3.97.66 \(\int \frac {e^x (-15 x^2+5 x^3+x^4-x^5)+e^{x^2} (60 x^2-44 x^4+8 x^6)+(e^x (4 x^3-2 x^4)+e^{x^2} (-16 x^3+16 x^5)) \log (\log (3))}{e^{x+x^2} (-1000+400 x^2-40 x^4)+e^{2 x} (125-50 x^2+5 x^4)+e^{2 x^2} (2000-800 x^2+80 x^4)+(e^{x+x^2} (800 x-160 x^3)+e^{2 x} (-100 x+20 x^3)+e^{2 x^2} (-1600 x+320 x^3)) \log (\log (3))+(20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2) \log ^2(\log (3))} \, dx\)

Optimal. Leaf size=34 \[ \frac {x^2}{5 \left (e^x-4 e^{x^2}\right ) \left (-\frac {5}{x}+x+2 \log (\log (3))\right )} \]

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Rubi [F]  time = 9.43, 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^x \left (-15 x^2+5 x^3+x^4-x^5\right )+e^{x^2} \left (60 x^2-44 x^4+8 x^6\right )+\left (e^x \left (4 x^3-2 x^4\right )+e^{x^2} \left (-16 x^3+16 x^5\right )\right ) \log (\log (3))}{e^{x+x^2} \left (-1000+400 x^2-40 x^4\right )+e^{2 x} \left (125-50 x^2+5 x^4\right )+e^{2 x^2} \left (2000-800 x^2+80 x^4\right )+\left (e^{x+x^2} \left (800 x-160 x^3\right )+e^{2 x} \left (-100 x+20 x^3\right )+e^{2 x^2} \left (-1600 x+320 x^3\right )\right ) \log (\log (3))+\left (20 e^{2 x} x^2+320 e^{2 x^2} x^2-160 e^{x+x^2} x^2\right ) \log ^2(\log (3))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-15*x^2 + 5*x^3 + x^4 - x^5) + E^x^2*(60*x^2 - 44*x^4 + 8*x^6) + (E^x*(4*x^3 - 2*x^4) + E^x^2*(-16*x
^3 + 16*x^5))*Log[Log[3]])/(E^(x + x^2)*(-1000 + 400*x^2 - 40*x^4) + E^(2*x)*(125 - 50*x^2 + 5*x^4) + E^(2*x^2
)*(2000 - 800*x^2 + 80*x^4) + (E^(x + x^2)*(800*x - 160*x^3) + E^(2*x)*(-100*x + 20*x^3) + E^(2*x^2)*(-1600*x
+ 320*x^3))*Log[Log[3]] + (20*E^(2*x)*x^2 + 320*E^(2*x^2)*x^2 - 160*E^(x + x^2)*x^2)*Log[Log[3]]^2),x]

[Out]

(2*(5 + Log[Log[3]] + 4*Log[Log[3]]^2)*Defer[Int][E^x/(E^x - 4*E^x^2)^2, x])/5 - ((9 + 8*Log[Log[3]]^2)*Defer[
Int][(E^x - 4*E^x^2)^(-1), x])/5 - ((1 + 4*Log[Log[3]])*Defer[Int][(E^x*x)/(E^x - 4*E^x^2)^2, x])/5 + (4*Log[L
og[3]]*Defer[Int][x/(E^x - 4*E^x^2), x])/5 + (2*Defer[Int][(E^x*x^2)/(E^x - 4*E^x^2)^2, x])/5 - (2*Defer[Int][
x^2/(E^x - 4*E^x^2), x])/5 - 2*(5 + 2*Log[Log[3]]^2)*Defer[Int][1/((E^x - 4*E^x^2)*(-5 + x^2 + 2*x*Log[Log[3]]
)^2), x] + (2*Log[Log[3]]*(15 + 4*Log[Log[3]]^2)*Defer[Int][x/((E^x - 4*E^x^2)*(-5 + x^2 + 2*x*Log[Log[3]])^2)
, x])/5 - ((5 + 40*Log[Log[3]] + 4*Log[Log[3]]^2 + 16*Log[Log[3]]^3)*(1 - Log[Log[3]]/Sqrt[5 + Log[Log[3]]^2])
*Defer[Int][E^x/((E^x - 4*E^x^2)^2*(2*x + 2*Log[Log[3]] - 2*Sqrt[5 + Log[Log[3]]^2])), x])/5 + (8*Log[Log[3]]*
(5 + 2*Log[Log[3]]^2)*(1 - Log[Log[3]]/Sqrt[5 + Log[Log[3]]^2])*Defer[Int][1/((E^x - 4*E^x^2)*(2*x + 2*Log[Log
[3]] - 2*Sqrt[5 + Log[Log[3]]^2])), x])/5 - (2*(5 + Log[Log[3]] + 4*Log[Log[3]]^2)*Defer[Int][E^x/((E^x - 4*E^
x^2)^2*(-2*x - 2*Log[Log[3]] + 2*Sqrt[5 + Log[Log[3]]^2])), x])/Sqrt[5 + Log[Log[3]]^2] + (11*(5 + 4*Log[Log[3
]]^2)*Defer[Int][1/((E^x - 4*E^x^2)*(-2*x - 2*Log[Log[3]] + 2*Sqrt[5 + Log[Log[3]]^2])), x])/(5*Sqrt[5 + Log[L
og[3]]^2]) - (2*(5 + Log[Log[3]] + 4*Log[Log[3]]^2)*Defer[Int][E^x/((E^x - 4*E^x^2)^2*(2*x + 2*Log[Log[3]] + 2
*Sqrt[5 + Log[Log[3]]^2])), x])/Sqrt[5 + Log[Log[3]]^2] - ((5 + 40*Log[Log[3]] + 4*Log[Log[3]]^2 + 16*Log[Log[
3]]^3)*(1 + Log[Log[3]]/Sqrt[5 + Log[Log[3]]^2])*Defer[Int][E^x/((E^x - 4*E^x^2)^2*(2*x + 2*Log[Log[3]] + 2*Sq
rt[5 + Log[Log[3]]^2])), x])/5 + (11*(5 + 4*Log[Log[3]]^2)*Defer[Int][1/((E^x - 4*E^x^2)*(2*x + 2*Log[Log[3]]
+ 2*Sqrt[5 + Log[Log[3]]^2])), x])/(5*Sqrt[5 + Log[Log[3]]^2]) + (8*Log[Log[3]]*(5 + 2*Log[Log[3]]^2)*(1 + Log
[Log[3]]/Sqrt[5 + Log[Log[3]]^2])*Defer[Int][1/((E^x - 4*E^x^2)*(2*x + 2*Log[Log[3]] + 2*Sqrt[5 + Log[Log[3]]^
2])), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2 \left (4 e^{x^2} \left (15-11 x^2+2 x^4-4 x \log (\log (3))+4 x^3 \log (\log (3))\right )+e^x \left (-15-x^3+x^2 (1-2 \log (\log (3)))+x (5+4 \log (\log (3)))\right )\right )}{5 \left (e^x-4 e^{x^2}\right )^2 \left (5-x^2-2 x \log (\log (3))\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {x^2 \left (4 e^{x^2} \left (15-11 x^2+2 x^4-4 x \log (\log (3))+4 x^3 \log (\log (3))\right )+e^x \left (-15-x^3+x^2 (1-2 \log (\log (3)))+x (5+4 \log (\log (3)))\right )\right )}{\left (e^x-4 e^{x^2}\right )^2 \left (5-x^2-2 x \log (\log (3))\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {e^x x^3 (-1+2 x)}{\left (e^x-4 e^{x^2}\right )^2 \left (-5+x^2+2 x \log (\log (3))\right )}-\frac {x^2 \left (15-11 x^2+2 x^4-4 x \log (\log (3))+4 x^3 \log (\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {e^x x^3 (-1+2 x)}{\left (e^x-4 e^{x^2}\right )^2 \left (-5+x^2+2 x \log (\log (3))\right )} \, dx-\frac {1}{5} \int \frac {x^2 \left (15-11 x^2+2 x^4-4 x \log (\log (3))+4 x^3 \log (\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )^2} \, dx\\ &=-\left (\frac {1}{5} \int \left (\frac {2 x^2}{e^x-4 e^{x^2}}-\frac {4 x \log (\log (3))}{e^x-4 e^{x^2}}+\frac {9 \left (1+\frac {8}{9} \log ^2(\log (3))\right )}{e^x-4 e^{x^2}}+\frac {8 x \log (\log (3)) \left (5+2 \log ^2(\log (3))\right )-11 \left (5+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (5-x^2-2 x \log (\log (3))\right )}+\frac {2 \left (5 \left (5+2 \log ^2(\log (3))\right )-x \log (\log (3)) \left (15+4 \log ^2(\log (3))\right )\right )}{\left (e^x-4 e^{x^2}\right ) \left (5-x^2-2 x \log (\log (3))\right )^2}\right ) \, dx\right )+\frac {1}{5} \int \left (\frac {2 e^x x^2}{\left (e^x-4 e^{x^2}\right )^2}-\frac {e^x x (1+4 \log (\log (3)))}{\left (e^x-4 e^{x^2}\right )^2}+\frac {2 e^x \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right )^2}+\frac {e^x \left (-10 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )+x \left (5+40 \log (\log (3))+4 \log ^2(\log (3))+16 \log ^3(\log (3))\right )\right )}{\left (e^x-4 e^{x^2}\right )^2 \left (5-x^2-2 x \log (\log (3))\right )}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {8 x \log (\log (3)) \left (5+2 \log ^2(\log (3))\right )-11 \left (5+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (5-x^2-2 x \log (\log (3))\right )} \, dx\right )+\frac {1}{5} \int \frac {e^x \left (-10 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )+x \left (5+40 \log (\log (3))+4 \log ^2(\log (3))+16 \log ^3(\log (3))\right )\right )}{\left (e^x-4 e^{x^2}\right )^2 \left (5-x^2-2 x \log (\log (3))\right )} \, dx+\frac {2}{5} \int \frac {e^x x^2}{\left (e^x-4 e^{x^2}\right )^2} \, dx-\frac {2}{5} \int \frac {x^2}{e^x-4 e^{x^2}} \, dx-\frac {2}{5} \int \frac {5 \left (5+2 \log ^2(\log (3))\right )-x \log (\log (3)) \left (15+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (5-x^2-2 x \log (\log (3))\right )^2} \, dx+\frac {1}{5} (-1-4 \log (\log (3))) \int \frac {e^x x}{\left (e^x-4 e^{x^2}\right )^2} \, dx+\frac {1}{5} (4 \log (\log (3))) \int \frac {x}{e^x-4 e^{x^2}} \, dx+\frac {1}{5} \left (2 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2} \, dx-\frac {1}{5} \left (9+8 \log ^2(\log (3))\right ) \int \frac {1}{e^x-4 e^{x^2}} \, dx\\ &=-\left (\frac {1}{5} \int \left (-\frac {8 x \log (\log (3)) \left (5+2 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )}+\frac {11 \left (5+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )}\right ) \, dx\right )+\frac {1}{5} \int \left (\frac {10 e^x \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right )^2 \left (-5+x^2+2 x \log (\log (3))\right )}-\frac {e^x x \left (5+40 \log (\log (3))+4 \log ^2(\log (3))+16 \log ^3(\log (3))\right )}{\left (e^x-4 e^{x^2}\right )^2 \left (-5+x^2+2 x \log (\log (3))\right )}\right ) \, dx+\frac {2}{5} \int \frac {e^x x^2}{\left (e^x-4 e^{x^2}\right )^2} \, dx-\frac {2}{5} \int \frac {x^2}{e^x-4 e^{x^2}} \, dx-\frac {2}{5} \int \left (\frac {5 \left (5+2 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )^2}-\frac {x \log (\log (3)) \left (15+4 \log ^2(\log (3))\right )}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )^2}\right ) \, dx+\frac {1}{5} (-1-4 \log (\log (3))) \int \frac {e^x x}{\left (e^x-4 e^{x^2}\right )^2} \, dx+\frac {1}{5} (4 \log (\log (3))) \int \frac {x}{e^x-4 e^{x^2}} \, dx+\frac {1}{5} \left (2 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2} \, dx-\frac {1}{5} \left (9+8 \log ^2(\log (3))\right ) \int \frac {1}{e^x-4 e^{x^2}} \, dx\\ &=\frac {2}{5} \int \frac {e^x x^2}{\left (e^x-4 e^{x^2}\right )^2} \, dx-\frac {2}{5} \int \frac {x^2}{e^x-4 e^{x^2}} \, dx+\frac {1}{5} (-1-4 \log (\log (3))) \int \frac {e^x x}{\left (e^x-4 e^{x^2}\right )^2} \, dx+\frac {1}{5} (4 \log (\log (3))) \int \frac {x}{e^x-4 e^{x^2}} \, dx-\left (2 \left (5+2 \log ^2(\log (3))\right )\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )^2} \, dx+\frac {1}{5} \left (8 \log (\log (3)) \left (5+2 \log ^2(\log (3))\right )\right ) \int \frac {x}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )} \, dx-\frac {1}{5} \left (11 \left (5+4 \log ^2(\log (3))\right )\right ) \int \frac {1}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )} \, dx+\frac {1}{5} \left (2 \log (\log (3)) \left (15+4 \log ^2(\log (3))\right )\right ) \int \frac {x}{\left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )^2} \, dx+\frac {1}{5} \left (2 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2} \, dx+\left (2 \left (5+\log (\log (3))+4 \log ^2(\log (3))\right )\right ) \int \frac {e^x}{\left (e^x-4 e^{x^2}\right )^2 \left (-5+x^2+2 x \log (\log (3))\right )} \, dx-\frac {1}{5} \left (9+8 \log ^2(\log (3))\right ) \int \frac {1}{e^x-4 e^{x^2}} \, dx+\frac {1}{5} \left (-5-40 \log (\log (3))-4 \log ^2(\log (3))-16 \log ^3(\log (3))\right ) \int \frac {e^x x}{\left (e^x-4 e^{x^2}\right )^2 \left (-5+x^2+2 x \log (\log (3))\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 33, normalized size = 0.97 \begin {gather*} \frac {x^3}{5 \left (e^x-4 e^{x^2}\right ) \left (-5+x^2+2 x \log (\log (3))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-15*x^2 + 5*x^3 + x^4 - x^5) + E^x^2*(60*x^2 - 44*x^4 + 8*x^6) + (E^x*(4*x^3 - 2*x^4) + E^x^2*
(-16*x^3 + 16*x^5))*Log[Log[3]])/(E^(x + x^2)*(-1000 + 400*x^2 - 40*x^4) + E^(2*x)*(125 - 50*x^2 + 5*x^4) + E^
(2*x^2)*(2000 - 800*x^2 + 80*x^4) + (E^(x + x^2)*(800*x - 160*x^3) + E^(2*x)*(-100*x + 20*x^3) + E^(2*x^2)*(-1
600*x + 320*x^3))*Log[Log[3]] + (20*E^(2*x)*x^2 + 320*E^(2*x^2)*x^2 - 160*E^(x + x^2)*x^2)*Log[Log[3]]^2),x]

[Out]

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

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fricas [A]  time = 0.53, size = 62, normalized size = 1.82 \begin {gather*} -\frac {x^{3} e^{\left (x^{2}\right )}}{5 \, {\left (4 \, {\left (x^{2} - 5\right )} e^{\left (2 \, x^{2}\right )} - {\left (x^{2} - 5\right )} e^{\left (x^{2} + x\right )} + 2 \, {\left (4 \, x e^{\left (2 \, x^{2}\right )} - x e^{\left (x^{2} + x\right )}\right )} \log \left (\log \relax (3)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x
^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3
-1600*x)*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2)*log(log(3))+(80*x^4-800*x^2+2000
)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp(x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x, algorithm="fricas")

[Out]

-1/5*x^3*e^(x^2)/(4*(x^2 - 5)*e^(2*x^2) - (x^2 - 5)*e^(x^2 + x) + 2*(4*x*e^(2*x^2) - x*e^(x^2 + x))*log(log(3)
))

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giac [A]  time = 0.19, size = 52, normalized size = 1.53 \begin {gather*} -\frac {x^{3}}{5 \, {\left (4 \, x^{2} e^{\left (x^{2}\right )} - x^{2} e^{x} + 8 \, x e^{\left (x^{2}\right )} \log \left (\log \relax (3)\right ) - 2 \, x e^{x} \log \left (\log \relax (3)\right ) - 20 \, e^{\left (x^{2}\right )} + 5 \, e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x
^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3
-1600*x)*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2)*log(log(3))+(80*x^4-800*x^2+2000
)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp(x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x, algorithm="giac")

[Out]

-1/5*x^3/(4*x^2*e^(x^2) - x^2*e^x + 8*x*e^(x^2)*log(log(3)) - 2*x*e^x*log(log(3)) - 20*e^(x^2) + 5*e^x)

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maple [A]  time = 0.08, size = 30, normalized size = 0.88

method	result	size
risch	\(\frac {x^{3}}{5 \left (2 \ln \left (\ln \relax (3)\right ) x +x^{2}-5\right ) \left ({\mathrm e}^{x}-4 \,{\mathrm e}^{x^{2}}\right )}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*ln(ln(3))+(8*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x^4+5*x^3
-15*x^2)*exp(x))/((320*x^2*exp(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*ln(ln(3))^2+((320*x^3-1600*x)*e
xp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2)*ln(ln(3))+(80*x^4-800*x^2+2000)*exp(x^2)^2
+(-40*x^4+400*x^2-1000)*exp(x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.52, size = 40, normalized size = 1.18 \begin {gather*} -\frac {x^{3}}{5 \, {\left (4 \, {\left (x^{2} + 2 \, x \log \left (\log \relax (3)\right ) - 5\right )} e^{\left (x^{2}\right )} - {\left (x^{2} + 2 \, x \log \left (\log \relax (3)\right ) - 5\right )} e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^5-16*x^3)*exp(x^2)+(-2*x^4+4*x^3)*exp(x))*log(log(3))+(8*x^6-44*x^4+60*x^2)*exp(x^2)+(-x^5+x
^4+5*x^3-15*x^2)*exp(x))/((320*x^2*exp(x^2)^2-160*x^2*exp(x)*exp(x^2)+20*exp(x)^2*x^2)*log(log(3))^2+((320*x^3
-1600*x)*exp(x^2)^2+(-160*x^3+800*x)*exp(x)*exp(x^2)+(20*x^3-100*x)*exp(x)^2)*log(log(3))+(80*x^4-800*x^2+2000
)*exp(x^2)^2+(-40*x^4+400*x^2-1000)*exp(x)*exp(x^2)+(5*x^4-50*x^2+125)*exp(x)^2),x, algorithm="maxima")

[Out]

-1/5*x^3/(4*(x^2 + 2*x*log(log(3)) - 5)*e^(x^2) - (x^2 + 2*x*log(log(3)) - 5)*e^x)

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mupad [B]  time = 6.06, size = 100, normalized size = 2.94 \begin {gather*} -\frac {{\mathrm {e}}^x\,\left (\frac {2\,x^4\,\ln \left (\ln \relax (3)\right )}{5}-\frac {4\,x^5\,\ln \left (\ln \relax (3)\right )}{5}-x^3+2\,x^4+\frac {x^5}{5}-\frac {2\,x^6}{5}\right )}{\left (4\,{\mathrm {e}}^{x^2}-{\mathrm {e}}^x\right )\,\left ({\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x\right )\,\left (4\,x^3\,\ln \left (\ln \relax (3)\right )+4\,x^2\,{\ln \left (\ln \relax (3)\right )}^2-20\,x\,\ln \left (\ln \relax (3)\right )-10\,x^2+x^4+25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(3))*(exp(x)*(4*x^3 - 2*x^4) - exp(x^2)*(16*x^3 - 16*x^5)) - exp(x)*(15*x^2 - 5*x^3 - x^4 + x^5) +
exp(x^2)*(60*x^2 - 44*x^4 + 8*x^6))/(exp(2*x)*(5*x^4 - 50*x^2 + 125) + log(log(3))^2*(20*x^2*exp(2*x) + 320*x
^2*exp(2*x^2) - 160*x^2*exp(x^2)*exp(x)) + exp(2*x^2)*(80*x^4 - 800*x^2 + 2000) - log(log(3))*(exp(2*x)*(100*x
- 20*x^3) + exp(2*x^2)*(1600*x - 320*x^3) - exp(x^2)*exp(x)*(800*x - 160*x^3)) - exp(x^2)*exp(x)*(40*x^4 - 40
0*x^2 + 1000)),x)

[Out]

-(exp(x)*((2*x^4*log(log(3)))/5 - (4*x^5*log(log(3)))/5 - x^3 + 2*x^4 + x^5/5 - (2*x^6)/5))/((4*exp(x^2) - exp
(x))*(exp(x) - 2*x*exp(x))*(4*x^3*log(log(3)) + 4*x^2*log(log(3))^2 - 20*x*log(log(3)) - 10*x^2 + x^4 + 25))

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sympy [A]  time = 0.30, size = 49, normalized size = 1.44 \begin {gather*} - \frac {x^{3}}{- 5 x^{2} e^{x} - 10 x e^{x} \log {\left (\log {\relax (3 )} \right )} + \left (20 x^{2} + 40 x \log {\left (\log {\relax (3 )} \right )} - 100\right ) e^{x^{2}} + 25 e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**5-16*x**3)*exp(x**2)+(-2*x**4+4*x**3)*exp(x))*ln(ln(3))+(8*x**6-44*x**4+60*x**2)*exp(x**2)+
(-x**5+x**4+5*x**3-15*x**2)*exp(x))/((320*x**2*exp(x**2)**2-160*x**2*exp(x)*exp(x**2)+20*exp(x)**2*x**2)*ln(ln
(3))**2+((320*x**3-1600*x)*exp(x**2)**2+(-160*x**3+800*x)*exp(x)*exp(x**2)+(20*x**3-100*x)*exp(x)**2)*ln(ln(3)
)+(80*x**4-800*x**2+2000)*exp(x**2)**2+(-40*x**4+400*x**2-1000)*exp(x)*exp(x**2)+(5*x**4-50*x**2+125)*exp(x)**
2),x)

[Out]

-x**3/(-5*x**2*exp(x) - 10*x*exp(x)*log(log(3)) + (20*x**2 + 40*x*log(log(3)) - 100)*exp(x**2) + 25*exp(x))

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