3.54.29 \(\int \frac {-1+(1+2 x+48 x^2) \log (x)-2 x \log ^2(x)-2 x \log ^3(x)+\log (x) \log (\frac {2 x}{\log (x)})}{\log (x)} \, dx\)

Optimal. Leaf size=29 \[ x \left (-2 x+16 x^2+x \left (3-\log ^2(x)\right )+\log \left (\frac {2 x}{\log (x)}\right )\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 11, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6742, 2298, 2304, 2305, 2549} \begin {gather*} 16 x^3+x^2-x^2 \log ^2(x)+x \log \left (\frac {2 x}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + (1 + 2*x + 48*x^2)*Log[x] - 2*x*Log[x]^2 - 2*x*Log[x]^3 + Log[x]*Log[(2*x)/Log[x]])/Log[x],x]

[Out]

x^2 + 16*x^3 - x^2*Log[x]^2 + x*Log[(2*x)/Log[x]]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 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]

Rule 2549

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[
u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {-1+\log (x)+2 x \log (x)+48 x^2 \log (x)-2 x \log ^2(x)-2 x \log ^3(x)}{\log (x)}+\log \left (\frac {2 x}{\log (x)}\right )\right ) \, dx\\ &=\int \frac {-1+\log (x)+2 x \log (x)+48 x^2 \log (x)-2 x \log ^2(x)-2 x \log ^3(x)}{\log (x)} \, dx+\int \log \left (\frac {2 x}{\log (x)}\right ) \, dx\\ &=x \log \left (\frac {2 x}{\log (x)}\right )-\int \left (1-\frac {1}{\log (x)}\right ) \, dx+\int \left (1+2 x+48 x^2-\frac {1}{\log (x)}-2 x \log (x)-2 x \log ^2(x)\right ) \, dx\\ &=x^2+16 x^3+x \log \left (\frac {2 x}{\log (x)}\right )-2 \int x \log (x) \, dx-2 \int x \log ^2(x) \, dx\\ &=\frac {3 x^2}{2}+16 x^3-x^2 \log (x)-x^2 \log ^2(x)+x \log \left (\frac {2 x}{\log (x)}\right )+2 \int x \log (x) \, dx\\ &=x^2+16 x^3-x^2 \log ^2(x)+x \log \left (\frac {2 x}{\log (x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.97 \begin {gather*} x^2+16 x^3-x^2 \log ^2(x)+x \log \left (\frac {2 x}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + (1 + 2*x + 48*x^2)*Log[x] - 2*x*Log[x]^2 - 2*x*Log[x]^3 + Log[x]*Log[(2*x)/Log[x]])/Log[x],x]

[Out]

x^2 + 16*x^3 - x^2*Log[x]^2 + x*Log[(2*x)/Log[x]]

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fricas [A]  time = 0.69, size = 28, normalized size = 0.97 \begin {gather*} -x^{2} \log \relax (x)^{2} + 16 \, x^{3} + x^{2} + x \log \left (\frac {2 \, x}{\log \relax (x)}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)*log(2*x/log(x))-2*x*log(x)^3-2*x*log(x)^2+(48*x^2+2*x+1)*log(x)-1)/log(x),x, algorithm="fric
as")

[Out]

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

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giac [A]  time = 1.01, size = 32, normalized size = 1.10 \begin {gather*} -x^{2} \log \relax (x)^{2} + 16 \, x^{3} + x^{2} + x \log \relax (2) + x \log \relax (x) - x \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)*log(2*x/log(x))-2*x*log(x)^3-2*x*log(x)^2+(48*x^2+2*x+1)*log(x)-1)/log(x),x, algorithm="giac
")

[Out]

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

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maple [A]  time = 0.09, size = 32, normalized size = 1.10

method	result	size
default	\(16 x^{3}+x^{2}-x^{2} \ln \relax (x )^{2}+x \ln \relax (2)+\ln \left (\frac {x}{\ln \relax (x )}\right ) x\)	\(32\)
risch	\(-x \ln \left (\ln \relax (x )\right )-x^{2} \ln \relax (x )^{2}+x \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) x}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} x}{2}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{2} x}{2}-\frac {i \pi \mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )^{3} x}{2}+16 x^{3}+x \ln \relax (2)+x^{2}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(x)*ln(2*x/ln(x))-2*x*ln(x)^3-2*x*ln(x)^2+(48*x^2+2*x+1)*ln(x)-1)/ln(x),x,method=_RETURNVERBOSE)

[Out]

16*x^3+x^2-x^2*ln(x)^2+x*ln(2)+ln(x/ln(x))*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {1}{2} \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + 16 \, x^{3} - x^{2} \log \relax (x) + \frac {3}{2} \, x^{2} + x {\left (\log \relax (2) - 1\right )} + x \log \relax (x) - x \log \left (\log \relax (x)\right ) + x - {\rm Ei}\left (\log \relax (x)\right ) + \int \frac {1}{\log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(x)*log(2*x/log(x))-2*x*log(x)^3-2*x*log(x)^2+(48*x^2+2*x+1)*log(x)-1)/log(x),x, algorithm="maxi
ma")

[Out]

-1/2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 16*x^3 - x^2*log(x) + 3/2*x^2 + x*(log(2) - 1) + x*log(x) - x*log(log(x
)) + x - Ei(log(x)) + integrate(1/log(x), x)

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mupad [B]  time = 3.65, size = 24, normalized size = 0.83 \begin {gather*} x\,\left (x+\ln \left (\frac {2\,x}{\ln \relax (x)}\right )-x\,{\ln \relax (x)}^2+16\,x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x*log(x)^2 + 2*x*log(x)^3 - log(x)*(2*x + 48*x^2 + 1) - log((2*x)/log(x))*log(x) + 1)/log(x),x)

[Out]

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

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sympy [A]  time = 0.33, size = 26, normalized size = 0.90 \begin {gather*} 16 x^{3} - x^{2} \log {\relax (x )}^{2} + x^{2} + x \log {\left (\frac {2 x}{\log {\relax (x )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(x)*ln(2*x/ln(x))-2*x*ln(x)**3-2*x*ln(x)**2+(48*x**2+2*x+1)*ln(x)-1)/ln(x),x)

[Out]

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

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