3.50.69 \(\int \frac {-18-30 x+6 x^2+20 x^3+4 x^4+e^{2 e^x} (-2 x^3+e^x (2 x^3+2 x^4))+(-6 x-16 x^2-2 x^3+4 x^4) \log (x)-2 x^3 \log ^2(x)+e^{e^x} (-6 x-16 x^2-2 x^3+4 x^4+e^x (6 x^2+6 x^3-4 x^4-4 x^5)+(-4 x^3+e^x (2 x^3+2 x^4)) \log (x))}{9 x^3+27 x^4+27 x^5+9 x^6} \, dx\)

Optimal. Leaf size=29 \[ \frac {\left (3+x \left (e^{e^x}-2 x+\log (x)\right )\right )^2}{\left (3 x+3 x^2\right )^2} \]

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Rubi [F]  time = 6.72, 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 {-18-30 x+6 x^2+20 x^3+4 x^4+e^{2 e^x} \left (-2 x^3+e^x \left (2 x^3+2 x^4\right )\right )+\left (-6 x-16 x^2-2 x^3+4 x^4\right ) \log (x)-2 x^3 \log ^2(x)+e^{e^x} \left (-6 x-16 x^2-2 x^3+4 x^4+e^x \left (6 x^2+6 x^3-4 x^4-4 x^5\right )+\left (-4 x^3+e^x \left (2 x^3+2 x^4\right )\right ) \log (x)\right )}{9 x^3+27 x^4+27 x^5+9 x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-18 - 30*x + 6*x^2 + 20*x^3 + 4*x^4 + E^(2*E^x)*(-2*x^3 + E^x*(2*x^3 + 2*x^4)) + (-6*x - 16*x^2 - 2*x^3 +
4*x^4)*Log[x] - 2*x^3*Log[x]^2 + E^E^x*(-6*x - 16*x^2 - 2*x^3 + 4*x^4 + E^x*(6*x^2 + 6*x^3 - 4*x^4 - 4*x^5) +
(-4*x^3 + E^x*(2*x^3 + 2*x^4))*Log[x]))/(9*x^3 + 27*x^4 + 27*x^5 + 9*x^6),x]

[Out]

x^(-2) - 2/x - 1/(9*(1 + x)^2) + (2*x^2)/(9*(1 + x)^2) + 14/(9*(1 + x)) - (8*Log[x])/9 + (2*Log[x])/(3*x) - (4
*Log[x])/(9*(1 + x)^2) + (2*x^2*Log[x])/(9*(1 + x)^2) + (2*x*Log[x])/(3*(1 + x)) + Log[x]^2/(9*(1 + x)^2) - (1
6*Defer[Int][E^E^x/(-1 - x), x])/9 - (2*Defer[Int][E^E^x/x^2, x])/3 + (2*Defer[Int][E^E^x/x, x])/9 + (2*Defer[
Int][E^(E^x + x)/x, x])/3 + (4*Defer[Int][E^E^x/(1 + x)^3, x])/9 - (4*Log[x]*Defer[Int][E^E^x/(1 + x)^3, x])/9
- (2*Defer[Int][E^(2*E^x)/(1 + x)^3, x])/9 + (8*Defer[Int][E^E^x/(1 + x)^2, x])/9 - (2*Defer[Int][E^(E^x + x)
/(1 + x)^2, x])/9 + (2*Log[x]*Defer[Int][E^(E^x + x)/(1 + x)^2, x])/9 + (2*Defer[Int][E^(2*E^x + x)/(1 + x)^2,
x])/9 - 2*Defer[Int][E^E^x/(1 + x), x] - (10*Defer[Int][E^(E^x + x)/(1 + x), x])/9 + (4*Defer[Int][Defer[Int]
[E^E^x/(1 + x)^3, x]/x, x])/9 - (2*Defer[Int][Defer[Int][E^(E^x + x)/(1 + x)^2, x]/x, x])/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (3+e^{e^x} x-2 x^2+x \log (x)\right ) \left (-3-5 x+\left (-1-e^{e^x}+e^{e^x+x}\right ) x^2+e^{e^x+x} x^3-x^2 \log (x)\right )}{9 x^3 (1+x)^3} \, dx\\ &=\frac {2}{9} \int \frac {\left (3+e^{e^x} x-2 x^2+x \log (x)\right ) \left (-3-5 x+\left (-1-e^{e^x}+e^{e^x+x}\right ) x^2+e^{e^x+x} x^3-x^2 \log (x)\right )}{x^3 (1+x)^3} \, dx\\ &=\frac {2}{9} \int \left (\frac {3 \left (-3-e^{e^x} x+2 x^2-x \log (x)\right )}{x^3 (1+x)^3}+\frac {5 \left (-3-e^{e^x} x+2 x^2-x \log (x)\right )}{x^2 (1+x)^3}+\frac {-3-e^{e^x} x+2 x^2-x \log (x)}{x (1+x)^3}+\frac {\log (x) \left (-3-e^{e^x} x+2 x^2-x \log (x)\right )}{x (1+x)^3}-\frac {e^{e^x} \left (3+e^{e^x} x-2 x^2+x \log (x)\right )}{x (1+x)^3}+\frac {e^{e^x+x} \left (3+e^{e^x} x-2 x^2+x \log (x)\right )}{x (1+x)^2}\right ) \, dx\\ &=\frac {2}{9} \int \frac {-3-e^{e^x} x+2 x^2-x \log (x)}{x (1+x)^3} \, dx+\frac {2}{9} \int \frac {\log (x) \left (-3-e^{e^x} x+2 x^2-x \log (x)\right )}{x (1+x)^3} \, dx-\frac {2}{9} \int \frac {e^{e^x} \left (3+e^{e^x} x-2 x^2+x \log (x)\right )}{x (1+x)^3} \, dx+\frac {2}{9} \int \frac {e^{e^x+x} \left (3+e^{e^x} x-2 x^2+x \log (x)\right )}{x (1+x)^2} \, dx+\frac {2}{3} \int \frac {-3-e^{e^x} x+2 x^2-x \log (x)}{x^3 (1+x)^3} \, dx+\frac {10}{9} \int \frac {-3-e^{e^x} x+2 x^2-x \log (x)}{x^2 (1+x)^3} \, dx\\ &=\frac {2}{9} \int \left (-\frac {e^{e^x}}{(1+x)^3}-\frac {3}{x (1+x)^3}+\frac {2 x}{(1+x)^3}-\frac {\log (x)}{(1+x)^3}\right ) \, dx-\frac {2}{9} \int \left (\frac {e^{2 e^x}}{(1+x)^3}+\frac {3 e^{e^x}}{x (1+x)^3}-\frac {2 e^{e^x} x}{(1+x)^3}+\frac {e^{e^x} \log (x)}{(1+x)^3}\right ) \, dx+\frac {2}{9} \int \left (-\frac {e^{e^x} \log (x)}{(1+x)^3}-\frac {3 \log (x)}{x (1+x)^3}+\frac {2 x \log (x)}{(1+x)^3}-\frac {\log ^2(x)}{(1+x)^3}\right ) \, dx+\frac {2}{9} \int \left (\frac {e^{2 e^x+x}}{(1+x)^2}+\frac {e^{e^x+x} \left (3-2 x^2+x \log (x)\right )}{x (1+x)^2}\right ) \, dx+\frac {2}{3} \int \left (-\frac {3}{x^3 (1+x)^3}-\frac {e^{e^x}}{x^2 (1+x)^3}+\frac {2}{x (1+x)^3}-\frac {\log (x)}{x^2 (1+x)^3}\right ) \, dx+\frac {10}{9} \int \left (\frac {2}{(1+x)^3}-\frac {3}{x^2 (1+x)^3}-\frac {e^{e^x}}{x (1+x)^3}-\frac {\log (x)}{x (1+x)^3}\right ) \, dx\\ &=-\frac {10}{9 (1+x)^2}-\frac {2}{9} \int \frac {e^{e^x}}{(1+x)^3} \, dx-\frac {2}{9} \int \frac {e^{2 e^x}}{(1+x)^3} \, dx+\frac {2}{9} \int \frac {e^{2 e^x+x}}{(1+x)^2} \, dx-\frac {2}{9} \int \frac {\log (x)}{(1+x)^3} \, dx-2 \left (\frac {2}{9} \int \frac {e^{e^x} \log (x)}{(1+x)^3} \, dx\right )-\frac {2}{9} \int \frac {\log ^2(x)}{(1+x)^3} \, dx+\frac {2}{9} \int \frac {e^{e^x+x} \left (3-2 x^2+x \log (x)\right )}{x (1+x)^2} \, dx+\frac {4}{9} \int \frac {x}{(1+x)^3} \, dx+\frac {4}{9} \int \frac {e^{e^x} x}{(1+x)^3} \, dx+\frac {4}{9} \int \frac {x \log (x)}{(1+x)^3} \, dx-\frac {2}{3} \int \frac {e^{e^x}}{x^2 (1+x)^3} \, dx-\frac {2}{3} \int \frac {1}{x (1+x)^3} \, dx-\frac {2}{3} \int \frac {e^{e^x}}{x (1+x)^3} \, dx-\frac {2}{3} \int \frac {\log (x)}{x^2 (1+x)^3} \, dx-\frac {2}{3} \int \frac {\log (x)}{x (1+x)^3} \, dx-\frac {10}{9} \int \frac {e^{e^x}}{x (1+x)^3} \, dx-\frac {10}{9} \int \frac {\log (x)}{x (1+x)^3} \, dx+\frac {4}{3} \int \frac {1}{x (1+x)^3} \, dx-2 \int \frac {1}{x^3 (1+x)^3} \, dx-\frac {10}{3} \int \frac {1}{x^2 (1+x)^3} \, dx\\ &=-\frac {10}{9 (1+x)^2}+\frac {2 x^2}{9 (1+x)^2}+\frac {\log (x)}{9 (1+x)^2}+\frac {2 x^2 \log (x)}{9 (1+x)^2}+\frac {\log ^2(x)}{9 (1+x)^2}-\frac {1}{9} \int \frac {1}{x (1+x)^2} \, dx-\frac {2}{9} \int \frac {e^{e^x}}{(1+x)^3} \, dx-\frac {2}{9} \int \frac {e^{2 e^x}}{(1+x)^3} \, dx+\frac {2}{9} \int \frac {e^{2 e^x+x}}{(1+x)^2} \, dx-\frac {2}{9} \int \frac {x}{(1+x)^2} \, dx-\frac {2}{9} \int \frac {\log (x)}{x (1+x)^2} \, dx+\frac {2}{9} \int \left (\frac {e^{e^x+x} \left (3-2 x^2\right )}{x (1+x)^2}+\frac {e^{e^x+x} \log (x)}{(1+x)^2}\right ) \, dx+\frac {4}{9} \int \left (-\frac {e^{e^x}}{(1+x)^3}+\frac {e^{e^x}}{(1+x)^2}\right ) \, dx-\frac {2}{3} \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^3}-\frac {1}{(1+x)^2}\right ) \, dx-\frac {2}{3} \int \left (\frac {e^{e^x}}{-1-x}+\frac {e^{e^x}}{x}-\frac {e^{e^x}}{(1+x)^3}-\frac {e^{e^x}}{(1+x)^2}\right ) \, dx-\frac {2}{3} \int \left (\frac {e^{e^x}}{x^2}-\frac {3 e^{e^x}}{x}+\frac {e^{e^x}}{(1+x)^3}+\frac {2 e^{e^x}}{(1+x)^2}+\frac {3 e^{e^x}}{1+x}\right ) \, dx+\frac {2}{3} \int \frac {\log (x)}{(1+x)^3} \, dx-\frac {2}{3} \int \frac {\log (x)}{x (1+x)^2} \, dx-\frac {2}{3} \int \left (\frac {\log (x)}{x^2}-\frac {3 \log (x)}{x}+\frac {\log (x)}{(1+x)^3}+\frac {2 \log (x)}{(1+x)^2}+\frac {3 \log (x)}{1+x}\right ) \, dx-\frac {10}{9} \int \left (\frac {e^{e^x}}{-1-x}+\frac {e^{e^x}}{x}-\frac {e^{e^x}}{(1+x)^3}-\frac {e^{e^x}}{(1+x)^2}\right ) \, dx+\frac {10}{9} \int \frac {\log (x)}{(1+x)^3} \, dx-\frac {10}{9} \int \frac {\log (x)}{x (1+x)^2} \, dx+\frac {4}{3} \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^3}-\frac {1}{(1+x)^2}\right ) \, dx-2 \int \left (\frac {1}{x^3}-\frac {3}{x^2}+\frac {6}{x}-\frac {1}{(1+x)^3}-\frac {3}{(1+x)^2}-\frac {6}{1+x}\right ) \, dx-\frac {10}{3} \int \left (\frac {1}{x^2}-\frac {3}{x}+\frac {1}{(1+x)^3}+\frac {2}{(1+x)^2}+\frac {3}{1+x}\right ) \, dx-2 \left (-\left (\frac {2}{9} \int \frac {\int \frac {e^{e^x}}{(1+x)^3} \, dx}{x} \, dx\right )+\frac {1}{9} (2 \log (x)) \int \frac {e^{e^x}}{(1+x)^3} \, dx\right )\\ &=\frac {1}{x^2}-\frac {8}{3 x}-\frac {1}{9 (1+x)^2}+\frac {2 x^2}{9 (1+x)^2}+\frac {4}{3 (1+x)}-\frac {4 \log (x)}{3}-\frac {7 \log (x)}{9 (1+x)^2}+\frac {2 x^2 \log (x)}{9 (1+x)^2}+\frac {\log ^2(x)}{9 (1+x)^2}+\frac {4}{3} \log (1+x)-\frac {1}{9} \int \left (\frac {1}{-1-x}+\frac {1}{x}-\frac {1}{(1+x)^2}\right ) \, dx-\frac {2}{9} \int \frac {e^{e^x}}{(1+x)^3} \, dx-\frac {2}{9} \int \frac {e^{2 e^x}}{(1+x)^3} \, dx+\frac {2}{9} \int \frac {e^{2 e^x+x}}{(1+x)^2} \, dx+\frac {2}{9} \int \frac {e^{e^x+x} \left (3-2 x^2\right )}{x (1+x)^2} \, dx-\frac {2}{9} \int \left (-\frac {1}{(1+x)^2}+\frac {1}{1+x}\right ) \, dx+\frac {2}{9} \int \frac {\log (x)}{(1+x)^2} \, dx+\frac {2}{9} \int \frac {e^{e^x+x} \log (x)}{(1+x)^2} \, dx-\frac {2}{9} \int \frac {\log (x)}{x (1+x)} \, dx+\frac {1}{3} \int \frac {1}{x (1+x)^2} \, dx-\frac {4}{9} \int \frac {e^{e^x}}{(1+x)^3} \, dx+\frac {4}{9} \int \frac {e^{e^x}}{(1+x)^2} \, dx+\frac {5}{9} \int \frac {1}{x (1+x)^2} \, dx-\frac {2}{3} \int \frac {e^{e^x}}{-1-x} \, dx-\frac {2}{3} \int \frac {e^{e^x}}{x^2} \, dx-\frac {2}{3} \int \frac {e^{e^x}}{x} \, dx+\frac {2}{3} \int \frac {e^{e^x}}{(1+x)^2} \, dx-\frac {2}{3} \int \frac {\log (x)}{x^2} \, dx-\frac {2}{3} \int \frac {\log (x)}{(1+x)^3} \, dx+\frac {2}{3} \int \frac {\log (x)}{(1+x)^2} \, dx-\frac {2}{3} \int \frac {\log (x)}{x (1+x)} \, dx-\frac {10}{9} \int \frac {e^{e^x}}{-1-x} \, dx-\frac {10}{9} \int \frac {e^{e^x}}{x} \, dx+\frac {10}{9} \int \frac {e^{e^x}}{(1+x)^3} \, dx+\frac {10}{9} \int \frac {e^{e^x}}{(1+x)^2} \, dx+\frac {10}{9} \int \frac {\log (x)}{(1+x)^2} \, dx-\frac {10}{9} \int \frac {\log (x)}{x (1+x)} \, dx-\frac {4}{3} \int \frac {e^{e^x}}{(1+x)^2} \, dx-\frac {4}{3} \int \frac {\log (x)}{(1+x)^2} \, dx+2 \int \frac {e^{e^x}}{x} \, dx-2 \int \frac {e^{e^x}}{1+x} \, dx+2 \int \frac {\log (x)}{x} \, dx-2 \int \frac {\log (x)}{1+x} \, dx-2 \left (-\left (\frac {2}{9} \int \frac {\int \frac {e^{e^x}}{(1+x)^3} \, dx}{x} \, dx\right )+\frac {1}{9} (2 \log (x)) \int \frac {e^{e^x}}{(1+x)^3} \, dx\right )\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [B]  time = 5.15, size = 78, normalized size = 2.69 \begin {gather*} \frac {9-16 x^2+e^{2 e^x} x^2-8 x^3+e^{e^x} \left (6 x-4 x^3\right )-2 x \left (-3-e^{e^x} x+2 x^2\right ) \log (x)+x^2 \log ^2(x)}{9 x^2 (1+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18 - 30*x + 6*x^2 + 20*x^3 + 4*x^4 + E^(2*E^x)*(-2*x^3 + E^x*(2*x^3 + 2*x^4)) + (-6*x - 16*x^2 - 2
*x^3 + 4*x^4)*Log[x] - 2*x^3*Log[x]^2 + E^E^x*(-6*x - 16*x^2 - 2*x^3 + 4*x^4 + E^x*(6*x^2 + 6*x^3 - 4*x^4 - 4*
x^5) + (-4*x^3 + E^x*(2*x^3 + 2*x^4))*Log[x]))/(9*x^3 + 27*x^4 + 27*x^5 + 9*x^6),x]

[Out]

(9 - 16*x^2 + E^(2*E^x)*x^2 - 8*x^3 + E^E^x*(6*x - 4*x^3) - 2*x*(-3 - E^E^x*x + 2*x^2)*Log[x] + x^2*Log[x]^2)/
(9*x^2*(1 + x)^2)

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fricas [B]  time = 0.65, size = 79, normalized size = 2.72 \begin {gather*} \frac {x^{2} \log \relax (x)^{2} - 8 \, x^{3} + x^{2} e^{\left (2 \, e^{x}\right )} - 16 \, x^{2} - 2 \, {\left (2 \, x^{3} - x^{2} \log \relax (x) - 3 \, x\right )} e^{\left (e^{x}\right )} - 2 \, {\left (2 \, x^{3} - 3 \, x\right )} \log \relax (x) + 9}{9 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^4+2*x^3)*exp(x)-2*x^3)*exp(exp(x))^2+(((2*x^4+2*x^3)*exp(x)-4*x^3)*log(x)+(-4*x^5-4*x^4+6*x^3
+6*x^2)*exp(x)+4*x^4-2*x^3-16*x^2-6*x)*exp(exp(x))-2*x^3*log(x)^2+(4*x^4-2*x^3-16*x^2-6*x)*log(x)+4*x^4+20*x^3
+6*x^2-30*x-18)/(9*x^6+27*x^5+27*x^4+9*x^3),x, algorithm="fricas")

[Out]

1/9*(x^2*log(x)^2 - 8*x^3 + x^2*e^(2*e^x) - 16*x^2 - 2*(2*x^3 - x^2*log(x) - 3*x)*e^(e^x) - 2*(2*x^3 - 3*x)*lo
g(x) + 9)/(x^4 + 2*x^3 + x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (x^{3} \log \relax (x)^{2} - 2 \, x^{4} - 10 \, x^{3} - 3 \, x^{2} + {\left (x^{3} - {\left (x^{4} + x^{3}\right )} e^{x}\right )} e^{\left (2 \, e^{x}\right )} - {\left (2 \, x^{4} - x^{3} - 8 \, x^{2} - {\left (2 \, x^{5} + 2 \, x^{4} - 3 \, x^{3} - 3 \, x^{2}\right )} e^{x} - {\left (2 \, x^{3} - {\left (x^{4} + x^{3}\right )} e^{x}\right )} \log \relax (x) - 3 \, x\right )} e^{\left (e^{x}\right )} - {\left (2 \, x^{4} - x^{3} - 8 \, x^{2} - 3 \, x\right )} \log \relax (x) + 15 \, x + 9\right )}}{9 \, {\left (x^{6} + 3 \, x^{5} + 3 \, x^{4} + x^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^4+2*x^3)*exp(x)-2*x^3)*exp(exp(x))^2+(((2*x^4+2*x^3)*exp(x)-4*x^3)*log(x)+(-4*x^5-4*x^4+6*x^3
+6*x^2)*exp(x)+4*x^4-2*x^3-16*x^2-6*x)*exp(exp(x))-2*x^3*log(x)^2+(4*x^4-2*x^3-16*x^2-6*x)*log(x)+4*x^4+20*x^3
+6*x^2-30*x-18)/(9*x^6+27*x^5+27*x^4+9*x^3),x, algorithm="giac")

[Out]

integrate(-2/9*(x^3*log(x)^2 - 2*x^4 - 10*x^3 - 3*x^2 + (x^3 - (x^4 + x^3)*e^x)*e^(2*e^x) - (2*x^4 - x^3 - 8*x
^2 - (2*x^5 + 2*x^4 - 3*x^3 - 3*x^2)*e^x - (2*x^3 - (x^4 + x^3)*e^x)*log(x) - 3*x)*e^(e^x) - (2*x^4 - x^3 - 8*
x^2 - 3*x)*log(x) + 15*x + 9)/(x^6 + 3*x^5 + 3*x^4 + x^3), x)

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maple [B]  time = 0.08, size = 116, normalized size = 4.00

method	result	size
risch	\(\frac {\ln \relax (x )^{2}}{9 x^{2}+18 x +9}-\frac {2 \left (2 x^{2}-3\right ) \ln \relax (x )}{9 x \left (x^{2}+2 x +1\right )}-\frac {8 x^{3}+16 x^{2}-9}{9 \left (x^{2}+2 x +1\right ) x^{2}}+\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{9 x^{2}+18 x +9}-\frac {2 \left (2 x^{2}-x \ln \relax (x )-3\right ) {\mathrm e}^{{\mathrm e}^{x}}}{9 x \left (x^{2}+2 x +1\right )}\)	\(116\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^4+2*x^3)*exp(x)-2*x^3)*exp(exp(x))^2+(((2*x^4+2*x^3)*exp(x)-4*x^3)*ln(x)+(-4*x^5-4*x^4+6*x^3+6*x^2)
*exp(x)+4*x^4-2*x^3-16*x^2-6*x)*exp(exp(x))-2*x^3*ln(x)^2+(4*x^4-2*x^3-16*x^2-6*x)*ln(x)+4*x^4+20*x^3+6*x^2-30
*x-18)/(9*x^6+27*x^5+27*x^4+9*x^3),x,method=_RETURNVERBOSE)

[Out]

1/9/(x^2+2*x+1)*ln(x)^2-2/9*(2*x^2-3)/x/(x^2+2*x+1)*ln(x)-1/9*(8*x^3+16*x^2-9)/(x^2+2*x+1)/x^2+1/9/(x^2+2*x+1)
*exp(2*exp(x))-2/9*(2*x^2-x*ln(x)-3)/x/(x^2+2*x+1)*exp(exp(x))

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maxima [B]  time = 0.41, size = 167, normalized size = 5.76 \begin {gather*} -\frac {12 \, x^{3} + 18 \, x^{2} + 4 \, x - 1}{x^{4} + 2 \, x^{3} + x^{2}} + \frac {x \log \relax (x)^{2} + 8 \, x^{2} + x e^{\left (2 \, e^{x}\right )} - 2 \, {\left (2 \, x^{2} - x \log \relax (x) - 3\right )} e^{\left (e^{x}\right )} - 2 \, {\left (2 \, x^{2} - 3\right )} \log \relax (x) + 14 \, x + 6}{9 \, {\left (x^{3} + 2 \, x^{2} + x\right )}} + \frac {5 \, {\left (6 \, x^{2} + 9 \, x + 2\right )}}{3 \, {\left (x^{3} + 2 \, x^{2} + x\right )}} + \frac {2 \, x + 3}{3 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {2 \, {\left (2 \, x + 1\right )}}{9 \, {\left (x^{2} + 2 \, x + 1\right )}} - \frac {10}{9 \, {\left (x^{2} + 2 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^4+2*x^3)*exp(x)-2*x^3)*exp(exp(x))^2+(((2*x^4+2*x^3)*exp(x)-4*x^3)*log(x)+(-4*x^5-4*x^4+6*x^3
+6*x^2)*exp(x)+4*x^4-2*x^3-16*x^2-6*x)*exp(exp(x))-2*x^3*log(x)^2+(4*x^4-2*x^3-16*x^2-6*x)*log(x)+4*x^4+20*x^3
+6*x^2-30*x-18)/(9*x^6+27*x^5+27*x^4+9*x^3),x, algorithm="maxima")

[Out]

-(12*x^3 + 18*x^2 + 4*x - 1)/(x^4 + 2*x^3 + x^2) + 1/9*(x*log(x)^2 + 8*x^2 + x*e^(2*e^x) - 2*(2*x^2 - x*log(x)
- 3)*e^(e^x) - 2*(2*x^2 - 3)*log(x) + 14*x + 6)/(x^3 + 2*x^2 + x) + 5/3*(6*x^2 + 9*x + 2)/(x^3 + 2*x^2 + x) +
1/3*(2*x + 3)/(x^2 + 2*x + 1) - 2/9*(2*x + 1)/(x^2 + 2*x + 1) - 10/9/(x^2 + 2*x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {30\,x+\ln \relax (x)\,\left (-4\,x^4+2\,x^3+16\,x^2+6\,x\right )+2\,x^3\,{\ln \relax (x)}^2-{\mathrm {e}}^{2\,{\mathrm {e}}^x}\,\left ({\mathrm {e}}^x\,\left (2\,x^4+2\,x^3\right )-2\,x^3\right )-6\,x^2-20\,x^3-4\,x^4+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (6\,x-{\mathrm {e}}^x\,\left (-4\,x^5-4\,x^4+6\,x^3+6\,x^2\right )+16\,x^2+2\,x^3-4\,x^4-\ln \relax (x)\,\left ({\mathrm {e}}^x\,\left (2\,x^4+2\,x^3\right )-4\,x^3\right )\right )+18}{9\,x^6+27\,x^5+27\,x^4+9\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*x + log(x)*(6*x + 16*x^2 + 2*x^3 - 4*x^4) + 2*x^3*log(x)^2 - exp(2*exp(x))*(exp(x)*(2*x^3 + 2*x^4) -
2*x^3) - 6*x^2 - 20*x^3 - 4*x^4 + exp(exp(x))*(6*x - exp(x)*(6*x^2 + 6*x^3 - 4*x^4 - 4*x^5) + 16*x^2 + 2*x^3 -
4*x^4 - log(x)*(exp(x)*(2*x^3 + 2*x^4) - 4*x^3)) + 18)/(9*x^3 + 27*x^4 + 27*x^5 + 9*x^6),x)

[Out]

int(-(30*x + log(x)*(6*x + 16*x^2 + 2*x^3 - 4*x^4) + 2*x^3*log(x)^2 - exp(2*exp(x))*(exp(x)*(2*x^3 + 2*x^4) -
2*x^3) - 6*x^2 - 20*x^3 - 4*x^4 + exp(exp(x))*(6*x - exp(x)*(6*x^2 + 6*x^3 - 4*x^4 - 4*x^5) + 16*x^2 + 2*x^3 -
4*x^4 - log(x)*(exp(x)*(2*x^3 + 2*x^4) - 4*x^3)) + 18)/(9*x^3 + 27*x^4 + 27*x^5 + 9*x^6), x)

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sympy [B]  time = 0.65, size = 158, normalized size = 5.45 \begin {gather*} \frac {\left (6 - 4 x^{2}\right ) \log {\relax (x )}}{9 x^{3} + 18 x^{2} + 9 x} + \frac {\left (9 x^{3} + 18 x^{2} + 9 x\right ) e^{2 e^{x}} + \left (- 36 x^{4} + 18 x^{3} \log {\relax (x )} - 72 x^{3} + 36 x^{2} \log {\relax (x )} + 18 x^{2} + 18 x \log {\relax (x )} + 108 x + 54\right ) e^{e^{x}}}{81 x^{5} + 324 x^{4} + 486 x^{3} + 324 x^{2} + 81 x} + \frac {- 8 x^{3} - 16 x^{2} + 9}{9 x^{4} + 18 x^{3} + 9 x^{2}} + \frac {\log {\relax (x )}^{2}}{9 x^{2} + 18 x + 9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**4+2*x**3)*exp(x)-2*x**3)*exp(exp(x))**2+(((2*x**4+2*x**3)*exp(x)-4*x**3)*ln(x)+(-4*x**5-4*x*
*4+6*x**3+6*x**2)*exp(x)+4*x**4-2*x**3-16*x**2-6*x)*exp(exp(x))-2*x**3*ln(x)**2+(4*x**4-2*x**3-16*x**2-6*x)*ln
(x)+4*x**4+20*x**3+6*x**2-30*x-18)/(9*x**6+27*x**5+27*x**4+9*x**3),x)

[Out]

(6 - 4*x**2)*log(x)/(9*x**3 + 18*x**2 + 9*x) + ((9*x**3 + 18*x**2 + 9*x)*exp(2*exp(x)) + (-36*x**4 + 18*x**3*l
og(x) - 72*x**3 + 36*x**2*log(x) + 18*x**2 + 18*x*log(x) + 108*x + 54)*exp(exp(x)))/(81*x**5 + 324*x**4 + 486*
x**3 + 324*x**2 + 81*x) + (-8*x**3 - 16*x**2 + 9)/(9*x**4 + 18*x**3 + 9*x**2) + log(x)**2/(9*x**2 + 18*x + 9)

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