3.46.96 \(\int \frac {-3 x \log (3)+e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} (6 x^2-6 x \log (3))+(6 e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} \log (3)+6 x \log (3)) \log (x)}{36 e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} x \log (3)+36 x^2 \log (3)+(-12 e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} x \log (3)-12 x^2 \log (3)) \log ^2(x)+(e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} x \log (3)+x^2 \log (3)) \log ^4(x)+(12 e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} x \log (3)+12 x^2 \log (3)+(-2 e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} x \log (3)-2 x^2 \log (3)) \log ^2(x)) \log (e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}}+x)+(e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}} x \log (3)+x^2 \log (3)) \log ^2(e^{\frac {-x^2+(-2+2 x) \log (3)}{\log (3)}}+x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {3}{6-\log ^2(x)+\log \left (e^{-2+2 x-\frac {x^2}{\log (3)}}+x\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 1.35, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 343, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {6688, 12, 6686} \begin {gather*} \frac {3}{\log \left (e^{-\frac {x^2}{\log (3)}+2 x-2}+x\right )-\log ^2(x)+6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3*x*Log[3] + E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*(6*x^2 - 6*x*Log[3]) + (6*E^((-x^2 + (-2 + 2*x)*Log[3
])/Log[3])*Log[3] + 6*x*Log[3])*Log[x])/(36*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] + 36*x^2*Log[3] + (
-12*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] - 12*x^2*Log[3])*Log[x]^2 + (E^((-x^2 + (-2 + 2*x)*Log[3])/
Log[3])*x*Log[3] + x^2*Log[3])*Log[x]^4 + (12*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] + 12*x^2*Log[3] +
 (-2*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] - 2*x^2*Log[3])*Log[x]^2)*Log[E^((-x^2 + (-2 + 2*x)*Log[3]
)/Log[3]) + x] + (E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] + x^2*Log[3])*Log[E^((-x^2 + (-2 + 2*x)*Log[3
])/Log[3]) + x]^2),x]

[Out]

3/(6 - Log[x]^2 + Log[E^(-2 + 2*x - x^2/Log[3]) + x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 e^{2 x} x (x-\log (3))-3 e^{2+\frac {x^2}{\log (3)}} x \log (3)+6 \left (e^{2 x}+e^{2+\frac {x^2}{\log (3)}} x\right ) \log (3) \log (x)}{x \left (e^{2 x}+e^{2+\frac {x^2}{\log (3)}} x\right ) \log (3) \left (6-\log ^2(x)+\log \left (e^{-2+2 x-\frac {x^2}{\log (3)}}+x\right )\right )^2} \, dx\\ &=\frac {\int \frac {6 e^{2 x} x (x-\log (3))-3 e^{2+\frac {x^2}{\log (3)}} x \log (3)+6 \left (e^{2 x}+e^{2+\frac {x^2}{\log (3)}} x\right ) \log (3) \log (x)}{x \left (e^{2 x}+e^{2+\frac {x^2}{\log (3)}} x\right ) \left (6-\log ^2(x)+\log \left (e^{-2+2 x-\frac {x^2}{\log (3)}}+x\right )\right )^2} \, dx}{\log (3)}\\ &=\frac {3}{6-\log ^2(x)+\log \left (e^{-2+2 x-\frac {x^2}{\log (3)}}+x\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.31, size = 31, normalized size = 1.00 \begin {gather*} -\frac {3}{-6+\log ^2(x)-\log \left (e^{-2+2 x-\frac {x^2}{\log (3)}}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x*Log[3] + E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*(6*x^2 - 6*x*Log[3]) + (6*E^((-x^2 + (-2 + 2*x)
*Log[3])/Log[3])*Log[3] + 6*x*Log[3])*Log[x])/(36*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] + 36*x^2*Log[
3] + (-12*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] - 12*x^2*Log[3])*Log[x]^2 + (E^((-x^2 + (-2 + 2*x)*Lo
g[3])/Log[3])*x*Log[3] + x^2*Log[3])*Log[x]^4 + (12*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] + 12*x^2*Lo
g[3] + (-2*E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] - 2*x^2*Log[3])*Log[x]^2)*Log[E^((-x^2 + (-2 + 2*x)*
Log[3])/Log[3]) + x] + (E^((-x^2 + (-2 + 2*x)*Log[3])/Log[3])*x*Log[3] + x^2*Log[3])*Log[E^((-x^2 + (-2 + 2*x)
*Log[3])/Log[3]) + x]^2),x]

[Out]

-3/(-6 + Log[x]^2 - Log[E^(-2 + 2*x - x^2/Log[3]) + x])

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 33, normalized size = 1.06 \begin {gather*} -\frac {3}{\log \relax (x)^{2} - \log \left (x + e^{\left (-\frac {x^{2} - 2 \, {\left (x - 1\right )} \log \relax (3)}{\log \relax (3)}\right )}\right ) - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))+6*x*log(3))*log(x)+(-6*x*log(3)+6*x^2)*exp(((2*x-2)*log(
3)-x^2)/log(3))-3*x*log(3))/((x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))+x^2*log(3))*log(exp(((2*x-2)*log(3)-x^
2)/log(3))+x)^2+((-2*x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))-2*x^2*log(3))*log(x)^2+12*x*log(3)*exp(((2*x-2)
*log(3)-x^2)/log(3))+12*x^2*log(3))*log(exp(((2*x-2)*log(3)-x^2)/log(3))+x)+(x*log(3)*exp(((2*x-2)*log(3)-x^2)
/log(3))+x^2*log(3))*log(x)^4+(-12*x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))-12*x^2*log(3))*log(x)^2+36*x*log(
3)*exp(((2*x-2)*log(3)-x^2)/log(3))+36*x^2*log(3)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 1.42, size = 34, normalized size = 1.10 \begin {gather*} -\frac {3}{\log \relax (x)^{2} - \log \left (x e^{2} + e^{\left (-\frac {x^{2} - 2 \, x \log \relax (3)}{\log \relax (3)}\right )}\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))+6*x*log(3))*log(x)+(-6*x*log(3)+6*x^2)*exp(((2*x-2)*log(
3)-x^2)/log(3))-3*x*log(3))/((x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))+x^2*log(3))*log(exp(((2*x-2)*log(3)-x^
2)/log(3))+x)^2+((-2*x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))-2*x^2*log(3))*log(x)^2+12*x*log(3)*exp(((2*x-2)
*log(3)-x^2)/log(3))+12*x^2*log(3))*log(exp(((2*x-2)*log(3)-x^2)/log(3))+x)+(x*log(3)*exp(((2*x-2)*log(3)-x^2)
/log(3))+x^2*log(3))*log(x)^4+(-12*x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))-12*x^2*log(3))*log(x)^2+36*x*log(
3)*exp(((2*x-2)*log(3)-x^2)/log(3))+36*x^2*log(3)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 37, normalized size = 1.19




method result size



risch \(-\frac {3}{\ln \relax (x )^{2}-\ln \left ({\mathrm e}^{\frac {2 x \ln \relax (3)-x^{2}-2 \ln \relax (3)}{\ln \relax (3)}}+x \right )-6}\) \(37\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*ln(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))+6*x*ln(3))*ln(x)+(-6*x*ln(3)+6*x^2)*exp(((2*x-2)*ln(3)-x^2)/ln(3)
)-3*x*ln(3))/((x*ln(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))+x^2*ln(3))*ln(exp(((2*x-2)*ln(3)-x^2)/ln(3))+x)^2+((-2*x
*ln(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))-2*x^2*ln(3))*ln(x)^2+12*x*ln(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))+12*x^2*ln
(3))*ln(exp(((2*x-2)*ln(3)-x^2)/ln(3))+x)+(x*ln(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))+x^2*ln(3))*ln(x)^4+(-12*x*ln
(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))-12*x^2*ln(3))*ln(x)^2+36*x*ln(3)*exp(((2*x-2)*ln(3)-x^2)/ln(3))+36*x^2*ln(3
)),x,method=_RETURNVERBOSE)

[Out]

-3/(ln(x)^2-ln(exp((2*x*ln(3)-x^2-2*ln(3))/ln(3))+x)-6)

________________________________________________________________________________________

maxima [A]  time = 0.60, size = 44, normalized size = 1.42 \begin {gather*} -\frac {3 \, \log \relax (3)}{\log \relax (3) \log \relax (x)^{2} + x^{2} - \log \relax (3) \log \left (x e^{\left (\frac {x^{2}}{\log \relax (3)} + 2\right )} + e^{\left (2 \, x\right )}\right ) - 4 \, \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))+6*x*log(3))*log(x)+(-6*x*log(3)+6*x^2)*exp(((2*x-2)*log(
3)-x^2)/log(3))-3*x*log(3))/((x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))+x^2*log(3))*log(exp(((2*x-2)*log(3)-x^
2)/log(3))+x)^2+((-2*x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))-2*x^2*log(3))*log(x)^2+12*x*log(3)*exp(((2*x-2)
*log(3)-x^2)/log(3))+12*x^2*log(3))*log(exp(((2*x-2)*log(3)-x^2)/log(3))+x)+(x*log(3)*exp(((2*x-2)*log(3)-x^2)
/log(3))+x^2*log(3))*log(x)^4+(-12*x*log(3)*exp(((2*x-2)*log(3)-x^2)/log(3))-12*x^2*log(3))*log(x)^2+36*x*log(
3)*exp(((2*x-2)*log(3)-x^2)/log(3))+36*x^2*log(3)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.03, size = 32, normalized size = 1.03 \begin {gather*} \frac {3}{-{\ln \relax (x)}^2+\ln \left (x+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {x^2}{\ln \relax (3)}}\right )+6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x*log(3) + exp((log(3)*(2*x - 2) - x^2)/log(3))*(6*x*log(3) - 6*x^2) - log(x)*(6*x*log(3) + 6*exp((log
(3)*(2*x - 2) - x^2)/log(3))*log(3)))/(log(x)^4*(x^2*log(3) + x*exp((log(3)*(2*x - 2) - x^2)/log(3))*log(3)) -
 log(x)^2*(12*x^2*log(3) + 12*x*exp((log(3)*(2*x - 2) - x^2)/log(3))*log(3)) + log(x + exp((log(3)*(2*x - 2) -
 x^2)/log(3)))*(12*x^2*log(3) - log(x)^2*(2*x^2*log(3) + 2*x*exp((log(3)*(2*x - 2) - x^2)/log(3))*log(3)) + 12
*x*exp((log(3)*(2*x - 2) - x^2)/log(3))*log(3)) + log(x + exp((log(3)*(2*x - 2) - x^2)/log(3)))^2*(x^2*log(3)
+ x*exp((log(3)*(2*x - 2) - x^2)/log(3))*log(3)) + 36*x^2*log(3) + 36*x*exp((log(3)*(2*x - 2) - x^2)/log(3))*l
og(3)),x)

[Out]

3/(log(x + exp(2*x)*exp(-2)*exp(-x^2/log(3))) - log(x)^2 + 6)

________________________________________________________________________________________

sympy [A]  time = 1.06, size = 27, normalized size = 0.87 \begin {gather*} \frac {3}{- \log {\relax (x )}^{2} + \log {\left (x + e^{\frac {- x^{2} + \left (2 x - 2\right ) \log {\relax (3 )}}{\log {\relax (3 )}}} \right )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*ln(3)*exp(((2*x-2)*ln(3)-x**2)/ln(3))+6*x*ln(3))*ln(x)+(-6*x*ln(3)+6*x**2)*exp(((2*x-2)*ln(3)-x*
*2)/ln(3))-3*x*ln(3))/((x*ln(3)*exp(((2*x-2)*ln(3)-x**2)/ln(3))+x**2*ln(3))*ln(exp(((2*x-2)*ln(3)-x**2)/ln(3))
+x)**2+((-2*x*ln(3)*exp(((2*x-2)*ln(3)-x**2)/ln(3))-2*x**2*ln(3))*ln(x)**2+12*x*ln(3)*exp(((2*x-2)*ln(3)-x**2)
/ln(3))+12*x**2*ln(3))*ln(exp(((2*x-2)*ln(3)-x**2)/ln(3))+x)+(x*ln(3)*exp(((2*x-2)*ln(3)-x**2)/ln(3))+x**2*ln(
3))*ln(x)**4+(-12*x*ln(3)*exp(((2*x-2)*ln(3)-x**2)/ln(3))-12*x**2*ln(3))*ln(x)**2+36*x*ln(3)*exp(((2*x-2)*ln(3
)-x**2)/ln(3))+36*x**2*ln(3)),x)

[Out]

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

________________________________________________________________________________________