3.4.58 \(\int \frac {4 e^{e^5} x+2 x^2+(-10 x^2-2 x^3-5 x^4-x^5+(-2 x^2-x^4) \log (3)+e^{2 e^5} (-20 x^2-4 x^3-4 x^2 \log (3))+e^{e^5} (-20 x^3-4 x^4-4 x^3 \log (3))) \log (5+x+\log (3))+(10 x^3+2 x^4+2 x^3 \log (3)+e^{2 e^5} (40 x+8 x^2+8 x \log (3))+e^{e^5} (-20-4 x+40 x^2+8 x^3+(-4+8 x^2) \log (3))) \log (5+x+\log (3)) \log (\log (5+x+\log (3)))+(-5 x^2-x^3+e^{2 e^5} (-20-4 x-4 \log (3))-x^2 \log (3)+e^{e^5} (-20 x-4 x^2-4 x \log (3))) \log (5+x+\log (3)) \log ^2(\log (5+x+\log (3)))}{(5 x^4+x^5+x^4 \log (3)+e^{2 e^5} (20 x^2+4 x^3+4 x^2 \log (3))+e^{e^5} (20 x^3+4 x^4+4 x^3 \log (3))) \log (5+x+\log (3))+(-10 x^3-2 x^4-2 x^3 \log (3)+e^{2 e^5} (-40 x-8 x^2-8 x \log (3))+e^{e^5} (-40 x^2-8 x^3-8 x^2 \log (3))) \log (5+x+\log (3)) \log (\log (5+x+\log (3)))+(5 x^2+x^3+x^2 \log (3)+e^{2 e^5} (20+4 x+4 \log (3))+e^{e^5} (20 x+4 x^2+4 x \log (3))) \log (5+x+\log (3)) \log ^2(\log (5+x+\log (3)))} \, dx\)

Optimal. Leaf size=33 \[ -1-x+\frac {x}{\left (e^{e^5}+\frac {x}{2}\right ) (x-\log (\log (5+x+\log (3))))} \]

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Rubi [F]  time = 13.55, 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 {4 e^{e^5} x+2 x^2+\left (-10 x^2-2 x^3-5 x^4-x^5+\left (-2 x^2-x^4\right ) \log (3)+e^{2 e^5} \left (-20 x^2-4 x^3-4 x^2 \log (3)\right )+e^{e^5} \left (-20 x^3-4 x^4-4 x^3 \log (3)\right )\right ) \log (5+x+\log (3))+\left (10 x^3+2 x^4+2 x^3 \log (3)+e^{2 e^5} \left (40 x+8 x^2+8 x \log (3)\right )+e^{e^5} \left (-20-4 x+40 x^2+8 x^3+\left (-4+8 x^2\right ) \log (3)\right )\right ) \log (5+x+\log (3)) \log (\log (5+x+\log (3)))+\left (-5 x^2-x^3+e^{2 e^5} (-20-4 x-4 \log (3))-x^2 \log (3)+e^{e^5} \left (-20 x-4 x^2-4 x \log (3)\right )\right ) \log (5+x+\log (3)) \log ^2(\log (5+x+\log (3)))}{\left (5 x^4+x^5+x^4 \log (3)+e^{2 e^5} \left (20 x^2+4 x^3+4 x^2 \log (3)\right )+e^{e^5} \left (20 x^3+4 x^4+4 x^3 \log (3)\right )\right ) \log (5+x+\log (3))+\left (-10 x^3-2 x^4-2 x^3 \log (3)+e^{2 e^5} \left (-40 x-8 x^2-8 x \log (3)\right )+e^{e^5} \left (-40 x^2-8 x^3-8 x^2 \log (3)\right )\right ) \log (5+x+\log (3)) \log (\log (5+x+\log (3)))+\left (5 x^2+x^3+x^2 \log (3)+e^{2 e^5} (20+4 x+4 \log (3))+e^{e^5} \left (20 x+4 x^2+4 x \log (3)\right )\right ) \log (5+x+\log (3)) \log ^2(\log (5+x+\log (3)))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-x + (2*(5 + Log[3]))/((5 - 2*E^E^5 + Log[3])*(x - Log[Log[5 + x + Log[3]]])) + (4*E^E^5*Defer[Int][(x - Log[L
og[5 + x + Log[3]]])^(-2), x])/(5 - 2*E^E^5 + Log[3]) - (8*E^(2*E^5)*Defer[Int][1/((2*E^E^5 + x)*(x - Log[Log[
5 + x + Log[3]]])^2), x])/(5 - 2*E^E^5 + Log[3]) + (4*E^E^5*(5 + Log[3])*Defer[Int][1/((2*E^E^5 + x)*(x - Log[
Log[5 + x + Log[3]]])^2), x])/(5 - 2*E^E^5 + Log[3]) - (4*E^E^5*Defer[Int][1/((2*E^E^5 + x)*Log[5 + x + Log[3]
]*(x - Log[Log[5 + x + Log[3]]])^2), x])/(5 - 2*E^E^5 + Log[3]) + 4*E^E^5*Defer[Int][1/((2*E^E^5 + x)^2*(x - L
og[Log[5 + x + Log[3]]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x \left (2 e^{e^5}+x\right )-\log (5+x+\log (3)) \left (x^2 \left (10+2 x+x^3+x^2 (5+\log (3))+\log (9)+e^{2 e^5} (20+4 x+\log (81))+e^{e^5} x (20+4 x+\log (81))\right )-\left (8 e^{2 e^5} x (5+x+\log (3))+4 e^{e^5} \left (-1+2 x^2\right ) (5+x+\log (3))+x^3 (10+2 x+\log (9))\right ) \log (\log (5+x+\log (3)))+\left (2 e^{e^5}+x\right )^2 (5+x+\log (3)) \log ^2(\log (5+x+\log (3)))\right )}{\left (2 e^{e^5}+x\right )^2 (5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx\\ &=\int \left (-1+\frac {x \left (4 e^{e^5}+2 x-2 x^2 \log (5+x+\log (3))-20 e^{e^5} \left (1+\frac {\log (3)}{5}\right ) \log (5+x+\log (3))-10 x \left (1+\frac {1}{5} \left (2 e^{e^5}+\log (3)\right )\right ) \log (5+x+\log (3))\right )}{\left (2 e^{e^5}+x\right )^2 (5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2}+\frac {4 e^{e^5}}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))}\right ) \, dx\\ &=-x+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\int \frac {x \left (4 e^{e^5}+2 x-2 x^2 \log (5+x+\log (3))-20 e^{e^5} \left (1+\frac {\log (3)}{5}\right ) \log (5+x+\log (3))-10 x \left (1+\frac {1}{5} \left (2 e^{e^5}+\log (3)\right )\right ) \log (5+x+\log (3))\right )}{\left (2 e^{e^5}+x\right )^2 (5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx\\ &=-x+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\int \frac {2 x (1-(5+x+\log (3)) \log (5+x+\log (3)))}{\left (2 e^{e^5}+x\right ) (5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx\\ &=-x+2 \int \frac {x (1-(5+x+\log (3)) \log (5+x+\log (3)))}{\left (2 e^{e^5}+x\right ) (5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx\\ &=-x+2 \int \left (\frac {(5+\log (3)) \left (1-x \log (5+x+\log (3))-5 \left (1+\frac {\log (3)}{5}\right ) \log (5+x+\log (3))\right )}{\left (5-2 e^{e^5}+\log (3)\right ) (5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2}+\frac {2 e^{e^5} \left (-1+x \log (5+x+\log (3))+5 \left (1+\frac {\log (3)}{5}\right ) \log (5+x+\log (3))\right )}{\left (2 e^{e^5}+x\right ) \left (5-2 e^{e^5}+\log (3)\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2}\right ) \, dx+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx\\ &=-x+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\frac {\left (4 e^{e^5}\right ) \int \frac {-1+x \log (5+x+\log (3))+5 \left (1+\frac {\log (3)}{5}\right ) \log (5+x+\log (3))}{\left (2 e^{e^5}+x\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}+\frac {(2 (5+\log (3))) \int \frac {1-x \log (5+x+\log (3))-5 \left (1+\frac {\log (3)}{5}\right ) \log (5+x+\log (3))}{(5+x+\log (3)) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}\\ &=-x+\frac {2 (5+\log (3))}{\left (5-2 e^{e^5}+\log (3)\right ) (x-\log (\log (5+x+\log (3))))}+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\frac {\left (4 e^{e^5}\right ) \int \frac {-1+(5+x+\log (3)) \log (5+x+\log (3))}{\left (2 e^{e^5}+x\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}\\ &=-x+\frac {2 (5+\log (3))}{\left (5-2 e^{e^5}+\log (3)\right ) (x-\log (\log (5+x+\log (3))))}+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\frac {\left (4 e^{e^5}\right ) \int \left (\frac {x}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2}+\frac {5 \left (1+\frac {\log (3)}{5}\right )}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2}-\frac {1}{\left (2 e^{e^5}+x\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2}\right ) \, dx}{5-2 e^{e^5}+\log (3)}\\ &=-x+\frac {2 (5+\log (3))}{\left (5-2 e^{e^5}+\log (3)\right ) (x-\log (\log (5+x+\log (3))))}+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\frac {\left (4 e^{e^5}\right ) \int \frac {x}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}-\frac {\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}+\frac {\left (4 e^{e^5} (5+\log (3))\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}\\ &=-x+\frac {2 (5+\log (3))}{\left (5-2 e^{e^5}+\log (3)\right ) (x-\log (\log (5+x+\log (3))))}+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\frac {\left (4 e^{e^5}\right ) \int \left (\frac {1}{(x-\log (\log (5+x+\log (3))))^2}-\frac {2 e^{e^5}}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2}\right ) \, dx}{5-2 e^{e^5}+\log (3)}-\frac {\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}+\frac {\left (4 e^{e^5} (5+\log (3))\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}\\ &=-x+\frac {2 (5+\log (3))}{\left (5-2 e^{e^5}+\log (3)\right ) (x-\log (\log (5+x+\log (3))))}+\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right )^2 (x-\log (\log (5+x+\log (3))))} \, dx+\frac {\left (4 e^{e^5}\right ) \int \frac {1}{(x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}-\frac {\left (4 e^{e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) \log (5+x+\log (3)) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}-\frac {\left (8 e^{2 e^5}\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}+\frac {\left (4 e^{e^5} (5+\log (3))\right ) \int \frac {1}{\left (2 e^{e^5}+x\right ) (x-\log (\log (5+x+\log (3))))^2} \, dx}{5-2 e^{e^5}+\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.49, size = 87, normalized size = 2.64 \begin {gather*} x \left (-1+\frac {-2 \left (2 e^{e^5}+x\right )+\left (x (10+2 x+\log (9))+e^{e^5} (20+4 x+\log (81))\right ) \log (5+x+\log (3))}{\left (2 e^{e^5}+x\right )^2 (-1+(5+x+\log (3)) \log (5+x+\log (3))) (x-\log (\log (5+x+\log (3))))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^E^5*x + 2*x^2 + (-10*x^2 - 2*x^3 - 5*x^4 - x^5 + (-2*x^2 - x^4)*Log[3] + E^(2*E^5)*(-20*x^2 - 4
*x^3 - 4*x^2*Log[3]) + E^E^5*(-20*x^3 - 4*x^4 - 4*x^3*Log[3]))*Log[5 + x + Log[3]] + (10*x^3 + 2*x^4 + 2*x^3*L
og[3] + E^(2*E^5)*(40*x + 8*x^2 + 8*x*Log[3]) + E^E^5*(-20 - 4*x + 40*x^2 + 8*x^3 + (-4 + 8*x^2)*Log[3]))*Log[
5 + x + Log[3]]*Log[Log[5 + x + Log[3]]] + (-5*x^2 - x^3 + E^(2*E^5)*(-20 - 4*x - 4*Log[3]) - x^2*Log[3] + E^E
^5*(-20*x - 4*x^2 - 4*x*Log[3]))*Log[5 + x + Log[3]]*Log[Log[5 + x + Log[3]]]^2)/((5*x^4 + x^5 + x^4*Log[3] +
E^(2*E^5)*(20*x^2 + 4*x^3 + 4*x^2*Log[3]) + E^E^5*(20*x^3 + 4*x^4 + 4*x^3*Log[3]))*Log[5 + x + Log[3]] + (-10*
x^3 - 2*x^4 - 2*x^3*Log[3] + E^(2*E^5)*(-40*x - 8*x^2 - 8*x*Log[3]) + E^E^5*(-40*x^2 - 8*x^3 - 8*x^2*Log[3]))*
Log[5 + x + Log[3]]*Log[Log[5 + x + Log[3]]] + (5*x^2 + x^3 + x^2*Log[3] + E^(2*E^5)*(20 + 4*x + 4*Log[3]) + E
^E^5*(20*x + 4*x^2 + 4*x*Log[3]))*Log[5 + x + Log[3]]*Log[Log[5 + x + Log[3]]]^2),x]

[Out]

x*(-1 + (-2*(2*E^E^5 + x) + (x*(10 + 2*x + Log[9]) + E^E^5*(20 + 4*x + Log[81]))*Log[5 + x + Log[3]])/((2*E^E^
5 + x)^2*(-1 + (5 + x + Log[3])*Log[5 + x + Log[3]])*(x - Log[Log[5 + x + Log[3]]])))

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fricas [B]  time = 0.70, size = 64, normalized size = 1.94 \begin {gather*} -\frac {x^{3} + 2 \, x^{2} e^{\left (e^{5}\right )} - {\left (x^{2} + 2 \, x e^{\left (e^{5}\right )}\right )} \log \left (\log \left (x + \log \relax (3) + 5\right )\right ) - 2 \, x}{x^{2} + 2 \, x e^{\left (e^{5}\right )} - {\left (x + 2 \, e^{\left (e^{5}\right )}\right )} \log \left (\log \left (x + \log \relax (3) + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*log(3)-4*x-20)*exp(exp(5))^2+(-4*x*log(3)-4*x^2-20*x)*exp(exp(5))-x^2*log(3)-x^3-5*x^2)*log(lo
g(3)+5+x)*log(log(log(3)+5+x))^2+((8*x*log(3)+8*x^2+40*x)*exp(exp(5))^2+((8*x^2-4)*log(3)+8*x^3+40*x^2-4*x-20)
*exp(exp(5))+2*x^3*log(3)+2*x^4+10*x^3)*log(log(3)+5+x)*log(log(log(3)+5+x))+((-4*x^2*log(3)-4*x^3-20*x^2)*exp
(exp(5))^2+(-4*x^3*log(3)-4*x^4-20*x^3)*exp(exp(5))+(-x^4-2*x^2)*log(3)-x^5-5*x^4-2*x^3-10*x^2)*log(log(3)+5+x
)+4*x*exp(exp(5))+2*x^2)/(((4*log(3)+20+4*x)*exp(exp(5))^2+(4*x*log(3)+4*x^2+20*x)*exp(exp(5))+x^2*log(3)+x^3+
5*x^2)*log(log(3)+5+x)*log(log(log(3)+5+x))^2+((-8*x*log(3)-8*x^2-40*x)*exp(exp(5))^2+(-8*x^2*log(3)-8*x^3-40*
x^2)*exp(exp(5))-2*x^3*log(3)-2*x^4-10*x^3)*log(log(3)+5+x)*log(log(log(3)+5+x))+((4*x^2*log(3)+4*x^3+20*x^2)*
exp(exp(5))^2+(4*x^3*log(3)+4*x^4+20*x^3)*exp(exp(5))+x^4*log(3)+x^5+5*x^4)*log(log(3)+5+x)),x, algorithm="fri
cas")

[Out]

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

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giac [B]  time = 2.37, size = 76, normalized size = 2.30 \begin {gather*} -\frac {x^{3} + 2 \, x^{2} e^{\left (e^{5}\right )} - x^{2} \log \left (\log \left (x + \log \relax (3) + 5\right )\right ) - 2 \, x e^{\left (e^{5}\right )} \log \left (\log \left (x + \log \relax (3) + 5\right )\right ) - 2 \, x}{x^{2} + 2 \, x e^{\left (e^{5}\right )} - x \log \left (\log \left (x + \log \relax (3) + 5\right )\right ) - 2 \, e^{\left (e^{5}\right )} \log \left (\log \left (x + \log \relax (3) + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*log(3)-4*x-20)*exp(exp(5))^2+(-4*x*log(3)-4*x^2-20*x)*exp(exp(5))-x^2*log(3)-x^3-5*x^2)*log(lo
g(3)+5+x)*log(log(log(3)+5+x))^2+((8*x*log(3)+8*x^2+40*x)*exp(exp(5))^2+((8*x^2-4)*log(3)+8*x^3+40*x^2-4*x-20)
*exp(exp(5))+2*x^3*log(3)+2*x^4+10*x^3)*log(log(3)+5+x)*log(log(log(3)+5+x))+((-4*x^2*log(3)-4*x^3-20*x^2)*exp
(exp(5))^2+(-4*x^3*log(3)-4*x^4-20*x^3)*exp(exp(5))+(-x^4-2*x^2)*log(3)-x^5-5*x^4-2*x^3-10*x^2)*log(log(3)+5+x
)+4*x*exp(exp(5))+2*x^2)/(((4*log(3)+20+4*x)*exp(exp(5))^2+(4*x*log(3)+4*x^2+20*x)*exp(exp(5))+x^2*log(3)+x^3+
5*x^2)*log(log(3)+5+x)*log(log(log(3)+5+x))^2+((-8*x*log(3)-8*x^2-40*x)*exp(exp(5))^2+(-8*x^2*log(3)-8*x^3-40*
x^2)*exp(exp(5))-2*x^3*log(3)-2*x^4-10*x^3)*log(log(3)+5+x)*log(log(log(3)+5+x))+((4*x^2*log(3)+4*x^3+20*x^2)*
exp(exp(5))^2+(4*x^3*log(3)+4*x^4+20*x^3)*exp(exp(5))+x^4*log(3)+x^5+5*x^4)*log(log(3)+5+x)),x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.84, size = 30, normalized size = 0.91

method	result	size
risch	\(-x +\frac {2 x}{\left (2 \,{\mathrm e}^{{\mathrm e}^{5}}+x \right ) \left (x -\ln \left (\ln \left (\ln \relax (3)+5+x \right )\right )\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*ln(3)-4*x-20)*exp(exp(5))^2+(-4*x*ln(3)-4*x^2-20*x)*exp(exp(5))-x^2*ln(3)-x^3-5*x^2)*ln(ln(3)+5+x)*l
n(ln(ln(3)+5+x))^2+((8*x*ln(3)+8*x^2+40*x)*exp(exp(5))^2+((8*x^2-4)*ln(3)+8*x^3+40*x^2-4*x-20)*exp(exp(5))+2*x
^3*ln(3)+2*x^4+10*x^3)*ln(ln(3)+5+x)*ln(ln(ln(3)+5+x))+((-4*x^2*ln(3)-4*x^3-20*x^2)*exp(exp(5))^2+(-4*x^3*ln(3
)-4*x^4-20*x^3)*exp(exp(5))+(-x^4-2*x^2)*ln(3)-x^5-5*x^4-2*x^3-10*x^2)*ln(ln(3)+5+x)+4*x*exp(exp(5))+2*x^2)/((
(4*ln(3)+20+4*x)*exp(exp(5))^2+(4*x*ln(3)+4*x^2+20*x)*exp(exp(5))+x^2*ln(3)+x^3+5*x^2)*ln(ln(3)+5+x)*ln(ln(ln(
3)+5+x))^2+((-8*x*ln(3)-8*x^2-40*x)*exp(exp(5))^2+(-8*x^2*ln(3)-8*x^3-40*x^2)*exp(exp(5))-2*x^3*ln(3)-2*x^4-10
*x^3)*ln(ln(3)+5+x)*ln(ln(ln(3)+5+x))+((4*x^2*ln(3)+4*x^3+20*x^2)*exp(exp(5))^2+(4*x^3*ln(3)+4*x^4+20*x^3)*exp
(exp(5))+x^4*ln(3)+x^5+5*x^4)*ln(ln(3)+5+x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.72, size = 64, normalized size = 1.94 \begin {gather*} -\frac {x^{3} + 2 \, x^{2} e^{\left (e^{5}\right )} - {\left (x^{2} + 2 \, x e^{\left (e^{5}\right )}\right )} \log \left (\log \left (x + \log \relax (3) + 5\right )\right ) - 2 \, x}{x^{2} + 2 \, x e^{\left (e^{5}\right )} - {\left (x + 2 \, e^{\left (e^{5}\right )}\right )} \log \left (\log \left (x + \log \relax (3) + 5\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*log(3)-4*x-20)*exp(exp(5))^2+(-4*x*log(3)-4*x^2-20*x)*exp(exp(5))-x^2*log(3)-x^3-5*x^2)*log(lo
g(3)+5+x)*log(log(log(3)+5+x))^2+((8*x*log(3)+8*x^2+40*x)*exp(exp(5))^2+((8*x^2-4)*log(3)+8*x^3+40*x^2-4*x-20)
*exp(exp(5))+2*x^3*log(3)+2*x^4+10*x^3)*log(log(3)+5+x)*log(log(log(3)+5+x))+((-4*x^2*log(3)-4*x^3-20*x^2)*exp
(exp(5))^2+(-4*x^3*log(3)-4*x^4-20*x^3)*exp(exp(5))+(-x^4-2*x^2)*log(3)-x^5-5*x^4-2*x^3-10*x^2)*log(log(3)+5+x
)+4*x*exp(exp(5))+2*x^2)/(((4*log(3)+20+4*x)*exp(exp(5))^2+(4*x*log(3)+4*x^2+20*x)*exp(exp(5))+x^2*log(3)+x^3+
5*x^2)*log(log(3)+5+x)*log(log(log(3)+5+x))^2+((-8*x*log(3)-8*x^2-40*x)*exp(exp(5))^2+(-8*x^2*log(3)-8*x^3-40*
x^2)*exp(exp(5))-2*x^3*log(3)-2*x^4-10*x^3)*log(log(3)+5+x)*log(log(log(3)+5+x))+((4*x^2*log(3)+4*x^3+20*x^2)*
exp(exp(5))^2+(4*x^3*log(3)+4*x^4+20*x^3)*exp(exp(5))+x^4*log(3)+x^5+5*x^4)*log(log(3)+5+x)),x, algorithm="max
ima")

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*exp(exp(5)) - log(x + log(3) + 5)*(log(3)*(2*x^2 + x^4) + exp(exp(5))*(4*x^3*log(3) + 20*x^3 + 4*x^4)
+ 10*x^2 + 2*x^3 + 5*x^4 + x^5 + exp(2*exp(5))*(4*x^2*log(3) + 20*x^2 + 4*x^3)) + 2*x^2 - log(log(x + log(3)
+ 5))^2*log(x + log(3) + 5)*(exp(2*exp(5))*(4*x + 4*log(3) + 20) + x^2*log(3) + 5*x^2 + x^3 + exp(exp(5))*(20*
x + 4*x*log(3) + 4*x^2)) + log(log(x + log(3) + 5))*log(x + log(3) + 5)*(exp(exp(5))*(log(3)*(8*x^2 - 4) - 4*x
+ 40*x^2 + 8*x^3 - 20) + exp(2*exp(5))*(40*x + 8*x*log(3) + 8*x^2) + 2*x^3*log(3) + 10*x^3 + 2*x^4))/(log(x +
log(3) + 5)*(exp(exp(5))*(4*x^3*log(3) + 20*x^3 + 4*x^4) + x^4*log(3) + 5*x^4 + x^5 + exp(2*exp(5))*(4*x^2*lo
g(3) + 20*x^2 + 4*x^3)) + log(log(x + log(3) + 5))^2*log(x + log(3) + 5)*(exp(2*exp(5))*(4*x + 4*log(3) + 20)
+ x^2*log(3) + 5*x^2 + x^3 + exp(exp(5))*(20*x + 4*x*log(3) + 4*x^2)) - log(log(x + log(3) + 5))*log(x + log(3
) + 5)*(exp(2*exp(5))*(40*x + 8*x*log(3) + 8*x^2) + exp(exp(5))*(8*x^2*log(3) + 40*x^2 + 8*x^3) + 2*x^3*log(3)
+ 10*x^3 + 2*x^4)),x)

[Out]

\text{Hanged}

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sympy [A]  time = 0.68, size = 36, normalized size = 1.09 \begin {gather*} - x - \frac {2 x}{- x^{2} - 2 x e^{e^{5}} + \left (x + 2 e^{e^{5}}\right ) \log {\left (\log {\left (x + \log {\relax (3 )} + 5 \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*ln(3)-4*x-20)*exp(exp(5))**2+(-4*x*ln(3)-4*x**2-20*x)*exp(exp(5))-x**2*ln(3)-x**3-5*x**2)*ln(l
n(3)+5+x)*ln(ln(ln(3)+5+x))**2+((8*x*ln(3)+8*x**2+40*x)*exp(exp(5))**2+((8*x**2-4)*ln(3)+8*x**3+40*x**2-4*x-20
)*exp(exp(5))+2*x**3*ln(3)+2*x**4+10*x**3)*ln(ln(3)+5+x)*ln(ln(ln(3)+5+x))+((-4*x**2*ln(3)-4*x**3-20*x**2)*exp
(exp(5))**2+(-4*x**3*ln(3)-4*x**4-20*x**3)*exp(exp(5))+(-x**4-2*x**2)*ln(3)-x**5-5*x**4-2*x**3-10*x**2)*ln(ln(
3)+5+x)+4*x*exp(exp(5))+2*x**2)/(((4*ln(3)+20+4*x)*exp(exp(5))**2+(4*x*ln(3)+4*x**2+20*x)*exp(exp(5))+x**2*ln(
3)+x**3+5*x**2)*ln(ln(3)+5+x)*ln(ln(ln(3)+5+x))**2+((-8*x*ln(3)-8*x**2-40*x)*exp(exp(5))**2+(-8*x**2*ln(3)-8*x
**3-40*x**2)*exp(exp(5))-2*x**3*ln(3)-2*x**4-10*x**3)*ln(ln(3)+5+x)*ln(ln(ln(3)+5+x))+((4*x**2*ln(3)+4*x**3+20
*x**2)*exp(exp(5))**2+(4*x**3*ln(3)+4*x**4+20*x**3)*exp(exp(5))+x**4*ln(3)+x**5+5*x**4)*ln(ln(3)+5+x)),x)

[Out]

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

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