3.11.1 \(\int \frac {2 x^2+4 x^3+(2 x+2 x^2+2 x^3) \log (1+x+x^2)+(e^3 (-1-x-x^2)+(-2 x-2 x^2-2 x^3) \log (1+x+x^2)) \log (e^3+2 x \log (1+x+x^2))+(e^3 (-1-x-x^2)+(-2 x-2 x^2-2 x^3) \log (1+x+x^2)) \log (e^3+2 x \log (1+x+x^2)) \log (\log (e^3+2 x \log (1+x+x^2)))}{(e^3 (x^2+x^3+x^4)+(2 x^3+2 x^4+2 x^5) \log (1+x+x^2)) \log (e^3+2 x \log (1+x+x^2))} \, dx\)

Optimal. Leaf size=22 \[ \frac {1+\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x} \]

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Rubi [F]  time = 31.31, 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 {2 x^2+4 x^3+\left (2 x+2 x^2+2 x^3\right ) \log \left (1+x+x^2\right )+\left (e^3 \left (-1-x-x^2\right )+\left (-2 x-2 x^2-2 x^3\right ) \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )+\left (e^3 \left (-1-x-x^2\right )+\left (-2 x-2 x^2-2 x^3\right ) \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{\left (e^3 \left (x^2+x^3+x^4\right )+\left (2 x^3+2 x^4+2 x^5\right ) \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x^2+4 x^3+2 x \left (1+x+x^2\right ) \log \left (1+x+x^2\right )-\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2 \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx\\ &=\int \left (\frac {2 x^2+4 x^3+2 x \log \left (1+x+x^2\right )+2 x^2 \log \left (1+x+x^2\right )+2 x^3 \log \left (1+x+x^2\right )-e^3 \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-e^3 x \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-e^3 x^2 \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-2 x \log \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-2 x^2 \log \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-2 x^3 \log \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}{x^2 \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}-\frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2}\right ) \, dx\\ &=\int \frac {2 x^2+4 x^3+2 x \log \left (1+x+x^2\right )+2 x^2 \log \left (1+x+x^2\right )+2 x^3 \log \left (1+x+x^2\right )-e^3 \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-e^3 x \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-e^3 x^2 \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-2 x \log \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-2 x^2 \log \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )-2 x^3 \log \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}{x^2 \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\int \frac {2 x^2 (1+2 x)-2 x \left (1+x+x^2\right ) \log \left (1+x+x^2\right ) \left (-1+\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )-e^3 \left (1+x+x^2\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}{x^2 \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\int \left (-\frac {1}{x^2}+\frac {2 \left (x+2 x^2+\log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+x^2 \log \left (1+x+x^2\right )\right )}{x \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}\right ) \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \frac {x+2 x^2+\log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+x^2 \log \left (1+x+x^2\right )}{x \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \frac {x (1+2 x)+\left (1+x+x^2\right ) \log \left (1+x+x^2\right )}{x \left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \left (\frac {x+2 x^2+\log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+x^2 \log \left (1+x+x^2\right )}{x \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}-\frac {(1+x) \left (x+2 x^2+\log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+x^2 \log \left (1+x+x^2\right )\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}\right ) \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \frac {x+2 x^2+\log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+x^2 \log \left (1+x+x^2\right )}{x \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-2 \int \frac {(1+x) \left (x+2 x^2+\log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+x^2 \log \left (1+x+x^2\right )\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \frac {x (1+2 x)+\left (1+x+x^2\right ) \log \left (1+x+x^2\right )}{x \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-2 \int \frac {(1+x) \left (x (1+2 x)+\left (1+x+x^2\right ) \log \left (1+x+x^2\right )\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \left (\frac {1}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {2 x}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {\log \left (1+x+x^2\right )}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {\log \left (1+x+x^2\right )}{x \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {x \log \left (1+x+x^2\right )}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}\right ) \, dx-2 \int \left (\frac {x}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {3 x^2}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {2 x^3}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {\log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {2 x \log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {2 x^2 \log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}+\frac {x^3 \log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )}\right ) \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\frac {1}{x}+2 \int \frac {1}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-2 \int \frac {x}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx+2 \int \frac {\log \left (1+x+x^2\right )}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx+2 \int \frac {\log \left (1+x+x^2\right )}{x \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx+2 \int \frac {x \log \left (1+x+x^2\right )}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-2 \int \frac {\log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-2 \int \frac {x^3 \log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx+4 \int \frac {x}{\left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-4 \int \frac {x^3}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-4 \int \frac {x \log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-4 \int \frac {x^2 \log \left (1+x+x^2\right )}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-6 \int \frac {x^2}{\left (1+x+x^2\right ) \left (e^3+2 x \log \left (1+x+x^2\right )\right ) \log \left (e^3+2 x \log \left (1+x+x^2\right )\right )} \, dx-\int \frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 24, normalized size = 1.09 \begin {gather*} \frac {1}{x}+\frac {\log \left (\log \left (e^3+2 x \log \left (1+x+x^2\right )\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 21, normalized size = 0.95 \begin {gather*} \frac {\log \left (\log \left (2 \, x \log \left (x^{2} + x + 1\right ) + e^{3}\right )\right ) + 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^3-2*x^2-2*x)*log(x^2+x+1)+(-x^2-x-1)*exp(3))*log(2*x*log(x^2+x+1)+exp(3))*log(log(2*x*log(x^
2+x+1)+exp(3)))+((-2*x^3-2*x^2-2*x)*log(x^2+x+1)+(-x^2-x-1)*exp(3))*log(2*x*log(x^2+x+1)+exp(3))+(2*x^3+2*x^2+
2*x)*log(x^2+x+1)+4*x^3+2*x^2)/((2*x^5+2*x^4+2*x^3)*log(x^2+x+1)+(x^4+x^3+x^2)*exp(3))/log(2*x*log(x^2+x+1)+ex
p(3)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {4 \, x^{3} - {\left ({\left (x^{2} + x + 1\right )} e^{3} + 2 \, {\left (x^{3} + x^{2} + x\right )} \log \left (x^{2} + x + 1\right )\right )} \log \left (2 \, x \log \left (x^{2} + x + 1\right ) + e^{3}\right ) \log \left (\log \left (2 \, x \log \left (x^{2} + x + 1\right ) + e^{3}\right )\right ) + 2 \, x^{2} + 2 \, {\left (x^{3} + x^{2} + x\right )} \log \left (x^{2} + x + 1\right ) - {\left ({\left (x^{2} + x + 1\right )} e^{3} + 2 \, {\left (x^{3} + x^{2} + x\right )} \log \left (x^{2} + x + 1\right )\right )} \log \left (2 \, x \log \left (x^{2} + x + 1\right ) + e^{3}\right )}{{\left ({\left (x^{4} + x^{3} + x^{2}\right )} e^{3} + 2 \, {\left (x^{5} + x^{4} + x^{3}\right )} \log \left (x^{2} + x + 1\right )\right )} \log \left (2 \, x \log \left (x^{2} + x + 1\right ) + e^{3}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^3-2*x^2-2*x)*log(x^2+x+1)+(-x^2-x-1)*exp(3))*log(2*x*log(x^2+x+1)+exp(3))*log(log(2*x*log(x^
2+x+1)+exp(3)))+((-2*x^3-2*x^2-2*x)*log(x^2+x+1)+(-x^2-x-1)*exp(3))*log(2*x*log(x^2+x+1)+exp(3))+(2*x^3+2*x^2+
2*x)*log(x^2+x+1)+4*x^3+2*x^2)/((2*x^5+2*x^4+2*x^3)*log(x^2+x+1)+(x^4+x^3+x^2)*exp(3))/log(2*x*log(x^2+x+1)+ex
p(3)),x, algorithm="giac")

[Out]

integrate((4*x^3 - ((x^2 + x + 1)*e^3 + 2*(x^3 + x^2 + x)*log(x^2 + x + 1))*log(2*x*log(x^2 + x + 1) + e^3)*lo
g(log(2*x*log(x^2 + x + 1) + e^3)) + 2*x^2 + 2*(x^3 + x^2 + x)*log(x^2 + x + 1) - ((x^2 + x + 1)*e^3 + 2*(x^3
+ x^2 + x)*log(x^2 + x + 1))*log(2*x*log(x^2 + x + 1) + e^3))/(((x^4 + x^3 + x^2)*e^3 + 2*(x^5 + x^4 + x^3)*lo
g(x^2 + x + 1))*log(2*x*log(x^2 + x + 1) + e^3)), x)

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maple [A]  time = 0.05, size = 24, normalized size = 1.09

method	result	size
risch	\(\frac {\ln \left (\ln \left (2 x \ln \left (x^{2}+x +1\right )+{\mathrm e}^{3}\right )\right )}{x}+\frac {1}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x^3-2*x^2-2*x)*ln(x^2+x+1)+(-x^2-x-1)*exp(3))*ln(2*x*ln(x^2+x+1)+exp(3))*ln(ln(2*x*ln(x^2+x+1)+exp(3
)))+((-2*x^3-2*x^2-2*x)*ln(x^2+x+1)+(-x^2-x-1)*exp(3))*ln(2*x*ln(x^2+x+1)+exp(3))+(2*x^3+2*x^2+2*x)*ln(x^2+x+1
)+4*x^3+2*x^2)/((2*x^5+2*x^4+2*x^3)*ln(x^2+x+1)+(x^4+x^3+x^2)*exp(3))/ln(2*x*ln(x^2+x+1)+exp(3)),x,method=_RET
URNVERBOSE)

[Out]

1/x*ln(ln(2*x*ln(x^2+x+1)+exp(3)))+1/x

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maxima [A]  time = 0.62, size = 21, normalized size = 0.95 \begin {gather*} \frac {\log \left (\log \left (2 \, x \log \left (x^{2} + x + 1\right ) + e^{3}\right )\right ) + 1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x^3-2*x^2-2*x)*log(x^2+x+1)+(-x^2-x-1)*exp(3))*log(2*x*log(x^2+x+1)+exp(3))*log(log(2*x*log(x^
2+x+1)+exp(3)))+((-2*x^3-2*x^2-2*x)*log(x^2+x+1)+(-x^2-x-1)*exp(3))*log(2*x*log(x^2+x+1)+exp(3))+(2*x^3+2*x^2+
2*x)*log(x^2+x+1)+4*x^3+2*x^2)/((2*x^5+2*x^4+2*x^3)*log(x^2+x+1)+(x^4+x^3+x^2)*exp(3))/log(2*x*log(x^2+x+1)+ex
p(3)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.72, size = 21, normalized size = 0.95 \begin {gather*} \frac {\ln \left (\ln \left ({\mathrm {e}}^3+2\,x\,\ln \left (x^2+x+1\right )\right )\right )+1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + x^2 + 1)*(2*x + 2*x^2 + 2*x^3) - log(exp(3) + 2*x*log(x + x^2 + 1))*(log(x + x^2 + 1)*(2*x + 2*x^
2 + 2*x^3) + exp(3)*(x + x^2 + 1)) + 2*x^2 + 4*x^3 - log(exp(3) + 2*x*log(x + x^2 + 1))*log(log(exp(3) + 2*x*l
og(x + x^2 + 1)))*(log(x + x^2 + 1)*(2*x + 2*x^2 + 2*x^3) + exp(3)*(x + x^2 + 1)))/(log(exp(3) + 2*x*log(x + x
^2 + 1))*(log(x + x^2 + 1)*(2*x^3 + 2*x^4 + 2*x^5) + exp(3)*(x^2 + x^3 + x^4))),x)

[Out]

(log(log(exp(3) + 2*x*log(x + x^2 + 1))) + 1)/x

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sympy [A]  time = 30.43, size = 22, normalized size = 1.00 \begin {gather*} \frac {\log {\left (\log {\left (2 x \log {\left (x^{2} + x + 1 \right )} + e^{3} \right )} \right )}}{x} + \frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x**3-2*x**2-2*x)*ln(x**2+x+1)+(-x**2-x-1)*exp(3))*ln(2*x*ln(x**2+x+1)+exp(3))*ln(ln(2*x*ln(x**
2+x+1)+exp(3)))+((-2*x**3-2*x**2-2*x)*ln(x**2+x+1)+(-x**2-x-1)*exp(3))*ln(2*x*ln(x**2+x+1)+exp(3))+(2*x**3+2*x
**2+2*x)*ln(x**2+x+1)+4*x**3+2*x**2)/((2*x**5+2*x**4+2*x**3)*ln(x**2+x+1)+(x**4+x**3+x**2)*exp(3))/ln(2*x*ln(x
**2+x+1)+exp(3)),x)

[Out]

log(log(2*x*log(x**2 + x + 1) + exp(3)))/x + 1/x

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