3.54.74 \(\int \frac {(-2 x-2 x \log (x)) \log (e^{-8-2 x} \log ^2(3))+(-2-\log (x)) \log ^2(e^{-8-2 x} \log ^2(3))}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx\)

Optimal. Leaf size=28 \[ \frac {\log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{4 x^2 (1+\log (x))^2} \]

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Rubi [F]  time = 1.19, 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 {(-2 x-2 x \log (x)) \log \left (e^{-8-2 x} \log ^2(3)\right )+(-2-\log (x)) \log ^2\left (e^{-8-2 x} \log ^2(3)\right )}{2 x^3+6 x^3 \log (x)+6 x^3 \log ^2(x)+2 x^3 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-2*x - 2*x*Log[x])*Log[E^(-8 - 2*x)*Log[3]^2] + (-2 - Log[x])*Log[E^(-8 - 2*x)*Log[3]^2]^2)/(2*x^3 + 6*x
^3*Log[x] + 6*x^3*Log[x]^2 + 2*x^3*Log[x]^3),x]

[Out]

-Defer[Int][Log[E^(-8 - 2*x)*Log[3]^2]/(x^2*(1 + Log[x])^2), x] + Defer[Int][Log[E^(-8 - 2*x)*Log[3]^2]^2/(x^3
*(-1 - Log[x])^3), x] + Defer[Int][(Log[x]*Log[E^(-8 - 2*x)*Log[3]^2]^2)/(x^3*(-1 - Log[x])^3), x]/2

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 28, normalized size = 1.00 \begin {gather*} \frac {\log ^2\left (e^{-2 (4+x)} \log ^2(3)\right )}{4 x^2 (1+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-2*x - 2*x*Log[x])*Log[E^(-8 - 2*x)*Log[3]^2] + (-2 - Log[x])*Log[E^(-8 - 2*x)*Log[3]^2]^2)/(2*x^3
+ 6*x^3*Log[x] + 6*x^3*Log[x]^2 + 2*x^3*Log[x]^3),x]

[Out]

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

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fricas [A]  time = 0.57, size = 50, normalized size = 1.79 \begin {gather*} \frac {4 \, x^{2} - 4 \, {\left (x + 4\right )} \log \left (\log \relax (3)^{2}\right ) + \log \left (\log \relax (3)^{2}\right )^{2} + 32 \, x + 64}{4 \, {\left (x^{2} \log \relax (x)^{2} + 2 \, x^{2} \log \relax (x) + x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x^2 - 4*(x + 4)*log(log(3)^2) + log(log(3)^2)^2 + 32*x + 64)/(x^2*log(x)^2 + 2*x^2*log(x) + x^2)

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giac [A]  time = 0.13, size = 46, normalized size = 1.64 \begin {gather*} \frac {x^{2} - 2 \, x \log \left (\log \relax (3)\right ) + \log \left (\log \relax (3)\right )^{2} + 8 \, x - 8 \, \log \left (\log \relax (3)\right ) + 16}{x^{2} \log \relax (x)^{2} + 2 \, x^{2} \log \relax (x) + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.25, size = 104, normalized size = 3.71

method	result	size
default	\(\frac {\left (-2 \ln \left (\ln \relax (3)^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )-4 x \right ) x +2 x^{2}+\frac {\left (\ln \left (\ln \relax (3)^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )^{2}}{2}-2 \left (\ln \left ({\mathrm e}^{x}\right )-x \right ) \left (\ln \left (\ln \relax (3)^{2} {\mathrm e}^{-8} {\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )+2 \left (\ln \left ({\mathrm e}^{x}\right )-x \right )^{2}}{2 x^{2} \left (\ln \relax (x )+1\right )^{2}}\)	\(104\)
risch	\(\frac {\ln \left ({\mathrm e}^{x}\right )^{2}}{x^{2} \left (\ln \relax (x )+1\right )^{2}}-\frac {\left (-16+i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-2 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+4 \ln \left (\ln \relax (3)\right )\right ) \ln \left ({\mathrm e}^{x}\right )}{2 x^{2} \left (\ln \relax (x )+1\right )^{2}}+\frac {256-128 \ln \left (\ln \relax (3)\right )+16 \ln \left (\ln \relax (3)\right )^{2}-32 i \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+64 i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+8 i \ln \left (\ln \relax (3)\right ) \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+8 i \ln \left (\ln \relax (3)\right ) \pi \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-16 i \ln \left (\ln \relax (3)\right ) \pi \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}-\pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{4} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+4 \pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{3} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-6 \pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{4}+4 \pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{5}-32 i \pi \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-\pi ^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{6}}{16 x^{2} \left (\ln \relax (x )+1\right )^{2}}\)	\(339\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((-2*ln(ln(3)^2/exp(4)^2/exp(x)^2)-4*x)*x+2*x^2+1/2*(ln(ln(3)^2/exp(4)^2/exp(x)^2)+2*ln(exp(x)))^2-2*(ln(e
xp(x))-x)*(ln(ln(3)^2/exp(4)^2/exp(x)^2)+2*ln(exp(x)))+2*(ln(exp(x))-x)^2)/x^2/(ln(x)+1)^2

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maxima [A]  time = 0.50, size = 45, normalized size = 1.61 \begin {gather*} \frac {x^{2} - 2 \, x {\left (\log \left (\log \relax (3)\right ) - 4\right )} + \log \left (\log \relax (3)\right )^{2} - 8 \, \log \left (\log \relax (3)\right ) + 16}{x^{2} \log \relax (x)^{2} + 2 \, x^{2} \log \relax (x) + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.72, size = 52, normalized size = 1.86 \begin {gather*} -\frac {{\ln \relax (x)}^2+2\,\ln \relax (x)}{{\left (\ln \relax (x)+1\right )}^2}-\frac {8\,\ln \left (\ln \relax (3)\right )-{\ln \left (\ln \relax (3)\right )}^2+x\,\left (2\,\ln \left (\ln \relax (3)\right )-8\right )-16}{x^2\,{\left (\ln \relax (x)+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(exp(-2*x)*exp(-8)*log(3)^2)^2*(log(x) + 2) + log(exp(-2*x)*exp(-8)*log(3)^2)*(2*x + 2*x*log(x)))/(6*
x^3*log(x) + 6*x^3*log(x)^2 + 2*x^3*log(x)^3 + 2*x^3),x)

[Out]

- (2*log(x) + log(x)^2)/(log(x) + 1)^2 - (8*log(log(3)) - log(log(3))^2 + x*(2*log(log(3)) - 8) - 16)/(x^2*(lo
g(x) + 1)^2)

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sympy [A]  time = 0.32, size = 49, normalized size = 1.75 \begin {gather*} \frac {x^{2} - 2 x \log {\left (\log {\relax (3 )} \right )} + 8 x - 8 \log {\left (\log {\relax (3 )} \right )} + \log {\left (\log {\relax (3 )} \right )}^{2} + 16}{x^{2} \log {\relax (x )}^{2} + 2 x^{2} \log {\relax (x )} + x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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