3.39.24 \(\int \frac {3-x-x \log (x)+(3+x-5 x^2+(-2+x-10 x^2) \log (x)-\log (5 x^2)) \log (\log (4))}{x} \, dx\)

Optimal. Leaf size=29 \[ \log (x) \left (3-x-\left (-3-x+5 x^2+\log \left (5 x^2\right )\right ) \log (\log (4))\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 56, normalized size of antiderivative = 1.93, number of steps used = 12, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {14, 2301, 2357, 2295, 2304} \begin {gather*} -\frac {1}{4} \log (\log (4)) \log ^2\left (5 x^2\right )-5 x^2 \log (\log (4)) \log (x)-\log (\log (4)) \log ^2(x)-x (1-\log (\log (4))) \log (x)+3 (1+\log (\log (4))) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - x - x*Log[x] + (3 + x - 5*x^2 + (-2 + x - 10*x^2)*Log[x] - Log[5*x^2])*Log[Log[4]])/x,x]

[Out]

-(x*Log[x]*(1 - Log[Log[4]])) - 5*x^2*Log[x]*Log[Log[4]] - Log[x]^2*Log[Log[4]] - (Log[5*x^2]^2*Log[Log[4]])/4
+ 3*Log[x]*(1 + Log[Log[4]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {\log \left (5 x^2\right ) \log (\log (4))}{x}+\frac {-x (1-\log (\log (4)))-x \log (x) (1-\log (\log (4)))-5 x^2 \log (\log (4))-2 \log (x) \log (\log (4))-10 x^2 \log (x) \log (\log (4))+3 (1+\log (\log (4)))}{x}\right ) \, dx\\ &=-\left (\log (\log (4)) \int \frac {\log \left (5 x^2\right )}{x} \, dx\right )+\int \frac {-x (1-\log (\log (4)))-x \log (x) (1-\log (\log (4)))-5 x^2 \log (\log (4))-2 \log (x) \log (\log (4))-10 x^2 \log (x) \log (\log (4))+3 (1+\log (\log (4)))}{x} \, dx\\ &=-\frac {1}{4} \log ^2\left (5 x^2\right ) \log (\log (4))+\int \left (\frac {\log (x) \left (-x (1-\log (\log (4)))-2 \log (\log (4))-10 x^2 \log (\log (4))\right )}{x}+\frac {-x (1-\log (\log (4)))-5 x^2 \log (\log (4))+3 (1+\log (\log (4)))}{x}\right ) \, dx\\ &=-\frac {1}{4} \log ^2\left (5 x^2\right ) \log (\log (4))+\int \frac {\log (x) \left (-x (1-\log (\log (4)))-2 \log (\log (4))-10 x^2 \log (\log (4))\right )}{x} \, dx+\int \frac {-x (1-\log (\log (4)))-5 x^2 \log (\log (4))+3 (1+\log (\log (4)))}{x} \, dx\\ &=-\frac {1}{4} \log ^2\left (5 x^2\right ) \log (\log (4))+\int \left (\log (x) (-1+\log (\log (4)))-\frac {2 \log (x) \log (\log (4))}{x}-10 x \log (x) \log (\log (4))\right ) \, dx+\int \left (-1+\log (\log (4))-5 x \log (\log (4))+\frac {3 (1+\log (\log (4)))}{x}\right ) \, dx\\ &=-x (1-\log (\log (4)))-\frac {5}{2} x^2 \log (\log (4))-\frac {1}{4} \log ^2\left (5 x^2\right ) \log (\log (4))+3 \log (x) (1+\log (\log (4)))+(-1+\log (\log (4))) \int \log (x) \, dx-(2 \log (\log (4))) \int \frac {\log (x)}{x} \, dx-(10 \log (\log (4))) \int x \log (x) \, dx\\ &=-x \log (x) (1-\log (\log (4)))-5 x^2 \log (x) \log (\log (4))-\log ^2(x) \log (\log (4))-\frac {1}{4} \log ^2\left (5 x^2\right ) \log (\log (4))+3 \log (x) (1+\log (\log (4)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 58, normalized size = 2.00 \begin {gather*} 3 \log (x)-x \log (x)+3 \log (x) \log (\log (4))+x \log (x) \log (\log (4))-5 x^2 \log (x) \log (\log (4))-\log ^2(x) \log (\log (4))-\frac {1}{4} \log ^2\left (5 x^2\right ) \log (\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - x - x*Log[x] + (3 + x - 5*x^2 + (-2 + x - 10*x^2)*Log[x] - Log[5*x^2])*Log[Log[4]])/x,x]

[Out]

3*Log[x] - x*Log[x] + 3*Log[x]*Log[Log[4]] + x*Log[x]*Log[Log[4]] - 5*x^2*Log[x]*Log[Log[4]] - Log[x]^2*Log[Lo
g[4]] - (Log[5*x^2]^2*Log[Log[4]])/4

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fricas [A]  time = 0.59, size = 37, normalized size = 1.28 \begin {gather*} -{\left (x - 3\right )} \log \relax (x) - {\left ({\left (5 \, x^{2} - x + \log \relax (5) - 3\right )} \log \relax (x) + 2 \, \log \relax (x)^{2}\right )} \log \left (2 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5*x^2)+(-10*x^2+x-2)*log(x)-5*x^2+x+3)*log(2*log(2))-x*log(x)+3-x)/x,x, algorithm="fricas")

[Out]

-(x - 3)*log(x) - ((5*x^2 - x + log(5) - 3)*log(x) + 2*log(x)^2)*log(2*log(2))

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giac [B]  time = 0.19, size = 65, normalized size = 2.24 \begin {gather*} -2 \, {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} \log \relax (x)^{2} - {\left (5 \, x^{2} {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} - x {\left (\log \relax (2) + \log \left (\log \relax (2)\right ) - 1\right )}\right )} \log \relax (x) - {\left (\log \relax (5) \log \relax (2) + \log \relax (5) \log \left (\log \relax (2)\right ) - 3 \, \log \relax (2) - 3 \, \log \left (\log \relax (2)\right ) - 3\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5*x^2)+(-10*x^2+x-2)*log(x)-5*x^2+x+3)*log(2*log(2))-x*log(x)+3-x)/x,x, algorithm="giac")

[Out]

-2*(log(2) + log(log(2)))*log(x)^2 - (5*x^2*(log(2) + log(log(2))) - x*(log(2) + log(log(2)) - 1))*log(x) - (l
og(5)*log(2) + log(5)*log(log(2)) - 3*log(2) - 3*log(log(2)) - 3)*log(x)

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maple [B]  time = 0.11, size = 66, normalized size = 2.28

method	result	size
norman	\(\left (\frac {3}{2}+\frac {3 \ln \relax (2)}{2}+\frac {3 \ln \left (\ln \relax (2)\right )}{2}\right ) \ln \left (5 x^{2}\right )+\left (-5 \ln \relax (2)-5 \ln \left (\ln \relax (2)\right )\right ) x^{2} \ln \relax (x )+\left (-\ln \relax (2)-\ln \left (\ln \relax (2)\right )\right ) \ln \relax (x ) \ln \left (5 x^{2}\right )+\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )-1\right ) x \ln \relax (x )\)	\(66\)
default	\(-5 x^{2} \ln \relax (x ) \ln \left (\ln \relax (2)\right )+x \ln \relax (x ) \ln \left (\ln \relax (2)\right )-\ln \relax (x )^{2} \ln \left (\ln \relax (2)\right )-x \ln \relax (x )+3 \ln \relax (2) \ln \relax (x )-\ln \relax (2) \ln \relax (5) \ln \relax (x )-\frac {\ln \relax (2) \ln \left (x^{2}\right )^{2}}{4}-5 x^{2} \ln \relax (2) \ln \relax (x )+x \ln \relax (2) \ln \relax (x )-\ln \relax (2) \ln \relax (x )^{2}+3 \ln \relax (x )-\ln \left (\ln \relax (2)\right ) \ln \relax (5) \ln \relax (x )-\frac {\ln \left (\ln \relax (2)\right ) \ln \left (x^{2}\right )^{2}}{4}+3 \ln \relax (x ) \ln \left (\ln \relax (2)\right )\)	\(111\)
risch	\(-2 \ln \relax (x )^{2} \ln \left (\ln \relax (2)\right )-2 \ln \relax (2) \ln \relax (x )^{2}+\left (-5 x^{2} \ln \left (\ln \relax (2)\right )-5 x^{2} \ln \relax (2)+x \ln \left (\ln \relax (2)\right )+x \ln \relax (2)-x \right ) \ln \relax (x )-\frac {\left (-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} \ln \left (\ln \relax (2)\right )-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3} \ln \relax (2)+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} \ln \left (\ln \relax (2)\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} \ln \relax (2)-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) \ln \left (\ln \relax (2)\right )-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) \ln \relax (2)-6+2 \ln \left (\ln \relax (2)\right ) \ln \relax (5)+2 \ln \relax (2) \ln \relax (5)-6 \ln \left (\ln \relax (2)\right )-6 \ln \relax (2)\right ) \ln \relax (x )}{2}\)	\(191\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-ln(5*x^2)+(-10*x^2+x-2)*ln(x)-5*x^2+x+3)*ln(2*ln(2))-x*ln(x)+3-x)/x,x,method=_RETURNVERBOSE)

[Out]

(3/2+3/2*ln(2)+3/2*ln(ln(2)))*ln(5*x^2)+(-5*ln(2)-5*ln(ln(2)))*x^2*ln(x)+(-ln(2)-ln(ln(2)))*ln(x)*ln(5*x^2)+(l
n(2)+ln(ln(2))-1)*x*ln(x)

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maxima [B]  time = 0.35, size = 96, normalized size = 3.31 \begin {gather*} -\frac {5}{2} \, x^{2} \log \left (2 \, \log \relax (2)\right ) - \frac {1}{4} \, \log \left (5 \, x^{2}\right )^{2} \log \left (2 \, \log \relax (2)\right ) - \log \relax (x)^{2} \log \left (2 \, \log \relax (2)\right ) - x \log \relax (x) - \frac {5}{2} \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \left (2 \, \log \relax (2)\right ) + {\left (x \log \relax (x) - x\right )} \log \left (2 \, \log \relax (2)\right ) + x \log \left (2 \, \log \relax (2)\right ) + 3 \, \log \relax (x) \log \left (2 \, \log \relax (2)\right ) + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5*x^2)+(-10*x^2+x-2)*log(x)-5*x^2+x+3)*log(2*log(2))-x*log(x)+3-x)/x,x, algorithm="maxima")

[Out]

-5/2*x^2*log(2*log(2)) - 1/4*log(5*x^2)^2*log(2*log(2)) - log(x)^2*log(2*log(2)) - x*log(x) - 5/2*(2*x^2*log(x
) - x^2)*log(2*log(2)) + (x*log(x) - x)*log(2*log(2)) + x*log(2*log(2)) + 3*log(x)*log(2*log(2)) + 3*log(x)

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mupad [B]  time = 2.50, size = 42, normalized size = 1.45 \begin {gather*} -\ln \relax (x)\,\left (x-3\,\ln \left (2\,\ln \relax (2)\right )+\ln \left (2\,\ln \relax (2)\right )\,\ln \left (5\,x^2\right )+5\,x^2\,\ln \left (2\,\ln \relax (2)\right )-x\,\ln \left (\ln \relax (4)\right )-3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + log(2*log(2))*(log(5*x^2) - x + log(x)*(10*x^2 - x + 2) + 5*x^2 - 3) + x*log(x) - 3)/x,x)

[Out]

-log(x)*(x - 3*log(2*log(2)) + log(2*log(2))*log(5*x^2) + 5*x^2*log(2*log(2)) - x*log(log(4)) - 3)

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sympy [B]  time = 0.37, size = 85, normalized size = 2.93 \begin {gather*} \left (- 5 x^{2} \log {\relax (2 )} - 5 x^{2} \log {\left (\log {\relax (2 )} \right )} - x + x \log {\left (\log {\relax (2 )} \right )} + x \log {\relax (2 )}\right ) \log {\relax (x )} + \left (- 2 \log {\relax (2 )} - 2 \log {\left (\log {\relax (2 )} \right )}\right ) \log {\relax (x )}^{2} + \left (- \log {\relax (2 )} \log {\relax (5 )} + 3 \log {\left (\log {\relax (2 )} \right )} - \log {\relax (5 )} \log {\left (\log {\relax (2 )} \right )} + 3 \log {\relax (2 )} + 3\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-ln(5*x**2)+(-10*x**2+x-2)*ln(x)-5*x**2+x+3)*ln(2*ln(2))-x*ln(x)+3-x)/x,x)

[Out]

(-5*x**2*log(2) - 5*x**2*log(log(2)) - x + x*log(log(2)) + x*log(2))*log(x) + (-2*log(2) - 2*log(log(2)))*log(
x)**2 + (-log(2)*log(5) + 3*log(log(2)) - log(5)*log(log(2)) + 3*log(2) + 3)*log(x)

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