3.57.18 \(\int \frac {(8 x+4 x^2+(-2 x-x^2) \log (x)) \log (4+2 x)+\log (e^{2 e^3} (3-x)) (12 x-4 x^2+(-3 x+x^2) \log (x)+(-6-x+x^2) \log (4+2 x))}{(-6 x-x^2+x^3) \log ^2(e^{2 e^3} (3-x))} \, dx\) [5618]

Optimal. Leaf size=27 \[ \frac {(-4+\log (x)) \log (4+2 x)}{\log \left (e^{2 e^3} (3-x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [F]
time = 7.38, 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 {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{\left (-6 x-x^2+x^3\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[Int][(-4 + Log[x])/((2 + x)*(2*E^3 + Log[3 - x])), x] + 2*E^3*Defer[Int][Log[4 + 2*x]/(x*(2*E^3 + Log[3
- x])^2), x] + Defer[Int][(Log[3 - x]*Log[4 + 2*x])/((3 - x)*(2*E^3 + Log[3 - x])^2), x] + Defer[Int][(Log[3 -
x]*Log[4 + 2*x])/((-3 + x)*(2*E^3 + Log[3 - x])^2), x] + Defer[Int][(Log[3 - x]*Log[4 + 2*x])/(x*(2*E^3 + Log
[3 - x])^2), x] + Defer[Int][(Log[x]*Log[4 + 2*x])/((3 - x)*(2*E^3 + Log[3 - x])^2), x] - 2*E^3*Defer[Subst][D
efer[Int][Log[10 - 2*x]/(x*(2*E^3 + Log[x])^2), x], x, 3 - x] + 2*(2 + E^3)*Defer[Subst][Defer[Int][Log[10 - 2
*x]/(x*(2*E^3 + Log[x])^2), x], x, 3 - x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (8 x+4 x^2+\left (-2 x-x^2\right ) \log (x)\right ) \log (4+2 x)+\log \left (e^{2 e^3} (3-x)\right ) \left (12 x-4 x^2+\left (-3 x+x^2\right ) \log (x)+\left (-6-x+x^2\right ) \log (4+2 x)\right )}{x \left (-6-x+x^2\right ) \log ^2\left (e^{2 e^3} (3-x)\right )} \, dx\\ &=\int \frac {x (2+x) (-4+\log (x)) \log (2 (2+x))-(-3+x) \left (2 e^3+\log (3-x)\right ) (-4 x+x \log (x)+(2+x) \log (2 (2+x)))}{x \left (6+x-x^2\right ) \left (2 e^3+\log (3-x)\right )^2} \, dx\\ &=\int \left (\frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )}+\frac {\left (6 e^3-4 \left (1+\frac {e^3}{2}\right ) x+3 \log (3-x)-x \log (3-x)+x \log (x)\right ) \log (4+2 x)}{(3-x) x \left (2 e^3+\log (3-x)\right )^2}\right ) \, dx\\ &=\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx+\int \frac {\left (6 e^3-4 \left (1+\frac {e^3}{2}\right ) x+3 \log (3-x)-x \log (3-x)+x \log (x)\right ) \log (4+2 x)}{(3-x) x \left (2 e^3+\log (3-x)\right )^2} \, dx\\ &=\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx+\int \left (\frac {\left (6 e^3-4 \left (1+\frac {e^3}{2}\right ) x+3 \log (3-x)-x \log (3-x)+x \log (x)\right ) \log (4+2 x)}{3 (3-x) \left (2 e^3+\log (3-x)\right )^2}+\frac {\left (6 e^3-4 \left (1+\frac {e^3}{2}\right ) x+3 \log (3-x)-x \log (3-x)+x \log (x)\right ) \log (4+2 x)}{3 x \left (2 e^3+\log (3-x)\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\left (6 e^3-4 \left (1+\frac {e^3}{2}\right ) x+3 \log (3-x)-x \log (3-x)+x \log (x)\right ) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\frac {1}{3} \int \frac {\left (6 e^3-4 \left (1+\frac {e^3}{2}\right ) x+3 \log (3-x)-x \log (3-x)+x \log (x)\right ) \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {2 \left (-2-e^3\right ) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2}+\frac {6 e^3 \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2}-\frac {\log (3-x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2}+\frac {3 \log (3-x) \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2}+\frac {\log (x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2}\right ) \, dx+\frac {1}{3} \int \left (\frac {6 e^3 \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2}+\frac {2 \left (-2-e^3\right ) x \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2}+\frac {3 \log (3-x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2}+\frac {x \log (3-x) \log (4+2 x)}{(-3+x) \left (2 e^3+\log (3-x)\right )^2}+\frac {x \log (x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2}\right ) \, dx+\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx\\ &=-\left (\frac {1}{3} \int \frac {\log (3-x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2} \, dx\right )+\frac {1}{3} \int \frac {x \log (3-x) \log (4+2 x)}{(-3+x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\frac {1}{3} \int \frac {\log (x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2} \, dx+\frac {1}{3} \int \frac {x \log (x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\left (2 e^3\right ) \int \frac {\log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\left (2 e^3\right ) \int \frac {\log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx-\frac {1}{3} \left (2 \left (2+e^3\right )\right ) \int \frac {\log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2} \, dx-\frac {1}{3} \left (2 \left (2+e^3\right )\right ) \int \frac {x \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx\\ &=-\left (\frac {1}{3} \int \frac {\log (3-x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2} \, dx\right )+\frac {1}{3} \int \frac {\log (x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2} \, dx+\frac {1}{3} \int \left (\frac {\log (3-x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2}+\frac {3 \log (3-x) \log (4+2 x)}{(-3+x) \left (2 e^3+\log (3-x)\right )^2}\right ) \, dx+\frac {1}{3} \int \left (-\frac {\log (x) \log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2}+\frac {3 \log (x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2}\right ) \, dx+\left (2 e^3\right ) \int \frac {\log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx-\left (2 e^3\right ) \text {Subst}\left (\int \frac {\log (10-2 x)}{x \left (2 e^3+\log (x)\right )^2} \, dx,x,3-x\right )-\frac {1}{3} \left (2 \left (2+e^3\right )\right ) \int \frac {\log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2} \, dx-\frac {1}{3} \left (2 \left (2+e^3\right )\right ) \int \left (-\frac {\log (4+2 x)}{\left (2 e^3+\log (3-x)\right )^2}+\frac {3 \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2}\right ) \, dx+\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx\\ &=\left (2 e^3\right ) \int \frac {\log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx-\left (2 e^3\right ) \text {Subst}\left (\int \frac {\log (10-2 x)}{x \left (2 e^3+\log (x)\right )^2} \, dx,x,3-x\right )-\left (2 \left (2+e^3\right )\right ) \int \frac {\log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{(-3+x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx\\ &=\left (2 e^3\right ) \int \frac {\log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx-\left (2 e^3\right ) \text {Subst}\left (\int \frac {\log (10-2 x)}{x \left (2 e^3+\log (x)\right )^2} \, dx,x,3-x\right )+\left (2 \left (2+e^3\right )\right ) \text {Subst}\left (\int \frac {\log (10-2 x)}{x \left (2 e^3+\log (x)\right )^2} \, dx,x,3-x\right )+\int \frac {-4+\log (x)}{(2+x) \left (2 e^3+\log (3-x)\right )} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{(-3+x) \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (3-x) \log (4+2 x)}{x \left (2 e^3+\log (3-x)\right )^2} \, dx+\int \frac {\log (x) \log (4+2 x)}{(3-x) \left (2 e^3+\log (3-x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]
time = 0.26, size = 25, normalized size = 0.93 \begin {gather*} \frac {(-4+\log (x)) \log (2 (2+x))}{2 e^3+\log (3-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 17.44, size = 26, normalized size = 0.96

method	result	size
risch	\(\frac {\left (\ln \left (x \right )-4\right ) \ln \left (2 x +4\right )}{\ln \left (\left (3-x \right ) {\mathrm e}^{2 \,{\mathrm e}^{3}}\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.67, size = 33, normalized size = 1.22 \begin {gather*} \frac {{\left (\log \left (x\right ) - 4\right )} \log \left (x + 2\right ) + \log \left (2\right ) \log \left (x\right ) - 4 \, \log \left (2\right )}{2 \, e^{3} + \log \left (-x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 24, normalized size = 0.89 \begin {gather*} \frac {{\left (\log \left (x\right ) - 4\right )} \log \left (2 \, x + 4\right )}{\log \left (-{\left (x - 3\right )} e^{\left (2 \, e^{3}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]
time = 0.14, size = 29, normalized size = 1.07 \begin {gather*} \frac {\log {\left (x \right )} \log {\left (2 x + 4 \right )} - 4 \log {\left (2 x + 4 \right )}}{\log {\left (\left (3 - x\right ) e^{2 e^{3}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x**2-x-6)*ln(2*x+4)+(x**2-3*x)*ln(x)-4*x**2+12*x)*ln((3-x)*exp(exp(3))**2)+((-x**2-2*x)*ln(x)+4*x
**2+8*x)*ln(2*x+4))/(x**3-x**2-6*x)/ln((3-x)*exp(exp(3))**2)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 32, normalized size = 1.19 \begin {gather*} \frac {\log \left (2 \, x + 4\right ) \log \left (x\right ) - 4 \, \log \left (2 \, x + 4\right )}{2 \, e^{3} + \log \left (-x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 3.76, size = 108, normalized size = 4.00 \begin {gather*} \ln \left (x\,\left (x+2\right )\right )+\frac {20}{x+2}-\frac {5\,\ln \left (x\right )}{x+2}+\frac {\ln \left (2\,x+4\right )\,\left (\ln \left (x\right )-4\right )-\frac {\ln \left (-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (x-3\right )\right )\,\left (x-3\right )\,\left (2\,\ln \left (2\,x+4\right )-4\,x+x\,\ln \left (2\,x+4\right )+x\,\ln \left (x\right )\right )}{x\,\left (x+2\right )}}{\ln \left (-{\mathrm {e}}^{2\,{\mathrm {e}}^3}\,\left (x-3\right )\right )}-\frac {3\,\ln \left (2\,x+4\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x*(x + 2)) + 20/(x + 2) - (5*log(x))/(x + 2) + (log(2*x + 4)*(log(x) - 4) - (log(-exp(2*exp(3))*(x - 3))*(
x - 3)*(2*log(2*x + 4) - 4*x + x*log(2*x + 4) + x*log(x)))/(x*(x + 2)))/log(-exp(2*exp(3))*(x - 3)) - (3*log(2
*x + 4))/x

________________________________________________________________________________________