3.67.71 \(\int \frac {e^{\frac {-30000 \log ^4(\frac {x}{5})+x^5 \log ^4(x)}{2500 \log ^4(\frac {x}{5})}} (4 x^4 \log (\frac {x}{5}) \log ^3(x)+(-4 x^4+5 x^4 \log (\frac {x}{5})) \log ^4(x))}{2500 \log ^5(\frac {x}{5})} \, dx\)

Optimal. Leaf size=23 \[ e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} \]

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Rubi [F]  time = 2.04, 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 {-30000 \log ^4\left (\frac {x}{5}\right )+x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} \left (4 x^4 \log \left (\frac {x}{5}\right ) \log ^3(x)+\left (-4 x^4+5 x^4 \log \left (\frac {x}{5}\right )\right ) \log ^4(x)\right )}{2500 \log ^5\left (\frac {x}{5}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-30000*Log[x/5]^4 + x^5*Log[x]^4)/(2500*Log[x/5]^4))*(4*x^4*Log[x/5]*Log[x]^3 + (-4*x^4 + 5*x^4*Log[x
/5])*Log[x]^4))/(2500*Log[x/5]^5),x]

[Out]

Defer[Int][(E^(-12 + (x^5*Log[x]^4)/(2500*Log[x/5]^4))*x^4*Log[x]^3)/Log[x/5]^4, x]/625 - Defer[Int][(E^(-12 +
(x^5*Log[x]^4)/(2500*Log[x/5]^4))*x^4*Log[x]^4)/Log[x/5]^5, x]/625 + Defer[Int][(E^(-12 + (x^5*Log[x]^4)/(250
0*Log[x/5]^4))*x^4*Log[x]^4)/Log[x/5]^4, x]/500

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^{\frac {-30000 \log ^4\left (\frac {x}{5}\right )+x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} \left (4 x^4 \log \left (\frac {x}{5}\right ) \log ^3(x)+\left (-4 x^4+5 x^4 \log \left (\frac {x}{5}\right )\right ) \log ^4(x)\right )}{\log ^5\left (\frac {x}{5}\right )} \, dx}{2500}\\ &=\frac {\int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^3(x) \left (-4 \log (x)+\log \left (\frac {x}{5}\right ) (4+5 \log (x))\right )}{\log ^5\left (\frac {x}{5}\right )} \, dx}{2500}\\ &=\frac {\int \left (\frac {4 e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^3(x)}{\log ^4\left (\frac {x}{5}\right )}+\frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \left (-4+5 \log \left (\frac {x}{5}\right )\right ) \log ^4(x)}{\log ^5\left (\frac {x}{5}\right )}\right ) \, dx}{2500}\\ &=\frac {\int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \left (-4+5 \log \left (\frac {x}{5}\right )\right ) \log ^4(x)}{\log ^5\left (\frac {x}{5}\right )} \, dx}{2500}+\frac {1}{625} \int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^3(x)}{\log ^4\left (\frac {x}{5}\right )} \, dx\\ &=\frac {\int \left (-\frac {4 e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^4(x)}{\log ^5\left (\frac {x}{5}\right )}+\frac {5 e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^4(x)}{\log ^4\left (\frac {x}{5}\right )}\right ) \, dx}{2500}+\frac {1}{625} \int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^3(x)}{\log ^4\left (\frac {x}{5}\right )} \, dx\\ &=\frac {1}{625} \int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^3(x)}{\log ^4\left (\frac {x}{5}\right )} \, dx-\frac {1}{625} \int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^4(x)}{\log ^5\left (\frac {x}{5}\right )} \, dx+\frac {1}{500} \int \frac {e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} x^4 \log ^4(x)}{\log ^4\left (\frac {x}{5}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 23, normalized size = 1.00 \begin {gather*} e^{-12+\frac {x^5 \log ^4(x)}{2500 \log ^4\left (\frac {x}{5}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-30000*Log[x/5]^4 + x^5*Log[x]^4)/(2500*Log[x/5]^4))*(4*x^4*Log[x/5]*Log[x]^3 + (-4*x^4 + 5*x^4
*Log[x/5])*Log[x]^4))/(2500*Log[x/5]^5),x]

[Out]

E^(-12 + (x^5*Log[x]^4)/(2500*Log[x/5]^4))

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fricas [B]  time = 0.61, size = 71, normalized size = 3.09 \begin {gather*} e^{\left (\frac {x^{5} \log \relax (5)^{4} + 4 \, x^{5} \log \relax (5)^{3} \log \left (\frac {1}{5} \, x\right ) + 6 \, x^{5} \log \relax (5)^{2} \log \left (\frac {1}{5} \, x\right )^{2} + 4 \, x^{5} \log \relax (5) \log \left (\frac {1}{5} \, x\right )^{3} + {\left (x^{5} - 30000\right )} \log \left (\frac {1}{5} \, x\right )^{4}}{2500 \, \log \left (\frac {1}{5} \, x\right )^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2500*((5*x^4*log(1/5*x)-4*x^4)*log(x)^4+4*x^4*log(1/5*x)*log(x)^3)*exp(1/10000*(x^5*log(x)^4-30000
*log(1/5*x)^4)/log(1/5*x)^4)^4/log(1/5*x)^5,x, algorithm="fricas")

[Out]

e^(1/2500*(x^5*log(5)^4 + 4*x^5*log(5)^3*log(1/5*x) + 6*x^5*log(5)^2*log(1/5*x)^2 + 4*x^5*log(5)*log(1/5*x)^3
+ (x^5 - 30000)*log(1/5*x)^4)/log(1/5*x)^4)

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giac [A]  time = 3.05, size = 18, normalized size = 0.78 \begin {gather*} e^{\left (\frac {x^{5} \log \relax (x)^{4}}{2500 \, \log \left (\frac {1}{5} \, x\right )^{4}} - 12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2500*((5*x^4*log(1/5*x)-4*x^4)*log(x)^4+4*x^4*log(1/5*x)*log(x)^3)*exp(1/10000*(x^5*log(x)^4-30000
*log(1/5*x)^4)/log(1/5*x)^4)^4/log(1/5*x)^5,x, algorithm="giac")

[Out]

e^(1/2500*x^5*log(x)^4/log(1/5*x)^4 - 12)

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maple [B]  time = 1.68, size = 61, normalized size = 2.65

method	result	size
risch	\({\mathrm e}^{-\frac {-x^{5} \ln \relax (x )^{4}+30000 \ln \relax (x )^{4}-120000 \ln \relax (x )^{3} \ln \relax (5)+180000 \ln \relax (x )^{2} \ln \relax (5)^{2}-120000 \ln \relax (x ) \ln \relax (5)^{3}+30000 \ln \relax (5)^{4}}{2500 \left (\ln \relax (5)-\ln \relax (x )\right )^{4}}}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2500*((5*x^4*ln(1/5*x)-4*x^4)*ln(x)^4+4*x^4*ln(1/5*x)*ln(x)^3)*exp(1/10000*(x^5*ln(x)^4-30000*ln(1/5*x)^
4)/ln(1/5*x)^4)^4/ln(1/5*x)^5,x,method=_RETURNVERBOSE)

[Out]

exp(-1/2500*(-x^5*ln(x)^4+30000*ln(x)^4-120000*ln(x)^3*ln(5)+180000*ln(x)^2*ln(5)^2-120000*ln(x)*ln(5)^3+30000
*ln(5)^4)/(ln(5)-ln(x))^4)

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maxima [B]  time = 0.69, size = 49, normalized size = 2.13 \begin {gather*} e^{\left (\frac {x^{5} \log \relax (x)^{4}}{2500 \, {\left (\log \relax (5)^{4} - 4 \, \log \relax (5)^{3} \log \relax (x) + 6 \, \log \relax (5)^{2} \log \relax (x)^{2} - 4 \, \log \relax (5) \log \relax (x)^{3} + \log \relax (x)^{4}\right )}} - 12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2500*((5*x^4*log(1/5*x)-4*x^4)*log(x)^4+4*x^4*log(1/5*x)*log(x)^3)*exp(1/10000*(x^5*log(x)^4-30000
*log(1/5*x)^4)/log(1/5*x)^4)^4/log(1/5*x)^5,x, algorithm="maxima")

[Out]

e^(1/2500*x^5*log(x)^4/(log(5)^4 - 4*log(5)^3*log(x) + 6*log(5)^2*log(x)^2 - 4*log(5)*log(x)^3 + log(x)^4) - 1
2)

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mupad [B]  time = 4.52, size = 277, normalized size = 12.04 \begin {gather*} 5^{\frac {48\,{\ln \relax (x)}^3}{{\ln \relax (x)}^4-4\,\ln \relax (5)\,{\ln \relax (x)}^3+6\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2-4\,{\ln \relax (5)}^3\,\ln \relax (x)+{\ln \relax (5)}^4}}\,x^{\frac {48\,{\ln \relax (5)}^3}{{\ln \relax (x)}^4-4\,\ln \relax (5)\,{\ln \relax (x)}^3+6\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2-4\,{\ln \relax (5)}^3\,\ln \relax (x)+{\ln \relax (5)}^4}}\,{\mathrm {e}}^{-\frac {72\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2}{{\ln \relax (x)}^4-4\,\ln \relax (5)\,{\ln \relax (x)}^3+6\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2-4\,{\ln \relax (5)}^3\,\ln \relax (x)+{\ln \relax (5)}^4}}\,{\mathrm {e}}^{\frac {x^5\,{\ln \relax (x)}^4}{2500\,\left ({\ln \relax (x)}^4-4\,\ln \relax (5)\,{\ln \relax (x)}^3+6\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2-4\,{\ln \relax (5)}^3\,\ln \relax (x)+{\ln \relax (5)}^4\right )}}\,{\mathrm {e}}^{-\frac {12\,{\ln \relax (x)}^4}{{\ln \relax (x)}^4-4\,\ln \relax (5)\,{\ln \relax (x)}^3+6\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2-4\,{\ln \relax (5)}^3\,\ln \relax (x)+{\ln \relax (5)}^4}}\,{\mathrm {e}}^{-\frac {12\,{\ln \relax (5)}^4}{{\ln \relax (x)}^4-4\,\ln \relax (5)\,{\ln \relax (x)}^3+6\,{\ln \relax (5)}^2\,{\ln \relax (x)}^2-4\,{\ln \relax (5)}^3\,\ln \relax (x)+{\ln \relax (5)}^4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((4*((x^5*log(x)^4)/10000 - 3*log(x/5)^4))/log(x/5)^4)*(log(x)^4*(5*x^4*log(x/5) - 4*x^4) + 4*x^4*log(
x/5)*log(x)^3))/(2500*log(x/5)^5),x)

[Out]

5^((48*log(x)^3)/(log(x)^4 - 4*log(5)^3*log(x) - 4*log(5)*log(x)^3 + 6*log(5)^2*log(x)^2 + log(5)^4))*x^((48*l
og(5)^3)/(log(x)^4 - 4*log(5)^3*log(x) - 4*log(5)*log(x)^3 + 6*log(5)^2*log(x)^2 + log(5)^4))*exp(-(72*log(5)^
2*log(x)^2)/(log(x)^4 - 4*log(5)^3*log(x) - 4*log(5)*log(x)^3 + 6*log(5)^2*log(x)^2 + log(5)^4))*exp((x^5*log(
x)^4)/(2500*(log(x)^4 - 4*log(5)^3*log(x) - 4*log(5)*log(x)^3 + 6*log(5)^2*log(x)^2 + log(5)^4)))*exp(-(12*log
(x)^4)/(log(x)^4 - 4*log(5)^3*log(x) - 4*log(5)*log(x)^3 + 6*log(5)^2*log(x)^2 + log(5)^4))*exp(-(12*log(5)^4)
/(log(x)^4 - 4*log(5)^3*log(x) - 4*log(5)*log(x)^3 + 6*log(5)^2*log(x)^2 + log(5)^4))

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sympy [A]  time = 0.85, size = 29, normalized size = 1.26 \begin {gather*} e^{\frac {4 \left (\frac {x^{5} \log {\relax (x )}^{4}}{10000} - 3 \left (\log {\relax (x )} - \log {\relax (5 )}\right )^{4}\right )}{\left (\log {\relax (x )} - \log {\relax (5 )}\right )^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2500*((5*x**4*ln(1/5*x)-4*x**4)*ln(x)**4+4*x**4*ln(1/5*x)*ln(x)**3)*exp(1/10000*(x**5*ln(x)**4-300
00*ln(1/5*x)**4)/ln(1/5*x)**4)**4/ln(1/5*x)**5,x)

[Out]

exp(4*(x**5*log(x)**4/10000 - 3*(log(x) - log(5))**4)/(log(x) - log(5))**4)

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