\(\int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} (-11 x-x^2+(-96+16 x+2 x^2+(-24-2 x) \log (12+x)) \log (-4+x-\log (12+x)))}{48 x^3-8 x^4-x^5+(12 x^3+x^4) \log (12+x)} \, dx\) [6415]

Optimal result

Integrand size = 101, antiderivative size = 15 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} \]

[Out]

exp(exp(ln(-ln(x+12)+x-4)/x^2))

Rubi [F]

\[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=\int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx \]

[In]

Int[(E^(-4 + x - Log[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^x^(-2)*(-11*x - x^2 + (-96 + 16*x + 2*x^2 + (-24 -
2*x)*Log[12 + x])*Log[-4 + x - Log[12 + x]]))/(48*x^3 - 8*x^4 - x^5 + (12*x^3 + x^4)*Log[12 + x]),x]

[Out]

(11*Defer[Int][(E^(-4 + x - Log[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^(-1 + x^(-2)))/x^2, x])/12 + Defer[Int]
[(E^(-4 + x - Log[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^(-1 + x^(-2)))/x, x]/144 - Defer[Int][(E^(-4 + x - Lo
g[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^(-1 + x^(-2)))/(12 + x), x]/144 - 2*Defer[Int][(E^(-4 + x - Log[12 +
x])^x^(-2)*(-4 + x - Log[12 + x])^x^(-2)*Log[-4 + x - Log[12 + x]])/x^3, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}} (x (11+x)-2 (12+x) (-4+x-\log (12+x)) \log (-4+x-\log (12+x)))}{x^3 (12+x)} \, dx \\ & = \int \left (\frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (11+x) (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{x^2 (12+x)}-\frac {2 e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \log (-4+x-\log (12+x))}{x^3}\right ) \, dx \\ & = -\left (2 \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \log (-4+x-\log (12+x))}{x^3} \, dx\right )+\int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (11+x) (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{x^2 (12+x)} \, dx \\ & = -\left (2 \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \log (-4+x-\log (12+x))}{x^3} \, dx\right )+\int \left (\frac {11 e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{12 x^2}+\frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{144 x}-\frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{144 (12+x)}\right ) \, dx \\ & = \frac {1}{144} \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{x} \, dx-\frac {1}{144} \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{12+x} \, dx+\frac {11}{12} \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{-1+\frac {1}{x^2}}}{x^2} \, dx-2 \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \log (-4+x-\log (12+x))}{x^3} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 5.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} \]

[In]

Integrate[(E^(-4 + x - Log[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^x^(-2)*(-11*x - x^2 + (-96 + 16*x + 2*x^2 +
(-24 - 2*x)*Log[12 + x])*Log[-4 + x - Log[12 + x]]))/(48*x^3 - 8*x^4 - x^5 + (12*x^3 + x^4)*Log[12 + x]),x]

[Out]

E^(-4 + x - Log[12 + x])^x^(-2)

Maple [A] (verified)

Time = 247.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00

method	result	size
risch	\({\mathrm e}^{\left (-\ln \left (x +12\right )+x -4\right )^{\frac {1}{x^{2}}}}\)	\(15\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {\ln \left (-\ln \left (x +12\right )+x -4\right )}{x^{2}}}}\)	\(17\)

[In]

int((((-2*x-24)*ln(x+12)+2*x^2+16*x-96)*ln(-ln(x+12)+x-4)-x^2-11*x)*exp(ln(-ln(x+12)+x-4)/x^2)*exp(exp(ln(-ln(
x+12)+x-4)/x^2))/((x^4+12*x^3)*ln(x+12)-x^5-8*x^4+48*x^3),x,method=_RETURNVERBOSE)

[Out]

exp((-ln(x+12)+x-4)^(1/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{\left ({\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )}\right )} \]

[In]

integrate((((-2*x-24)*log(x+12)+2*x^2+16*x-96)*log(-log(x+12)+x-4)-x^2-11*x)*exp(log(-log(x+12)+x-4)/x^2)*exp(
exp(log(-log(x+12)+x-4)/x^2))/((x^4+12*x^3)*log(x+12)-x^5-8*x^4+48*x^3),x, algorithm="fricas")

[Out]

e^((x - log(x + 12) - 4)^(x^(-2)))

Sympy [A] (verification not implemented)

Time = 171.53 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{e^{\frac {\log {\left (x - \log {\left (x + 12 \right )} - 4 \right )}}{x^{2}}}} \]

[In]

integrate((((-2*x-24)*ln(x+12)+2*x**2+16*x-96)*ln(-ln(x+12)+x-4)-x**2-11*x)*exp(ln(-ln(x+12)+x-4)/x**2)*exp(ex
p(ln(-ln(x+12)+x-4)/x**2))/((x**4+12*x**3)*ln(x+12)-x**5-8*x**4+48*x**3),x)

[Out]

exp(exp(log(x - log(x + 12) - 4)/x**2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{\left ({\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )}\right )} \]

[In]

integrate((((-2*x-24)*log(x+12)+2*x^2+16*x-96)*log(-log(x+12)+x-4)-x^2-11*x)*exp(log(-log(x+12)+x-4)/x^2)*exp(
exp(log(-log(x+12)+x-4)/x^2))/((x^4+12*x^3)*log(x+12)-x^5-8*x^4+48*x^3),x, algorithm="maxima")

[Out]

e^((x - log(x + 12) - 4)^(x^(-2)))

Giac [F]

\[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=\int { \frac {{\left (x^{2} - 2 \, {\left (x^{2} - {\left (x + 12\right )} \log \left (x + 12\right ) + 8 \, x - 48\right )} \log \left (x - \log \left (x + 12\right ) - 4\right ) + 11 \, x\right )} {\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )} e^{\left ({\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )}\right )}}{x^{5} + 8 \, x^{4} - 48 \, x^{3} - {\left (x^{4} + 12 \, x^{3}\right )} \log \left (x + 12\right )} \,d x } \]

[In]

integrate((((-2*x-24)*log(x+12)+2*x^2+16*x-96)*log(-log(x+12)+x-4)-x^2-11*x)*exp(log(-log(x+12)+x-4)/x^2)*exp(
exp(log(-log(x+12)+x-4)/x^2))/((x^4+12*x^3)*log(x+12)-x^5-8*x^4+48*x^3),x, algorithm="giac")

[Out]

integrate((x^2 - 2*(x^2 - (x + 12)*log(x + 12) + 8*x - 48)*log(x - log(x + 12) - 4) + 11*x)*(x - log(x + 12) -
4)^(x^(-2))*e^((x - log(x + 12) - 4)^(x^(-2)))/(x^5 + 8*x^4 - 48*x^3 - (x^4 + 12*x^3)*log(x + 12)), x)

Mupad [B] (verification not implemented)

Time = 14.47 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx={\mathrm {e}}^{{\left (x-\ln \left (x+12\right )-4\right )}^{\frac {1}{x^2}}} \]

[In]

int(-(exp(log(x - log(x + 12) - 4)/x^2)*exp(exp(log(x - log(x + 12) - 4)/x^2))*(11*x - log(x - log(x + 12) - 4
)*(16*x + 2*x^2 - log(x + 12)*(2*x + 24) - 96) + x^2))/(log(x + 12)*(12*x^3 + x^4) + 48*x^3 - 8*x^4 - x^5),x)

[Out]

exp((x - log(x + 12) - 4)^(1/x^2))