3.39.33 \(\int \frac {e^{\frac {\log (5) \log (\frac {x}{3})}{-x+(2+x) \log (\frac {x}{3})}} (-\log (5)+\log (5) \log (\frac {x}{3})-\log (5) \log ^2(\frac {x}{3}))}{x^2+(-4 x-2 x^2) \log (\frac {x}{3})+(4+4 x+x^2) \log ^2(\frac {x}{3})} \, dx\)

Optimal. Leaf size=24 \[ 5^{\frac {1}{2+x-\frac {x}{\log \left (\frac {x}{3}\right )}}}-e^5 \]

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Rubi [F]  time = 3.82, 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 {\log (5) \log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-\log (5)+\log (5) \log \left (\frac {x}{3}\right )-\log (5) \log ^2\left (\frac {x}{3}\right )\right )}{x^2+\left (-4 x-2 x^2\right ) \log \left (\frac {x}{3}\right )+\left (4+4 x+x^2\right ) \log ^2\left (\frac {x}{3}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-3*Log[5]*Defer[Subst][Defer[Int][E^((Log[5]*Log[x])/(-3*x + (2 + 3*x)*Log[x]))/(2 + 3*x)^2, x], x, x/3] - 3*(
2 + Log[3])*Log[5]*Defer[Subst][Defer[Int][E^((Log[5]*Log[x])/(-3*x + (2 + 3*x)*Log[x]))/(-3*x + 2*Log[x] + 3*
x*Log[x])^2, x], x, x/3] - 12*Log[5]*Defer[Subst][Defer[Int][E^((Log[5]*Log[x])/(-3*x + (2 + 3*x)*Log[x]))/((2
+ 3*x)^2*(-3*x + 2*Log[x] + 3*x*Log[x])^2), x], x, x/3] + 12*Log[5]*Defer[Subst][Defer[Int][E^((Log[5]*Log[x]
)/(-3*x + (2 + 3*x)*Log[x]))/((2 + 3*x)*(-3*x + 2*Log[x] + 3*x*Log[x])^2), x], x, x/3] + 12*Log[5]*Defer[Subst
][Defer[Int][E^((Log[5]*Log[x])/(-3*x + (2 + 3*x)*Log[x]))/((2 + 3*x)^2*(-3*x + 2*Log[x] + 3*x*Log[x])), x], x
, x/3] - 6*Log[5]*Defer[Subst][Defer[Int][E^((Log[5]*Log[x])/(-3*x + (2 + 3*x)*Log[x]))/((2 + 3*x)*(-3*x + 2*L
og[x] + 3*x*Log[x])), x], x, x/3] + 3*Log[5]*Defer[Subst][Defer[Int][(E^((Log[5]*Log[x])/(-3*x + (2 + 3*x)*Log
[x]))*Log[3*x])/(-3*x + 2*Log[x] + 3*x*Log[x])^2, x], x, x/3]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5^{\frac {\log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \log (5) \left (-1-\log (3)-\log ^2\left (\frac {x}{3}\right )+\log (x)\right )}{\left (x-(2+x) \log \left (\frac {x}{3}\right )\right )^2} \, dx\\ &=\log (5) \int \frac {5^{\frac {\log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \left (-1-\log (3)-\log ^2\left (\frac {x}{3}\right )+\log (x)\right )}{\left (x-(2+x) \log \left (\frac {x}{3}\right )\right )^2} \, dx\\ &=(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \left (-1-\log (3)-\log ^2(x)+\log (3 x)\right )}{(3 x-(2+3 x) \log (x))^2} \, dx,x,\frac {x}{3}\right )\\ &=(3 \log (5)) \operatorname {Subst}\left (\int \left (\frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \left (-1-\log (3)-\log ^2(x)\right )}{(-3 x+2 \log (x)+3 x \log (x))^2}+\frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \log (3 x)}{(-3 x+2 \log (x)+3 x \log (x))^2}\right ) \, dx,x,\frac {x}{3}\right )\\ &=(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \left (-1-\log (3)-\log ^2(x)\right )}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )+(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \log (3 x)}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )\\ &=(3 \log (5)) \operatorname {Subst}\left (\int \left (-\frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2}+\frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \left (-4-12 x (1+\log (3))-9 x^2 (2+\log (3))-\log (81)\right )}{(2+3 x)^2 (3 x-2 \log (x)-3 x \log (x))^2}-\frac {6 e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} x}{(2+3 x)^2 (-3 x+2 \log (x)+3 x \log (x))}\right ) \, dx,x,\frac {x}{3}\right )+(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \log (3 x)}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )\\ &=-\left ((3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2} \, dx,x,\frac {x}{3}\right )\right )+(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \left (-4-12 x (1+\log (3))-9 x^2 (2+\log (3))-\log (81)\right )}{(2+3 x)^2 (3 x-2 \log (x)-3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )+(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \log (3 x)}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )-(18 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} x}{(2+3 x)^2 (-3 x+2 \log (x)+3 x \log (x))} \, dx,x,\frac {x}{3}\right )\\ &=-\left ((3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2} \, dx,x,\frac {x}{3}\right )\right )+(3 \log (5)) \operatorname {Subst}\left (\int \left (-\frac {4 e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2 (-3 x+2 \log (x)+3 x \log (x))^2}+\frac {4 e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x) (-3 x+2 \log (x)+3 x \log (x))^2}-\frac {2 e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \left (1+\frac {\log (3)}{2}\right )}{(-3 x+2 \log (x)+3 x \log (x))^2}\right ) \, dx,x,\frac {x}{3}\right )+(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \log (3 x)}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )-(18 \log (5)) \operatorname {Subst}\left (\int \left (-\frac {2 e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{3 (2+3 x)^2 (-3 x+2 \log (x)+3 x \log (x))}+\frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{3 (2+3 x) (-3 x+2 \log (x)+3 x \log (x))}\right ) \, dx,x,\frac {x}{3}\right )\\ &=-\left ((3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2} \, dx,x,\frac {x}{3}\right )\right )+(3 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}} \log (3 x)}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )-(6 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x) (-3 x+2 \log (x)+3 x \log (x))} \, dx,x,\frac {x}{3}\right )-(12 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2 (-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )+(12 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x) (-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )+(12 \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(2+3 x)^2 (-3 x+2 \log (x)+3 x \log (x))} \, dx,x,\frac {x}{3}\right )-(3 (2+\log (3)) \log (5)) \operatorname {Subst}\left (\int \frac {e^{\frac {\log (5) \log (x)}{-3 x+(2+3 x) \log (x)}}}{(-3 x+2 \log (x)+3 x \log (x))^2} \, dx,x,\frac {x}{3}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 25, normalized size = 1.04 \begin {gather*} 5^{\frac {\log \left (\frac {x}{3}\right )}{-x+(2+x) \log \left (\frac {x}{3}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5^(Log[x/3]/(-x + (2 + x)*Log[x/3]))

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fricas [A]  time = 0.53, size = 22, normalized size = 0.92 \begin {gather*} e^{\left (\frac {\log \relax (5) \log \left (\frac {1}{3} \, x\right )}{{\left (x + 2\right )} \log \left (\frac {1}{3} \, x\right ) - x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1/3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*
x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3*x)+x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.11, size = 26, normalized size = 1.08 \begin {gather*} e^{\left (\frac {\log \relax (5) \log \left (\frac {1}{3} \, x\right )}{x \log \left (\frac {1}{3} \, x\right ) - x + 2 \, \log \left (\frac {1}{3} \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1/3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*
x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3*x)+x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.20, size = 31, normalized size = 1.29

method	result	size
risch	\(\left (\frac {x}{3}\right )^{\frac {\ln \relax (5)}{x \left (\ln \relax (x )-\ln \relax (3)\right )+2 \ln \relax (x )-2 \ln \relax (3)-x}}\)	\(31\)
norman	\(\frac {x \ln \left (\frac {x}{3}\right ) {\mathrm e}^{\frac {\ln \relax (5) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}-x \,{\mathrm e}^{\frac {\ln \relax (5) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}+2 \ln \left (\frac {x}{3}\right ) {\mathrm e}^{\frac {\ln \relax (5) \ln \left (\frac {x}{3}\right )}{\left (2+x \right ) \ln \left (\frac {x}{3}\right )-x}}}{x \ln \left (\frac {x}{3}\right )+2 \ln \left (\frac {x}{3}\right )-x}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(5)*ln(1/3*x)^2+ln(5)*ln(1/3*x)-ln(5))*exp(ln(5)*ln(1/3*x)/((2+x)*ln(1/3*x)-x))/((x^2+4*x+4)*ln(1/3*x)
^2+(-2*x^2-4*x)*ln(1/3*x)+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.52, size = 53, normalized size = 2.21 \begin {gather*} e^{\left (\frac {\log \relax (5) \log \relax (3)}{x {\left (\log \relax (3) + 1\right )} - {\left (x + 2\right )} \log \relax (x) + 2 \, \log \relax (3)} - \frac {\log \relax (5) \log \relax (x)}{x {\left (\log \relax (3) + 1\right )} - {\left (x + 2\right )} \log \relax (x) + 2 \, \log \relax (3)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(5)*log(1/3*x)^2+log(5)*log(1/3*x)-log(5))*exp(log(5)*log(1/3*x)/((2+x)*log(1/3*x)-x))/((x^2+4*
x+4)*log(1/3*x)^2+(-2*x^2-4*x)*log(1/3*x)+x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.66, size = 56, normalized size = 2.33 \begin {gather*} {\mathrm {e}}^{-\frac {\ln \relax (5)\,\ln \relax (x)}{x+2\,\ln \relax (3)-2\,\ln \relax (x)+x\,\ln \relax (3)-x\,\ln \relax (x)}}\,{\mathrm {e}}^{\frac {\ln \relax (3)\,\ln \relax (5)}{x+2\,\ln \relax (3)-2\,\ln \relax (x)+x\,\ln \relax (3)-x\,\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.67, size = 19, normalized size = 0.79 \begin {gather*} e^{\frac {\log {\relax (5 )} \log {\left (\frac {x}{3} \right )}}{- x + \left (x + 2\right ) \log {\left (\frac {x}{3} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(5)*ln(1/3*x)**2+ln(5)*ln(1/3*x)-ln(5))*exp(ln(5)*ln(1/3*x)/((2+x)*ln(1/3*x)-x))/((x**2+4*x+4)*l
n(1/3*x)**2+(-2*x**2-4*x)*ln(1/3*x)+x**2),x)

[Out]

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

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