3.77.20 \(\int \frac {-12-9 x+3 x^2+e^4 (3 x+3 x^2)+e^2 (-6 x+6 x^2)+(6 x-6 x^2+e^2 (-6 x-6 x^2)) \log (x)+(3 x+3 x^2) \log ^2(x)}{20 x^3-5 x^4-5 e^4 x^4+e^2 (20 x^3-10 x^4)+(-4 x^3+x^4+e^4 x^4+e^2 (-4 x^3+2 x^4)) \log (3)+(-20 x^3+10 x^4+10 e^2 x^4+(4 x^3-2 x^4-2 e^2 x^4) \log (3)) \log (x)+(-5 x^4+x^4 \log (3)) \log ^2(x)+(40 x^2-10 x^3-10 e^4 x^3+e^2 (40 x^2-20 x^3)+(-8 x^2+2 x^3+2 e^4 x^3+e^2 (-8 x^2+4 x^3)) \log (3)+(-40 x^2+20 x^3+20 e^2 x^3+(8 x^2-4 x^3-4 e^2 x^3) \log (3)) \log (x)+(-10 x^3+2 x^3 \log (3)) \log ^2(x)) \log (\frac {4-x-e^2 x+x \log (x)}{-1-e^2+\log (x)})+(20 x-5 x^2-5 e^4 x^2+e^2 (20 x-10 x^2)+(-4 x+x^2+e^4 x^2+e^2 (-4 x+2 x^2)) \log (3)+(-20 x+10 x^2+10 e^2 x^2+(4 x-2 x^2-2 e^2 x^2) \log (3)) \log (x)+(-5 x^2+x^2 \log (3)) \log ^2(x)) \log ^2(\frac {4-x-e^2 x+x \log (x)}{-1-e^2+\log (x)})} \, dx\)

Optimal. Leaf size=30 \[ \frac {3}{(5-\log (3)) \left (x+\log \left (x-\frac {4}{1+e^2-\log (x)}\right )\right )} \]

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Rubi [F]  time = 7.37, 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 {-12-9 x+3 x^2+e^4 \left (3 x+3 x^2\right )+e^2 \left (-6 x+6 x^2\right )+\left (6 x-6 x^2+e^2 \left (-6 x-6 x^2\right )\right ) \log (x)+\left (3 x+3 x^2\right ) \log ^2(x)}{20 x^3-5 x^4-5 e^4 x^4+e^2 \left (20 x^3-10 x^4\right )+\left (-4 x^3+x^4+e^4 x^4+e^2 \left (-4 x^3+2 x^4\right )\right ) \log (3)+\left (-20 x^3+10 x^4+10 e^2 x^4+\left (4 x^3-2 x^4-2 e^2 x^4\right ) \log (3)\right ) \log (x)+\left (-5 x^4+x^4 \log (3)\right ) \log ^2(x)+\left (40 x^2-10 x^3-10 e^4 x^3+e^2 \left (40 x^2-20 x^3\right )+\left (-8 x^2+2 x^3+2 e^4 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (3)+\left (-40 x^2+20 x^3+20 e^2 x^3+\left (8 x^2-4 x^3-4 e^2 x^3\right ) \log (3)\right ) \log (x)+\left (-10 x^3+2 x^3 \log (3)\right ) \log ^2(x)\right ) \log \left (\frac {4-x-e^2 x+x \log (x)}{-1-e^2+\log (x)}\right )+\left (20 x-5 x^2-5 e^4 x^2+e^2 \left (20 x-10 x^2\right )+\left (-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )\right ) \log (3)+\left (-20 x+10 x^2+10 e^2 x^2+\left (4 x-2 x^2-2 e^2 x^2\right ) \log (3)\right ) \log (x)+\left (-5 x^2+x^2 \log (3)\right ) \log ^2(x)\right ) \log ^2\left (\frac {4-x-e^2 x+x \log (x)}{-1-e^2+\log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-12 - 9*x + 3*x^2 + E^4*(3*x + 3*x^2) + E^2*(-6*x + 6*x^2) + (6*x - 6*x^2 + E^2*(-6*x - 6*x^2))*Log[x] +
(3*x + 3*x^2)*Log[x]^2)/(20*x^3 - 5*x^4 - 5*E^4*x^4 + E^2*(20*x^3 - 10*x^4) + (-4*x^3 + x^4 + E^4*x^4 + E^2*(-
4*x^3 + 2*x^4))*Log[3] + (-20*x^3 + 10*x^4 + 10*E^2*x^4 + (4*x^3 - 2*x^4 - 2*E^2*x^4)*Log[3])*Log[x] + (-5*x^4
+ x^4*Log[3])*Log[x]^2 + (40*x^2 - 10*x^3 - 10*E^4*x^3 + E^2*(40*x^2 - 20*x^3) + (-8*x^2 + 2*x^3 + 2*E^4*x^3
+ E^2*(-8*x^2 + 4*x^3))*Log[3] + (-40*x^2 + 20*x^3 + 20*E^2*x^3 + (8*x^2 - 4*x^3 - 4*E^2*x^3)*Log[3])*Log[x] +
(-10*x^3 + 2*x^3*Log[3])*Log[x]^2)*Log[(4 - x - E^2*x + x*Log[x])/(-1 - E^2 + Log[x])] + (20*x - 5*x^2 - 5*E^
4*x^2 + E^2*(20*x - 10*x^2) + (-4*x + x^2 + E^4*x^2 + E^2*(-4*x + 2*x^2))*Log[3] + (-20*x + 10*x^2 + 10*E^2*x^
2 + (4*x - 2*x^2 - 2*E^2*x^2)*Log[3])*Log[x] + (-5*x^2 + x^2*Log[3])*Log[x]^2)*Log[(4 - x - E^2*x + x*Log[x])/
(-1 - E^2 + Log[x])]^2),x]

[Out]

(-3*(3 + 2*E^2 - E^4)*Defer[Int][1/((1 + E^2 - Log[x])*(4 - (1 + E^2)*x + x*Log[x])*(x + Log[(-4 + x + E^2*x -
x*Log[x])/(1 + E^2 - Log[x])])^2), x])/(5 - Log[3]) + (3*(1 + E^2)^2*Defer[Int][x/((1 + E^2 - Log[x])*(4 - (1
+ E^2)*x + x*Log[x])*(x + Log[(-4 + x + E^2*x - x*Log[x])/(1 + E^2 - Log[x])])^2), x])/(5 - Log[3]) + (6*(1 -
E^2)*Defer[Int][Log[x]/((1 + E^2 - Log[x])*(4 - (1 + E^2)*x + x*Log[x])*(x + Log[(-4 + x + E^2*x - x*Log[x])/
(1 + E^2 - Log[x])])^2), x])/(5 - Log[3]) + (3*Defer[Int][Log[x]^2/((1 + E^2 - Log[x])*(4 - (1 + E^2)*x + x*Lo
g[x])*(x + Log[(-4 + x + E^2*x - x*Log[x])/(1 + E^2 - Log[x])])^2), x])/(5 - Log[3]) + (3*Defer[Int][(x*Log[x]
^2)/((1 + E^2 - Log[x])*(4 - (1 + E^2)*x + x*Log[x])*(x + Log[(-4 + x + E^2*x - x*Log[x])/(1 + E^2 - Log[x])])
^2), x])/(5 - Log[3]) + (12*Defer[Int][1/(x*(-1 - E^2 + Log[x])*(4 - (1 + E^2)*x + x*Log[x])*(x + Log[(-4 + x
+ E^2*x - x*Log[x])/(1 + E^2 - Log[x])])^2), x])/(5 - Log[3]) + (6*(1 + E^2)*Defer[Int][(x*Log[x])/((-1 - E^2
+ Log[x])*(4 - (1 + E^2)*x + x*Log[x])*(x + Log[(-4 + x + E^2*x - x*Log[x])/(1 + E^2 - Log[x])])^2), x])/(5 -
Log[3])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-12-9 x+3 x^2+e^4 \left (3 x+3 x^2\right )+e^2 \left (-6 x+6 x^2\right )+\left (6 x-6 x^2+e^2 \left (-6 x-6 x^2\right )\right ) \log (x)+\left (3 x+3 x^2\right ) \log ^2(x)}{20 x^3+\left (-5-5 e^4\right ) x^4+e^2 \left (20 x^3-10 x^4\right )+\left (-4 x^3+x^4+e^4 x^4+e^2 \left (-4 x^3+2 x^4\right )\right ) \log (3)+\left (-20 x^3+10 x^4+10 e^2 x^4+\left (4 x^3-2 x^4-2 e^2 x^4\right ) \log (3)\right ) \log (x)+\left (-5 x^4+x^4 \log (3)\right ) \log ^2(x)+\left (40 x^2-10 x^3-10 e^4 x^3+e^2 \left (40 x^2-20 x^3\right )+\left (-8 x^2+2 x^3+2 e^4 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (3)+\left (-40 x^2+20 x^3+20 e^2 x^3+\left (8 x^2-4 x^3-4 e^2 x^3\right ) \log (3)\right ) \log (x)+\left (-10 x^3+2 x^3 \log (3)\right ) \log ^2(x)\right ) \log \left (\frac {4-x-e^2 x+x \log (x)}{-1-e^2+\log (x)}\right )+\left (20 x-5 x^2-5 e^4 x^2+e^2 \left (20 x-10 x^2\right )+\left (-4 x+x^2+e^4 x^2+e^2 \left (-4 x+2 x^2\right )\right ) \log (3)+\left (-20 x+10 x^2+10 e^2 x^2+\left (4 x-2 x^2-2 e^2 x^2\right ) \log (3)\right ) \log (x)+\left (-5 x^2+x^2 \log (3)\right ) \log ^2(x)\right ) \log ^2\left (\frac {4-x-e^2 x+x \log (x)}{-1-e^2+\log (x)}\right )} \, dx\\ &=\int \frac {3 \left (-4+\left (-3-2 e^2+e^4\right ) x+\left (1+e^2\right )^2 x^2-2 x \left (-1+x+e^2 (1+x)\right ) \log (x)+x (1+x) \log ^2(x)\right )}{x (5-\log (3)) \left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx\\ &=\frac {3 \int \frac {-4+\left (-3-2 e^2+e^4\right ) x+\left (1+e^2\right )^2 x^2-2 x \left (-1+x+e^2 (1+x)\right ) \log (x)+x (1+x) \log ^2(x)}{x \left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}\\ &=\frac {3 \int \left (\frac {-3-2 e^2+e^4}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}+\frac {\left (1+e^2\right )^2 x}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}+\frac {2 \left (1-e^2\right ) \log (x)}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}+\frac {\log ^2(x)}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}+\frac {x \log ^2(x)}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}+\frac {4}{x \left (-1-e^2+\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}+\frac {2 \left (1+e^2\right ) x \log (x)}{\left (-1-e^2+\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2}\right ) \, dx}{5-\log (3)}\\ &=\frac {3 \int \frac {\log ^2(x)}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}+\frac {3 \int \frac {x \log ^2(x)}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}+\frac {12 \int \frac {1}{x \left (-1-e^2+\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}+\frac {\left (6 \left (1-e^2\right )\right ) \int \frac {\log (x)}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}+\frac {\left (6 \left (1+e^2\right )\right ) \int \frac {x \log (x)}{\left (-1-e^2+\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}+\frac {\left (3 \left (1+e^2\right )^2\right ) \int \frac {x}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}+\frac {\left (3 \left (-3-2 e^2+e^4\right )\right ) \int \frac {1}{\left (1+e^2-\log (x)\right ) \left (4-\left (1+e^2\right ) x+x \log (x)\right ) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )^2} \, dx}{5-\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 38, normalized size = 1.27 \begin {gather*} -\frac {3}{(-5+\log (3)) \left (x+\log \left (\frac {-4+x+e^2 x-x \log (x)}{1+e^2-\log (x)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 - 9*x + 3*x^2 + E^4*(3*x + 3*x^2) + E^2*(-6*x + 6*x^2) + (6*x - 6*x^2 + E^2*(-6*x - 6*x^2))*Log
[x] + (3*x + 3*x^2)*Log[x]^2)/(20*x^3 - 5*x^4 - 5*E^4*x^4 + E^2*(20*x^3 - 10*x^4) + (-4*x^3 + x^4 + E^4*x^4 +
E^2*(-4*x^3 + 2*x^4))*Log[3] + (-20*x^3 + 10*x^4 + 10*E^2*x^4 + (4*x^3 - 2*x^4 - 2*E^2*x^4)*Log[3])*Log[x] + (
-5*x^4 + x^4*Log[3])*Log[x]^2 + (40*x^2 - 10*x^3 - 10*E^4*x^3 + E^2*(40*x^2 - 20*x^3) + (-8*x^2 + 2*x^3 + 2*E^
4*x^3 + E^2*(-8*x^2 + 4*x^3))*Log[3] + (-40*x^2 + 20*x^3 + 20*E^2*x^3 + (8*x^2 - 4*x^3 - 4*E^2*x^3)*Log[3])*Lo
g[x] + (-10*x^3 + 2*x^3*Log[3])*Log[x]^2)*Log[(4 - x - E^2*x + x*Log[x])/(-1 - E^2 + Log[x])] + (20*x - 5*x^2
- 5*E^4*x^2 + E^2*(20*x - 10*x^2) + (-4*x + x^2 + E^4*x^2 + E^2*(-4*x + 2*x^2))*Log[3] + (-20*x + 10*x^2 + 10*
E^2*x^2 + (4*x - 2*x^2 - 2*E^2*x^2)*Log[3])*Log[x] + (-5*x^2 + x^2*Log[3])*Log[x]^2)*Log[(4 - x - E^2*x + x*Lo
g[x])/(-1 - E^2 + Log[x])]^2),x]

[Out]

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

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fricas [A]  time = 1.67, size = 41, normalized size = 1.37 \begin {gather*} -\frac {3}{x \log \relax (3) + {\left (\log \relax (3) - 5\right )} \log \left (\frac {x e^{2} - x \log \relax (x) + x - 4}{e^{2} - \log \relax (x) + 1}\right ) - 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+3*x)*log(x)^2+((-6*x^2-6*x)*exp(2)-6*x^2+6*x)*log(x)+(3*x^2+3*x)*exp(2)^2+(6*x^2-6*x)*exp(2)
+3*x^2-9*x-12)/(((x^2*log(3)-5*x^2)*log(x)^2+((-2*x^2*exp(2)-2*x^2+4*x)*log(3)+10*x^2*exp(2)+10*x^2-20*x)*log(
x)+(x^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x)*log(3)-5*x^2*exp(2)^2+(-10*x^2+20*x)*exp(2)-5*x^2+20*x)*log((x*lo
g(x)-exp(2)*x-x+4)/(log(x)-1-exp(2)))^2+((2*x^3*log(3)-10*x^3)*log(x)^2+((-4*x^3*exp(2)-4*x^3+8*x^2)*log(3)+20
*x^3*exp(2)+20*x^3-40*x^2)*log(x)+(2*x^3*exp(2)^2+(4*x^3-8*x^2)*exp(2)+2*x^3-8*x^2)*log(3)-10*x^3*exp(2)^2+(-2
0*x^3+40*x^2)*exp(2)-10*x^3+40*x^2)*log((x*log(x)-exp(2)*x-x+4)/(log(x)-1-exp(2)))+(x^4*log(3)-5*x^4)*log(x)^2
+((-2*x^4*exp(2)-2*x^4+4*x^3)*log(3)+10*x^4*exp(2)+10*x^4-20*x^3)*log(x)+(x^4*exp(2)^2+(2*x^4-4*x^3)*exp(2)+x^
4-4*x^3)*log(3)-5*x^4*exp(2)^2+(-10*x^4+20*x^3)*exp(2)-5*x^4+20*x^3),x, algorithm="fricas")

[Out]

-3/(x*log(3) + (log(3) - 5)*log((x*e^2 - x*log(x) + x - 4)/(e^2 - log(x) + 1)) - 5*x)

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giac [B]  time = 4.92, size = 67, normalized size = 2.23 \begin {gather*} -\frac {3}{x \log \relax (3) + \log \relax (3) \log \left (x e^{2} - x \log \relax (x) + x - 4\right ) - \log \relax (3) \log \left (e^{2} - \log \relax (x) + 1\right ) - 5 \, x - 5 \, \log \left (x e^{2} - x \log \relax (x) + x - 4\right ) + 5 \, \log \left (e^{2} - \log \relax (x) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+3*x)*log(x)^2+((-6*x^2-6*x)*exp(2)-6*x^2+6*x)*log(x)+(3*x^2+3*x)*exp(2)^2+(6*x^2-6*x)*exp(2)
+3*x^2-9*x-12)/(((x^2*log(3)-5*x^2)*log(x)^2+((-2*x^2*exp(2)-2*x^2+4*x)*log(3)+10*x^2*exp(2)+10*x^2-20*x)*log(
x)+(x^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x)*log(3)-5*x^2*exp(2)^2+(-10*x^2+20*x)*exp(2)-5*x^2+20*x)*log((x*lo
g(x)-exp(2)*x-x+4)/(log(x)-1-exp(2)))^2+((2*x^3*log(3)-10*x^3)*log(x)^2+((-4*x^3*exp(2)-4*x^3+8*x^2)*log(3)+20
*x^3*exp(2)+20*x^3-40*x^2)*log(x)+(2*x^3*exp(2)^2+(4*x^3-8*x^2)*exp(2)+2*x^3-8*x^2)*log(3)-10*x^3*exp(2)^2+(-2
0*x^3+40*x^2)*exp(2)-10*x^3+40*x^2)*log((x*log(x)-exp(2)*x-x+4)/(log(x)-1-exp(2)))+(x^4*log(3)-5*x^4)*log(x)^2
+((-2*x^4*exp(2)-2*x^4+4*x^3)*log(3)+10*x^4*exp(2)+10*x^4-20*x^3)*log(x)+(x^4*exp(2)^2+(2*x^4-4*x^3)*exp(2)+x^
4-4*x^3)*log(3)-5*x^4*exp(2)^2+(-10*x^4+20*x^3)*exp(2)-5*x^4+20*x^3),x, algorithm="giac")

[Out]

-3/(x*log(3) + log(3)*log(x*e^2 - x*log(x) + x - 4) - log(3)*log(e^2 - log(x) + 1) - 5*x - 5*log(x*e^2 - x*log
(x) + x - 4) + 5*log(e^2 - log(x) + 1))

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maple [C]  time = 0.28, size = 239, normalized size = 7.97

method	result	size
risch	\(-\frac {6 i}{\left (\ln \relax (3)-5\right ) \left (\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2}-\ln \relax (x )+1}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )}{{\mathrm e}^{2}-\ln \relax (x )+1}\right )-\pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )}{{\mathrm e}^{2}-\ln \relax (x )+1}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2}-\ln \relax (x )+1}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )}{{\mathrm e}^{2}-\ln \relax (x )+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )}{{\mathrm e}^{2}-\ln \relax (x )+1}\right )^{3}+2 i x -2 i \ln \left ({\mathrm e}^{2}-\ln \relax (x )+1\right )+2 i \ln \left ({\mathrm e}^{2} x -4+x \left (1-\ln \relax (x )\right )\right )\right )}\)	\(239\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2+3*x)*ln(x)^2+((-6*x^2-6*x)*exp(2)-6*x^2+6*x)*ln(x)+(3*x^2+3*x)*exp(2)^2+(6*x^2-6*x)*exp(2)+3*x^2-9
*x-12)/(((x^2*ln(3)-5*x^2)*ln(x)^2+((-2*x^2*exp(2)-2*x^2+4*x)*ln(3)+10*x^2*exp(2)+10*x^2-20*x)*ln(x)+(x^2*exp(
2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x)*ln(3)-5*x^2*exp(2)^2+(-10*x^2+20*x)*exp(2)-5*x^2+20*x)*ln((x*ln(x)-exp(2)*x-x
+4)/(ln(x)-1-exp(2)))^2+((2*x^3*ln(3)-10*x^3)*ln(x)^2+((-4*x^3*exp(2)-4*x^3+8*x^2)*ln(3)+20*x^3*exp(2)+20*x^3-
40*x^2)*ln(x)+(2*x^3*exp(2)^2+(4*x^3-8*x^2)*exp(2)+2*x^3-8*x^2)*ln(3)-10*x^3*exp(2)^2+(-20*x^3+40*x^2)*exp(2)-
10*x^3+40*x^2)*ln((x*ln(x)-exp(2)*x-x+4)/(ln(x)-1-exp(2)))+(x^4*ln(3)-5*x^4)*ln(x)^2+((-2*x^4*exp(2)-2*x^4+4*x
^3)*ln(3)+10*x^4*exp(2)+10*x^4-20*x^3)*ln(x)+(x^4*exp(2)^2+(2*x^4-4*x^3)*exp(2)+x^4-4*x^3)*ln(3)-5*x^4*exp(2)^
2+(-10*x^4+20*x^3)*exp(2)-5*x^4+20*x^3),x,method=_RETURNVERBOSE)

[Out]

-6*I/(ln(3)-5)/(Pi*csgn(I*(exp(2)*x-4+x*(1-ln(x))))*csgn(I/(exp(2)-ln(x)+1))*csgn(I/(exp(2)-ln(x)+1)*(exp(2)*x
-4+x*(1-ln(x))))-Pi*csgn(I*(exp(2)*x-4+x*(1-ln(x))))*csgn(I/(exp(2)-ln(x)+1)*(exp(2)*x-4+x*(1-ln(x))))^2-Pi*cs
gn(I/(exp(2)-ln(x)+1))*csgn(I/(exp(2)-ln(x)+1)*(exp(2)*x-4+x*(1-ln(x))))^2+Pi*csgn(I/(exp(2)-ln(x)+1)*(exp(2)*
x-4+x*(1-ln(x))))^3+2*I*x-2*I*ln(exp(2)-ln(x)+1)+2*I*ln(exp(2)*x-4+x*(1-ln(x))))

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maxima [A]  time = 0.75, size = 45, normalized size = 1.50 \begin {gather*} -\frac {3}{x {\left (\log \relax (3) - 5\right )} + {\left (\log \relax (3) - 5\right )} \log \left (-x {\left (e^{2} + 1\right )} + x \log \relax (x) + 4\right ) - {\left (\log \relax (3) - 5\right )} \log \left (-e^{2} + \log \relax (x) - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+3*x)*log(x)^2+((-6*x^2-6*x)*exp(2)-6*x^2+6*x)*log(x)+(3*x^2+3*x)*exp(2)^2+(6*x^2-6*x)*exp(2)
+3*x^2-9*x-12)/(((x^2*log(3)-5*x^2)*log(x)^2+((-2*x^2*exp(2)-2*x^2+4*x)*log(3)+10*x^2*exp(2)+10*x^2-20*x)*log(
x)+(x^2*exp(2)^2+(2*x^2-4*x)*exp(2)+x^2-4*x)*log(3)-5*x^2*exp(2)^2+(-10*x^2+20*x)*exp(2)-5*x^2+20*x)*log((x*lo
g(x)-exp(2)*x-x+4)/(log(x)-1-exp(2)))^2+((2*x^3*log(3)-10*x^3)*log(x)^2+((-4*x^3*exp(2)-4*x^3+8*x^2)*log(3)+20
*x^3*exp(2)+20*x^3-40*x^2)*log(x)+(2*x^3*exp(2)^2+(4*x^3-8*x^2)*exp(2)+2*x^3-8*x^2)*log(3)-10*x^3*exp(2)^2+(-2
0*x^3+40*x^2)*exp(2)-10*x^3+40*x^2)*log((x*log(x)-exp(2)*x-x+4)/(log(x)-1-exp(2)))+(x^4*log(3)-5*x^4)*log(x)^2
+((-2*x^4*exp(2)-2*x^4+4*x^3)*log(3)+10*x^4*exp(2)+10*x^4-20*x^3)*log(x)+(x^4*exp(2)^2+(2*x^4-4*x^3)*exp(2)+x^
4-4*x^3)*log(3)-5*x^4*exp(2)^2+(-10*x^4+20*x^3)*exp(2)-5*x^4+20*x^3),x, algorithm="maxima")

[Out]

-3/(x*(log(3) - 5) + (log(3) - 5)*log(-x*(e^2 + 1) + x*log(x) + 4) - (log(3) - 5)*log(-e^2 + log(x) - 1))

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mupad [B]  time = 7.41, size = 36, normalized size = 1.20 \begin {gather*} -\frac {3}{\left (\ln \relax (3)-5\right )\,\left (x+\ln \left (\frac {x+x\,{\mathrm {e}}^2-x\,\ln \relax (x)-4}{{\mathrm {e}}^2-\ln \relax (x)+1}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*x - log(x)^2*(3*x + 3*x^2) - exp(4)*(3*x + 3*x^2) + exp(2)*(6*x - 6*x^2) - 3*x^2 + log(x)*(exp(2)*(6*x
+ 6*x^2) - 6*x + 6*x^2) + 12)/(log((x + x*exp(2) - x*log(x) - 4)/(exp(2) - log(x) + 1))^2*(log(3)*(4*x + exp(2
)*(4*x - 2*x^2) - x^2*exp(4) - x^2) - 20*x + log(x)*(20*x - 10*x^2*exp(2) + log(3)*(2*x^2*exp(2) - 4*x + 2*x^2
) - 10*x^2) - exp(2)*(20*x - 10*x^2) + 5*x^2*exp(4) + 5*x^2 - log(x)^2*(x^2*log(3) - 5*x^2)) - exp(2)*(20*x^3
- 10*x^4) + 5*x^4*exp(4) + log(3)*(exp(2)*(4*x^3 - 2*x^4) - x^4*exp(4) + 4*x^3 - x^4) - 20*x^3 + 5*x^4 - log(x
)^2*(x^4*log(3) - 5*x^4) + log(x)*(log(3)*(2*x^4*exp(2) - 4*x^3 + 2*x^4) - 10*x^4*exp(2) + 20*x^3 - 10*x^4) +
log((x + x*exp(2) - x*log(x) - 4)/(exp(2) - log(x) + 1))*(10*x^3*exp(4) - exp(2)*(40*x^2 - 20*x^3) + log(3)*(e
xp(2)*(8*x^2 - 4*x^3) - 2*x^3*exp(4) + 8*x^2 - 2*x^3) - 40*x^2 + 10*x^3 - log(x)^2*(2*x^3*log(3) - 10*x^3) + l
og(x)*(log(3)*(4*x^3*exp(2) - 8*x^2 + 4*x^3) - 20*x^3*exp(2) + 40*x^2 - 20*x^3))),x)

[Out]

-3/((log(3) - 5)*(x + log((x + x*exp(2) - x*log(x) - 4)/(exp(2) - log(x) + 1))))

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sympy [A]  time = 1.25, size = 39, normalized size = 1.30 \begin {gather*} - \frac {3}{- 5 x + x \log {\relax (3 )} + \left (-5 + \log {\relax (3 )}\right ) \log {\left (\frac {x \log {\relax (x )} - x e^{2} - x + 4}{\log {\relax (x )} - e^{2} - 1} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2+3*x)*ln(x)**2+((-6*x**2-6*x)*exp(2)-6*x**2+6*x)*ln(x)+(3*x**2+3*x)*exp(2)**2+(6*x**2-6*x)*e
xp(2)+3*x**2-9*x-12)/(((x**2*ln(3)-5*x**2)*ln(x)**2+((-2*x**2*exp(2)-2*x**2+4*x)*ln(3)+10*x**2*exp(2)+10*x**2-
20*x)*ln(x)+(x**2*exp(2)**2+(2*x**2-4*x)*exp(2)+x**2-4*x)*ln(3)-5*x**2*exp(2)**2+(-10*x**2+20*x)*exp(2)-5*x**2
+20*x)*ln((x*ln(x)-exp(2)*x-x+4)/(ln(x)-1-exp(2)))**2+((2*x**3*ln(3)-10*x**3)*ln(x)**2+((-4*x**3*exp(2)-4*x**3
+8*x**2)*ln(3)+20*x**3*exp(2)+20*x**3-40*x**2)*ln(x)+(2*x**3*exp(2)**2+(4*x**3-8*x**2)*exp(2)+2*x**3-8*x**2)*l
n(3)-10*x**3*exp(2)**2+(-20*x**3+40*x**2)*exp(2)-10*x**3+40*x**2)*ln((x*ln(x)-exp(2)*x-x+4)/(ln(x)-1-exp(2)))+
(x**4*ln(3)-5*x**4)*ln(x)**2+((-2*x**4*exp(2)-2*x**4+4*x**3)*ln(3)+10*x**4*exp(2)+10*x**4-20*x**3)*ln(x)+(x**4
*exp(2)**2+(2*x**4-4*x**3)*exp(2)+x**4-4*x**3)*ln(3)-5*x**4*exp(2)**2+(-10*x**4+20*x**3)*exp(2)-5*x**4+20*x**3
),x)

[Out]

-3/(-5*x + x*log(3) + (-5 + log(3))*log((x*log(x) - x*exp(2) - x + 4)/(log(x) - exp(2) - 1)))

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