3.28.99 \(\int \frac {-18+36 x-18 x^2+(-18 x+18 x^2) \log (x)+2 x^2 \log ^3(x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=17 \[ -2+x^2+\frac {(-3+3 x)^2}{\log ^2(x)} \]

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Rubi [B]  time = 0.26, antiderivative size = 54, normalized size of antiderivative = 3.18, number of steps used = 21, number of rules used = 12, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6688, 12, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2320, 2330} \begin {gather*} x^2+\frac {9 x^2}{\log ^2(x)}+\frac {18 x^2}{\log (x)}-\frac {18 x}{\log ^2(x)}+\frac {9}{\log ^2(x)}+\frac {18 (1-x) x}{\log (x)}-\frac {18 x}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-18 + 36*x - 18*x^2 + (-18*x + 18*x^2)*Log[x] + 2*x^2*Log[x]^3)/(x*Log[x]^3),x]

[Out]

x^2 + 9/Log[x]^2 - (18*x)/Log[x]^2 + (9*x^2)/Log[x]^2 - (18*x)/Log[x] + (18*(1 - x)*x)/Log[x] + (18*x^2)/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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 2306

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log
[c*x^n])^(p + 1))/(b*d*n*(p + 1)), x] - Dist[(m + 1)/(b*n*(p + 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] && LtQ[p, -1]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2320

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(x*(d + e*x)^q*(a
+ b*Log[c*x^n])^(p + 1))/(b*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), Int[(d + e*x)^q*(a + b*Log[c*x^n])^
(p + 1), x], x] + Dist[(d*q)/(b*n*(p + 1)), Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^(p + 1), x], x]) /; FreeQ
[{a, b, c, d, e, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 2330

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

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 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} &=\int 2 \left (x-\frac {9 (-1+x)^2}{x \log ^3(x)}+\frac {9 (-1+x)}{\log ^2(x)}\right ) \, dx\\ &=2 \int \left (x-\frac {9 (-1+x)^2}{x \log ^3(x)}+\frac {9 (-1+x)}{\log ^2(x)}\right ) \, dx\\ &=x^2-18 \int \frac {(-1+x)^2}{x \log ^3(x)} \, dx+18 \int \frac {-1+x}{\log ^2(x)} \, dx\\ &=x^2+\frac {18 (1-x) x}{\log (x)}-18 \int \left (-\frac {2}{\log ^3(x)}+\frac {1}{x \log ^3(x)}+\frac {x}{\log ^3(x)}\right ) \, dx+18 \int \frac {1}{\log (x)} \, dx+36 \int \frac {-1+x}{\log (x)} \, dx\\ &=x^2+\frac {18 (1-x) x}{\log (x)}+18 \text {li}(x)-18 \int \frac {1}{x \log ^3(x)} \, dx-18 \int \frac {x}{\log ^3(x)} \, dx+36 \int \left (-\frac {1}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx+36 \int \frac {1}{\log ^3(x)} \, dx\\ &=x^2-\frac {18 x}{\log ^2(x)}+\frac {9 x^2}{\log ^2(x)}+\frac {18 (1-x) x}{\log (x)}+18 \text {li}(x)+18 \int \frac {1}{\log ^2(x)} \, dx-18 \int \frac {x}{\log ^2(x)} \, dx-18 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )-36 \int \frac {1}{\log (x)} \, dx+36 \int \frac {x}{\log (x)} \, dx\\ &=x^2+\frac {9}{\log ^2(x)}-\frac {18 x}{\log ^2(x)}+\frac {9 x^2}{\log ^2(x)}-\frac {18 x}{\log (x)}+\frac {18 (1-x) x}{\log (x)}+\frac {18 x^2}{\log (x)}-18 \text {li}(x)+18 \int \frac {1}{\log (x)} \, dx-36 \int \frac {x}{\log (x)} \, dx+36 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=x^2+36 \text {Ei}(2 \log (x))+\frac {9}{\log ^2(x)}-\frac {18 x}{\log ^2(x)}+\frac {9 x^2}{\log ^2(x)}-\frac {18 x}{\log (x)}+\frac {18 (1-x) x}{\log (x)}+\frac {18 x^2}{\log (x)}-36 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=x^2+\frac {9}{\log ^2(x)}-\frac {18 x}{\log ^2(x)}+\frac {9 x^2}{\log ^2(x)}-\frac {18 x}{\log (x)}+\frac {18 (1-x) x}{\log (x)}+\frac {18 x^2}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.53 \begin {gather*} x^2+\frac {9}{\log ^2(x)}-\frac {18 x}{\log ^2(x)}+\frac {9 x^2}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-18 + 36*x - 18*x^2 + (-18*x + 18*x^2)*Log[x] + 2*x^2*Log[x]^3)/(x*Log[x]^3),x]

[Out]

x^2 + 9/Log[x]^2 - (18*x)/Log[x]^2 + (9*x^2)/Log[x]^2

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fricas [A]  time = 0.59, size = 23, normalized size = 1.35 \begin {gather*} \frac {x^{2} \log \relax (x)^{2} + 9 \, x^{2} - 18 \, x + 9}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^3+(18*x^2-18*x)*log(x)-18*x^2+36*x-18)/x/log(x)^3,x, algorithm="fricas")

[Out]

(x^2*log(x)^2 + 9*x^2 - 18*x + 9)/log(x)^2

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giac [A]  time = 0.24, size = 18, normalized size = 1.06 \begin {gather*} x^{2} + \frac {9 \, {\left (x^{2} - 2 \, x + 1\right )}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^3+(18*x^2-18*x)*log(x)-18*x^2+36*x-18)/x/log(x)^3,x, algorithm="giac")

[Out]

x^2 + 9*(x^2 - 2*x + 1)/log(x)^2

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maple [A]  time = 0.03, size = 19, normalized size = 1.12

method	result	size
risch	\(x^{2}+\frac {9 x^{2}-18 x +9}{\ln \relax (x )^{2}}\)	\(19\)
norman	\(\frac {9+x^{2} \ln \relax (x )^{2}-18 x +9 x^{2}}{\ln \relax (x )^{2}}\)	\(24\)
default	\(x^{2}+\frac {9 x^{2}}{\ln \relax (x )^{2}}-\frac {18 x}{\ln \relax (x )^{2}}+\frac {9}{\ln \relax (x )^{2}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2*ln(x)^3+(18*x^2-18*x)*ln(x)-18*x^2+36*x-18)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

x^2+9*(x^2-2*x+1)/ln(x)^2

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maxima [C]  time = 0.45, size = 42, normalized size = 2.47 \begin {gather*} x^{2} + \frac {9}{\log \relax (x)^{2}} - 18 \, \Gamma \left (-1, -\log \relax (x)\right ) + 36 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 36 \, \Gamma \left (-2, -\log \relax (x)\right ) + 72 \, \Gamma \left (-2, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2*log(x)^3+(18*x^2-18*x)*log(x)-18*x^2+36*x-18)/x/log(x)^3,x, algorithm="maxima")

[Out]

x^2 + 9/log(x)^2 - 18*gamma(-1, -log(x)) + 36*gamma(-1, -2*log(x)) - 36*gamma(-2, -log(x)) + 72*gamma(-2, -2*l
og(x))

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mupad [B]  time = 1.94, size = 15, normalized size = 0.88 \begin {gather*} x^2+\frac {9\,{\left (x-1\right )}^2}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(18*x - 18*x^2) - 2*x^2*log(x)^3 - 36*x + 18*x^2 + 18)/(x*log(x)^3),x)

[Out]

x^2 + (9*(x - 1)^2)/log(x)^2

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sympy [A]  time = 0.09, size = 17, normalized size = 1.00 \begin {gather*} x^{2} + \frac {9 x^{2} - 18 x + 9}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2*ln(x)**3+(18*x**2-18*x)*ln(x)-18*x**2+36*x-18)/x/ln(x)**3,x)

[Out]

x**2 + (9*x**2 - 18*x + 9)/log(x)**2

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