3.103.3 \(\int \frac {e^{\frac {1}{9} (225 e \log ^2(x^2)-150 e \log (x^2) \log (\log (\frac {4-x}{4}))+25 e \log ^2(\log (\frac {4-x}{4})))} ((-150 e x+e (-3600+900 x) \log (\frac {4-x}{4})) \log (x^2)+(50 e x+e (1200-300 x) \log (\frac {4-x}{4})) \log (\log (\frac {4-x}{4})))}{(-36 x+9 x^2) \log (\frac {4-x}{4})} \, dx\)

Optimal. Leaf size=29 \[ e^{25 e \left (-\log \left (x^2\right )+\frac {1}{3} \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )^2} \]

________________________________________________________________________________________

Rubi [A]  time = 4.06, antiderivative size = 27, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, integrand size = 136, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {1593, 6741, 12, 6706} \begin {gather*} e^{\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((225*E*Log[x^2]^2 - 150*E*Log[x^2]*Log[Log[(4 - x)/4]] + 25*E*Log[Log[(4 - x)/4]]^2)/9)*((-150*E*x + E
*(-3600 + 900*x)*Log[(4 - x)/4])*Log[x^2] + (50*E*x + E*(1200 - 300*x)*Log[(4 - x)/4])*Log[Log[(4 - x)/4]]))/(
(-36*x + 9*x^2)*Log[(4 - x)/4]),x]

[Out]

E^((25*E*(3*Log[x^2] - Log[Log[1 - x/4]])^2)/9)

Rule 12

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {1}{9} \left (225 e \log ^2\left (x^2\right )-150 e \log \left (x^2\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )+25 e \log ^2\left (\log \left (\frac {4-x}{4}\right )\right )\right )\right ) \left (\left (-150 e x+e (-3600+900 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (x^2\right )+\left (50 e x+e (1200-300 x) \log \left (\frac {4-x}{4}\right )\right ) \log \left (\log \left (\frac {4-x}{4}\right )\right )\right )}{x (-36+9 x) \log \left (\frac {4-x}{4}\right )} \, dx\\ &=\int \frac {50 e^{1+\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2} \left (x+24 \log \left (1-\frac {x}{4}\right )-6 x \log \left (1-\frac {x}{4}\right )\right ) \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )}{(36-9 x) x \log \left (1-\frac {x}{4}\right )} \, dx\\ &=50 \int \frac {e^{1+\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2} \left (x+24 \log \left (1-\frac {x}{4}\right )-6 x \log \left (1-\frac {x}{4}\right )\right ) \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )}{(36-9 x) x \log \left (1-\frac {x}{4}\right )} \, dx\\ &=e^{\frac {25}{9} e \left (3 \log \left (x^2\right )-\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 25, normalized size = 0.86 \begin {gather*} e^{\frac {25}{9} e \left (-3 \log \left (x^2\right )+\log \left (\log \left (1-\frac {x}{4}\right )\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((225*E*Log[x^2]^2 - 150*E*Log[x^2]*Log[Log[(4 - x)/4]] + 25*E*Log[Log[(4 - x)/4]]^2)/9)*((-150*E
*x + E*(-3600 + 900*x)*Log[(4 - x)/4])*Log[x^2] + (50*E*x + E*(1200 - 300*x)*Log[(4 - x)/4])*Log[Log[(4 - x)/4
]]))/((-36*x + 9*x^2)*Log[(4 - x)/4]),x]

[Out]

E^((25*E*(-3*Log[x^2] + Log[Log[1 - x/4]])^2)/9)

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 40, normalized size = 1.38 \begin {gather*} e^{\left (25 \, e \log \left (x^{2}\right )^{2} - \frac {50}{3} \, e \log \left (x^{2}\right ) \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right ) + \frac {25}{9} \, e \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*x*exp(1))*log(log(-1/4*x+1))+((900*x-3600)*exp(1)*log(-1/4*x
+1)-150*x*exp(1))*log(x^2))*exp(25/9*exp(1)*log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*ex
p(1)*log(x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x, algorithm="fricas")

[Out]

e^(25*e*log(x^2)^2 - 50/3*e*log(x^2)*log(log(-1/4*x + 1)) + 25/9*e*log(log(-1/4*x + 1))^2)

________________________________________________________________________________________

giac [A]  time = 9.01, size = 40, normalized size = 1.38 \begin {gather*} e^{\left (25 \, e \log \left (x^{2}\right )^{2} - \frac {50}{3} \, e \log \left (x^{2}\right ) \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right ) + \frac {25}{9} \, e \log \left (\log \left (-\frac {1}{4} \, x + 1\right )\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*x*exp(1))*log(log(-1/4*x+1))+((900*x-3600)*exp(1)*log(-1/4*x
+1)-150*x*exp(1))*log(x^2))*exp(25/9*exp(1)*log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*ex
p(1)*log(x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x, algorithm="giac")

[Out]

e^(25*e*log(x^2)^2 - 50/3*e*log(x^2)*log(log(-1/4*x + 1)) + 25/9*e*log(log(-1/4*x + 1))^2)

________________________________________________________________________________________

maple [C]  time = 0.25, size = 187, normalized size = 6.45

method	result	size
risch	\(\ln \left (-\frac {x}{4}+1\right )^{-\frac {50 \,{\mathrm e} \left (-i \pi \,\mathrm {csgn}\left (i x^{2}\right )+i \pi \,\mathrm {csgn}\left (i x \right )+2 \ln \relax (x )\right )}{3}} x^{-100 i \pi \,{\mathrm e} \,\mathrm {csgn}\left (i x^{2}\right )} x^{100 i \pi \,{\mathrm e} \,\mathrm {csgn}\left (i x \right )} {\mathrm e}^{\frac {25 \,{\mathrm e} \left (-9 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{6}+36 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{5} \mathrm {csgn}\left (i x \right )-54 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{4} \mathrm {csgn}\left (i x \right )^{2}+36 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{3} \mathrm {csgn}\left (i x \right )^{3}-9 \pi ^{2} \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )^{4}+144 \ln \relax (x )^{2}+4 \ln \left (\ln \left (-\frac {x}{4}+1\right )\right )^{2}\right )}{36}}\)	\(187\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-300*x+1200)*exp(1)*ln(-1/4*x+1)+50*x*exp(1))*ln(ln(-1/4*x+1))+((900*x-3600)*exp(1)*ln(-1/4*x+1)-150*x*
exp(1))*ln(x^2))*exp(25/9*exp(1)*ln(ln(-1/4*x+1))^2-50/3*exp(1)*ln(x^2)*ln(ln(-1/4*x+1))+25*exp(1)*ln(x^2)^2)/
(9*x^2-36*x)/ln(-1/4*x+1),x,method=_RETURNVERBOSE)

[Out]

ln(-1/4*x+1)^(-50/3*exp(1)*(-I*Pi*csgn(I*x^2)+I*Pi*csgn(I*x)+2*ln(x)))*(x^(-50*I*Pi*exp(1)*csgn(I*x^2)))^2*x^(
100*I*Pi*exp(1)*csgn(I*x))*exp(25/36*exp(1)*(-9*Pi^2*csgn(I*x^2)^6+36*Pi^2*csgn(I*x^2)^5*csgn(I*x)-54*Pi^2*csg
n(I*x^2)^4*csgn(I*x)^2+36*Pi^2*csgn(I*x^2)^3*csgn(I*x)^3-9*Pi^2*csgn(I*x^2)^2*csgn(I*x)^4+144*ln(x)^2+4*ln(ln(
-1/4*x+1))^2))

________________________________________________________________________________________

maxima [A]  time = 0.90, size = 46, normalized size = 1.59 \begin {gather*} e^{\left (100 \, e \log \relax (x)^{2} - \frac {100}{3} \, e \log \relax (x) \log \left (-2 \, \log \relax (2) + \log \left (-x + 4\right )\right ) + \frac {25}{9} \, e \log \left (-2 \, \log \relax (2) + \log \left (-x + 4\right )\right )^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-300*x+1200)*exp(1)*log(-1/4*x+1)+50*x*exp(1))*log(log(-1/4*x+1))+((900*x-3600)*exp(1)*log(-1/4*x
+1)-150*x*exp(1))*log(x^2))*exp(25/9*exp(1)*log(log(-1/4*x+1))^2-50/3*exp(1)*log(x^2)*log(log(-1/4*x+1))+25*ex
p(1)*log(x^2)^2)/(9*x^2-36*x)/log(-1/4*x+1),x, algorithm="maxima")

[Out]

e^(100*e*log(x)^2 - 100/3*e*log(x)*log(-2*log(2) + log(-x + 4)) + 25/9*e*log(-2*log(2) + log(-x + 4))^2)

________________________________________________________________________________________

mupad [B]  time = 7.58, size = 42, normalized size = 1.45 \begin {gather*} {\mathrm {e}}^{-\frac {50\,\ln \left (\ln \left (1-\frac {x}{4}\right )\right )\,\ln \left (x^2\right )\,\mathrm {e}}{3}}\,{\mathrm {e}}^{\frac {25\,{\ln \left (\ln \left (1-\frac {x}{4}\right )\right )}^2\,\mathrm {e}}{9}}\,{\mathrm {e}}^{25\,{\ln \left (x^2\right )}^2\,\mathrm {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((25*log(log(1 - x/4))^2*exp(1))/9 + 25*log(x^2)^2*exp(1) - (50*log(log(1 - x/4))*log(x^2)*exp(1))/3)
*(log(log(1 - x/4))*(50*x*exp(1) - exp(1)*log(1 - x/4)*(300*x - 1200)) - log(x^2)*(150*x*exp(1) - exp(1)*log(1
- x/4)*(900*x - 3600))))/(log(1 - x/4)*(36*x - 9*x^2)),x)

[Out]

exp(-(50*log(log(1 - x/4))*log(x^2)*exp(1))/3)*exp((25*log(log(1 - x/4))^2*exp(1))/9)*exp(25*log(x^2)^2*exp(1)
)

________________________________________________________________________________________

sympy [B]  time = 3.08, size = 49, normalized size = 1.69 \begin {gather*} e^{25 e \log {\left (x^{2} \right )}^{2} - \frac {50 e \log {\left (x^{2} \right )} \log {\left (\log {\left (1 - \frac {x}{4} \right )} \right )}}{3} + \frac {25 e \log {\left (\log {\left (1 - \frac {x}{4} \right )} \right )}^{2}}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-300*x+1200)*exp(1)*ln(-1/4*x+1)+50*x*exp(1))*ln(ln(-1/4*x+1))+((900*x-3600)*exp(1)*ln(-1/4*x+1)-
150*x*exp(1))*ln(x**2))*exp(25/9*exp(1)*ln(ln(-1/4*x+1))**2-50/3*exp(1)*ln(x**2)*ln(ln(-1/4*x+1))+25*exp(1)*ln
(x**2)**2)/(9*x**2-36*x)/ln(-1/4*x+1),x)

[Out]

exp(25*E*log(x**2)**2 - 50*E*log(x**2)*log(log(1 - x/4))/3 + 25*E*log(log(1 - x/4))**2/9)

________________________________________________________________________________________