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

Optimal. Leaf size=20 \[ 2+2 \left (4+2 x+\frac {\log ^2(4)}{\log ^2(x)}+\log (x)\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6742, 43, 2302, 30} \begin {gather*} 4 x+\frac {2 \log ^2(4)}{\log ^2(x)}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

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 {2 (1+2 x)}{x}-\frac {4 \log ^2(4)}{x \log ^3(x)}\right ) \, dx\\ &=2 \int \frac {1+2 x}{x} \, dx-\left (4 \log ^2(4)\right ) \int \frac {1}{x \log ^3(x)} \, dx\\ &=2 \int \left (2+\frac {1}{x}\right ) \, dx-\left (4 \log ^2(4)\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )\\ &=4 x+\frac {2 \log ^2(4)}{\log ^2(x)}+2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.90 \begin {gather*} 4 x+\frac {2 \log ^2(4)}{\log ^2(x)}+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 24, normalized size = 1.20 \begin {gather*} \frac {2 \, {\left (2 \, x \log \relax (x)^{2} + \log \relax (x)^{3} + 4 \, \log \relax (2)^{2}\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*x*log(x)^2 + log(x)^3 + 4*log(2)^2)/log(x)^2

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giac [A]  time = 0.41, size = 18, normalized size = 0.90 \begin {gather*} 4 \, x + \frac {8 \, \log \relax (2)^{2}}{\log \relax (x)^{2}} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x + 8*log(2)^2/log(x)^2 + 2*log(x)

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maple [A]  time = 0.03, size = 19, normalized size = 0.95

method	result	size
default	\(4 x +2 \ln \relax (x )+\frac {8 \ln \relax (2)^{2}}{\ln \relax (x )^{2}}\)	\(19\)
risch	\(4 x +2 \ln \relax (x )+\frac {8 \ln \relax (2)^{2}}{\ln \relax (x )^{2}}\)	\(19\)
norman	\(\frac {2 \ln \relax (x )^{3}+8 \ln \relax (2)^{2}+4 x \ln \relax (x )^{2}}{\ln \relax (x )^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x+2*ln(x)+8*ln(2)^2/ln(x)^2

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maxima [A]  time = 0.61, size = 18, normalized size = 0.90 \begin {gather*} 4 \, x + \frac {8 \, \log \relax (2)^{2}}{\log \relax (x)^{2}} + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x + 8*log(2)^2/log(x)^2 + 2*log(x)

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mupad [B]  time = 0.67, size = 18, normalized size = 0.90 \begin {gather*} 4\,x+2\,\ln \relax (x)+\frac {8\,{\ln \relax (2)}^2}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x + 2*log(x) + (8*log(2)^2)/log(x)^2

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sympy [A]  time = 0.10, size = 19, normalized size = 0.95 \begin {gather*} 4 x + 2 \log {\relax (x )} + \frac {8 \log {\relax (2 )}^{2}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x + 2*log(x) + 8*log(2)**2/log(x)**2

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