3.55.23 \(\int (-1+(-16 x^3 \log ^3(x)-16 x^3 \log ^4(x)) \log ^2(\frac {3 \log (2)}{e^2})) \, dx\)

Optimal. Leaf size=24 \[ 5-x-4 x^4 \log ^4(x) \log ^2\left (\frac {3 \log (2)}{e^2}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2305, 2304} \begin {gather*} -4 x^4 (2-\log (\log (8)))^2 \log ^4(x)-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-1 + (-16*x^3*Log[x]^3 - 16*x^3*Log[x]^4)*Log[(3*Log[2])/E^2]^2,x]

[Out]

-x - 4*x^4*Log[x]^4*(2 - Log[Log[8]])^2

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.83 \begin {gather*} -x-4 x^4 \log ^4(x) (-2+\log (\log (8)))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-1 + (-16*x^3*Log[x]^3 - 16*x^3*Log[x]^4)*Log[(3*Log[2])/E^2]^2,x]

[Out]

-x - 4*x^4*Log[x]^4*(-2 + Log[Log[8]])^2

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fricas [A]  time = 0.75, size = 22, normalized size = 0.92 \begin {gather*} -4 \, x^{4} \log \left (3 \, e^{\left (-2\right )} \log \relax (2)\right )^{2} \log \relax (x)^{4} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^3*log(x)^4-16*x^3*log(x)^3)*log(3*log(2)/exp(2))^2-1,x, algorithm="fricas")

[Out]

-4*x^4*log(3*e^(-2)*log(2))^2*log(x)^4 - x

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giac [A]  time = 0.21, size = 22, normalized size = 0.92 \begin {gather*} -4 \, x^{4} \log \left (3 \, e^{\left (-2\right )} \log \relax (2)\right )^{2} \log \relax (x)^{4} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^3*log(x)^4-16*x^3*log(x)^3)*log(3*log(2)/exp(2))^2-1,x, algorithm="giac")

[Out]

-4*x^4*log(3*e^(-2)*log(2))^2*log(x)^4 - x

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maple [A]  time = 0.06, size = 23, normalized size = 0.96

method	result	size
risch	\(-4 \left (\ln \relax (3)+\ln \left (\ln \relax (2)\right )-2\right )^{2} x^{4} \ln \relax (x )^{4}-x\)	\(23\)
default	\(-x -4 \ln \relax (x )^{4} x^{4} \ln \left (3 \ln \relax (2) {\mathrm e}^{-2}\right )^{2}\)	\(25\)
norman	\(\left (-4 \ln \relax (3)^{2}-8 \ln \relax (3) \ln \left (\ln \relax (2)\right )-4 \ln \left (\ln \relax (2)\right )^{2}+16 \ln \relax (3)+16 \ln \left (\ln \relax (2)\right )-16\right ) x^{4} \ln \relax (x )^{4}-x\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x^3*ln(x)^4-16*x^3*ln(x)^3)*ln(3*ln(2)/exp(2))^2-1,x,method=_RETURNVERBOSE)

[Out]

-4*(ln(3)+ln(ln(2))-2)^2*x^4*ln(x)^4-x

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maxima [B]  time = 0.40, size = 66, normalized size = 2.75 \begin {gather*} -\frac {1}{8} \, {\left ({\left (32 \, \log \relax (x)^{4} - 32 \, \log \relax (x)^{3} + 24 \, \log \relax (x)^{2} - 12 \, \log \relax (x) + 3\right )} x^{4} + {\left (32 \, \log \relax (x)^{3} - 24 \, \log \relax (x)^{2} + 12 \, \log \relax (x) - 3\right )} x^{4}\right )} \log \left (3 \, e^{\left (-2\right )} \log \relax (2)\right )^{2} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x^3*log(x)^4-16*x^3*log(x)^3)*log(3*log(2)/exp(2))^2-1,x, algorithm="maxima")

[Out]

-1/8*((32*log(x)^4 - 32*log(x)^3 + 24*log(x)^2 - 12*log(x) + 3)*x^4 + (32*log(x)^3 - 24*log(x)^2 + 12*log(x) -
3)*x^4)*log(3*e^(-2)*log(2))^2 - x

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mupad [B]  time = 3.41, size = 22, normalized size = 0.92 \begin {gather*} -4\,{\ln \left (3\,{\mathrm {e}}^{-2}\,\ln \relax (2)\right )}^2\,x^4\,{\ln \relax (x)}^4-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(- log(3*exp(-2)*log(2))^2*(16*x^3*log(x)^3 + 16*x^3*log(x)^4) - 1,x)

[Out]

- x - 4*x^4*log(3*exp(-2)*log(2))^2*log(x)^4

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sympy [B]  time = 0.17, size = 65, normalized size = 2.71 \begin {gather*} - x + \left (- 16 x^{4} + 16 x^{4} \log {\left (\log {\relax (2 )} \right )} - 4 x^{4} \log {\relax (3 )}^{2} - 4 x^{4} \log {\left (\log {\relax (2 )} \right )}^{2} - 8 x^{4} \log {\relax (3 )} \log {\left (\log {\relax (2 )} \right )} + 16 x^{4} \log {\relax (3 )}\right ) \log {\relax (x )}^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x**3*ln(x)**4-16*x**3*ln(x)**3)*ln(3*ln(2)/exp(2))**2-1,x)

[Out]

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

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