3.26.54 \(\int \frac {4+(8 x+x^2) \log ^2(x)}{4 x \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ 7+2 x+\frac {x^2}{8}-\frac {x+\frac {x}{\log (x)}}{x} \]

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Rubi [A]  time = 0.08, antiderivative size = 17, normalized size of antiderivative = 0.68, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {12, 6688, 2302, 30} \begin {gather*} \frac {x^2}{8}+2 x-\frac {1}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 + (8*x + x^2)*Log[x]^2)/(4*x*Log[x]^2),x]

[Out]

2*x + x^2/8 - Log[x]^(-1)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

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

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {4+\left (8 x+x^2\right ) \log ^2(x)}{x \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (8+x+\frac {4}{x \log ^2(x)}\right ) \, dx\\ &=2 x+\frac {x^2}{8}+\int \frac {1}{x \log ^2(x)} \, dx\\ &=2 x+\frac {x^2}{8}+\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=2 x+\frac {x^2}{8}-\frac {1}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 17, normalized size = 0.68 \begin {gather*} 2 x+\frac {x^2}{8}-\frac {1}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + (8*x + x^2)*Log[x]^2)/(4*x*Log[x]^2),x]

[Out]

2*x + x^2/8 - Log[x]^(-1)

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fricas [A]  time = 1.01, size = 18, normalized size = 0.72 \begin {gather*} \frac {{\left (x^{2} + 16 \, x\right )} \log \relax (x) - 8}{8 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^2+8*x)*log(x)^2+4)/x/log(x)^2,x, algorithm="fricas")

[Out]

1/8*((x^2 + 16*x)*log(x) - 8)/log(x)

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giac [A]  time = 0.26, size = 15, normalized size = 0.60 \begin {gather*} \frac {1}{8} \, x^{2} + 2 \, x - \frac {1}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^2+8*x)*log(x)^2+4)/x/log(x)^2,x, algorithm="giac")

[Out]

1/8*x^2 + 2*x - 1/log(x)

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maple [A]  time = 0.01, size = 16, normalized size = 0.64

method	result	size
default	\(\frac {x^{2}}{8}+2 x -\frac {1}{\ln \relax (x )}\)	\(16\)
risch	\(\frac {x^{2}}{8}+2 x -\frac {1}{\ln \relax (x )}\)	\(16\)
norman	\(\frac {-1+2 x \ln \relax (x )+\frac {x^{2} \ln \relax (x )}{8}}{\ln \relax (x )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((x^2+8*x)*ln(x)^2+4)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/8*x^2+2*x-1/ln(x)

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maxima [A]  time = 0.44, size = 15, normalized size = 0.60 \begin {gather*} \frac {1}{8} \, x^{2} + 2 \, x - \frac {1}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x^2+8*x)*log(x)^2+4)/x/log(x)^2,x, algorithm="maxima")

[Out]

1/8*x^2 + 2*x - 1/log(x)

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mupad [B]  time = 1.42, size = 15, normalized size = 0.60 \begin {gather*} 2\,x-\frac {1}{\ln \relax (x)}+\frac {x^2}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x)^2*(8*x + x^2))/4 + 1)/(x*log(x)^2),x)

[Out]

2*x - 1/log(x) + x^2/8

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sympy [A]  time = 0.08, size = 12, normalized size = 0.48 \begin {gather*} \frac {x^{2}}{8} + 2 x - \frac {1}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((x**2+8*x)*ln(x)**2+4)/x/ln(x)**2,x)

[Out]

x**2/8 + 2*x - 1/log(x)

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