^6*Defer[Int][Log[x]/(x*
(1 + 4*E^3*Log[x])^2*(x - Log[-((-3 + x)*x)])), x])/5 - (2*E^6*Defer[Int][Log[-x + Log[-((-3 + x)*x)]]/(x*(1 +
4*E^3*Log[x])^2), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (-3+5 x-x^2-4 e^3 \left (3-5 x+x^2\right ) \log (x)+4 e^3 (-3+x) (x-\log (-((-3+x) x))) \log (-x+\log (-((-3+x) x)))\right )}{10 (3-x) x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx\\ &=\frac {1}{10} e^3 \int \frac {-3+5 x-x^2-4 e^3 \left (3-5 x+x^2\right ) \log (x)+4 e^3 (-3+x) (x-\log (-((-3+x) x))) \log (-x+\log (-((-3+x) x)))}{(3-x) x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx\\ &=\frac {1}{10} e^3 \int \left (-\frac {5}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}+\frac {3}{(-3+x) x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}+\frac {x}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}+\frac {4 e^3 \left (3-5 x+x^2\right ) \log (x)}{(-3+x) x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}-\frac {4 e^3 \log (-x+\log (-((-3+x) x)))}{x \left (1+4 e^3 \log (x)\right )^2}\right ) \, dx\\ &=\frac {1}{10} e^3 \int \frac {x}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx+\frac {1}{10} \left (3 e^3\right ) \int \frac {1}{(-3+x) x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{2} e^3 \int \frac {1}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx+\frac {1}{5} \left (2 e^6\right ) \int \frac {\left (3-5 x+x^2\right ) \log (x)}{(-3+x) x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{5} \left (2 e^6\right ) \int \frac {\log (-x+\log (-((-3+x) x)))}{x \left (1+4 e^3 \log (x)\right )^2} \, dx\\ &=\frac {1}{10} e^3 \int \left (\frac {1}{\left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}+\frac {3}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}\right ) \, dx+\frac {1}{10} \left (3 e^3\right ) \int \left (\frac {1}{3 (-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}-\frac {1}{3 x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}\right ) \, dx-\frac {1}{2} e^3 \int \frac {1}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx+\frac {1}{5} \left (2 e^6\right ) \int \left (\frac {\log (x)}{\left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}-\frac {\log (x)}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}-\frac {\log (x)}{x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))}\right ) \, dx-\frac {1}{5} \left (2 e^6\right ) \int \frac {\log (-x+\log (-((-3+x) x)))}{x \left (1+4 e^3 \log (x)\right )^2} \, dx\\ &=\frac {1}{10} e^3 \int \frac {1}{\left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx+\frac {1}{10} e^3 \int \frac {1}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{10} e^3 \int \frac {1}{x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx+\frac {1}{10} \left (3 e^3\right ) \int \frac {1}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{2} e^3 \int \frac {1}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx+\frac {1}{5} \left (2 e^6\right ) \int \frac {\log (x)}{\left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{5} \left (2 e^6\right ) \int \frac {\log (x)}{(-3+x) \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{5} \left (2 e^6\right ) \int \frac {\log (x)}{x \left (1+4 e^3 \log (x)\right )^2 (x-\log (-((-3+x) x)))} \, dx-\frac {1}{5} \left (2 e^6\right ) \int \frac {\log (-x+\log (-((-3+x) x)))}{x \left (1+4 e^3 \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 30, normalized size = 1.03 \begin {gather*} \frac {e^3 \log (-x+\log (-((-3+x) x)))}{10 \left (1+4 e^3 \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(-3 + 5*x - x^2) + E^6*(-12 + 20*x - 4*x^2)*Log[x] + (E^6*(-12*x + 4*x^2) + E^6*(12 - 4*x)*Log[
3*x - x^2])*Log[-x + Log[3*x - x^2]])/(30*x^2 - 10*x^3 + (-30*x + 10*x^2)*Log[3*x - x^2] + Log[x]*(E^3*(240*x^
2 - 80*x^3) + E^3*(-240*x + 80*x^2)*Log[3*x - x^2]) + Log[x]^2*(E^6*(480*x^2 - 160*x^3) + E^6*(-480*x + 160*x^
2)*Log[3*x - x^2])),x]

[Out]

(E^3*Log[-x + Log[-((-3 + x)*x)]])/(10*(1 + 4*E^3*Log[x]))

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fricas [A]  time = 1.01, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{3} \log \left (-x + \log \left (-x^{2} + 3 \, x\right )\right )}{10 \, {\left (4 \, e^{3} \log \relax (x) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+12)*exp(3)^2*log(-x^2+3*x)+(4*x^2-12*x)*exp(3)^2)*log(log(-x^2+3*x)-x)+(-4*x^2+20*x-12)*exp(
3)^2*log(x)+(-x^2+5*x-3)*exp(3))/(((160*x^2-480*x)*exp(3)^2*log(-x^2+3*x)+(-160*x^3+480*x^2)*exp(3)^2)*log(x)^
2+((80*x^2-240*x)*exp(3)*log(-x^2+3*x)+(-80*x^3+240*x^2)*exp(3))*log(x)+(10*x^2-30*x)*log(-x^2+3*x)-10*x^3+30*
x^2),x, algorithm="fricas")

[Out]

1/10*e^3*log(-x + log(-x^2 + 3*x))/(4*e^3*log(x) + 1)

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giac [A]  time = 0.71, size = 29, normalized size = 1.00 \begin {gather*} \frac {e^{3} \log \left (-x + \log \left (-x^{2} + 3 \, x\right )\right )}{10 \, {\left (4 \, e^{3} \log \relax (x) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+12)*exp(3)^2*log(-x^2+3*x)+(4*x^2-12*x)*exp(3)^2)*log(log(-x^2+3*x)-x)+(-4*x^2+20*x-12)*exp(
3)^2*log(x)+(-x^2+5*x-3)*exp(3))/(((160*x^2-480*x)*exp(3)^2*log(-x^2+3*x)+(-160*x^3+480*x^2)*exp(3)^2)*log(x)^
2+((80*x^2-240*x)*exp(3)*log(-x^2+3*x)+(-80*x^3+240*x^2)*exp(3))*log(x)+(10*x^2-30*x)*log(-x^2+3*x)-10*x^3+30*
x^2),x, algorithm="giac")

[Out]

1/10*e^3*log(-x + log(-x^2 + 3*x))/(4*e^3*log(x) + 1)

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maple [C]  time = 0.25, size = 100, normalized size = 3.45

method	result	size
risch	\(\frac {{\mathrm e}^{3} \ln \left (i \pi +\ln \relax (x )+\ln \left (x -3\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \left (x -3\right )\right ) \left (-\mathrm {csgn}\left (i x \left (x -3\right )\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \left (x -3\right )\right )+\mathrm {csgn}\left (i \left (x -3\right )\right )\right )}{2}+i \pi \mathrm {csgn}\left (i x \left (x -3\right )\right )^{2} \left (\mathrm {csgn}\left (i x \left (x -3\right )\right )-1\right )-x \right )}{40 \ln \relax (x ) {\mathrm e}^{3}+10}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x+12)*exp(3)^2*ln(-x^2+3*x)+(4*x^2-12*x)*exp(3)^2)*ln(ln(-x^2+3*x)-x)+(-4*x^2+20*x-12)*exp(3)^2*ln(x
)+(-x^2+5*x-3)*exp(3))/(((160*x^2-480*x)*exp(3)^2*ln(-x^2+3*x)+(-160*x^3+480*x^2)*exp(3)^2)*ln(x)^2+((80*x^2-2
40*x)*exp(3)*ln(-x^2+3*x)+(-80*x^3+240*x^2)*exp(3))*ln(x)+(10*x^2-30*x)*ln(-x^2+3*x)-10*x^3+30*x^2),x,method=_
RETURNVERBOSE)

[Out]

1/10*exp(3)/(4*ln(x)*exp(3)+1)*ln(I*Pi+ln(x)+ln(x-3)-1/2*I*Pi*csgn(I*x*(x-3))*(-csgn(I*x*(x-3))+csgn(I*x))*(-c
sgn(I*x*(x-3))+csgn(I*(x-3)))+I*Pi*csgn(I*x*(x-3))^2*(csgn(I*x*(x-3))-1)-x)

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maxima [A]  time = 0.43, size = 27, normalized size = 0.93 \begin {gather*} \frac {e^{3} \log \left (-x + \log \relax (x) + \log \left (-x + 3\right )\right )}{10 \, {\left (4 \, e^{3} \log \relax (x) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+12)*exp(3)^2*log(-x^2+3*x)+(4*x^2-12*x)*exp(3)^2)*log(log(-x^2+3*x)-x)+(-4*x^2+20*x-12)*exp(
3)^2*log(x)+(-x^2+5*x-3)*exp(3))/(((160*x^2-480*x)*exp(3)^2*log(-x^2+3*x)+(-160*x^3+480*x^2)*exp(3)^2)*log(x)^
2+((80*x^2-240*x)*exp(3)*log(-x^2+3*x)+(-80*x^3+240*x^2)*exp(3))*log(x)+(10*x^2-30*x)*log(-x^2+3*x)-10*x^3+30*
x^2),x, algorithm="maxima")

[Out]

1/10*e^3*log(-x + log(x) + log(-x + 3))/(4*e^3*log(x) + 1)

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mupad [B]  time = 1.00, size = 28, normalized size = 0.97 \begin {gather*} \frac {{\mathrm {e}}^3\,\ln \left (\ln \left (3\,x-x^2\right )-x\right )}{40\,{\mathrm {e}}^3\,\ln \relax (x)+10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(3*x - x^2) - x)*(exp(6)*(12*x - 4*x^2) + exp(6)*log(3*x - x^2)*(4*x - 12)) + exp(3)*(x^2 - 5*x +
3) + exp(6)*log(x)*(4*x^2 - 20*x + 12))/(log(x)*(exp(3)*(240*x^2 - 80*x^3) - exp(3)*log(3*x - x^2)*(240*x - 8
0*x^2)) + log(x)^2*(exp(6)*(480*x^2 - 160*x^3) - exp(6)*log(3*x - x^2)*(480*x - 160*x^2)) - log(3*x - x^2)*(30
*x - 10*x^2) + 30*x^2 - 10*x^3),x)

[Out]

(exp(3)*log(log(3*x - x^2) - x))/(40*exp(3)*log(x) + 10)

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sympy [A]  time = 1.10, size = 24, normalized size = 0.83 \begin {gather*} \frac {e^{3} \log {\left (- x + \log {\left (- x^{2} + 3 x \right )} \right )}}{40 e^{3} \log {\relax (x )} + 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x+12)*exp(3)**2*ln(-x**2+3*x)+(4*x**2-12*x)*exp(3)**2)*ln(ln(-x**2+3*x)-x)+(-4*x**2+20*x-12)*e
xp(3)**2*ln(x)+(-x**2+5*x-3)*exp(3))/(((160*x**2-480*x)*exp(3)**2*ln(-x**2+3*x)+(-160*x**3+480*x**2)*exp(3)**2
)*ln(x)**2+((80*x**2-240*x)*exp(3)*ln(-x**2+3*x)+(-80*x**3+240*x**2)*exp(3))*ln(x)+(10*x**2-30*x)*ln(-x**2+3*x
)-10*x**3+30*x**2),x)

[Out]

exp(3)*log(-x + log(-x**2 + 3*x))/(40*exp(3)*log(x) + 10)

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