3.66.18 \(\int \frac {-5 x^2-20 x^5-15 x^4 \log ^2(3)+e^{\frac {16-8 x+2 x^2}{x^2}} (160 x-40 x^2-10 x^3+(160-40 x-5 x^2) \log ^2(3))}{x^4+2 x^7+x^{10}+(2 x^6+2 x^9) \log ^2(3)+x^8 \log ^4(3)+e^{\frac {2 (16-8 x+2 x^2)}{x^2}} (x^6+2 x^5 \log ^2(3)+x^4 \log ^4(3))+e^{\frac {16-8 x+2 x^2}{x^2}} (2 x^5+2 x^8+(2 x^4+4 x^7) \log ^2(3)+2 x^6 \log ^4(3))} \, dx\)

Optimal. Leaf size=31 \[ \frac {5}{x+x \left (e^{1+\frac {(-4+x)^2}{x^2}}+x^2\right ) \left (x+\log ^2(3)\right )} \]

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Rubi [F]  time = 43.06, 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 {-5 x^2-20 x^5-15 x^4 \log ^2(3)+e^{\frac {16-8 x+2 x^2}{x^2}} \left (160 x-40 x^2-10 x^3+\left (160-40 x-5 x^2\right ) \log ^2(3)\right )}{x^4+2 x^7+x^{10}+\left (2 x^6+2 x^9\right ) \log ^2(3)+x^8 \log ^4(3)+e^{\frac {2 \left (16-8 x+2 x^2\right )}{x^2}} \left (x^6+2 x^5 \log ^2(3)+x^4 \log ^4(3)\right )+e^{\frac {16-8 x+2 x^2}{x^2}} \left (2 x^5+2 x^8+\left (2 x^4+4 x^7\right ) \log ^2(3)+2 x^6 \log ^4(3)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*x^2 - 20*x^5 - 15*x^4*Log[3]^2 + E^((16 - 8*x + 2*x^2)/x^2)*(160*x - 40*x^2 - 10*x^3 + (160 - 40*x - 5
*x^2)*Log[3]^2))/(x^4 + 2*x^7 + x^10 + (2*x^6 + 2*x^9)*Log[3]^2 + x^8*Log[3]^4 + E^((2*(16 - 8*x + 2*x^2))/x^2
)*(x^6 + 2*x^5*Log[3]^2 + x^4*Log[3]^4) + E^((16 - 8*x + 2*x^2)/x^2)*(2*x^5 + 2*x^8 + (2*x^4 + 4*x^7)*Log[3]^2
+ 2*x^6*Log[3]^4)),x]

[Out]

160*Log[3]^2*Defer[Int][E^((2*(8 + 4*x + x^2))/x^2)/(x^4*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x
^2*Log[3]^2))^2), x] + 40*(4 - Log[3]^2)*Defer[Int][E^((2*(8 + 4*x + x^2))/x^2)/(x^3*(E^(2 + 16/x^2)*(x + Log[
3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))^2), x] - 40*Defer[Int][E^((2*(8 + 4*x + x^2))/x^2)/(x^2*(E^(2 + 16/x
^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))^2), x] - 5*Defer[Int][E^((2*(8 + 4*x + x^2))/x^2)/(x*(E
^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))^2), x] + 10*Log[3]^4*Defer[Int][E^((2*(8 + 4*
x + x^2))/x^2)/((1 + x^3 + x^2*Log[3]^2)*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))^2)
, x] + 25*Log[3]^2*Defer[Int][(E^((2*(8 + 4*x + x^2))/x^2)*x)/((1 + x^3 + x^2*Log[3]^2)*(E^(2 + 16/x^2)*(x + L
og[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))^2), x] + 15*Defer[Int][(E^((2*(8 + 4*x + x^2))/x^2)*x^2)/((1 + x^
3 + x^2*Log[3]^2)*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))^2), x] - 5*Defer[Int][E^(
8/x)/(x^2*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))), x] - 10*Log[3]^2*Defer[Int][E^(
8/x)/((1 + x^3 + x^2*Log[3]^2)*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2))), x] - 15*De
fer[Int][(E^(8/x)*x)/((1 + x^3 + x^2*Log[3]^2)*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^
2))), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 e^{8/x} \left (-e^{8/x} \left (x^2+4 x^5+3 x^4 \log ^2(3)\right )-e^{2+\frac {16}{x^2}} \left (2 x^3-32 \log ^2(3)+8 x \left (-4+\log ^2(3)\right )+x^2 \left (8+\log ^2(3)\right )\right )\right )}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ &=5 \int \frac {e^{8/x} \left (-e^{8/x} \left (x^2+4 x^5+3 x^4 \log ^2(3)\right )-e^{2+\frac {16}{x^2}} \left (2 x^3-32 \log ^2(3)+8 x \left (-4+\log ^2(3)\right )+x^2 \left (8+\log ^2(3)\right )\right )\right )}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ &=5 \int \left (-\frac {e^{8/x} \left (1+4 x^3+3 x^2 \log ^2(3)\right )}{x^2 \left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )}+\frac {e^{2+\frac {16}{x^2}+\frac {8}{x}} \left (2 x^6+32 \log ^2(3)-8 x^5 \left (1-\frac {\log ^2(3)}{2}\right )+32 x \left (1-\frac {\log ^2(3)}{4}\right )-8 x^2 \left (1-4 \log ^4(3)\right )+32 x^4 \left (1+\frac {1}{16} \log ^2(3) \left (-8+\log ^2(3)\right )\right )-x^3 \left (1+8 \log ^2(3) \left (-8+\log ^2(3)\right )\right )\right )}{x^4 \left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{8/x} \left (1+4 x^3+3 x^2 \log ^2(3)\right )}{x^2 \left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )} \, dx\right )+5 \int \frac {e^{2+\frac {16}{x^2}+\frac {8}{x}} \left (2 x^6+32 \log ^2(3)-8 x^5 \left (1-\frac {\log ^2(3)}{2}\right )+32 x \left (1-\frac {\log ^2(3)}{4}\right )-8 x^2 \left (1-4 \log ^4(3)\right )+32 x^4 \left (1+\frac {1}{16} \log ^2(3) \left (-8+\log ^2(3)\right )\right )-x^3 \left (1+8 \log ^2(3) \left (-8+\log ^2(3)\right )\right )\right )}{x^4 \left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{8/x} \left (1+4 x^3+3 x^2 \log ^2(3)\right )}{x^2 \left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx\right )+5 \int \left (-\frac {8 e^{2+\frac {16}{x^2}+\frac {8}{x}}}{x^2 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}-\frac {e^{2+\frac {16}{x^2}+\frac {8}{x}}}{x \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}+\frac {32 e^{2+\frac {16}{x^2}+\frac {8}{x}} \log ^2(3)}{x^4 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}-\frac {8 e^{2+\frac {16}{x^2}+\frac {8}{x}} \left (-4+\log ^2(3)\right )}{x^3 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}+\frac {e^{2+\frac {16}{x^2}+\frac {8}{x}} \left (3 x^2+5 x \log ^2(3)+2 \log ^4(3)\right )}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}\right ) \, dx\\ &=-\left (5 \int \frac {e^{2+\frac {16}{x^2}+\frac {8}{x}}}{x \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx\right )+5 \int \frac {e^{2+\frac {16}{x^2}+\frac {8}{x}} \left (3 x^2+5 x \log ^2(3)+2 \log ^4(3)\right )}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx-5 \int \left (\frac {e^{8/x}}{x^2 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )}+\frac {e^{8/x} \left (3 x+2 \log ^2(3)\right )}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )}\right ) \, dx-40 \int \frac {e^{2+\frac {16}{x^2}+\frac {8}{x}}}{x^2 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx+\left (160 \log ^2(3)\right ) \int \frac {e^{2+\frac {16}{x^2}+\frac {8}{x}}}{x^4 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx+\left (40 \left (4-\log ^2(3)\right )\right ) \int \frac {e^{2+\frac {16}{x^2}+\frac {8}{x}}}{x^3 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{8/x}}{x^2 \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )} \, dx\right )-5 \int \frac {e^{8/x} \left (3 x+2 \log ^2(3)\right )}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )} \, dx-5 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+5 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} \left (3 x^2+5 x \log ^2(3)+2 \log ^4(3)\right )}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx-40 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (160 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (40 \left (4-\log ^2(3)\right )\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^3 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\right )-5 \int \frac {e^{8/x}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx-5 \int \frac {e^{8/x} \left (3 x+2 \log ^2(3)\right )}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx+5 \int \left (\frac {3 e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x^2}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}+\frac {5 e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x \log ^2(3)}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}+\frac {2 e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} \log ^4(3)}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2}\right ) \, dx-40 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (160 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (40 \left (4-\log ^2(3)\right )\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^3 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\right )-5 \int \frac {e^{8/x}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx-5 \int \left (\frac {3 e^{8/x} x}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )}+\frac {2 e^{8/x} \log ^2(3)}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )}\right ) \, dx+15 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x^2}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx-40 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (25 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx+\left (160 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (10 \log ^4(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )^2} \, dx+\left (40 \left (4-\log ^2(3)\right )\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^3 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\right )-5 \int \frac {e^{8/x}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx-15 \int \frac {e^{8/x} x}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )} \, dx+15 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x^2}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx-40 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx-\left (10 \log ^2(3)\right ) \int \frac {e^{8/x}}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{8/x}+e^{2+\frac {16}{x^2}} x+e^{8/x} x^3+e^{2+\frac {16}{x^2}} \log ^2(3)+e^{8/x} x^2 \log ^2(3)\right )} \, dx+\left (25 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (160 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (10 \log ^4(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (40 \left (4-\log ^2(3)\right )\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^3 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ &=-\left (5 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\right )-5 \int \frac {e^{8/x}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx+15 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x^2}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx-15 \int \frac {e^{8/x} x}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx-40 \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^2 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx-\left (10 \log ^2(3)\right ) \int \frac {e^{8/x}}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \, dx+\left (25 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}} x}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (160 \log ^2(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^4 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (10 \log ^4(3)\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{\left (1+x^3+x^2 \log ^2(3)\right ) \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx+\left (40 \left (4-\log ^2(3)\right )\right ) \int \frac {e^{\frac {2 \left (8+4 x+x^2\right )}{x^2}}}{x^3 \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 52, normalized size = 1.68 \begin {gather*} \frac {5 e^{8/x}}{x \left (e^{2+\frac {16}{x^2}} \left (x+\log ^2(3)\right )+e^{8/x} \left (1+x^3+x^2 \log ^2(3)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x^2 - 20*x^5 - 15*x^4*Log[3]^2 + E^((16 - 8*x + 2*x^2)/x^2)*(160*x - 40*x^2 - 10*x^3 + (160 - 40
*x - 5*x^2)*Log[3]^2))/(x^4 + 2*x^7 + x^10 + (2*x^6 + 2*x^9)*Log[3]^2 + x^8*Log[3]^4 + E^((2*(16 - 8*x + 2*x^2
))/x^2)*(x^6 + 2*x^5*Log[3]^2 + x^4*Log[3]^4) + E^((16 - 8*x + 2*x^2)/x^2)*(2*x^5 + 2*x^8 + (2*x^4 + 4*x^7)*Lo
g[3]^2 + 2*x^6*Log[3]^4)),x]

[Out]

(5*E^(8/x))/(x*(E^(2 + 16/x^2)*(x + Log[3]^2) + E^(8/x)*(1 + x^3 + x^2*Log[3]^2)))

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fricas [A]  time = 0.64, size = 42, normalized size = 1.35 \begin {gather*} \frac {5}{x^{3} \log \relax (3)^{2} + x^{4} + {\left (x \log \relax (3)^{2} + x^{2}\right )} e^{\left (\frac {2 \, {\left (x^{2} - 4 \, x + 8\right )}}{x^{2}}\right )} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-40*x+160)*log(3)^2-10*x^3-40*x^2+160*x)*exp((2*x^2-8*x+16)/x^2)-15*x^4*log(3)^2-20*x^5-5*x
^2)/((x^4*log(3)^4+2*x^5*log(3)^2+x^6)*exp((2*x^2-8*x+16)/x^2)^2+(2*x^6*log(3)^4+(4*x^7+2*x^4)*log(3)^2+2*x^8+
2*x^5)*exp((2*x^2-8*x+16)/x^2)+x^8*log(3)^4+(2*x^9+2*x^6)*log(3)^2+x^10+2*x^7+x^4),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.36, size = 55, normalized size = 1.77 \begin {gather*} \frac {5}{x^{3} \log \relax (3)^{2} + x^{4} + x e^{\left (\frac {2 \, {\left (x^{2} - 4 \, x + 8\right )}}{x^{2}}\right )} \log \relax (3)^{2} + x^{2} e^{\left (\frac {2 \, {\left (x^{2} - 4 \, x + 8\right )}}{x^{2}}\right )} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-40*x+160)*log(3)^2-10*x^3-40*x^2+160*x)*exp((2*x^2-8*x+16)/x^2)-15*x^4*log(3)^2-20*x^5-5*x
^2)/((x^4*log(3)^4+2*x^5*log(3)^2+x^6)*exp((2*x^2-8*x+16)/x^2)^2+(2*x^6*log(3)^4+(4*x^7+2*x^4)*log(3)^2+2*x^8+
2*x^5)*exp((2*x^2-8*x+16)/x^2)+x^8*log(3)^4+(2*x^9+2*x^6)*log(3)^2+x^10+2*x^7+x^4),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.64, size = 56, normalized size = 1.81

method	result	size
risch	\(\frac {5}{x \left (x^{2} \ln \relax (3)^{2}+\ln \relax (3)^{2} {\mathrm e}^{\frac {2 x^{2}-8 x +16}{x^{2}}}+x^{3}+x \,{\mathrm e}^{\frac {2 x^{2}-8 x +16}{x^{2}}}+1\right )}\)	\(56\)
norman	\(\frac {5}{x \left (x^{2} \ln \relax (3)^{2}+\ln \relax (3)^{2} {\mathrm e}^{\frac {2 x^{2}-8 x +16}{x^{2}}}+x^{3}+x \,{\mathrm e}^{\frac {2 x^{2}-8 x +16}{x^{2}}}+1\right )}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2-40*x+160)*ln(3)^2-10*x^3-40*x^2+160*x)*exp((2*x^2-8*x+16)/x^2)-15*x^4*ln(3)^2-20*x^5-5*x^2)/((x^
4*ln(3)^4+2*x^5*ln(3)^2+x^6)*exp((2*x^2-8*x+16)/x^2)^2+(2*x^6*ln(3)^4+(4*x^7+2*x^4)*ln(3)^2+2*x^8+2*x^5)*exp((
2*x^2-8*x+16)/x^2)+x^8*ln(3)^4+(2*x^9+2*x^6)*ln(3)^2+x^10+2*x^7+x^4),x,method=_RETURNVERBOSE)

[Out]

5/x/(x^2*ln(3)^2+exp(2*(x^2-4*x+8)/x^2)*ln(3)^2+x^3+exp(2*(x^2-4*x+8)/x^2)*x+1)

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maxima [A]  time = 0.55, size = 53, normalized size = 1.71 \begin {gather*} \frac {5 \, e^{\frac {8}{x}}}{{\left (x^{3} \log \relax (3)^{2} + x^{4} + x\right )} e^{\frac {8}{x}} + {\left (x e^{2} \log \relax (3)^{2} + x^{2} e^{2}\right )} e^{\left (\frac {16}{x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-40*x+160)*log(3)^2-10*x^3-40*x^2+160*x)*exp((2*x^2-8*x+16)/x^2)-15*x^4*log(3)^2-20*x^5-5*x
^2)/((x^4*log(3)^4+2*x^5*log(3)^2+x^6)*exp((2*x^2-8*x+16)/x^2)^2+(2*x^6*log(3)^4+(4*x^7+2*x^4)*log(3)^2+2*x^8+
2*x^5)*exp((2*x^2-8*x+16)/x^2)+x^8*log(3)^4+(2*x^9+2*x^6)*log(3)^2+x^10+2*x^7+x^4),x, algorithm="maxima")

[Out]

5*e^(8/x)/((x^3*log(3)^2 + x^4 + x)*e^(8/x) + (x*e^2*log(3)^2 + x^2*e^2)*e^(16/x^2))

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mupad [B]  time = 5.12, size = 383, normalized size = 12.35 \begin {gather*} \frac {5\,{\left (x^5+2\,{\ln \relax (3)}^2\,x^4+{\ln \relax (3)}^4\,x^3\right )}^2\,\left (32\,x+64\,x^3\,{\ln \relax (3)}^2+32\,x^2\,{\ln \relax (3)}^4-16\,x^4\,{\ln \relax (3)}^2-8\,x^3\,{\ln \relax (3)}^4+4\,x^5\,{\ln \relax (3)}^2+2\,x^4\,{\ln \relax (3)}^4-8\,x\,{\ln \relax (3)}^2+32\,{\ln \relax (3)}^2-8\,x^2-x^3+32\,x^4-8\,x^5+2\,x^6\right )}{x^4\,\left ({\mathrm {e}}^{\frac {16}{x^2}-\frac {8}{x}+2}+\frac {x^3+{\ln \relax (3)}^2\,x^2+1}{x+{\ln \relax (3)}^2}\right )\,{\left (x+{\ln \relax (3)}^2\right )}^3\,\left (96\,x^5\,{\ln \relax (3)}^2+96\,x^4\,{\ln \relax (3)}^4-24\,x^6\,{\ln \relax (3)}^2+32\,x^3\,{\ln \relax (3)}^6-24\,x^5\,{\ln \relax (3)}^4-2\,x^7\,{\ln \relax (3)}^2-8\,x^4\,{\ln \relax (3)}^6-x^6\,{\ln \relax (3)}^4+128\,x^8\,{\ln \relax (3)}^2+192\,x^7\,{\ln \relax (3)}^4-32\,x^9\,{\ln \relax (3)}^2+128\,x^6\,{\ln \relax (3)}^6-48\,x^8\,{\ln \relax (3)}^4+8\,x^{10}\,{\ln \relax (3)}^2+32\,x^5\,{\ln \relax (3)}^8-32\,x^7\,{\ln \relax (3)}^6+12\,x^9\,{\ln \relax (3)}^4-8\,x^6\,{\ln \relax (3)}^8+8\,x^8\,{\ln \relax (3)}^6+2\,x^7\,{\ln \relax (3)}^8+32\,x^6-8\,x^7-x^8+32\,x^9-8\,x^{10}+2\,x^{11}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x^4*log(3)^2 + exp((2*x^2 - 8*x + 16)/x^2)*(log(3)^2*(40*x + 5*x^2 - 160) - 160*x + 40*x^2 + 10*x^3)
+ 5*x^2 + 20*x^5)/(x^8*log(3)^4 + exp((2*x^2 - 8*x + 16)/x^2)*(2*x^6*log(3)^4 + 2*x^5 + 2*x^8 + log(3)^2*(2*x^
4 + 4*x^7)) + exp((2*(2*x^2 - 8*x + 16))/x^2)*(2*x^5*log(3)^2 + x^4*log(3)^4 + x^6) + x^4 + 2*x^7 + x^10 + log
(3)^2*(2*x^6 + 2*x^9)),x)

[Out]

(5*(2*x^4*log(3)^2 + x^3*log(3)^4 + x^5)^2*(32*x + 64*x^3*log(3)^2 + 32*x^2*log(3)^4 - 16*x^4*log(3)^2 - 8*x^3
*log(3)^4 + 4*x^5*log(3)^2 + 2*x^4*log(3)^4 - 8*x*log(3)^2 + 32*log(3)^2 - 8*x^2 - x^3 + 32*x^4 - 8*x^5 + 2*x^
6))/(x^4*(exp(16/x^2 - 8/x + 2) + (x^2*log(3)^2 + x^3 + 1)/(x + log(3)^2))*(x + log(3)^2)^3*(96*x^5*log(3)^2 +
96*x^4*log(3)^4 - 24*x^6*log(3)^2 + 32*x^3*log(3)^6 - 24*x^5*log(3)^4 - 2*x^7*log(3)^2 - 8*x^4*log(3)^6 - x^6
*log(3)^4 + 128*x^8*log(3)^2 + 192*x^7*log(3)^4 - 32*x^9*log(3)^2 + 128*x^6*log(3)^6 - 48*x^8*log(3)^4 + 8*x^1
0*log(3)^2 + 32*x^5*log(3)^8 - 32*x^7*log(3)^6 + 12*x^9*log(3)^4 - 8*x^6*log(3)^8 + 8*x^8*log(3)^6 + 2*x^7*log
(3)^8 + 32*x^6 - 8*x^7 - x^8 + 32*x^9 - 8*x^10 + 2*x^11))

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sympy [A]  time = 0.45, size = 39, normalized size = 1.26 \begin {gather*} \frac {5}{x^{4} + x^{3} \log {\relax (3 )}^{2} + x + \left (x^{2} + x \log {\relax (3 )}^{2}\right ) e^{\frac {2 x^{2} - 8 x + 16}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2-40*x+160)*ln(3)**2-10*x**3-40*x**2+160*x)*exp((2*x**2-8*x+16)/x**2)-15*x**4*ln(3)**2-20*x
**5-5*x**2)/((x**4*ln(3)**4+2*x**5*ln(3)**2+x**6)*exp((2*x**2-8*x+16)/x**2)**2+(2*x**6*ln(3)**4+(4*x**7+2*x**4
)*ln(3)**2+2*x**8+2*x**5)*exp((2*x**2-8*x+16)/x**2)+x**8*ln(3)**4+(2*x**9+2*x**6)*ln(3)**2+x**10+2*x**7+x**4),
x)

[Out]

5/(x**4 + x**3*log(3)**2 + x + (x**2 + x*log(3)**2)*exp((2*x**2 - 8*x + 16)/x**2))

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