3.4.40 \(\int \frac {e^{\frac {-5 x-5 x^2-5 x^3+(-6+e^x (-2-x)-3 x) \log (\log (x))}{10 x^2+5 x^3}} (-12-12 x-3 x^2+e^x (-4-4 x-x^2)+(10 x+10 x^2-5 x^3) \log (x)+(24+24 x+6 x^2+e^x (8+4 x-2 x^2-x^3)) \log (x) \log (\log (x)))}{(20 x^3+20 x^4+5 x^5) \log (x)} \, dx\)

Optimal. Leaf size=34 \[ e^{\frac {-x+\frac {-1+x}{2+x}-\frac {\left (3+e^x\right ) \log (\log (x))}{5 x}}{x}} \]

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Rubi [F]  time = 53.61, 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 {\exp \left (\frac {-5 x-5 x^2-5 x^3+\left (-6+e^x (-2-x)-3 x\right ) \log (\log (x))}{10 x^2+5 x^3}\right ) \left (-12-12 x-3 x^2+e^x \left (-4-4 x-x^2\right )+\left (10 x+10 x^2-5 x^3\right ) \log (x)+\left (24+24 x+6 x^2+e^x \left (8+4 x-2 x^2-x^3\right )\right ) \log (x) \log (\log (x))\right )}{\left (20 x^3+20 x^4+5 x^5\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-5*x - 5*x^2 - 5*x^3 + (-6 + E^x*(-2 - x) - 3*x)*Log[Log[x]])/(10*x^2 + 5*x^3))*(-12 - 12*x - 3*x^2 +
E^x*(-4 - 4*x - x^2) + (10*x + 10*x^2 - 5*x^3)*Log[x] + (24 + 24*x + 6*x^2 + E^x*(8 + 4*x - 2*x^2 - x^3))*Log
[x]*Log[Log[x]]))/((20*x^3 + 20*x^4 + 5*x^5)*Log[x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-5 x-5 x^2-5 x^3+\left (-6+e^x (-2-x)-3 x\right ) \log (\log (x))}{10 x^2+5 x^3}\right ) \left (-12-12 x-3 x^2+e^x \left (-4-4 x-x^2\right )+\left (10 x+10 x^2-5 x^3\right ) \log (x)+\left (24+24 x+6 x^2+e^x \left (8+4 x-2 x^2-x^3\right )\right ) \log (x) \log (\log (x))\right )}{x^3 \left (20+20 x+5 x^2\right ) \log (x)} \, dx\\ &=\int \frac {\exp \left (\frac {-5 x-5 x^2-5 x^3+\left (-6+e^x (-2-x)-3 x\right ) \log (\log (x))}{10 x^2+5 x^3}\right ) \left (-12-12 x-3 x^2+e^x \left (-4-4 x-x^2\right )+\left (10 x+10 x^2-5 x^3\right ) \log (x)+\left (24+24 x+6 x^2+e^x \left (8+4 x-2 x^2-x^3\right )\right ) \log (x) \log (\log (x))\right )}{5 x^3 (2+x)^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \frac {\exp \left (\frac {-5 x-5 x^2-5 x^3+\left (-6+e^x (-2-x)-3 x\right ) \log (\log (x))}{10 x^2+5 x^3}\right ) \left (-12-12 x-3 x^2+e^x \left (-4-4 x-x^2\right )+\left (10 x+10 x^2-5 x^3\right ) \log (x)+\left (24+24 x+6 x^2+e^x \left (8+4 x-2 x^2-x^3\right )\right ) \log (x) \log (\log (x))\right )}{x^3 (2+x)^2 \log (x)} \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-\left (\left (3+e^x\right ) (2+x)^2\right )+\log (x) \left (-5 x \left (-2-2 x+x^2\right )-\left (-6+e^x (-2+x)\right ) (2+x)^2 \log (\log (x))\right )\right )}{x^3 (2+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {e^{x-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) (1-2 \log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{x^3}+\frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-12-12 x-3 x^2+10 x \log (x)+10 x^2 \log (x)-5 x^3 \log (x)+24 \log (x) \log (\log (x))+24 x \log (x) \log (\log (x))+6 x^2 \log (x) \log (\log (x))\right )}{x^3 (2+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{x-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) (1-2 \log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{x^3} \, dx\right )+\frac {1}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-12-12 x-3 x^2+10 x \log (x)+10 x^2 \log (x)-5 x^3 \log (x)+24 \log (x) \log (\log (x))+24 x \log (x) \log (\log (x))+6 x^2 \log (x) \log (\log (x))\right )}{x^3 (2+x)^2} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) (1-2 \log (x) \log (\log (x))+x \log (x) \log (\log (x)))}{x^3} \, dx\right )+\frac {1}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-3 (2+x)^2+\log (x) \left (-5 x \left (-2-2 x+x^2\right )+6 (2+x)^2 \log (\log (x))\right )\right )}{x^3 (2+x)^2} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) (1+(-2+x) \log (x) \log (\log (x)))}{x^3} \, dx\right )+\frac {1}{5} \int \left (\frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-12-12 x-3 x^2+10 x \log (x)+10 x^2 \log (x)-5 x^3 \log (x)\right )}{x^3 (2+x)^2}+\frac {6 e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{1-\frac {3+e^x+5 x^2}{5 x^2}}(x) \log (\log (x))}{x^3}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-12-12 x-3 x^2+10 x \log (x)+10 x^2 \log (x)-5 x^3 \log (x)\right )}{x^3 (2+x)^2} \, dx-\frac {1}{5} \int \left (\frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3}+\frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} (-2+x) \log ^{1-\frac {3+e^x+5 x^2}{5 x^2}}(x) \log (\log (x))}{x^3}\right ) \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{1-\frac {3+e^x+5 x^2}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx\right )+\frac {1}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x) \left (-3 (2+x)^2-5 x \left (-2-2 x+x^2\right ) \log (x)\right )}{x^3 (2+x)^2} \, dx-\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} (-2+x) \log ^{1-\frac {3+e^x+5 x^2}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx\right )+\frac {1}{5} \int \left (-\frac {3 e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3}-\frac {5 e^{-\frac {1+x+x^2}{x (2+x)}} \left (-2-2 x+x^2\right ) \log ^{1-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^2 (2+x)^2}\right ) \, dx-\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} (-2+x) \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx\right )-\frac {1}{5} \int \left (-\frac {2 e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3}+\frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^2}\right ) \, dx-\frac {3}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx-\int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \left (-2-2 x+x^2\right ) \log ^{1-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^2 (2+x)^2} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx\right )-\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^2} \, dx+\frac {2}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx-\frac {3}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx-\int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \left (-2-2 x+x^2\right ) \log ^{-\frac {3+e^x}{5 x^2}}(x)}{x^2 (2+x)^2} \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx\right )-\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^2} \, dx+\frac {2}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx-\frac {3}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx-\int \left (-\frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x)}{2 x^2}+\frac {3 e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x)}{2 (2+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx\right )-\frac {1}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^2} \, dx+\frac {2}{5} \int \frac {e^{\frac {(-1+x) (1+x)^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx+\frac {1}{2} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x)}{x^2} \, dx-\frac {3}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x+5 x^2}{5 x^2}}(x)}{x^3} \, dx+\frac {6}{5} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \log (\log (x))}{x^3} \, dx-\frac {3}{2} \int \frac {e^{-\frac {1+x+x^2}{x (2+x)}} \log ^{-\frac {3+e^x}{5 x^2}}(x)}{(2+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 35, normalized size = 1.03 \begin {gather*} e^{-\frac {1+x+x^2}{2 x+x^2}} \log ^{-\frac {3+e^x}{5 x^2}}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-5*x - 5*x^2 - 5*x^3 + (-6 + E^x*(-2 - x) - 3*x)*Log[Log[x]])/(10*x^2 + 5*x^3))*(-12 - 12*x - 3
*x^2 + E^x*(-4 - 4*x - x^2) + (10*x + 10*x^2 - 5*x^3)*Log[x] + (24 + 24*x + 6*x^2 + E^x*(8 + 4*x - 2*x^2 - x^3
))*Log[x]*Log[Log[x]]))/((20*x^3 + 20*x^4 + 5*x^5)*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.61, size = 43, normalized size = 1.26 \begin {gather*} e^{\left (-\frac {5 \, x^{3} + 5 \, x^{2} + {\left ({\left (x + 2\right )} e^{x} + 3 \, x + 6\right )} \log \left (\log \relax (x)\right ) + 5 \, x}{5 \, {\left (x^{3} + 2 \, x^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-2*x^2+4*x+8)*exp(x)+6*x^2+24*x+24)*log(x)*log(log(x))+(-5*x^3+10*x^2+10*x)*log(x)+(-x^2-4*x-
4)*exp(x)-3*x^2-12*x-12)*exp((((-x-2)*exp(x)-3*x-6)*log(log(x))-5*x^3-5*x^2-5*x)/(5*x^3+10*x^2))/(5*x^5+20*x^4
+20*x^3)/log(x),x, algorithm="fricas")

[Out]

e^(-1/5*(5*x^3 + 5*x^2 + ((x + 2)*e^x + 3*x + 6)*log(log(x)) + 5*x)/(x^3 + 2*x^2))

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giac [B]  time = 9.94, size = 118, normalized size = 3.47 \begin {gather*} e^{\left (-\frac {x^{3}}{x^{3} + 2 \, x^{2}} - \frac {x e^{x} \log \left (\log \relax (x)\right )}{5 \, {\left (x^{3} + 2 \, x^{2}\right )}} - \frac {x^{2}}{x^{3} + 2 \, x^{2}} - \frac {3 \, x \log \left (\log \relax (x)\right )}{5 \, {\left (x^{3} + 2 \, x^{2}\right )}} - \frac {2 \, e^{x} \log \left (\log \relax (x)\right )}{5 \, {\left (x^{3} + 2 \, x^{2}\right )}} - \frac {x}{x^{3} + 2 \, x^{2}} - \frac {6 \, \log \left (\log \relax (x)\right )}{5 \, {\left (x^{3} + 2 \, x^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-2*x^2+4*x+8)*exp(x)+6*x^2+24*x+24)*log(x)*log(log(x))+(-5*x^3+10*x^2+10*x)*log(x)+(-x^2-4*x-
4)*exp(x)-3*x^2-12*x-12)*exp((((-x-2)*exp(x)-3*x-6)*log(log(x))-5*x^3-5*x^2-5*x)/(5*x^3+10*x^2))/(5*x^5+20*x^4
+20*x^3)/log(x),x, algorithm="giac")

[Out]

e^(-x^3/(x^3 + 2*x^2) - 1/5*x*e^x*log(log(x))/(x^3 + 2*x^2) - x^2/(x^3 + 2*x^2) - 3/5*x*log(log(x))/(x^3 + 2*x
^2) - 2/5*e^x*log(log(x))/(x^3 + 2*x^2) - x/(x^3 + 2*x^2) - 6/5*log(log(x))/(x^3 + 2*x^2))

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maple [A]  time = 0.09, size = 51, normalized size = 1.50

method	result	size
risch	\({\mathrm e}^{-\frac {{\mathrm e}^{x} \ln \left (\ln \relax (x )\right ) x +5 x^{3}+2 \,{\mathrm e}^{x} \ln \left (\ln \relax (x )\right )+3 x \ln \left (\ln \relax (x )\right )+5 x^{2}+6 \ln \left (\ln \relax (x )\right )+5 x}{5 x^{2} \left (2+x \right )}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^3-2*x^2+4*x+8)*exp(x)+6*x^2+24*x+24)*ln(x)*ln(ln(x))+(-5*x^3+10*x^2+10*x)*ln(x)+(-x^2-4*x-4)*exp(x)-
3*x^2-12*x-12)*exp((((-x-2)*exp(x)-3*x-6)*ln(ln(x))-5*x^3-5*x^2-5*x)/(5*x^3+10*x^2))/(5*x^5+20*x^4+20*x^3)/ln(
x),x,method=_RETURNVERBOSE)

[Out]

exp(-1/5*(exp(x)*ln(ln(x))*x+5*x^3+2*exp(x)*ln(ln(x))+3*x*ln(ln(x))+5*x^2+6*ln(ln(x))+5*x)/x^2/(2+x))

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maxima [A]  time = 1.38, size = 33, normalized size = 0.97 \begin {gather*} e^{\left (-\frac {e^{x} \log \left (\log \relax (x)\right )}{5 \, x^{2}} + \frac {3}{2 \, {\left (x + 2\right )}} - \frac {1}{2 \, x} - \frac {3 \, \log \left (\log \relax (x)\right )}{5 \, x^{2}} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^3-2*x^2+4*x+8)*exp(x)+6*x^2+24*x+24)*log(x)*log(log(x))+(-5*x^3+10*x^2+10*x)*log(x)+(-x^2-4*x-
4)*exp(x)-3*x^2-12*x-12)*exp((((-x-2)*exp(x)-3*x-6)*log(log(x))-5*x^3-5*x^2-5*x)/(5*x^3+10*x^2))/(5*x^5+20*x^4
+20*x^3)/log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.81, size = 97, normalized size = 2.85 \begin {gather*} \frac {{\mathrm {e}}^{-\frac {5\,x}{5\,x^3+10\,x^2}}\,{\mathrm {e}}^{-\frac {5\,x^2}{5\,x^3+10\,x^2}}\,{\mathrm {e}}^{-\frac {5\,x^3}{5\,x^3+10\,x^2}}}{{\ln \relax (x)}^{\frac {{\mathrm {e}}^x+3}{5\,\left (x^2+2\,x\right )}+\frac {2\,\left ({\mathrm {e}}^x+3\right )}{5\,\left (x^3+2\,x^2\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x + 5*x^2 + 5*x^3 + log(log(x))*(3*x + exp(x)*(x + 2) + 6))/(10*x^2 + 5*x^3))*(12*x + exp(x)*(4*
x + x^2 + 4) + 3*x^2 - log(x)*(10*x + 10*x^2 - 5*x^3) - log(log(x))*log(x)*(24*x + 6*x^2 + exp(x)*(4*x - 2*x^2
- x^3 + 8) + 24) + 12))/(log(x)*(20*x^3 + 20*x^4 + 5*x^5)),x)

[Out]

(exp(-(5*x)/(10*x^2 + 5*x^3))*exp(-(5*x^2)/(10*x^2 + 5*x^3))*exp(-(5*x^3)/(10*x^2 + 5*x^3)))/log(x)^((exp(x) +
3)/(5*(2*x + x^2)) + (2*(exp(x) + 3))/(5*(2*x^2 + x^3)))

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sympy [A]  time = 3.98, size = 42, normalized size = 1.24 \begin {gather*} e^{\frac {- 5 x^{3} - 5 x^{2} - 5 x + \left (- 3 x + \left (- x - 2\right ) e^{x} - 6\right ) \log {\left (\log {\relax (x )} \right )}}{5 x^{3} + 10 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**3-2*x**2+4*x+8)*exp(x)+6*x**2+24*x+24)*ln(x)*ln(ln(x))+(-5*x**3+10*x**2+10*x)*ln(x)+(-x**2-4*
x-4)*exp(x)-3*x**2-12*x-12)*exp((((-x-2)*exp(x)-3*x-6)*ln(ln(x))-5*x**3-5*x**2-5*x)/(5*x**3+10*x**2))/(5*x**5+
20*x**4+20*x**3)/ln(x),x)

[Out]

exp((-5*x**3 - 5*x**2 - 5*x + (-3*x + (-x - 2)*exp(x) - 6)*log(log(x)))/(5*x**3 + 10*x**2))

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