3.3.11 \(\int \frac {3-3 x+(-12+3 x-3 \log (x)) \log (4-x+\log (x))+(64+e^2 (32-8 x)+e^4 (4-x)-16 x+(16+8 e^2+e^4) \log (x)) \log ^2(4-x+\log (x))}{(-12 x+3 x^2-3 x \log (x)) \log (4-x+\log (x))+(64 x-16 x^2+e^2 (32 x-8 x^2)+e^4 (4 x-x^2)+(16 x+8 e^2 x+e^4 x) \log (x)) \log ^2(4-x+\log (x))} \, dx\)

Optimal. Leaf size=26 \[ \log \left (\frac {1}{3} x \left (\left (4+e^2\right )^2-\frac {3}{\log (4-x+\log (x))}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 2.20, antiderivative size = 34, normalized size of antiderivative = 1.31, number of steps used = 5, number of rules used = 3, integrand size = 161, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6688, 6742, 6684} \begin {gather*} \log (x)-\log (\log (-x+\log (x)+4))+\log \left (3-\left (4+e^2\right )^2 \log (-x+\log (x)+4)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 3*x + (-12 + 3*x - 3*Log[x])*Log[4 - x + Log[x]] + (64 + E^2*(32 - 8*x) + E^4*(4 - x) - 16*x + (16 +
8*E^2 + E^4)*Log[x])*Log[4 - x + Log[x]]^2)/((-12*x + 3*x^2 - 3*x*Log[x])*Log[4 - x + Log[x]] + (64*x - 16*x^2
+ E^2*(32*x - 8*x^2) + E^4*(4*x - x^2) + (16*x + 8*E^2*x + E^4*x)*Log[x])*Log[4 - x + Log[x]]^2),x]

[Out]

Log[x] - Log[Log[4 - x + Log[x]]] + Log[3 - (4 + E^2)^2*Log[4 - x + Log[x]]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 (-1+x)-3 (-4+x-\log (x)) \log (4-x+\log (x))+\left (4+e^2\right )^2 (-4+x-\log (x)) \log ^2(4-x+\log (x))}{x (4-x+\log (x)) \log (4-x+\log (x)) \left (3-\left (4+e^2\right )^2 \log (4-x+\log (x))\right )} \, dx\\ &=\int \left (\frac {1}{x}+\frac {1-x}{x (-4+x-\log (x)) \log (4-x+\log (x))}+\frac {\left (4+e^2\right )^2 (-1+x)}{x (4-x+\log (x)) \left (3-16 \left (1+\frac {1}{16} e^2 \left (8+e^2\right )\right ) \log (4-x+\log (x))\right )}\right ) \, dx\\ &=\log (x)+\left (4+e^2\right )^2 \int \frac {-1+x}{x (4-x+\log (x)) \left (3-16 \left (1+\frac {1}{16} e^2 \left (8+e^2\right )\right ) \log (4-x+\log (x))\right )} \, dx+\int \frac {1-x}{x (-4+x-\log (x)) \log (4-x+\log (x))} \, dx\\ &=\log (x)-\log (\log (4-x+\log (x)))+\log \left (3-\left (4+e^2\right )^2 \log (4-x+\log (x))\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.08, size = 53, normalized size = 2.04 \begin {gather*} \log (x)-\log (\log (4-x+\log (x)))+\log \left (3-16 \log (4-x+\log (x))-8 e^2 \log (4-x+\log (x))-e^4 \log (4-x+\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 3*x + (-12 + 3*x - 3*Log[x])*Log[4 - x + Log[x]] + (64 + E^2*(32 - 8*x) + E^4*(4 - x) - 16*x +
(16 + 8*E^2 + E^4)*Log[x])*Log[4 - x + Log[x]]^2)/((-12*x + 3*x^2 - 3*x*Log[x])*Log[4 - x + Log[x]] + (64*x -
16*x^2 + E^2*(32*x - 8*x^2) + E^4*(4*x - x^2) + (16*x + 8*E^2*x + E^4*x)*Log[x])*Log[4 - x + Log[x]]^2),x]

[Out]

Log[x] - Log[Log[4 - x + Log[x]]] + Log[3 - 16*Log[4 - x + Log[x]] - 8*E^2*Log[4 - x + Log[x]] - E^4*Log[4 - x
+ Log[x]]]

________________________________________________________________________________________

fricas [A]  time = 0.63, size = 34, normalized size = 1.31 \begin {gather*} \log \left ({\left (e^{4} + 8 \, e^{2} + 16\right )} \log \left (-x + \log \relax (x) + 4\right ) - 3\right ) + \log \relax (x) - \log \left (\log \left (-x + \log \relax (x) + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(2)^2+8*exp(2)+16)*log(x)+(-x+4)*exp(2)^2+(-8*x+32)*exp(2)-16*x+64)*log(log(x)-x+4)^2+(-3*log(
x)+3*x-12)*log(log(x)-x+4)-3*x+3)/(((x*exp(2)^2+8*exp(2)*x+16*x)*log(x)+(-x^2+4*x)*exp(2)^2+(-8*x^2+32*x)*exp(
2)-16*x^2+64*x)*log(log(x)-x+4)^2+(-3*x*log(x)+3*x^2-12*x)*log(log(x)-x+4)),x, algorithm="fricas")

[Out]

log((e^4 + 8*e^2 + 16)*log(-x + log(x) + 4) - 3) + log(x) - log(log(-x + log(x) + 4))

________________________________________________________________________________________

giac [B]  time = 0.73, size = 51, normalized size = 1.96 \begin {gather*} \log \left (-e^{4} \log \left (-x + \log \relax (x) + 4\right ) - 8 \, e^{2} \log \left (-x + \log \relax (x) + 4\right ) - 16 \, \log \left (-x + \log \relax (x) + 4\right ) + 3\right ) + \log \relax (x) - \log \left (\log \left (-x + \log \relax (x) + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(2)^2+8*exp(2)+16)*log(x)+(-x+4)*exp(2)^2+(-8*x+32)*exp(2)-16*x+64)*log(log(x)-x+4)^2+(-3*log(
x)+3*x-12)*log(log(x)-x+4)-3*x+3)/(((x*exp(2)^2+8*exp(2)*x+16*x)*log(x)+(-x^2+4*x)*exp(2)^2+(-8*x^2+32*x)*exp(
2)-16*x^2+64*x)*log(log(x)-x+4)^2+(-3*x*log(x)+3*x^2-12*x)*log(log(x)-x+4)),x, algorithm="giac")

[Out]

log(-e^4*log(-x + log(x) + 4) - 8*e^2*log(-x + log(x) + 4) - 16*log(-x + log(x) + 4) + 3) + log(x) - log(log(-
x + log(x) + 4))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 37, normalized size = 1.42

method	result	size
risch	\(\ln \relax (x )-\ln \left (\ln \left (\ln \relax (x )-x +4\right )\right )+\ln \left (\ln \left (\ln \relax (x )-x +4\right )-\frac {3}{{\mathrm e}^{4}+8 \,{\mathrm e}^{2}+16}\right )\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((exp(2)^2+8*exp(2)+16)*ln(x)+(-x+4)*exp(2)^2+(-8*x+32)*exp(2)-16*x+64)*ln(ln(x)-x+4)^2+(-3*ln(x)+3*x-12)
*ln(ln(x)-x+4)-3*x+3)/(((x*exp(2)^2+8*exp(2)*x+16*x)*ln(x)+(-x^2+4*x)*exp(2)^2+(-8*x^2+32*x)*exp(2)-16*x^2+64*
x)*ln(ln(x)-x+4)^2+(-3*x*ln(x)+3*x^2-12*x)*ln(ln(x)-x+4)),x,method=_RETURNVERBOSE)

[Out]

ln(x)-ln(ln(ln(x)-x+4))+ln(ln(ln(x)-x+4)-3/(exp(4)+8*exp(2)+16))

________________________________________________________________________________________

maxima [A]  time = 0.75, size = 45, normalized size = 1.73 \begin {gather*} \log \relax (x) + \log \left (\frac {{\left (e^{4} + 8 \, e^{2} + 16\right )} \log \left (-x + \log \relax (x) + 4\right ) - 3}{e^{4} + 8 \, e^{2} + 16}\right ) - \log \left (\log \left (-x + \log \relax (x) + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(2)^2+8*exp(2)+16)*log(x)+(-x+4)*exp(2)^2+(-8*x+32)*exp(2)-16*x+64)*log(log(x)-x+4)^2+(-3*log(
x)+3*x-12)*log(log(x)-x+4)-3*x+3)/(((x*exp(2)^2+8*exp(2)*x+16*x)*log(x)+(-x^2+4*x)*exp(2)^2+(-8*x^2+32*x)*exp(
2)-16*x^2+64*x)*log(log(x)-x+4)^2+(-3*x*log(x)+3*x^2-12*x)*log(log(x)-x+4)),x, algorithm="maxima")

[Out]

log(x) + log(((e^4 + 8*e^2 + 16)*log(-x + log(x) + 4) - 3)/(e^4 + 8*e^2 + 16)) - log(log(-x + log(x) + 4))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\left (16\,x+{\mathrm {e}}^4\,\left (x-4\right )-\ln \relax (x)\,\left (8\,{\mathrm {e}}^2+{\mathrm {e}}^4+16\right )+{\mathrm {e}}^2\,\left (8\,x-32\right )-64\right )\,{\ln \left (\ln \relax (x)-x+4\right )}^2+\left (3\,\ln \relax (x)-3\,x+12\right )\,\ln \left (\ln \relax (x)-x+4\right )+3\,x-3}{{\ln \left (\ln \relax (x)-x+4\right )}^2\,\left (64\,x+{\mathrm {e}}^4\,\left (4\,x-x^2\right )+{\mathrm {e}}^2\,\left (32\,x-8\,x^2\right )-16\,x^2+\ln \relax (x)\,\left (16\,x+8\,x\,{\mathrm {e}}^2+x\,{\mathrm {e}}^4\right )\right )-\ln \left (\ln \relax (x)-x+4\right )\,\left (12\,x+3\,x\,\ln \relax (x)-3\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + log(log(x) - x + 4)*(3*log(x) - 3*x + 12) + log(log(x) - x + 4)^2*(16*x + exp(4)*(x - 4) - log(x)*
(8*exp(2) + exp(4) + 16) + exp(2)*(8*x - 32) - 64) - 3)/(log(log(x) - x + 4)^2*(64*x + exp(4)*(4*x - x^2) + ex
p(2)*(32*x - 8*x^2) - 16*x^2 + log(x)*(16*x + 8*x*exp(2) + x*exp(4))) - log(log(x) - x + 4)*(12*x + 3*x*log(x)
- 3*x^2)),x)

[Out]

int(-(3*x + log(log(x) - x + 4)*(3*log(x) - 3*x + 12) + log(log(x) - x + 4)^2*(16*x + exp(4)*(x - 4) - log(x)*
(8*exp(2) + exp(4) + 16) + exp(2)*(8*x - 32) - 64) - 3)/(log(log(x) - x + 4)^2*(64*x + exp(4)*(4*x - x^2) + ex
p(2)*(32*x - 8*x^2) - 16*x^2 + log(x)*(16*x + 8*x*exp(2) + x*exp(4))) - log(log(x) - x + 4)*(12*x + 3*x*log(x)
- 3*x^2)), x)

________________________________________________________________________________________

sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: PolynomialError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((exp(2)**2+8*exp(2)+16)*ln(x)+(-x+4)*exp(2)**2+(-8*x+32)*exp(2)-16*x+64)*ln(ln(x)-x+4)**2+(-3*ln(x
)+3*x-12)*ln(ln(x)-x+4)-3*x+3)/(((x*exp(2)**2+8*exp(2)*x+16*x)*ln(x)+(-x**2+4*x)*exp(2)**2+(-8*x**2+32*x)*exp(
2)-16*x**2+64*x)*ln(ln(x)-x+4)**2+(-3*x*ln(x)+3*x**2-12*x)*ln(ln(x)-x+4)),x)

[Out]

Exception raised: PolynomialError

________________________________________________________________________________________