3.53.89 \(\int \frac {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} (-2-7 x+6 x^2)+(3 e^4-3 x+e^{2 x} x) \log (\frac {1}{3} (15 e^4-15 x+5 e^{2 x} x))}{12 x^2-6 x^3+e^4 (-12 x+6 x^2)+e^{2 x} (-4 x^2+2 x^3)+(6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} (-2 x+x^2)) \log (\frac {1}{3} (15 e^4-15 x+5 e^{2 x} x))} \, dx\)

Optimal. Leaf size=31 \[ \log \left ((4-2 x) \left (2 x+\log \left (5 \left (e^4-x+\frac {1}{3} e^{2 x} x\right )\right )\right )\right ) \]

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Rubi [F]  time = 4.04, 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 {6+9 x-12 x^2+e^4 (-12+12 x)+e^{2 x} \left (-2-7 x+6 x^2\right )+\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )}{12 x^2-6 x^3+e^4 \left (-12 x+6 x^2\right )+e^{2 x} \left (-4 x^2+2 x^3\right )+\left (6 x-3 x^2+e^4 (-6+3 x)+e^{2 x} \left (-2 x+x^2\right )\right ) \log \left (\frac {1}{3} \left (15 e^4-15 x+5 e^{2 x} x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(6 + 9*x - 12*x^2 + E^4*(-12 + 12*x) + E^(2*x)*(-2 - 7*x + 6*x^2) + (3*E^4 - 3*x + E^(2*x)*x)*Log[(15*E^4
- 15*x + 5*E^(2*x)*x)/3])/(12*x^2 - 6*x^3 + E^4*(-12*x + 6*x^2) + E^(2*x)*(-4*x^2 + 2*x^3) + (6*x - 3*x^2 + E^
4*(-6 + 3*x) + E^(2*x)*(-2*x + x^2))*Log[(15*E^4 - 15*x + 5*E^(2*x)*x)/3]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6-12 e^4 (-1+x)-9 x+12 x^2-e^{2 x} \left (-2-7 x+6 x^2\right )-\left (3 e^4-3 x+e^{2 x} x\right ) \log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )}{(2-x) \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx\\ &=\int \left (\frac {3 \left (-e^4-2 e^4 x+2 x^2\right )}{x \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}+\frac {-2-7 x+6 x^2+x \log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )}{(-2+x) x \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}\right ) \, dx\\ &=3 \int \frac {-e^4-2 e^4 x+2 x^2}{x \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx+\int \frac {-2-7 x+6 x^2+x \log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )}{(-2+x) x \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx\\ &=3 \int \left (-\frac {2 e^4}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}-\frac {e^4}{x \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}+\frac {2 x}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}\right ) \, dx+\int \left (\frac {1}{-2+x}+\frac {1+4 x}{x \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}\right ) \, dx\\ &=\log (2-x)+6 \int \frac {x}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx-\left (3 e^4\right ) \int \frac {1}{x \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx-\left (6 e^4\right ) \int \frac {1}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx+\int \frac {1+4 x}{x \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx\\ &=\log (2-x)+6 \int \frac {x}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx-\left (3 e^4\right ) \int \frac {1}{x \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx-\left (6 e^4\right ) \int \frac {1}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx+\int \left (\frac {4}{2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )}+\frac {1}{x \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )}\right ) \, dx\\ &=\log (2-x)+4 \int \frac {1}{2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )} \, dx+6 \int \frac {x}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx-\left (3 e^4\right ) \int \frac {1}{x \left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx-\left (6 e^4\right ) \int \frac {1}{\left (3 e^4-3 x+e^{2 x} x\right ) \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx+\int \frac {1}{x \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 32, normalized size = 1.03 \begin {gather*} \log (2-x)+\log \left (2 x+\log \left (5 e^4-5 x+\frac {5}{3} e^{2 x} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + 9*x - 12*x^2 + E^4*(-12 + 12*x) + E^(2*x)*(-2 - 7*x + 6*x^2) + (3*E^4 - 3*x + E^(2*x)*x)*Log[(1
5*E^4 - 15*x + 5*E^(2*x)*x)/3])/(12*x^2 - 6*x^3 + E^4*(-12*x + 6*x^2) + E^(2*x)*(-4*x^2 + 2*x^3) + (6*x - 3*x^
2 + E^4*(-6 + 3*x) + E^(2*x)*(-2*x + x^2))*Log[(15*E^4 - 15*x + 5*E^(2*x)*x)/3]),x]

[Out]

Log[2 - x] + Log[2*x + Log[5*E^4 - 5*x + (5*E^(2*x)*x)/3]]

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fricas [A]  time = 0.53, size = 26, normalized size = 0.84 \begin {gather*} \log \left (2 \, x + \log \left (\frac {5}{3} \, x e^{\left (2 \, x\right )} - 5 \, x + 5 \, e^{4}\right )\right ) + \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-
12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(3*x-6)*exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*ex
p(2*x)+(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x, algorithm="fricas")

[Out]

log(2*x + log(5/3*x*e^(2*x) - 5*x + 5*e^4)) + log(x - 2)

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giac [A]  time = 0.38, size = 26, normalized size = 0.84 \begin {gather*} \log \left (2 \, x + \log \left (\frac {5}{3} \, x e^{\left (2 \, x\right )} - 5 \, x + 5 \, e^{4}\right )\right ) + \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-
12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(3*x-6)*exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*ex
p(2*x)+(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x, algorithm="giac")

[Out]

log(2*x + log(5/3*x*e^(2*x) - 5*x + 5*e^4)) + log(x - 2)

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maple [A]  time = 0.37, size = 27, normalized size = 0.87

method	result	size
norman	\(\ln \left (x -2\right )+\ln \left (2 x +\ln \left (\frac {5 x \,{\mathrm e}^{2 x}}{3}+5 \,{\mathrm e}^{4}-5 x \right )\right )\)	\(27\)
risch	\(\ln \left (x -2\right )+\ln \left (2 x +\ln \left (\frac {5 x \,{\mathrm e}^{2 x}}{3}+5 \,{\mathrm e}^{4}-5 x \right )\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(2*x)+3*exp(4)-3*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-12*x^2+
9*x+6)/(((x^2-2*x)*exp(2*x)+(3*x-6)*exp(4)-3*x^2+6*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*exp(2*x)+(
6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(x-2)+ln(2*x+ln(5/3*x*exp(2*x)+5*exp(4)-5*x))

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maxima [A]  time = 0.55, size = 31, normalized size = 1.00 \begin {gather*} \log \left (2 \, x + \log \relax (5) - \log \relax (3) + \log \left (x e^{\left (2 \, x\right )} - 3 \, x + 3 \, e^{4}\right )\right ) + \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2*x)+3*exp(4)-3*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x^2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-
12*x^2+9*x+6)/(((x^2-2*x)*exp(2*x)+(3*x-6)*exp(4)-3*x^2+6*x)*log(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x^3-4*x^2)*ex
p(2*x)+(6*x^2-12*x)*exp(4)-6*x^3+12*x^2),x, algorithm="maxima")

[Out]

log(2*x + log(5) - log(3) + log(x*e^(2*x) - 3*x + 3*e^4)) + log(x - 2)

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mupad [B]  time = 3.72, size = 26, normalized size = 0.84 \begin {gather*} \ln \left (2\,x+\ln \left (5\,{\mathrm {e}}^4-5\,x+\frac {5\,x\,{\mathrm {e}}^{2\,x}}{3}\right )\right )+\ln \left (x-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x - exp(2*x)*(7*x - 6*x^2 + 2) - 12*x^2 + log(5*exp(4) - 5*x + (5*x*exp(2*x))/3)*(3*exp(4) - 3*x + x*e
xp(2*x)) + exp(4)*(12*x - 12) + 6)/(exp(4)*(12*x - 6*x^2) - log(5*exp(4) - 5*x + (5*x*exp(2*x))/3)*(6*x - exp(
2*x)*(2*x - x^2) - 3*x^2 + exp(4)*(3*x - 6)) + exp(2*x)*(4*x^2 - 2*x^3) - 12*x^2 + 6*x^3),x)

[Out]

log(2*x + log(5*exp(4) - 5*x + (5*x*exp(2*x))/3)) + log(x - 2)

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sympy [A]  time = 0.48, size = 29, normalized size = 0.94 \begin {gather*} \log {\left (x - 2 \right )} + \log {\left (2 x + \log {\left (\frac {5 x e^{2 x}}{3} - 5 x + 5 e^{4} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(2*x)+3*exp(4)-3*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(6*x**2-7*x-2)*exp(2*x)+(12*x-12)*exp(4)-
12*x**2+9*x+6)/(((x**2-2*x)*exp(2*x)+(3*x-6)*exp(4)-3*x**2+6*x)*ln(5/3*x*exp(2*x)+5*exp(4)-5*x)+(2*x**3-4*x**2
)*exp(2*x)+(6*x**2-12*x)*exp(4)-6*x**3+12*x**2),x)

[Out]

log(x - 2) + log(2*x + log(5*x*exp(2*x)/3 - 5*x + 5*exp(4)))

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