3.67.29 \(\int \frac {e^{\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}} (8-2 x-4 \log (5 x))}{(48-24 x+3 x^2) \log ^3(5 x)} \, dx\)

Optimal. Leaf size=22 \[ 5 e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}} \]

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Rubi [F]  time = 1.49, 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 {\exp \left (\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}\right ) (8-2 x-4 \log (5 x))}{\left (48-24 x+3 x^2\right ) \log ^3(5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((x + (24 - 6*x + (-12 + 3*x)*Log[5])*Log[5*x]^2)/((-12 + 3*x)*Log[5*x]^2))*(8 - 2*x - 4*Log[5*x]))/((4
8 - 24*x + 3*x^2)*Log[5*x]^3),x]

[Out]

(-10*Defer[Int][E^(-2 + x/(3*(-4 + x)*Log[5*x]^2))/((-4 + x)*Log[5*x]^3), x])/3 - (20*Defer[Int][E^(-2 + x/(3*
(-4 + x)*Log[5*x]^2))/((-4 + x)^2*Log[5*x]^2), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}\right ) (8-2 x-4 \log (5 x))}{3 (-4+x)^2 \log ^3(5 x)} \, dx\\ &=\frac {1}{3} \int \frac {\exp \left (\frac {x+(24-6 x+(-12+3 x) \log (5)) \log ^2(5 x)}{(-12+3 x) \log ^2(5 x)}\right ) (8-2 x-4 \log (5 x))}{(-4+x)^2 \log ^3(5 x)} \, dx\\ &=\frac {1}{3} \int \frac {10 e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}} (4-x-2 \log (5 x))}{(4-x)^2 \log ^3(5 x)} \, dx\\ &=\frac {10}{3} \int \frac {e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}} (4-x-2 \log (5 x))}{(4-x)^2 \log ^3(5 x)} \, dx\\ &=\frac {10}{3} \int \left (-\frac {e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}}}{(-4+x) \log ^3(5 x)}-\frac {2 e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}}}{(-4+x)^2 \log ^2(5 x)}\right ) \, dx\\ &=-\left (\frac {10}{3} \int \frac {e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}}}{(-4+x) \log ^3(5 x)} \, dx\right )-\frac {20}{3} \int \frac {e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}}}{(-4+x)^2 \log ^2(5 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 22, normalized size = 1.00 \begin {gather*} 5 e^{-2+\frac {x}{3 (-4+x) \log ^2(5 x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x + (24 - 6*x + (-12 + 3*x)*Log[5])*Log[5*x]^2)/((-12 + 3*x)*Log[5*x]^2))*(8 - 2*x - 4*Log[5*x]
))/((48 - 24*x + 3*x^2)*Log[5*x]^3),x]

[Out]

5*E^(-2 + x/(3*(-4 + x)*Log[5*x]^2))

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fricas [A]  time = 0.53, size = 35, normalized size = 1.59 \begin {gather*} e^{\left (\frac {3 \, {\left ({\left (x - 4\right )} \log \relax (5) - 2 \, x + 8\right )} \log \left (5 \, x\right )^{2} + x}{3 \, {\left (x - 4\right )} \log \left (5 \, x\right )^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(5*x)-2*x+8)*exp((((3*x-12)*log(5)-6*x+24)*log(5*x)^2+x)/(3*x-12)/log(5*x)^2)/(3*x^2-24*x+48)
/log(5*x)^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 6.92, size = 137, normalized size = 6.23 \begin {gather*} e^{\left (\frac {x \log \relax (5) \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} - \frac {2 \, x \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} - \frac {4 \, \log \relax (5) \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} + \frac {8 \, \log \left (5 \, x\right )^{2}}{x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}} + \frac {x}{3 \, {\left (x \log \left (5 \, x\right )^{2} - 4 \, \log \left (5 \, x\right )^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(5*x)-2*x+8)*exp((((3*x-12)*log(5)-6*x+24)*log(5*x)^2+x)/(3*x-12)/log(5*x)^2)/(3*x^2-24*x+48)
/log(5*x)^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.10, size = 53, normalized size = 2.41

method	result	size
risch	\(5^{\frac {x}{x -4}} \left (\frac {1}{625}\right )^{\frac {1}{x -4}} {\mathrm e}^{-\frac {6 x \ln \left (5 x \right )^{2}-24 \ln \left (5 x \right )^{2}-x}{3 \left (x -4\right ) \ln \left (5 x \right )^{2}}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(5*x)-2*x+8)*exp((((3*x-12)*ln(5)-6*x+24)*ln(5*x)^2+x)/(3*x-12)/ln(5*x)^2)/(3*x^2-24*x+48)/ln(5*x)^3
,x,method=_RETURNVERBOSE)

[Out]

5^(x/(x-4))*(1/625)^(1/(x-4))*exp(-1/3*(6*x*ln(5*x)^2-24*ln(5*x)^2-x)/(x-4)/ln(5*x)^2)

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maxima [B]  time = 0.70, size = 62, normalized size = 2.82 \begin {gather*} 5 \, e^{\left (\frac {4}{3 \, {\left (x \log \relax (5)^{2} + {\left (x - 4\right )} \log \relax (x)^{2} - 4 \, \log \relax (5)^{2} + 2 \, {\left (x \log \relax (5) - 4 \, \log \relax (5)\right )} \log \relax (x)\right )}} + \frac {1}{3 \, {\left (\log \relax (5)^{2} + 2 \, \log \relax (5) \log \relax (x) + \log \relax (x)^{2}\right )}} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(5*x)-2*x+8)*exp((((3*x-12)*log(5)-6*x+24)*log(5*x)^2+x)/(3*x-12)/log(5*x)^2)/(3*x^2-24*x+48)
/log(5*x)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.00, size = 401, normalized size = 18.23 \begin {gather*} \frac {5^{\frac {{\ln \relax (x)}^2}{{\ln \relax (x)}^2+2\,\ln \relax (5)\,\ln \relax (x)+{\ln \relax (5)}^2}}\,{\mathrm {e}}^{\frac {3\,x\,{\ln \relax (5)}^3}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {6\,x\,{\ln \relax (5)}^2}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {24\,{\ln \relax (x)}^2}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {x}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {6\,x\,{\ln \relax (x)}^2}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}\,{\mathrm {e}}^{-\frac {12\,{\ln \relax (5)}^3}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {24\,{\ln \relax (5)}^2}{3\,x\,{\ln \relax (x)}^2-12\,{\ln \relax (x)}^2+3\,x\,{\ln \relax (5)}^2-24\,\ln \relax (5)\,\ln \relax (x)-12\,{\ln \relax (5)}^2+6\,x\,\ln \relax (5)\,\ln \relax (x)}}}{x^{\frac {2\,\left (2\,\ln \relax (5)-{\ln \relax (5)}^2\right )}{{\ln \relax (x)}^2+2\,\ln \relax (5)\,\ln \relax (x)+{\ln \relax (5)}^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x + log(5*x)^2*(log(5)*(3*x - 12) - 6*x + 24))/(log(5*x)^2*(3*x - 12)))*(2*x + 4*log(5*x) - 8))/(lo
g(5*x)^3*(3*x^2 - 24*x + 48)),x)

[Out]

(5^(log(x)^2/(log(x)^2 + 2*log(5)*log(x) + log(5)^2))*exp((3*x*log(5)^3)/(3*x*log(x)^2 - 12*log(x)^2 + 3*x*log
(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(-(6*x*log(5)^2)/(3*x*log(x)^2 - 12*log(x)^2 +
3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp((24*log(x)^2)/(3*x*log(x)^2 - 12*log(
x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(x/(3*x*log(x)^2 - 12*log(x)^2 +
3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(-(6*x*log(x)^2)/(3*x*log(x)^2 - 12*lo
g(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp(-(12*log(5)^3)/(3*x*log(x)^2
- 12*log(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x)))*exp((24*log(5)^2)/(3*x*log
(x)^2 - 12*log(x)^2 + 3*x*log(5)^2 - 24*log(5)*log(x) - 12*log(5)^2 + 6*x*log(5)*log(x))))/x^((2*(2*log(5) - l
og(5)^2))/(log(x)^2 + 2*log(5)*log(x) + log(5)^2))

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sympy [A]  time = 0.95, size = 34, normalized size = 1.55 \begin {gather*} e^{\frac {x + \left (- 6 x + \left (3 x - 12\right ) \log {\relax (5 )} + 24\right ) \log {\left (5 x \right )}^{2}}{\left (3 x - 12\right ) \log {\left (5 x \right )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(5*x)-2*x+8)*exp((((3*x-12)*ln(5)-6*x+24)*ln(5*x)**2+x)/(3*x-12)/ln(5*x)**2)/(3*x**2-24*x+48)/
ln(5*x)**3,x)

[Out]

exp((x + (-6*x + (3*x - 12)*log(5) + 24)*log(5*x)**2)/((3*x - 12)*log(5*x)**2))

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