3.45.2 \(\int \frac {e^{2 e^4} (36-24 e^{2 x}+4 e^{4 x})+36 x^2+3 x^3+(72 x+6 x^2) \log (3)+36 \log ^2(3)+e^{2 x} (-24 x^2-x^3+2 x^4+(-48 x-2 x^2+2 x^3) \log (3)-24 \log ^2(3))+e^{4 x} (4 x^2+8 x \log (3)+4 \log ^2(3))+e^{e^4} (-72 x-6 x^2+e^{4 x} (-8 x-8 \log (3))-72 \log (3)+e^{2 x} (48 x+2 x^2-2 x^3+48 \log (3)))}{9 x^3+e^{2 e^4} (9 x-6 e^{2 x} x+e^{4 x} x)+18 x^2 \log (3)+9 x \log ^2(3)+e^{2 x} (-6 x^3-12 x^2 \log (3)-6 x \log ^2(3))+e^{4 x} (x^3+2 x^2 \log (3)+x \log ^2(3))+e^{e^4} (-18 x^2-18 x \log (3)+e^{4 x} (-2 x^2-2 x \log (3))+e^{2 x} (12 x^2+12 x \log (3)))} \, dx\)

Optimal. Leaf size=33 \[ \frac {x^2}{\left (3-e^{2 x}\right ) \left (-e^{e^4}+x+\log (3)\right )}+4 \log (x) \]

________________________________________________________________________________________

Rubi [F]  time = 6.37, 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 {e^{2 e^4} \left (36-24 e^{2 x}+4 e^{4 x}\right )+36 x^2+3 x^3+\left (72 x+6 x^2\right ) \log (3)+36 \log ^2(3)+e^{2 x} \left (-24 x^2-x^3+2 x^4+\left (-48 x-2 x^2+2 x^3\right ) \log (3)-24 \log ^2(3)\right )+e^{4 x} \left (4 x^2+8 x \log (3)+4 \log ^2(3)\right )+e^{e^4} \left (-72 x-6 x^2+e^{4 x} (-8 x-8 \log (3))-72 \log (3)+e^{2 x} \left (48 x+2 x^2-2 x^3+48 \log (3)\right )\right )}{9 x^3+e^{2 e^4} \left (9 x-6 e^{2 x} x+e^{4 x} x\right )+18 x^2 \log (3)+9 x \log ^2(3)+e^{2 x} \left (-6 x^3-12 x^2 \log (3)-6 x \log ^2(3)\right )+e^{4 x} \left (x^3+2 x^2 \log (3)+x \log ^2(3)\right )+e^{e^4} \left (-18 x^2-18 x \log (3)+e^{4 x} \left (-2 x^2-2 x \log (3)\right )+e^{2 x} \left (12 x^2+12 x \log (3)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*E^4)*(36 - 24*E^(2*x) + 4*E^(4*x)) + 36*x^2 + 3*x^3 + (72*x + 6*x^2)*Log[3] + 36*Log[3]^2 + E^(2*x)*
(-24*x^2 - x^3 + 2*x^4 + (-48*x - 2*x^2 + 2*x^3)*Log[3] - 24*Log[3]^2) + E^(4*x)*(4*x^2 + 8*x*Log[3] + 4*Log[3
]^2) + E^E^4*(-72*x - 6*x^2 + E^(4*x)*(-8*x - 8*Log[3]) - 72*Log[3] + E^(2*x)*(48*x + 2*x^2 - 2*x^3 + 48*Log[3
])))/(9*x^3 + E^(2*E^4)*(9*x - 6*E^(2*x)*x + E^(4*x)*x) + 18*x^2*Log[3] + 9*x*Log[3]^2 + E^(2*x)*(-6*x^3 - 12*
x^2*Log[3] - 6*x*Log[3]^2) + E^(4*x)*(x^3 + 2*x^2*Log[3] + x*Log[3]^2) + E^E^4*(-18*x^2 - 18*x*Log[3] + E^(4*x
)*(-2*x^2 - 2*x*Log[3]) + E^(2*x)*(12*x^2 + 12*x*Log[3]))),x]

[Out]

-1/3*x + x/(3 - E^(2*x)) + (2*E^E^4 - Log[9])/(2*(3 - E^(2*x))) + (x*(2*E^E^4 - Log[9]))/3 + (x*(1 - 2*E^E^4 +
(2*Log[9]^2)/Log[81]))/3 + Log[3 - E^(2*x)]/6 - ((2*E^E^4 - Log[9])*Log[3 - E^(2*x)])/6 - ((1 - 2*E^E^4 + (2*
Log[9]^2)/Log[81])*Log[3 - E^(2*x)])/6 + 4*Log[x] + (E^E^4 - Log[3])^2*Defer[Int][1/((-3 + E^(2*x))*(E^E^4 - x
- Log[3])^2), x] - (2*E^(2*E^4) + 6*Log[3]^2 - Log[9]^2 - E^E^4*Log[81])*Defer[Int][1/((-3 + E^(2*x))*(E^E^4
- x - Log[3])), x] - 3*(2*E^E^4 - Log[9])^2*Defer[Int][1/((-3 + E^(2*x))^2*(2*E^E^4 - 2*x - Log[9])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {36 e^{2 e^4}-24 e^{2 \left (e^4+x\right )}+4 e^{2 e^4+4 x}-8 e^{e^4+4 x} (x+\log (3))+4 e^{4 x} (x+\log (3))^2-6 e^{e^4} \left (12 x+x^2+12 \log (3)\right )+e^{e^4+2 x} \left (48 x+2 x^2-2 x^3+48 \log (3)\right )+3 \left (x^3+24 x \log (3)+12 \log ^2(3)+x^2 (12+\log (9))\right )+e^{2 x} \left (2 x^4-48 x \log (3)-24 \log ^2(3)+x^3 (-1+\log (9))-x^2 (24+\log (9))\right )}{\left (3-e^{2 x}\right )^2 x \left (e^{e^4}-x-\log (3)\right )^2} \, dx\\ &=\int \left (\frac {4}{x}+\frac {12 x^2}{\left (3-e^{2 x}\right )^2 \left (-2 e^{e^4}+2 x+\log (9)\right )}+\frac {x \left (-2 e^{e^4}-2 x^2+x \left (1+2 e^{e^4}-\log (9)\right )+\log (9)\right )}{\left (3-e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2}\right ) \, dx\\ &=4 \log (x)+12 \int \frac {x^2}{\left (3-e^{2 x}\right )^2 \left (-2 e^{e^4}+2 x+\log (9)\right )} \, dx+\int \frac {x \left (-2 e^{e^4}-2 x^2+x \left (1+2 e^{e^4}-\log (9)\right )+\log (9)\right )}{\left (3-e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx\\ &=4 \log (x)+12 \int \left (\frac {x}{2 \left (-3+e^{2 x}\right )^2}+\frac {2 e^{e^4}-\log (9)}{4 \left (-3+e^{2 x}\right )^2}-\frac {\left (2 e^{e^4}-\log (9)\right )^2}{4 \left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )}\right ) \, dx+\int \left (\frac {2 x}{-3+e^{2 x}}+\frac {\left (e^{e^4}-\log (3)\right )^2}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2}-\frac {1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}}{-3+e^{2 x}}-\frac {2 e^{2 e^4}+6 \log ^2(3)-\log ^2(9)-e^{e^4} \log (81)}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )}\right ) \, dx\\ &=4 \log (x)+2 \int \frac {x}{-3+e^{2 x}} \, dx+6 \int \frac {x}{\left (-3+e^{2 x}\right )^2} \, dx+\left (e^{e^4}-\log (3)\right )^2 \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx+\left (3 \left (2 e^{e^4}-\log (9)\right )\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2} \, dx-\left (3 \left (2 e^{e^4}-\log (9)\right )^2\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )} \, dx+\left (-1+2 e^{e^4}-\frac {2 \log ^2(9)}{\log (81)}\right ) \int \frac {1}{-3+e^{2 x}} \, dx+\left (-2 e^{2 e^4}-6 \log ^2(3)+\log ^2(9)+e^{e^4} \log (81)\right ) \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )} \, dx\\ &=-\frac {x^2}{3}+4 \log (x)+\frac {2}{3} \int \frac {e^{2 x} x}{-3+e^{2 x}} \, dx+2 \int \frac {e^{2 x} x}{\left (-3+e^{2 x}\right )^2} \, dx-2 \int \frac {x}{-3+e^{2 x}} \, dx+\left (e^{e^4}-\log (3)\right )^2 \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx+\frac {1}{2} \left (3 \left (2 e^{e^4}-\log (9)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{(-3+x)^2 x} \, dx,x,e^{2 x}\right )-\left (3 \left (2 e^{e^4}-\log (9)\right )^2\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )} \, dx+\frac {1}{2} \left (-1+2 e^{e^4}-\frac {2 \log ^2(9)}{\log (81)}\right ) \operatorname {Subst}\left (\int \frac {1}{(-3+x) x} \, dx,x,e^{2 x}\right )+\left (-2 e^{2 e^4}-6 \log ^2(3)+\log ^2(9)+e^{e^4} \log (81)\right ) \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )} \, dx\\ &=\frac {x}{3-e^{2 x}}+\frac {1}{3} x \log \left (1-\frac {e^{2 x}}{3}\right )+4 \log (x)-\frac {1}{3} \int \log \left (1-\frac {e^{2 x}}{3}\right ) \, dx-\frac {2}{3} \int \frac {e^{2 x} x}{-3+e^{2 x}} \, dx+\left (e^{e^4}-\log (3)\right )^2 \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx+\frac {1}{2} \left (3 \left (2 e^{e^4}-\log (9)\right )\right ) \operatorname {Subst}\left (\int \left (\frac {1}{3 (-3+x)^2}-\frac {1}{9 (-3+x)}+\frac {1}{9 x}\right ) \, dx,x,e^{2 x}\right )-\left (3 \left (2 e^{e^4}-\log (9)\right )^2\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )} \, dx+\frac {1}{6} \left (-1+2 e^{e^4}-\frac {2 \log ^2(9)}{\log (81)}\right ) \operatorname {Subst}\left (\int \frac {1}{-3+x} \, dx,x,e^{2 x}\right )+\frac {1}{6} \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )+\left (-2 e^{2 e^4}-6 \log ^2(3)+\log ^2(9)+e^{e^4} \log (81)\right ) \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )} \, dx+\int \frac {1}{-3+e^{2 x}} \, dx\\ &=\frac {x}{3-e^{2 x}}+\frac {2 e^{e^4}-\log (9)}{2 \left (3-e^{2 x}\right )}+\frac {1}{3} x \left (2 e^{e^4}-\log (9)\right )+\frac {1}{3} x \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right )-\frac {1}{6} \left (2 e^{e^4}-\log (9)\right ) \log \left (3-e^{2 x}\right )-\frac {1}{6} \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right ) \log \left (3-e^{2 x}\right )+4 \log (x)-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3}\right )}{x} \, dx,x,e^{2 x}\right )+\frac {1}{3} \int \log \left (1-\frac {e^{2 x}}{3}\right ) \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(-3+x) x} \, dx,x,e^{2 x}\right )+\left (e^{e^4}-\log (3)\right )^2 \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx-\left (3 \left (2 e^{e^4}-\log (9)\right )^2\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )} \, dx+\left (-2 e^{2 e^4}-6 \log ^2(3)+\log ^2(9)+e^{e^4} \log (81)\right ) \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )} \, dx\\ &=\frac {x}{3-e^{2 x}}+\frac {2 e^{e^4}-\log (9)}{2 \left (3-e^{2 x}\right )}+\frac {1}{3} x \left (2 e^{e^4}-\log (9)\right )+\frac {1}{3} x \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right )-\frac {1}{6} \left (2 e^{e^4}-\log (9)\right ) \log \left (3-e^{2 x}\right )-\frac {1}{6} \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right ) \log \left (3-e^{2 x}\right )+4 \log (x)+\frac {1}{6} \text {Li}_2\left (\frac {e^{2 x}}{3}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{-3+x} \, dx,x,e^{2 x}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3}\right )}{x} \, dx,x,e^{2 x}\right )+\left (e^{e^4}-\log (3)\right )^2 \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx-\left (3 \left (2 e^{e^4}-\log (9)\right )^2\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )} \, dx+\left (-2 e^{2 e^4}-6 \log ^2(3)+\log ^2(9)+e^{e^4} \log (81)\right ) \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )} \, dx\\ &=-\frac {x}{3}+\frac {x}{3-e^{2 x}}+\frac {2 e^{e^4}-\log (9)}{2 \left (3-e^{2 x}\right )}+\frac {1}{3} x \left (2 e^{e^4}-\log (9)\right )+\frac {1}{3} x \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right )+\frac {1}{6} \log \left (3-e^{2 x}\right )-\frac {1}{6} \left (2 e^{e^4}-\log (9)\right ) \log \left (3-e^{2 x}\right )-\frac {1}{6} \left (1-2 e^{e^4}+\frac {2 \log ^2(9)}{\log (81)}\right ) \log \left (3-e^{2 x}\right )+4 \log (x)+\left (e^{e^4}-\log (3)\right )^2 \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )^2} \, dx-\left (3 \left (2 e^{e^4}-\log (9)\right )^2\right ) \int \frac {1}{\left (-3+e^{2 x}\right )^2 \left (2 e^{e^4}-2 x-\log (9)\right )} \, dx+\left (-2 e^{2 e^4}-6 \log ^2(3)+\log ^2(9)+e^{e^4} \log (81)\right ) \int \frac {1}{\left (-3+e^{2 x}\right ) \left (e^{e^4}-x-\log (3)\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 47, normalized size = 1.42 \begin {gather*} -\frac {x^2 \left (-2 e^{e^4}+2 x+\log (9)\right )}{2 \left (-3+e^{2 x}\right ) \left (-e^{e^4}+x+\log (3)\right )^2}+4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^4)*(36 - 24*E^(2*x) + 4*E^(4*x)) + 36*x^2 + 3*x^3 + (72*x + 6*x^2)*Log[3] + 36*Log[3]^2 + E^
(2*x)*(-24*x^2 - x^3 + 2*x^4 + (-48*x - 2*x^2 + 2*x^3)*Log[3] - 24*Log[3]^2) + E^(4*x)*(4*x^2 + 8*x*Log[3] + 4
*Log[3]^2) + E^E^4*(-72*x - 6*x^2 + E^(4*x)*(-8*x - 8*Log[3]) - 72*Log[3] + E^(2*x)*(48*x + 2*x^2 - 2*x^3 + 48
*Log[3])))/(9*x^3 + E^(2*E^4)*(9*x - 6*E^(2*x)*x + E^(4*x)*x) + 18*x^2*Log[3] + 9*x*Log[3]^2 + E^(2*x)*(-6*x^3
- 12*x^2*Log[3] - 6*x*Log[3]^2) + E^(4*x)*(x^3 + 2*x^2*Log[3] + x*Log[3]^2) + E^E^4*(-18*x^2 - 18*x*Log[3] +
E^(4*x)*(-2*x^2 - 2*x*Log[3]) + E^(2*x)*(12*x^2 + 12*x*Log[3]))),x]

[Out]

-1/2*(x^2*(-2*E^E^4 + 2*x + Log[9]))/((-3 + E^(2*x))*(-E^E^4 + x + Log[3])^2) + 4*Log[x]

________________________________________________________________________________________

fricas [B]  time = 0.64, size = 68, normalized size = 2.06 \begin {gather*} -\frac {x^{2} - 4 \, {\left ({\left (x + \log \relax (3)\right )} e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} - 3\right )} e^{\left (e^{4}\right )} - 3 \, x - 3 \, \log \relax (3)\right )} \log \relax (x)}{{\left (x + \log \relax (3)\right )} e^{\left (2 \, x\right )} - {\left (e^{\left (2 \, x\right )} - 3\right )} e^{\left (e^{4}\right )} - 3 \, x - 3 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2*x)^2-24*exp(2*x)+36)*exp(exp(4))^2+((-8*log(3)-8*x)*exp(2*x)^2+(48*log(3)-2*x^3+2*x^2+48*x
)*exp(2*x)-72*log(3)-6*x^2-72*x)*exp(exp(4))+(4*log(3)^2+8*x*log(3)+4*x^2)*exp(2*x)^2+(-24*log(3)^2+(2*x^3-2*x
^2-48*x)*log(3)+2*x^4-x^3-24*x^2)*exp(2*x)+36*log(3)^2+(6*x^2+72*x)*log(3)+3*x^3+36*x^2)/((x*exp(2*x)^2-6*x*ex
p(2*x)+9*x)*exp(exp(4))^2+((-2*x*log(3)-2*x^2)*exp(2*x)^2+(12*x*log(3)+12*x^2)*exp(2*x)-18*x*log(3)-18*x^2)*ex
p(exp(4))+(x*log(3)^2+2*x^2*log(3)+x^3)*exp(2*x)^2+(-6*x*log(3)^2-12*x^2*log(3)-6*x^3)*exp(2*x)+9*x*log(3)^2+1
8*x^2*log(3)+9*x^3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.30, size = 92, normalized size = 2.79 \begin {gather*} \frac {4 \, x e^{\left (2 \, x\right )} \log \relax (x) + 4 \, e^{\left (2 \, x\right )} \log \relax (3) \log \relax (x) - x^{2} - 12 \, x \log \relax (x) - 4 \, e^{\left (2 \, x + e^{4}\right )} \log \relax (x) + 12 \, e^{\left (e^{4}\right )} \log \relax (x) - 12 \, \log \relax (3) \log \relax (x)}{x e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \relax (3) - 3 \, x - e^{\left (2 \, x + e^{4}\right )} + 3 \, e^{\left (e^{4}\right )} - 3 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2*x)^2-24*exp(2*x)+36)*exp(exp(4))^2+((-8*log(3)-8*x)*exp(2*x)^2+(48*log(3)-2*x^3+2*x^2+48*x
)*exp(2*x)-72*log(3)-6*x^2-72*x)*exp(exp(4))+(4*log(3)^2+8*x*log(3)+4*x^2)*exp(2*x)^2+(-24*log(3)^2+(2*x^3-2*x
^2-48*x)*log(3)+2*x^4-x^3-24*x^2)*exp(2*x)+36*log(3)^2+(6*x^2+72*x)*log(3)+3*x^3+36*x^2)/((x*exp(2*x)^2-6*x*ex
p(2*x)+9*x)*exp(exp(4))^2+((-2*x*log(3)-2*x^2)*exp(2*x)^2+(12*x*log(3)+12*x^2)*exp(2*x)-18*x*log(3)-18*x^2)*ex
p(exp(4))+(x*log(3)^2+2*x^2*log(3)+x^3)*exp(2*x)^2+(-6*x*log(3)^2-12*x^2*log(3)-6*x^3)*exp(2*x)+9*x*log(3)^2+1
8*x^2*log(3)+9*x^3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.88, size = 30, normalized size = 0.91

method	result	size
norman	\(-\frac {x^{2}}{\left ({\mathrm e}^{2 x}-3\right ) \left (\ln \relax (3)+x -{\mathrm e}^{{\mathrm e}^{4}}\right )}+4 \ln \relax (x )\)	\(30\)
risch	\(-\frac {x^{2}}{\left ({\mathrm e}^{2 x}-3\right ) \left (\ln \relax (3)+x -{\mathrm e}^{{\mathrm e}^{4}}\right )}+4 \ln \relax (x )\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(2*x)^2-24*exp(2*x)+36)*exp(exp(4))^2+((-8*ln(3)-8*x)*exp(2*x)^2+(48*ln(3)-2*x^3+2*x^2+48*x)*exp(2*
x)-72*ln(3)-6*x^2-72*x)*exp(exp(4))+(4*ln(3)^2+8*x*ln(3)+4*x^2)*exp(2*x)^2+(-24*ln(3)^2+(2*x^3-2*x^2-48*x)*ln(
3)+2*x^4-x^3-24*x^2)*exp(2*x)+36*ln(3)^2+(6*x^2+72*x)*ln(3)+3*x^3+36*x^2)/((x*exp(2*x)^2-6*x*exp(2*x)+9*x)*exp
(exp(4))^2+((-2*x*ln(3)-2*x^2)*exp(2*x)^2+(12*x*ln(3)+12*x^2)*exp(2*x)-18*x*ln(3)-18*x^2)*exp(exp(4))+(x*ln(3)
^2+2*x^2*ln(3)+x^3)*exp(2*x)^2+(-6*x*ln(3)^2-12*x^2*ln(3)-6*x^3)*exp(2*x)+9*x*ln(3)^2+18*x^2*ln(3)+9*x^3),x,me
thod=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 39, normalized size = 1.18 \begin {gather*} -\frac {x^{2}}{{\left (x - e^{\left (e^{4}\right )} + \log \relax (3)\right )} e^{\left (2 \, x\right )} - 3 \, x + 3 \, e^{\left (e^{4}\right )} - 3 \, \log \relax (3)} + 4 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2*x)^2-24*exp(2*x)+36)*exp(exp(4))^2+((-8*log(3)-8*x)*exp(2*x)^2+(48*log(3)-2*x^3+2*x^2+48*x
)*exp(2*x)-72*log(3)-6*x^2-72*x)*exp(exp(4))+(4*log(3)^2+8*x*log(3)+4*x^2)*exp(2*x)^2+(-24*log(3)^2+(2*x^3-2*x
^2-48*x)*log(3)+2*x^4-x^3-24*x^2)*exp(2*x)+36*log(3)^2+(6*x^2+72*x)*log(3)+3*x^3+36*x^2)/((x*exp(2*x)^2-6*x*ex
p(2*x)+9*x)*exp(exp(4))^2+((-2*x*log(3)-2*x^2)*exp(2*x)^2+(12*x*log(3)+12*x^2)*exp(2*x)-18*x*log(3)-18*x^2)*ex
p(exp(4))+(x*log(3)^2+2*x^2*log(3)+x^3)*exp(2*x)^2+(-6*x*log(3)^2-12*x^2*log(3)-6*x^3)*exp(2*x)+9*x*log(3)^2+1
8*x^2*log(3)+9*x^3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.46, size = 44, normalized size = 1.33 \begin {gather*} 4\,\ln \relax (x)-\frac {x^2\,\ln \relax (3)-x^2\,{\mathrm {e}}^{{\mathrm {e}}^4}+x^3}{\left ({\mathrm {e}}^{2\,x}-3\right )\,{\left (x+\ln \relax (3)-{\mathrm {e}}^{{\mathrm {e}}^4}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(4))*(4*exp(4*x) - 24*exp(2*x) + 36) - exp(exp(4))*(72*x + 72*log(3) - exp(2*x)*(48*x + 48*log(3
) + 2*x^2 - 2*x^3) + exp(4*x)*(8*x + 8*log(3)) + 6*x^2) + log(3)*(72*x + 6*x^2) + exp(4*x)*(8*x*log(3) + 4*log
(3)^2 + 4*x^2) + 36*log(3)^2 + 36*x^2 + 3*x^3 - exp(2*x)*(log(3)*(48*x + 2*x^2 - 2*x^3) + 24*log(3)^2 + 24*x^2
+ x^3 - 2*x^4))/(exp(4*x)*(x*log(3)^2 + 2*x^2*log(3) + x^3) - exp(2*x)*(6*x*log(3)^2 + 12*x^2*log(3) + 6*x^3)
+ 9*x*log(3)^2 + 18*x^2*log(3) + 9*x^3 + exp(2*exp(4))*(9*x - 6*x*exp(2*x) + x*exp(4*x)) - exp(exp(4))*(18*x*
log(3) + exp(4*x)*(2*x*log(3) + 2*x^2) - exp(2*x)*(12*x*log(3) + 12*x^2) + 18*x^2)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.36, size = 37, normalized size = 1.12 \begin {gather*} - \frac {x^{2}}{- 3 x + \left (x - e^{e^{4}} + \log {\relax (3 )}\right ) e^{2 x} - 3 \log {\relax (3 )} + 3 e^{e^{4}}} + 4 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(2*x)**2-24*exp(2*x)+36)*exp(exp(4))**2+((-8*ln(3)-8*x)*exp(2*x)**2+(48*ln(3)-2*x**3+2*x**2+4
8*x)*exp(2*x)-72*ln(3)-6*x**2-72*x)*exp(exp(4))+(4*ln(3)**2+8*x*ln(3)+4*x**2)*exp(2*x)**2+(-24*ln(3)**2+(2*x**
3-2*x**2-48*x)*ln(3)+2*x**4-x**3-24*x**2)*exp(2*x)+36*ln(3)**2+(6*x**2+72*x)*ln(3)+3*x**3+36*x**2)/((x*exp(2*x
)**2-6*x*exp(2*x)+9*x)*exp(exp(4))**2+((-2*x*ln(3)-2*x**2)*exp(2*x)**2+(12*x*ln(3)+12*x**2)*exp(2*x)-18*x*ln(3
)-18*x**2)*exp(exp(4))+(x*ln(3)**2+2*x**2*ln(3)+x**3)*exp(2*x)**2+(-6*x*ln(3)**2-12*x**2*ln(3)-6*x**3)*exp(2*x
)+9*x*ln(3)**2+18*x**2*ln(3)+9*x**3),x)

[Out]

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

________________________________________________________________________________________