3.69.30 \(\int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log (\frac {16}{x^2})}} ((-2 e^{16-x} \log (5)+2 e^{16-x} \log (\frac {16}{x^2})) \log (x)+(e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log (\frac {16}{x^2})) \log ^2(x))}{x^2 \log ^2(5)-2 x^2 \log (5) \log (\frac {16}{x^2})+x^2 \log ^2(\frac {16}{x^2})} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {e^{16-x} \log ^2(x)}{x \left (-\log (5)+\log \left (\frac {16}{x^2}\right )\right )}} \]

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Rubi [F]  time = 37.81, 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^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((E^(16 - x)*Log[x]^2)/(-(x*Log[5]) + x*Log[16/x^2]))*((-2*E^(16 - x)*Log[5] + 2*E^(16 - x)*Log[16/x^2]
)*Log[x] + (E^(16 - x)*(2 + (1 + x)*Log[5]) + E^(16 - x)*(-1 - x)*Log[16/x^2])*Log[x]^2))/(x^2*Log[5]^2 - 2*x^
2*Log[5]*Log[16/x^2] + x^2*Log[16/x^2]^2),x]

[Out]

2*Defer[Int][(E^(16 - x + (E^(16 - x)*Log[x]^2)/(x*Log[16/(5*x^2)]))*Log[x])/(x^2*Log[16/(5*x^2)]), x] + (2 +
Log[5])*Defer[Int][(E^(16 - x + (E^(16 - x)*Log[x]^2)/(x*Log[16/(5*x^2)]))*Log[x]^2)/(x^2*Log[16/(5*x^2)]^2),
x] + Log[5]*Defer[Int][(E^(16 - x + (E^(16 - x)*Log[x]^2)/(x*Log[16/(5*x^2)]))*Log[x]^2)/(x*Log[16/(5*x^2)]^2)
, x] - Defer[Int][(E^(16 - x + (E^(16 - x)*Log[x]^2)/(x*Log[16/(5*x^2)]))*Log[16/x^2]*Log[x]^2)/(x^2*Log[16/(5
*x^2)]^2), x] - Defer[Int][(E^(16 - x + (E^(16 - x)*Log[x]^2)/(x*Log[16/(5*x^2)]))*Log[16/x^2]*Log[x]^2)/(x*Lo
g[16/(5*x^2)]^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log (x) \left (-2 \log (5)+(2+\log (5)+x \log (5)) \log (x)-\log \left (\frac {16}{x^2}\right ) (-2+(1+x) \log (x))\right )}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )} \, dx\\ &=\int \left (\frac {2 e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log (x)}{x^2 \log \left (\frac {16}{5 x^2}\right )}+\frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \left (2 \left (1+\frac {\log (5)}{2}\right )+x \log (5)-\log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )}\right ) \, dx\\ &=2 \int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log (x)}{x^2 \log \left (\frac {16}{5 x^2}\right )} \, dx+\int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \left (2 \left (1+\frac {\log (5)}{2}\right )+x \log (5)-\log \left (\frac {16}{x^2}\right )-x \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )} \, dx\\ &=2 \int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log (x)}{x^2 \log \left (\frac {16}{5 x^2}\right )} \, dx+\int \left (\frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log (5) \log ^2(x)}{x \log ^2\left (\frac {16}{5 x^2}\right )}+\frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} (2+\log (5)) \log ^2(x)}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )}-\frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log \left (\frac {16}{x^2}\right ) \log ^2(x)}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )}-\frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log \left (\frac {16}{x^2}\right ) \log ^2(x)}{x \log ^2\left (\frac {16}{5 x^2}\right )}\right ) \, dx\\ &=2 \int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log (x)}{x^2 \log \left (\frac {16}{5 x^2}\right )} \, dx+\log (5) \int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log ^2(x)}{x \log ^2\left (\frac {16}{5 x^2}\right )} \, dx+(2+\log (5)) \int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log ^2(x)}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )} \, dx-\int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log \left (\frac {16}{x^2}\right ) \log ^2(x)}{x^2 \log ^2\left (\frac {16}{5 x^2}\right )} \, dx-\int \frac {e^{16-x+\frac {e^{16-x} \log ^2(x)}{x \log \left (\frac {16}{5 x^2}\right )}} \log \left (\frac {16}{x^2}\right ) \log ^2(x)}{x \log ^2\left (\frac {16}{5 x^2}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 2.84, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {e^{16-x} \log ^2(x)}{-x \log (5)+x \log \left (\frac {16}{x^2}\right )}} \left (\left (-2 e^{16-x} \log (5)+2 e^{16-x} \log \left (\frac {16}{x^2}\right )\right ) \log (x)+\left (e^{16-x} (2+(1+x) \log (5))+e^{16-x} (-1-x) \log \left (\frac {16}{x^2}\right )\right ) \log ^2(x)\right )}{x^2 \log ^2(5)-2 x^2 \log (5) \log \left (\frac {16}{x^2}\right )+x^2 \log ^2\left (\frac {16}{x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((E^(16 - x)*Log[x]^2)/(-(x*Log[5]) + x*Log[16/x^2]))*((-2*E^(16 - x)*Log[5] + 2*E^(16 - x)*Log[1
6/x^2])*Log[x] + (E^(16 - x)*(2 + (1 + x)*Log[5]) + E^(16 - x)*(-1 - x)*Log[16/x^2])*Log[x]^2))/(x^2*Log[5]^2
- 2*x^2*Log[5]*Log[16/x^2] + x^2*Log[16/x^2]^2),x]

[Out]

Integrate[(E^((E^(16 - x)*Log[x]^2)/(-(x*Log[5]) + x*Log[16/x^2]))*((-2*E^(16 - x)*Log[5] + 2*E^(16 - x)*Log[1
6/x^2])*Log[x] + (E^(16 - x)*(2 + (1 + x)*Log[5]) + E^(16 - x)*(-1 - x)*Log[16/x^2])*Log[x]^2))/(x^2*Log[5]^2
- 2*x^2*Log[5]*Log[16/x^2] + x^2*Log[16/x^2]^2), x]

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fricas [B]  time = 0.94, size = 63, normalized size = 2.10 \begin {gather*} e^{\left (-\frac {16 \, e^{\left (-x + 16\right )} \log \relax (2)^{2} - 8 \, e^{\left (-x + 16\right )} \log \relax (2) \log \left (\frac {16}{x^{2}}\right ) + e^{\left (-x + 16\right )} \log \left (\frac {16}{x^{2}}\right )^{2}}{4 \, {\left (x \log \relax (5) - x \log \left (\frac {16}{x^{2}}\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(16-x)*log(16/x^2)+((x+1)*log(5)+2)*exp(16-x))*log(x)^2+(2*exp(16-x)*log(16/x^2)-2*log(5
)*exp(16-x))*log(x))*exp(exp(16-x)*log(x)^2/(x*log(16/x^2)-x*log(5)))/(x^2*log(16/x^2)^2-2*x^2*log(5)*log(16/x
^2)+x^2*log(5)^2),x, algorithm="fricas")

[Out]

e^(-1/4*(16*e^(-x + 16)*log(2)^2 - 8*e^(-x + 16)*log(2)*log(16/x^2) + e^(-x + 16)*log(16/x^2)^2)/(x*log(5) - x
*log(16/x^2)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left ({\left (x + 1\right )} e^{\left (-x + 16\right )} \log \left (\frac {16}{x^{2}}\right ) - {\left ({\left (x + 1\right )} \log \relax (5) + 2\right )} e^{\left (-x + 16\right )}\right )} \log \relax (x)^{2} + 2 \, {\left (e^{\left (-x + 16\right )} \log \relax (5) - e^{\left (-x + 16\right )} \log \left (\frac {16}{x^{2}}\right )\right )} \log \relax (x)\right )} e^{\left (-\frac {e^{\left (-x + 16\right )} \log \relax (x)^{2}}{x \log \relax (5) - x \log \left (\frac {16}{x^{2}}\right )}\right )}}{x^{2} \log \relax (5)^{2} - 2 \, x^{2} \log \relax (5) \log \left (\frac {16}{x^{2}}\right ) + x^{2} \log \left (\frac {16}{x^{2}}\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(16-x)*log(16/x^2)+((x+1)*log(5)+2)*exp(16-x))*log(x)^2+(2*exp(16-x)*log(16/x^2)-2*log(5
)*exp(16-x))*log(x))*exp(exp(16-x)*log(x)^2/(x*log(16/x^2)-x*log(5)))/(x^2*log(16/x^2)^2-2*x^2*log(5)*log(16/x
^2)+x^2*log(5)^2),x, algorithm="giac")

[Out]

integrate(-(((x + 1)*e^(-x + 16)*log(16/x^2) - ((x + 1)*log(5) + 2)*e^(-x + 16))*log(x)^2 + 2*(e^(-x + 16)*log
(5) - e^(-x + 16)*log(16/x^2))*log(x))*e^(-e^(-x + 16)*log(x)^2/(x*log(5) - x*log(16/x^2)))/(x^2*log(5)^2 - 2*
x^2*log(5)*log(16/x^2) + x^2*log(16/x^2)^2), x)

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maple [C]  time = 4.55, size = 81, normalized size = 2.70

method	result	size
risch	\({\mathrm e}^{\frac {2 \,{\mathrm e}^{16-x} \ln \relax (x )^{2}}{x \left (i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}-4 \ln \relax (x )-2 \ln \relax (5)+8 \ln \relax (2)\right )}}\)	\(81\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x-1)*exp(16-x)*ln(16/x^2)+((x+1)*ln(5)+2)*exp(16-x))*ln(x)^2+(2*exp(16-x)*ln(16/x^2)-2*ln(5)*exp(16-x)
)*ln(x))*exp(exp(16-x)*ln(x)^2/(x*ln(16/x^2)-x*ln(5)))/(x^2*ln(16/x^2)^2-2*x^2*ln(5)*ln(16/x^2)+x^2*ln(5)^2),x
,method=_RETURNVERBOSE)

[Out]

exp(2*exp(16-x)*ln(x)^2/x/(I*Pi*csgn(I*x^2)^3-2*I*Pi*csgn(I*x^2)^2*csgn(I*x)+I*Pi*csgn(I*x^2)*csgn(I*x)^2-4*ln
(x)-2*ln(5)+8*ln(2)))

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maxima [A]  time = 0.97, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (-\frac {e^{16} \log \relax (x)^{2}}{x {\left (\log \relax (5) - 4 \, \log \relax (2)\right )} e^{x} + 2 \, x e^{x} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(16-x)*log(16/x^2)+((x+1)*log(5)+2)*exp(16-x))*log(x)^2+(2*exp(16-x)*log(16/x^2)-2*log(5
)*exp(16-x))*log(x))*exp(exp(16-x)*log(x)^2/(x*log(16/x^2)-x*log(5)))/(x^2*log(16/x^2)^2-2*x^2*log(5)*log(16/x
^2)+x^2*log(5)^2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(exp(16 - x)*log(x)^2)/(x*log(5) - x*log(16/x^2)))*(log(x)^2*(exp(16 - x)*(log(5)*(x + 1) + 2) - exp
(16 - x)*log(16/x^2)*(x + 1)) - log(x)*(2*exp(16 - x)*log(5) - 2*exp(16 - x)*log(16/x^2))))/(x^2*log(5)^2 + x^
2*log(16/x^2)^2 - 2*x^2*log(5)*log(16/x^2)),x)

[Out]

-int(-(exp(-(exp(16 - x)*log(x)^2)/(x*log(5) - x*log(16/x^2)))*(log(x)^2*(exp(16 - x)*(log(5)*(x + 1) + 2) - e
xp(16 - x)*log(16/x^2)*(x + 1)) - log(x)*(2*exp(16 - x)*log(5) - 2*exp(16 - x)*log(16/x^2))))/(x^2*log(5)^2 +
x^2*log(16/x^2)^2 - 2*x^2*log(5)*log(16/x^2)), x)

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sympy [A]  time = 2.72, size = 26, normalized size = 0.87 \begin {gather*} e^{\frac {e^{16 - x} \log {\relax (x )}^{2}}{x \left (- 2 \log {\relax (x )} + \log {\left (16 \right )}\right ) - x \log {\relax (5 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-1)*exp(16-x)*ln(16/x**2)+((x+1)*ln(5)+2)*exp(16-x))*ln(x)**2+(2*exp(16-x)*ln(16/x**2)-2*ln(5)*
exp(16-x))*ln(x))*exp(exp(16-x)*ln(x)**2/(x*ln(16/x**2)-x*ln(5)))/(x**2*ln(16/x**2)**2-2*x**2*ln(5)*ln(16/x**2
)+x**2*ln(5)**2),x)

[Out]

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

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