3.19.39 \(\int \frac {e^{e^{-2+x}} (e^8 (25-10 x+x^2)+e^4 (5 x^2-x^3))+e^{e^{-2+x}} (e^8 (-25+10 x-x^2)+e^4 (-15 x^2+2 x^3)+e^{-2+x} (e^8 (25 x-10 x^2+x^3)+e^4 (5 x^3-x^4))) \log (x) \log (\log (x))}{(x^6+e^8 (25 x^2-10 x^3+x^4)+e^4 (10 x^4-2 x^5)) \log (x)} \, dx\)

Optimal. Leaf size=32 \[ \frac {e^{e^{-2+x}} \log (\log (x))}{x \left (1+\frac {x^2}{e^4 (5-x)}\right )} \]

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Rubi [F]  time = 22.87, 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^{e^{-2+x}} \left (e^8 \left (25-10 x+x^2\right )+e^4 \left (5 x^2-x^3\right )\right )+e^{e^{-2+x}} \left (e^8 \left (-25+10 x-x^2\right )+e^4 \left (-15 x^2+2 x^3\right )+e^{-2+x} \left (e^8 \left (25 x-10 x^2+x^3\right )+e^4 \left (5 x^3-x^4\right )\right )\right ) \log (x) \log (\log (x))}{\left (x^6+e^8 \left (25 x^2-10 x^3+x^4\right )+e^4 \left (10 x^4-2 x^5\right )\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^E^(-2 + x)*(E^8*(25 - 10*x + x^2) + E^4*(5*x^2 - x^3)) + E^E^(-2 + x)*(E^8*(-25 + 10*x - x^2) + E^4*(-1
5*x^2 + 2*x^3) + E^(-2 + x)*(E^8*(25*x - 10*x^2 + x^3) + E^4*(5*x^3 - x^4)))*Log[x]*Log[Log[x]])/((x^6 + E^8*(
25*x^2 - 10*x^3 + x^4) + E^4*(10*x^4 - 2*x^5))*Log[x]),x]

[Out]

(-2*Defer[Int][E^(-2 + E^(-2 + x))/((E^4 - E^2*Sqrt[-20 + E^4] - 2*x)*Log[x]), x])/Sqrt[-20 + E^4] + (2*Defer[
Int][E^(-2 + E^(-2 + x))/((E^4 + E^2*Sqrt[-20 + E^4] - 2*x)*Log[x]), x])/Sqrt[-20 + E^4] + Defer[Int][E^E^(-2
+ x)/(x^2*Log[x]), x] + (2*Defer[Int][(E^(-2 + E^(-2 + x))*Log[Log[x]])/(E^4 - E^2*Sqrt[-20 + E^4] - 2*x), x])
/Sqrt[-20 + E^4] + (1 - E^2/Sqrt[-20 + E^4])*Defer[Int][(E^(-2 + E^(-2 + x) + x)*Log[Log[x]])/(E^4 - E^2*Sqrt[
-20 + E^4] - 2*x), x] - (2*Defer[Int][(E^(-2 + E^(-2 + x))*Log[Log[x]])/(E^4 + E^2*Sqrt[-20 + E^4] - 2*x), x])
/Sqrt[-20 + E^4] + (1 + E^2/Sqrt[-20 + E^4])*Defer[Int][(E^(-2 + E^(-2 + x) + x)*Log[Log[x]])/(E^4 + E^2*Sqrt[
-20 + E^4] - 2*x), x] - Defer[Int][(E^E^(-2 + x)*Log[Log[x]])/x^2, x] + Defer[Int][(E^(-2 + E^(-2 + x) + x)*Lo
g[Log[x]])/x, x] - 10*Defer[Int][(E^(4 + E^(-2 + x))*Log[Log[x]])/(-5*E^4 + E^4*x - x^2)^2, x] + Defer[Int][(E
^(4 + E^(-2 + x))*x*Log[Log[x]])/(-5*E^4 + E^4*x - x^2)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2+e^{-2+x}} \left (-e^2 (-5+x) \left (-e^4 (-5+x)+x^2\right )-\left (e^6 (-5+x)^2-e^{4+x} (-5+x)^2 x+e^2 (15-2 x) x^2+e^x (-5+x) x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \left (5 e^4-e^4 x+x^2\right )^2 \log (x)} \, dx\\ &=\int \left (-\frac {e^{2+e^{-2+x}+x} (-5+x) \log (\log (x))}{x \left (5 e^4-e^4 x+x^2\right )}+\frac {e^{4+e^{-2+x}} \left (25 e^4-10 e^4 x+5 \left (1+\frac {e^4}{5}\right ) x^2-x^3-25 e^4 \log (x) \log (\log (x))+10 e^4 x \log (x) \log (\log (x))-15 \left (1+\frac {e^4}{15}\right ) x^2 \log (x) \log (\log (x))+2 x^3 \log (x) \log (\log (x))\right )}{x^2 \left (5 e^4-e^4 x+x^2\right )^2 \log (x)}\right ) \, dx\\ &=-\int \frac {e^{2+e^{-2+x}+x} (-5+x) \log (\log (x))}{x \left (5 e^4-e^4 x+x^2\right )} \, dx+\int \frac {e^{4+e^{-2+x}} \left (25 e^4-10 e^4 x+5 \left (1+\frac {e^4}{5}\right ) x^2-x^3-25 e^4 \log (x) \log (\log (x))+10 e^4 x \log (x) \log (\log (x))-15 \left (1+\frac {e^4}{15}\right ) x^2 \log (x) \log (\log (x))+2 x^3 \log (x) \log (\log (x))\right )}{x^2 \left (5 e^4-e^4 x+x^2\right )^2 \log (x)} \, dx\\ &=-\int \left (-\frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x}-\frac {e^{-2+e^{-2+x}+x} x \log (\log (x))}{-5 e^4+e^4 x-x^2}\right ) \, dx+\int \frac {e^{4+e^{-2+x}} \left (-\left ((-5+x) \left (-e^4 (-5+x)+x^2\right )\right )-\left (e^4 (-5+x)^2+(15-2 x) x^2\right ) \log (x) \log (\log (x))\right )}{x^2 \left (5 e^4-e^4 x+x^2\right )^2 \log (x)} \, dx\\ &=\int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x} \, dx+\int \frac {e^{-2+e^{-2+x}+x} x \log (\log (x))}{-5 e^4+e^4 x-x^2} \, dx+\int \left (\frac {e^{4+e^{-2+x}} (5-x)}{x^2 \left (5 e^4-e^4 x+x^2\right ) \log (x)}+\frac {e^{4+e^{-2+x}} \left (-25 e^4+10 e^4 x-\left (15+e^4\right ) x^2+2 x^3\right ) \log (\log (x))}{x^2 \left (5 e^4-e^4 x+x^2\right )^2}\right ) \, dx\\ &=\int \frac {e^{4+e^{-2+x}} (5-x)}{x^2 \left (5 e^4-e^4 x+x^2\right ) \log (x)} \, dx+\int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x} \, dx+\int \frac {e^{4+e^{-2+x}} \left (-25 e^4+10 e^4 x-\left (15+e^4\right ) x^2+2 x^3\right ) \log (\log (x))}{x^2 \left (5 e^4-e^4 x+x^2\right )^2} \, dx+\int \left (\frac {e^{-2+e^{-2+x}+x} \left (1-\frac {e^2}{\sqrt {-20+e^4}}\right ) \log (\log (x))}{e^4-e^2 \sqrt {-20+e^4}-2 x}+\frac {e^{-2+e^{-2+x}+x} \left (1+\frac {e^2}{\sqrt {-20+e^4}}\right ) \log (\log (x))}{e^4+e^2 \sqrt {-20+e^4}-2 x}\right ) \, dx\\ &=\left (1-\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4-e^2 \sqrt {-20+e^4}-2 x} \, dx+\left (1+\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4+e^2 \sqrt {-20+e^4}-2 x} \, dx+\int \left (\frac {e^{e^{-2+x}}}{x^2 \log (x)}+\frac {e^{e^{-2+x}}}{\left (-5 e^4+e^4 x-x^2\right ) \log (x)}\right ) \, dx+\int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x} \, dx+\int \left (-\frac {e^{e^{-2+x}} \log (\log (x))}{x^2}+\frac {e^{4+e^{-2+x}} (-10+x) \log (\log (x))}{\left (-5 e^4+e^4 x-x^2\right )^2}-\frac {e^{e^{-2+x}} \log (\log (x))}{-5 e^4+e^4 x-x^2}\right ) \, dx\\ &=\left (1-\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4-e^2 \sqrt {-20+e^4}-2 x} \, dx+\left (1+\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4+e^2 \sqrt {-20+e^4}-2 x} \, dx+\int \frac {e^{e^{-2+x}}}{x^2 \log (x)} \, dx+\int \frac {e^{e^{-2+x}}}{\left (-5 e^4+e^4 x-x^2\right ) \log (x)} \, dx-\int \frac {e^{e^{-2+x}} \log (\log (x))}{x^2} \, dx+\int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x} \, dx+\int \frac {e^{4+e^{-2+x}} (-10+x) \log (\log (x))}{\left (-5 e^4+e^4 x-x^2\right )^2} \, dx-\int \frac {e^{e^{-2+x}} \log (\log (x))}{-5 e^4+e^4 x-x^2} \, dx\\ &=\left (1-\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4-e^2 \sqrt {-20+e^4}-2 x} \, dx+\left (1+\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4+e^2 \sqrt {-20+e^4}-2 x} \, dx+\int \left (-\frac {2 e^{-2+e^{-2+x}}}{\sqrt {-20+e^4} \left (e^4-e^2 \sqrt {-20+e^4}-2 x\right ) \log (x)}+\frac {2 e^{-2+e^{-2+x}}}{\sqrt {-20+e^4} \left (e^4+e^2 \sqrt {-20+e^4}-2 x\right ) \log (x)}\right ) \, dx+\int \frac {e^{e^{-2+x}}}{x^2 \log (x)} \, dx-\int \frac {e^{e^{-2+x}} \log (\log (x))}{x^2} \, dx+\int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x} \, dx-\int \left (-\frac {2 e^{-2+e^{-2+x}} \log (\log (x))}{\sqrt {-20+e^4} \left (e^4-e^2 \sqrt {-20+e^4}-2 x\right )}+\frac {2 e^{-2+e^{-2+x}} \log (\log (x))}{\sqrt {-20+e^4} \left (e^4+e^2 \sqrt {-20+e^4}-2 x\right )}\right ) \, dx+\int \left (-\frac {10 e^{4+e^{-2+x}} \log (\log (x))}{\left (-5 e^4+e^4 x-x^2\right )^2}+\frac {e^{4+e^{-2+x}} x \log (\log (x))}{\left (-5 e^4+e^4 x-x^2\right )^2}\right ) \, dx\\ &=-\left (10 \int \frac {e^{4+e^{-2+x}} \log (\log (x))}{\left (-5 e^4+e^4 x-x^2\right )^2} \, dx\right )-\frac {2 \int \frac {e^{-2+e^{-2+x}}}{\left (e^4-e^2 \sqrt {-20+e^4}-2 x\right ) \log (x)} \, dx}{\sqrt {-20+e^4}}+\frac {2 \int \frac {e^{-2+e^{-2+x}}}{\left (e^4+e^2 \sqrt {-20+e^4}-2 x\right ) \log (x)} \, dx}{\sqrt {-20+e^4}}+\frac {2 \int \frac {e^{-2+e^{-2+x}} \log (\log (x))}{e^4-e^2 \sqrt {-20+e^4}-2 x} \, dx}{\sqrt {-20+e^4}}-\frac {2 \int \frac {e^{-2+e^{-2+x}} \log (\log (x))}{e^4+e^2 \sqrt {-20+e^4}-2 x} \, dx}{\sqrt {-20+e^4}}+\left (1-\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4-e^2 \sqrt {-20+e^4}-2 x} \, dx+\left (1+\frac {e^2}{\sqrt {-20+e^4}}\right ) \int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{e^4+e^2 \sqrt {-20+e^4}-2 x} \, dx+\int \frac {e^{e^{-2+x}}}{x^2 \log (x)} \, dx-\int \frac {e^{e^{-2+x}} \log (\log (x))}{x^2} \, dx+\int \frac {e^{-2+e^{-2+x}+x} \log (\log (x))}{x} \, dx+\int \frac {e^{4+e^{-2+x}} x \log (\log (x))}{\left (-5 e^4+e^4 x-x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 32, normalized size = 1.00 \begin {gather*} \frac {e^{4+e^{-2+x}} (-5+x) \log (\log (x))}{e^4 (-5+x) x-x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^(-2 + x)*(E^8*(25 - 10*x + x^2) + E^4*(5*x^2 - x^3)) + E^E^(-2 + x)*(E^8*(-25 + 10*x - x^2) + E
^4*(-15*x^2 + 2*x^3) + E^(-2 + x)*(E^8*(25*x - 10*x^2 + x^3) + E^4*(5*x^3 - x^4)))*Log[x]*Log[Log[x]])/((x^6 +
E^8*(25*x^2 - 10*x^3 + x^4) + E^4*(10*x^4 - 2*x^5))*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.67, size = 32, normalized size = 1.00 \begin {gather*} -\frac {{\left (x - 5\right )} e^{\left (e^{\left (x - 2\right )} + 4\right )} \log \left (\log \relax (x)\right )}{x^{3} - {\left (x^{2} - 5 \, x\right )} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^3-10*x^2+25*x)*exp(4)^2+(-x^4+5*x^3)*exp(4))*exp(x-2)+(-x^2+10*x-25)*exp(4)^2+(2*x^3-15*x^2)*e
xp(4))*log(x)*exp(exp(x-2))*log(log(x))+((x^2-10*x+25)*exp(4)^2+(-x^3+5*x^2)*exp(4))*exp(exp(x-2)))/((x^4-10*x
^3+25*x^2)*exp(4)^2+(-2*x^5+10*x^4)*exp(4)+x^6)/log(x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x^{2} - 10 \, x + 25\right )} e^{8} - {\left (2 \, x^{3} - 15 \, x^{2}\right )} e^{4} - {\left ({\left (x^{3} - 10 \, x^{2} + 25 \, x\right )} e^{8} - {\left (x^{4} - 5 \, x^{3}\right )} e^{4}\right )} e^{\left (x - 2\right )}\right )} e^{\left (e^{\left (x - 2\right )}\right )} \log \relax (x) \log \left (\log \relax (x)\right ) - {\left ({\left (x^{2} - 10 \, x + 25\right )} e^{8} - {\left (x^{3} - 5 \, x^{2}\right )} e^{4}\right )} e^{\left (e^{\left (x - 2\right )}\right )}}{{\left (x^{6} + {\left (x^{4} - 10 \, x^{3} + 25 \, x^{2}\right )} e^{8} - 2 \, {\left (x^{5} - 5 \, x^{4}\right )} e^{4}\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^3-10*x^2+25*x)*exp(4)^2+(-x^4+5*x^3)*exp(4))*exp(x-2)+(-x^2+10*x-25)*exp(4)^2+(2*x^3-15*x^2)*e
xp(4))*log(x)*exp(exp(x-2))*log(log(x))+((x^2-10*x+25)*exp(4)^2+(-x^3+5*x^2)*exp(4))*exp(exp(x-2)))/((x^4-10*x
^3+25*x^2)*exp(4)^2+(-2*x^5+10*x^4)*exp(4)+x^6)/log(x),x, algorithm="giac")

[Out]

integrate(-(((x^2 - 10*x + 25)*e^8 - (2*x^3 - 15*x^2)*e^4 - ((x^3 - 10*x^2 + 25*x)*e^8 - (x^4 - 5*x^3)*e^4)*e^
(x - 2))*e^(e^(x - 2))*log(x)*log(log(x)) - ((x^2 - 10*x + 25)*e^8 - (x^3 - 5*x^2)*e^4)*e^(e^(x - 2)))/((x^6 +
(x^4 - 10*x^3 + 25*x^2)*e^8 - 2*(x^5 - 5*x^4)*e^4)*log(x)), x)

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maple [A]  time = 0.12, size = 34, normalized size = 1.06

method	result	size
risch	\(\frac {\ln \left (\ln \relax (x )\right ) \left (x -5\right ) {\mathrm e}^{4+{\mathrm e}^{x -2}}}{\left (x \,{\mathrm e}^{4}-x^{2}-5 \,{\mathrm e}^{4}\right ) x}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x^3-10*x^2+25*x)*exp(4)^2+(-x^4+5*x^3)*exp(4))*exp(x-2)+(-x^2+10*x-25)*exp(4)^2+(2*x^3-15*x^2)*exp(4))
*ln(x)*exp(exp(x-2))*ln(ln(x))+((x^2-10*x+25)*exp(4)^2+(-x^3+5*x^2)*exp(4))*exp(exp(x-2)))/((x^4-10*x^3+25*x^2
)*exp(4)^2+(-2*x^5+10*x^4)*exp(4)+x^6)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.54, size = 37, normalized size = 1.16 \begin {gather*} -\frac {{\left (x e^{4} - 5 \, e^{4}\right )} e^{\left (e^{\left (x - 2\right )}\right )} \log \left (\log \relax (x)\right )}{x^{3} - x^{2} e^{4} + 5 \, x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^3-10*x^2+25*x)*exp(4)^2+(-x^4+5*x^3)*exp(4))*exp(x-2)+(-x^2+10*x-25)*exp(4)^2+(2*x^3-15*x^2)*e
xp(4))*log(x)*exp(exp(x-2))*log(log(x))+((x^2-10*x+25)*exp(4)^2+(-x^3+5*x^2)*exp(4))*exp(exp(x-2)))/((x^4-10*x
^3+25*x^2)*exp(4)^2+(-2*x^5+10*x^4)*exp(4)+x^6)/log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.93, size = 34, normalized size = 1.06 \begin {gather*} -\frac {\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,{\mathrm {e}}^x+4}\,\left (x-5\right )}{x\,\left (x^2-{\mathrm {e}}^4\,x+5\,{\mathrm {e}}^4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(x - 2))*(exp(4)*(5*x^2 - x^3) + exp(8)*(x^2 - 10*x + 25)) - log(log(x))*exp(exp(x - 2))*log(x)*(e
xp(4)*(15*x^2 - 2*x^3) - exp(x - 2)*(exp(8)*(25*x - 10*x^2 + x^3) + exp(4)*(5*x^3 - x^4)) + exp(8)*(x^2 - 10*x
+ 25)))/(log(x)*(exp(8)*(25*x^2 - 10*x^3 + x^4) + exp(4)*(10*x^4 - 2*x^5) + x^6)),x)

[Out]

-(log(log(x))*exp(exp(-2)*exp(x) + 4)*(x - 5))/(x*(5*exp(4) - x*exp(4) + x^2))

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sympy [A]  time = 0.67, size = 42, normalized size = 1.31 \begin {gather*} \frac {\left (- x e^{4} \log {\left (\log {\relax (x )} \right )} + 5 e^{4} \log {\left (\log {\relax (x )} \right )}\right ) e^{e^{x - 2}}}{x^{3} - x^{2} e^{4} + 5 x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x**3-10*x**2+25*x)*exp(4)**2+(-x**4+5*x**3)*exp(4))*exp(x-2)+(-x**2+10*x-25)*exp(4)**2+(2*x**3-1
5*x**2)*exp(4))*ln(x)*exp(exp(x-2))*ln(ln(x))+((x**2-10*x+25)*exp(4)**2+(-x**3+5*x**2)*exp(4))*exp(exp(x-2)))/
((x**4-10*x**3+25*x**2)*exp(4)**2+(-2*x**5+10*x**4)*exp(4)+x**6)/ln(x),x)

[Out]

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

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