3.66.93 \(\int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{9 x \log ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ 3+x^2+\frac {16 (-2+x)}{9 \log (x)} \]

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Rubi [A]  time = 0.27, antiderivative size = 21, normalized size of antiderivative = 1.31, number of steps used = 12, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {12, 6741, 6742, 2353, 2297, 2298, 2302, 30} \begin {gather*} x^2+\frac {16 x}{9 \log (x)}-\frac {32}{9 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 16*x + 16*x*Log[x] + 18*x^2*Log[x]^2)/(9*x*Log[x]^2),x]

[Out]

x^2 - 32/(9*Log[x]) + (16*x)/(9*Log[x])

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

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

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 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= 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} &=\frac {1}{9} \int \frac {32-16 x+16 x \log (x)+18 x^2 \log ^2(x)}{x \log ^2(x)} \, dx\\ &=\frac {1}{9} \int \frac {2 \left (16-8 x+8 x \log (x)+9 x^2 \log ^2(x)\right )}{x \log ^2(x)} \, dx\\ &=\frac {2}{9} \int \frac {16-8 x+8 x \log (x)+9 x^2 \log ^2(x)}{x \log ^2(x)} \, dx\\ &=\frac {2}{9} \int \left (9 x-\frac {8 (-2+x)}{x \log ^2(x)}+\frac {8}{\log (x)}\right ) \, dx\\ &=x^2-\frac {16}{9} \int \frac {-2+x}{x \log ^2(x)} \, dx+\frac {16}{9} \int \frac {1}{\log (x)} \, dx\\ &=x^2+\frac {16 \text {li}(x)}{9}-\frac {16}{9} \int \left (\frac {1}{\log ^2(x)}-\frac {2}{x \log ^2(x)}\right ) \, dx\\ &=x^2+\frac {16 \text {li}(x)}{9}-\frac {16}{9} \int \frac {1}{\log ^2(x)} \, dx+\frac {32}{9} \int \frac {1}{x \log ^2(x)} \, dx\\ &=x^2+\frac {16 x}{9 \log (x)}+\frac {16 \text {li}(x)}{9}-\frac {16}{9} \int \frac {1}{\log (x)} \, dx+\frac {32}{9} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=x^2-\frac {32}{9 \log (x)}+\frac {16 x}{9 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.31 \begin {gather*} x^2-\frac {32}{9 \log (x)}+\frac {16 x}{9 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 16*x + 16*x*Log[x] + 18*x^2*Log[x]^2)/(9*x*Log[x]^2),x]

[Out]

x^2 - 32/(9*Log[x]) + (16*x)/(9*Log[x])

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fricas [A]  time = 0.58, size = 18, normalized size = 1.12 \begin {gather*} \frac {9 \, x^{2} \log \relax (x) + 16 \, x - 32}{9 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(18*x^2*log(x)^2+16*x*log(x)-16*x+32)/x/log(x)^2,x, algorithm="fricas")

[Out]

1/9*(9*x^2*log(x) + 16*x - 32)/log(x)

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giac [A]  time = 2.20, size = 13, normalized size = 0.81 \begin {gather*} x^{2} + \frac {16 \, {\left (x - 2\right )}}{9 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(18*x^2*log(x)^2+16*x*log(x)-16*x+32)/x/log(x)^2,x, algorithm="giac")

[Out]

x^2 + 16/9*(x - 2)/log(x)

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maple [A]  time = 0.04, size = 14, normalized size = 0.88

method	result	size
risch	\(x^{2}+\frac {\frac {16 x}{9}-\frac {32}{9}}{\ln \relax (x )}\)	\(14\)
norman	\(\frac {-\frac {32}{9}+x^{2} \ln \relax (x )+\frac {16 x}{9}}{\ln \relax (x )}\)	\(17\)
default	\(x^{2}+\frac {16 x}{9 \ln \relax (x )}-\frac {32}{9 \ln \relax (x )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*(18*x^2*ln(x)^2+16*x*ln(x)-16*x+32)/x/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

x^2+16/9*(x-2)/ln(x)

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maxima [C]  time = 0.55, size = 23, normalized size = 1.44 \begin {gather*} x^{2} - \frac {32}{9 \, \log \relax (x)} + \frac {16}{9} \, {\rm Ei}\left (\log \relax (x)\right ) - \frac {16}{9} \, \Gamma \left (-1, -\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(18*x^2*log(x)^2+16*x*log(x)-16*x+32)/x/log(x)^2,x, algorithm="maxima")

[Out]

x^2 - 32/9/log(x) + 16/9*Ei(log(x)) - 16/9*gamma(-1, -log(x))

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mupad [B]  time = 4.15, size = 14, normalized size = 0.88 \begin {gather*} x^2+\frac {\frac {16\,x}{9}-\frac {32}{9}}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*log(x)^2 - (16*x)/9 + (16*x*log(x))/9 + 32/9)/(x*log(x)^2),x)

[Out]

x^2 + ((16*x)/9 - 32/9)/log(x)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.75 \begin {gather*} x^{2} + \frac {16 x - 32}{9 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*(18*x**2*ln(x)**2+16*x*ln(x)-16*x+32)/x/ln(x)**2,x)

[Out]

x**2 + (16*x - 32)/(9*log(x))

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