3.40.11 \(\int \frac {e^{\frac {-12-6 x-x^2+(4+x) \log (4 x^2)}{-3-x+\log (4 x^2)}} (4+6 x+x^2+(-5-2 x) \log (4 x^2)+\log ^2(4 x^2))}{9+6 x+x^2+(-6-2 x) \log (4 x^2)+\log ^2(4 x^2)} \, dx\)

Optimal. Leaf size=21 \[ e^{4+x-\frac {x}{3+x-\log \left (4 x^2\right )}} \]

________________________________________________________________________________________

Rubi [F]  time = 3.66, 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 {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) \left (4+6 x+x^2+(-5-2 x) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )\right )}{9+6 x+x^2+(-6-2 x) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-12 - 6*x - x^2 + (4 + x)*Log[4*x^2])/(-3 - x + Log[4*x^2]))*(4 + 6*x + x^2 + (-5 - 2*x)*Log[4*x^2] +
Log[4*x^2]^2))/(9 + 6*x + x^2 + (-6 - 2*x)*Log[4*x^2] + Log[4*x^2]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) \left (4+6 x+x^2+(-5-2 x) \log \left (4 x^2\right )+\log ^2\left (4 x^2\right )\right )}{\left (3+x-\log \left (4 x^2\right )\right )^2} \, dx\\ &=\int \left (\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )+\frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) (-2+x)}{\left (3+x-\log \left (4 x^2\right )\right )^2}+\frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )}{-3-x+\log \left (4 x^2\right )}\right ) \, dx\\ &=\int \exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) \, dx+\int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) (-2+x)}{\left (3+x-\log \left (4 x^2\right )\right )^2} \, dx+\int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )}{-3-x+\log \left (4 x^2\right )} \, dx\\ &=\int \exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) \, dx+\int \left (-\frac {2 \exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )}{\left (3+x-\log \left (4 x^2\right )\right )^2}+\frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) x}{\left (3+x-\log \left (4 x^2\right )\right )^2}\right ) \, dx+\int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )}{-3-x+\log \left (4 x^2\right )} \, dx\\ &=-\left (2 \int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )}{\left (3+x-\log \left (4 x^2\right )\right )^2} \, dx\right )+\int \exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) \, dx+\int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right ) x}{\left (3+x-\log \left (4 x^2\right )\right )^2} \, dx+\int \frac {\exp \left (\frac {-12-6 x-x^2+(4+x) \log \left (4 x^2\right )}{-3-x+\log \left (4 x^2\right )}\right )}{-3-x+\log \left (4 x^2\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.82, size = 20, normalized size = 0.95 \begin {gather*} e^{4+x+\frac {x}{-3-x+\log \left (4 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-12 - 6*x - x^2 + (4 + x)*Log[4*x^2])/(-3 - x + Log[4*x^2]))*(4 + 6*x + x^2 + (-5 - 2*x)*Log[4*
x^2] + Log[4*x^2]^2))/(9 + 6*x + x^2 + (-6 - 2*x)*Log[4*x^2] + Log[4*x^2]^2),x]

[Out]

E^(4 + x + x/(-3 - x + Log[4*x^2]))

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 34, normalized size = 1.62 \begin {gather*} e^{\left (\frac {x^{2} - {\left (x + 4\right )} \log \left (4 \, x^{2}\right ) + 6 \, x + 12}{x - \log \left (4 \, x^{2}\right ) + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(4*x^2)^2+(-2*x-5)*log(4*x^2)+x^2+6*x+4)*exp(((4+x)*log(4*x^2)-x^2-6*x-12)/(log(4*x^2)-3-x))/(lo
g(4*x^2)^2+(-2*x-6)*log(4*x^2)+x^2+6*x+9),x, algorithm="fricas")

[Out]

e^((x^2 - (x + 4)*log(4*x^2) + 6*x + 12)/(x - log(4*x^2) + 3))

________________________________________________________________________________________

giac [B]  time = 1.07, size = 93, normalized size = 4.43 \begin {gather*} e^{\left (\frac {x^{2}}{x - \log \left (4 \, x^{2}\right ) + 3} - \frac {x \log \left (4 \, x^{2}\right )}{x - \log \left (4 \, x^{2}\right ) + 3} + \frac {6 \, x}{x - \log \left (4 \, x^{2}\right ) + 3} - \frac {4 \, \log \left (4 \, x^{2}\right )}{x - \log \left (4 \, x^{2}\right ) + 3} + \frac {12}{x - \log \left (4 \, x^{2}\right ) + 3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(4*x^2)^2+(-2*x-5)*log(4*x^2)+x^2+6*x+4)*exp(((4+x)*log(4*x^2)-x^2-6*x-12)/(log(4*x^2)-3-x))/(lo
g(4*x^2)^2+(-2*x-6)*log(4*x^2)+x^2+6*x+9),x, algorithm="giac")

[Out]

e^(x^2/(x - log(4*x^2) + 3) - x*log(4*x^2)/(x - log(4*x^2) + 3) + 6*x/(x - log(4*x^2) + 3) - 4*log(4*x^2)/(x -
log(4*x^2) + 3) + 12/(x - log(4*x^2) + 3))

________________________________________________________________________________________

maple [A]  time = 0.30, size = 41, normalized size = 1.95

method	result	size
risch	\({\mathrm e}^{\frac {-x \ln \left (4 x^{2}\right )+x^{2}-4 \ln \left (4 x^{2}\right )+6 x +12}{x -\ln \left (4 x^{2}\right )+3}}\)	\(41\)
norman	\(\frac {x \,{\mathrm e}^{\frac {\left (4+x \right ) \ln \left (4 x^{2}\right )-x^{2}-6 x -12}{\ln \left (4 x^{2}\right )-3-x}}-\ln \left (4 x^{2}\right ) {\mathrm e}^{\frac {\left (4+x \right ) \ln \left (4 x^{2}\right )-x^{2}-6 x -12}{\ln \left (4 x^{2}\right )-3-x}}+3 \,{\mathrm e}^{\frac {\left (4+x \right ) \ln \left (4 x^{2}\right )-x^{2}-6 x -12}{\ln \left (4 x^{2}\right )-3-x}}}{x -\ln \left (4 x^{2}\right )+3}\)	\(133\)
default	\(\frac {x \,{\mathrm e}^{\frac {\left (4+x \right ) \ln \left (4 x^{2}\right )-x^{2}-6 x -12}{\ln \left (4 x^{2}\right )-3-x}}+\left (-\ln \left (4 x^{2}\right )+2 \ln \relax (x )+3\right ) {\mathrm e}^{\frac {\left (4+x \right ) \ln \left (4 x^{2}\right )-x^{2}-6 x -12}{\ln \left (4 x^{2}\right )-3-x}}-2 \ln \relax (x ) {\mathrm e}^{\frac {\left (4+x \right ) \ln \left (4 x^{2}\right )-x^{2}-6 x -12}{\ln \left (4 x^{2}\right )-3-x}}}{x -\ln \left (4 x^{2}\right )+3}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(4*x^2)^2+(-2*x-5)*ln(4*x^2)+x^2+6*x+4)*exp(((4+x)*ln(4*x^2)-x^2-6*x-12)/(ln(4*x^2)-3-x))/(ln(4*x^2)^2+
(-2*x-6)*ln(4*x^2)+x^2+6*x+9),x,method=_RETURNVERBOSE)

[Out]

exp((-x*ln(4*x^2)+x^2-4*ln(4*x^2)+6*x+12)/(x-ln(4*x^2)+3))

________________________________________________________________________________________

maxima [B]  time = 0.67, size = 53, normalized size = 2.52 \begin {gather*} e^{\left (x - \frac {2 \, \log \relax (2)}{x - 2 \, \log \relax (2) - 2 \, \log \relax (x) + 3} - \frac {2 \, \log \relax (x)}{x - 2 \, \log \relax (2) - 2 \, \log \relax (x) + 3} + \frac {3}{x - 2 \, \log \relax (2) - 2 \, \log \relax (x) + 3} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(4*x^2)^2+(-2*x-5)*log(4*x^2)+x^2+6*x+4)*exp(((4+x)*log(4*x^2)-x^2-6*x-12)/(log(4*x^2)-3-x))/(lo
g(4*x^2)^2+(-2*x-6)*log(4*x^2)+x^2+6*x+9),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.54, size = 83, normalized size = 3.95 \begin {gather*} {\mathrm {e}}^{\frac {x^2}{x-\ln \left (x^2\right )-2\,\ln \relax (2)+3}}\,{\mathrm {e}}^{\frac {12}{x-\ln \left (x^2\right )-2\,\ln \relax (2)+3}}\,{\mathrm {e}}^{\frac {6\,x}{x-\ln \left (x^2\right )-2\,\ln \relax (2)+3}}\,{\left (\frac {1}{4\,x^2}\right )}^{\frac {x+4}{x-\ln \left (x^2\right )-2\,\ln \relax (2)+3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((6*x + x^2 - log(4*x^2)*(x + 4) + 12)/(x - log(4*x^2) + 3))*(6*x + log(4*x^2)^2 - log(4*x^2)*(2*x + 5
) + x^2 + 4))/(6*x + log(4*x^2)^2 - log(4*x^2)*(2*x + 6) + x^2 + 9),x)

[Out]

exp(x^2/(x - log(x^2) - 2*log(2) + 3))*exp(12/(x - log(x^2) - 2*log(2) + 3))*exp((6*x)/(x - log(x^2) - 2*log(2
) + 3))*(1/(4*x^2))^((x + 4)/(x - log(x^2) - 2*log(2) + 3))

________________________________________________________________________________________

sympy [A]  time = 0.67, size = 29, normalized size = 1.38 \begin {gather*} e^{\frac {- x^{2} - 6 x + \left (x + 4\right ) \log {\left (4 x^{2} \right )} - 12}{- x + \log {\left (4 x^{2} \right )} - 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(4*x**2)**2+(-2*x-5)*ln(4*x**2)+x**2+6*x+4)*exp(((4+x)*ln(4*x**2)-x**2-6*x-12)/(ln(4*x**2)-3-x))/
(ln(4*x**2)**2+(-2*x-6)*ln(4*x**2)+x**2+6*x+9),x)

[Out]

exp((-x**2 - 6*x + (x + 4)*log(4*x**2) - 12)/(-x + log(4*x**2) - 3))

________________________________________________________________________________________