3.10.90 \(\int \frac {e^{-x} (-4 e^x x+(48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)) \log (16-8 x+x^2)+e^x (-8+2 x) \log (16-8 x+x^2) \log (\log (16-8 x+x^2)))}{(-4 x^2+x^3) \log (16-8 x+x^2)} \, dx\)

Optimal. Leaf size=37 \[ \frac {4 \left (3 e^{-x}+2 x+\frac {1}{2} \left (5-\log (x)-\log \left (\log \left ((4-x)^2\right )\right )\right )\right )}{x} \]

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Rubi [F]  time = 2.42, 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 {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{\left (-4 x^2+x^3\right ) \log \left (16-8 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4*E^x*x + (48 + E^x*(48 - 12*x) + 36*x - 12*x^2 + E^x*(-8 + 2*x)*Log[x])*Log[16 - 8*x + x^2] + E^x*(-8 +
2*x)*Log[16 - 8*x + x^2]*Log[Log[16 - 8*x + x^2]])/(E^x*(-4*x^2 + x^3)*Log[16 - 8*x + x^2]),x]

[Out]

10/x + 12/(E^x*x) - (2*Log[x])/x - Log[Log[(-4 + x)^2]]/2 + Defer[Int][1/(x*Log[(-4 + x)^2]), x] + 2*Defer[Int
][Log[Log[(-4 + x)^2]]/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-4 e^x x+\left (48+e^x (48-12 x)+36 x-12 x^2+e^x (-8+2 x) \log (x)\right ) \log \left (16-8 x+x^2\right )+e^x (-8+2 x) \log \left (16-8 x+x^2\right ) \log \left (\log \left (16-8 x+x^2\right )\right )\right )}{(-4+x) x^2 \log \left (16-8 x+x^2\right )} \, dx\\ &=\int \frac {2 \left (-6-6 e^{-x}-6 e^{-x} x-\frac {2 x}{(-4+x) \log \left ((-4+x)^2\right )}+\log (x)+\log \left (\log \left ((-4+x)^2\right )\right )\right )}{x^2} \, dx\\ &=2 \int \frac {-6-6 e^{-x}-6 e^{-x} x-\frac {2 x}{(-4+x) \log \left ((-4+x)^2\right )}+\log (x)+\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx\\ &=2 \int \left (-\frac {6 e^{-x} (1+x)}{x^2}+\frac {-2 x+24 \log \left ((-4+x)^2\right )-6 x \log \left ((-4+x)^2\right )-4 \log \left ((-4+x)^2\right ) \log (x)+x \log \left ((-4+x)^2\right ) \log (x)-4 \log \left ((-4+x)^2\right ) \log \left (\log \left ((-4+x)^2\right )\right )+x \log \left ((-4+x)^2\right ) \log \left (\log \left ((-4+x)^2\right )\right )}{(-4+x) x^2 \log \left ((-4+x)^2\right )}\right ) \, dx\\ &=2 \int \frac {-2 x+24 \log \left ((-4+x)^2\right )-6 x \log \left ((-4+x)^2\right )-4 \log \left ((-4+x)^2\right ) \log (x)+x \log \left ((-4+x)^2\right ) \log (x)-4 \log \left ((-4+x)^2\right ) \log \left (\log \left ((-4+x)^2\right )\right )+x \log \left ((-4+x)^2\right ) \log \left (\log \left ((-4+x)^2\right )\right )}{(-4+x) x^2 \log \left ((-4+x)^2\right )} \, dx-12 \int \frac {e^{-x} (1+x)}{x^2} \, dx\\ &=\frac {12 e^{-x}}{x}+2 \int \frac {-6-\frac {2 x}{(-4+x) \log \left ((-4+x)^2\right )}+\log (x)+\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx\\ &=\frac {12 e^{-x}}{x}+2 \int \left (\frac {-2 x+24 \log \left ((-4+x)^2\right )-6 x \log \left ((-4+x)^2\right )-4 \log \left ((-4+x)^2\right ) \log (x)+x \log \left ((-4+x)^2\right ) \log (x)}{(-4+x) x^2 \log \left ((-4+x)^2\right )}+\frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2}\right ) \, dx\\ &=\frac {12 e^{-x}}{x}+2 \int \frac {-2 x+24 \log \left ((-4+x)^2\right )-6 x \log \left ((-4+x)^2\right )-4 \log \left ((-4+x)^2\right ) \log (x)+x \log \left ((-4+x)^2\right ) \log (x)}{(-4+x) x^2 \log \left ((-4+x)^2\right )} \, dx+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx\\ &=\frac {12 e^{-x}}{x}+2 \int \frac {-6-\frac {2 x}{(-4+x) \log \left ((-4+x)^2\right )}+\log (x)}{x^2} \, dx+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx\\ &=\frac {12 e^{-x}}{x}+2 \int \left (-\frac {2 \left (x-12 \log \left ((-4+x)^2\right )+3 x \log \left ((-4+x)^2\right )\right )}{(-4+x) x^2 \log \left ((-4+x)^2\right )}+\frac {\log (x)}{x^2}\right ) \, dx+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx\\ &=\frac {12 e^{-x}}{x}+2 \int \frac {\log (x)}{x^2} \, dx+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx-4 \int \frac {x-12 \log \left ((-4+x)^2\right )+3 x \log \left ((-4+x)^2\right )}{(-4+x) x^2 \log \left ((-4+x)^2\right )} \, dx\\ &=-\frac {2}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx-4 \int \frac {3+\frac {x}{(-4+x) \log \left ((-4+x)^2\right )}}{x^2} \, dx\\ &=-\frac {2}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx-4 \int \left (\frac {3}{x^2}+\frac {1}{(-4+x) x \log \left ((-4+x)^2\right )}\right ) \, dx\\ &=\frac {10}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx-4 \int \frac {1}{(-4+x) x \log \left ((-4+x)^2\right )} \, dx\\ &=\frac {10}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx-4 \int \left (\frac {1}{4 (-4+x) \log \left ((-4+x)^2\right )}-\frac {1}{4 x \log \left ((-4+x)^2\right )}\right ) \, dx\\ &=\frac {10}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx-\int \frac {1}{(-4+x) \log \left ((-4+x)^2\right )} \, dx+\int \frac {1}{x \log \left ((-4+x)^2\right )} \, dx\\ &=\frac {10}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx+\int \frac {1}{x \log \left ((-4+x)^2\right )} \, dx-\operatorname {Subst}\left (\int \frac {1}{x \log \left (x^2\right )} \, dx,x,-4+x\right )\\ &=\frac {10}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left ((-4+x)^2\right )\right )+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx+\int \frac {1}{x \log \left ((-4+x)^2\right )} \, dx\\ &=\frac {10}{x}+\frac {12 e^{-x}}{x}-\frac {2 \log (x)}{x}-\frac {1}{2} \log \left (\log \left ((-4+x)^2\right )\right )+2 \int \frac {\log \left (\log \left ((-4+x)^2\right )\right )}{x^2} \, dx+\int \frac {1}{x \log \left ((-4+x)^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 23, normalized size = 0.62 \begin {gather*} -\frac {2 \left (-5-6 e^{-x}+\log (x)+\log \left (\log \left ((-4+x)^2\right )\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4*E^x*x + (48 + E^x*(48 - 12*x) + 36*x - 12*x^2 + E^x*(-8 + 2*x)*Log[x])*Log[16 - 8*x + x^2] + E^x
*(-8 + 2*x)*Log[16 - 8*x + x^2]*Log[Log[16 - 8*x + x^2]])/(E^x*(-4*x^2 + x^3)*Log[16 - 8*x + x^2]),x]

[Out]

(-2*(-5 - 6/E^x + Log[x] + Log[Log[(-4 + x)^2]]))/x

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fricas [A]  time = 0.77, size = 33, normalized size = 0.89 \begin {gather*} -\frac {2 \, {\left (e^{x} \log \relax (x) + e^{x} \log \left (\log \left (x^{2} - 8 \, x + 16\right )\right ) - 5 \, e^{x} - 6\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-8)*exp(x)*log(x^2-8*x+16)*log(log(x^2-8*x+16))+((2*x-8)*exp(x)*log(x)+(-12*x+48)*exp(x)-12*x^2
+36*x+48)*log(x^2-8*x+16)-4*exp(x)*x)/(x^3-4*x^2)/exp(x)/log(x^2-8*x+16),x, algorithm="fricas")

[Out]

-2*(e^x*log(x) + e^x*log(log(x^2 - 8*x + 16)) - 5*e^x - 6)*e^(-x)/x

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giac [A]  time = 0.71, size = 29, normalized size = 0.78 \begin {gather*} \frac {2 \, {\left (6 \, e^{\left (-x\right )} - \log \relax (x) - \log \left (\log \left (x^{2} - 8 \, x + 16\right )\right ) + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-8)*exp(x)*log(x^2-8*x+16)*log(log(x^2-8*x+16))+((2*x-8)*exp(x)*log(x)+(-12*x+48)*exp(x)-12*x^2
+36*x+48)*log(x^2-8*x+16)-4*exp(x)*x)/(x^3-4*x^2)/exp(x)/log(x^2-8*x+16),x, algorithm="giac")

[Out]

2*(6*e^(-x) - log(x) - log(log(x^2 - 8*x + 16)) + 5)/x

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maple [C]  time = 0.24, size = 69, normalized size = 1.86

method	result	size
risch	\(-\frac {2 \ln \left (2 \ln \left (x -4\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (x -4\right )^{2}\right ) \left (-\mathrm {csgn}\left (i \left (x -4\right )^{2}\right )+\mathrm {csgn}\left (i \left (x -4\right )\right )\right )^{2}}{2}\right )}{x}-\frac {2 \left ({\mathrm e}^{x} \ln \relax (x )-5 \,{\mathrm e}^{x}-6\right ) {\mathrm e}^{-x}}{x}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x-8)*exp(x)*ln(x^2-8*x+16)*ln(ln(x^2-8*x+16))+((2*x-8)*exp(x)*ln(x)+(-12*x+48)*exp(x)-12*x^2+36*x+48)*
ln(x^2-8*x+16)-4*exp(x)*x)/(x^3-4*x^2)/exp(x)/ln(x^2-8*x+16),x,method=_RETURNVERBOSE)

[Out]

-2/x*ln(2*ln(x-4)-1/2*I*Pi*csgn(I*(x-4)^2)*(-csgn(I*(x-4)^2)+csgn(I*(x-4)))^2)-2*(exp(x)*ln(x)-5*exp(x)-6)/x*e
xp(-x)

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maxima [A]  time = 0.79, size = 28, normalized size = 0.76 \begin {gather*} \frac {2 \, {\left (6 \, e^{\left (-x\right )} - \log \relax (2) - \log \relax (x) - \log \left (\log \left (x - 4\right )\right ) + 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-8)*exp(x)*log(x^2-8*x+16)*log(log(x^2-8*x+16))+((2*x-8)*exp(x)*log(x)+(-12*x+48)*exp(x)-12*x^2
+36*x+48)*log(x^2-8*x+16)-4*exp(x)*x)/(x^3-4*x^2)/exp(x)/log(x^2-8*x+16),x, algorithm="maxima")

[Out]

2*(6*e^(-x) - log(2) - log(x) - log(log(x - 4)) + 5)/x

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mupad [B]  time = 1.10, size = 50, normalized size = 1.35 \begin {gather*} \frac {12\,{\mathrm {e}}^{-x}}{x}-\frac {2\,\ln \relax (x)}{x}+\frac {10}{x}+\frac {\ln \left (\ln \left (x^2-8\,x+16\right )\right )\,\left (8\,x-2\,x^2\right )}{x^2\,\left (x-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(log(x^2 - 8*x + 16)*(36*x - exp(x)*(12*x - 48) - 12*x^2 + exp(x)*log(x)*(2*x - 8) + 48) - 4*x*e
xp(x) + exp(x)*log(x^2 - 8*x + 16)*log(log(x^2 - 8*x + 16))*(2*x - 8)))/(log(x^2 - 8*x + 16)*(4*x^2 - x^3)),x)

[Out]

(12*exp(-x))/x - (2*log(x))/x + 10/x + (log(log(x^2 - 8*x + 16))*(8*x - 2*x^2))/(x^2*(x - 4))

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sympy [A]  time = 0.88, size = 31, normalized size = 0.84 \begin {gather*} - \frac {2 \log {\relax (x )}}{x} - \frac {2 \log {\left (\log {\left (x^{2} - 8 x + 16 \right )} \right )}}{x} + \frac {10}{x} + \frac {12 e^{- x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x-8)*exp(x)*ln(x**2-8*x+16)*ln(ln(x**2-8*x+16))+((2*x-8)*exp(x)*ln(x)+(-12*x+48)*exp(x)-12*x**2+
36*x+48)*ln(x**2-8*x+16)-4*exp(x)*x)/(x**3-4*x**2)/exp(x)/ln(x**2-8*x+16),x)

[Out]

-2*log(x)/x - 2*log(log(x**2 - 8*x + 16))/x + 10/x + 12*exp(-x)/x

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