3.89.44 \(\int \frac {2+8 x+8 x^2+(-2-8 x-8 x^2) \log (x)+4 x \log ^2(x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=20 \[ \log (2)-\left (1-\frac {1+2 x}{\log (x)}\right )^2 \]

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Rubi [A]  time = 0.41, antiderivative size = 36, normalized size of antiderivative = 1.80, number of steps used = 23, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {6742, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178} \begin {gather*} -\frac {4 x^2}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {1}{\log ^2(x)}+\frac {4 x}{\log (x)}+\frac {2}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 8*x + 8*x^2 + (-2 - 8*x - 8*x^2)*Log[x] + 4*x*Log[x]^2)/(x*Log[x]^3),x]

[Out]

-Log[x]^(-2) - (4*x)/Log[x]^2 - (4*x^2)/Log[x]^2 + 2/Log[x] + (4*x)/Log[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 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 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)^2}{x \log ^3(x)}-\frac {2 (1+2 x)^2}{x \log ^2(x)}+\frac {4}{\log (x)}\right ) \, dx\\ &=2 \int \frac {(1+2 x)^2}{x \log ^3(x)} \, dx-2 \int \frac {(1+2 x)^2}{x \log ^2(x)} \, dx+4 \int \frac {1}{\log (x)} \, dx\\ &=4 \text {li}(x)+2 \int \left (\frac {4}{\log ^3(x)}+\frac {1}{x \log ^3(x)}+\frac {4 x}{\log ^3(x)}\right ) \, dx-2 \int \left (\frac {4}{\log ^2(x)}+\frac {1}{x \log ^2(x)}+\frac {4 x}{\log ^2(x)}\right ) \, dx\\ &=4 \text {li}(x)+2 \int \frac {1}{x \log ^3(x)} \, dx-2 \int \frac {1}{x \log ^2(x)} \, dx+8 \int \frac {1}{\log ^3(x)} \, dx+8 \int \frac {x}{\log ^3(x)} \, dx-8 \int \frac {1}{\log ^2(x)} \, dx-8 \int \frac {x}{\log ^2(x)} \, dx\\ &=-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {8 x}{\log (x)}+\frac {8 x^2}{\log (x)}+4 \text {li}(x)+2 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )-2 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )+4 \int \frac {1}{\log ^2(x)} \, dx+8 \int \frac {x}{\log ^2(x)} \, dx-8 \int \frac {1}{\log (x)} \, dx-16 \int \frac {x}{\log (x)} \, dx\\ &=-\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)}-4 \text {li}(x)+4 \int \frac {1}{\log (x)} \, dx+16 \int \frac {x}{\log (x)} \, dx-16 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-16 \text {Ei}(2 \log (x))-\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)}+16 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right )\\ &=-\frac {1}{\log ^2(x)}-\frac {4 x}{\log ^2(x)}-\frac {4 x^2}{\log ^2(x)}+\frac {2}{\log (x)}+\frac {4 x}{\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 20, normalized size = 1.00 \begin {gather*} -\frac {(1+2 x) (1+2 x-2 \log (x))}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 8*x + 8*x^2 + (-2 - 8*x - 8*x^2)*Log[x] + 4*x*Log[x]^2)/(x*Log[x]^3),x]

[Out]

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

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fricas [A]  time = 0.45, size = 25, normalized size = 1.25 \begin {gather*} -\frac {4 \, x^{2} - 2 \, {\left (2 \, x + 1\right )} \log \relax (x) + 4 \, x + 1}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+(-8*x^2-8*x-2)*log(x)+8*x^2+8*x+2)/x/log(x)^3,x, algorithm="fricas")

[Out]

-(4*x^2 - 2*(2*x + 1)*log(x) + 4*x + 1)/log(x)^2

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giac [A]  time = 0.14, size = 25, normalized size = 1.25 \begin {gather*} -\frac {4 \, x^{2} - 4 \, x \log \relax (x) + 4 \, x - 2 \, \log \relax (x) + 1}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+(-8*x^2-8*x-2)*log(x)+8*x^2+8*x+2)/x/log(x)^3,x, algorithm="giac")

[Out]

-(4*x^2 - 4*x*log(x) + 4*x - 2*log(x) + 1)/log(x)^2

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maple [A]  time = 0.03, size = 25, normalized size = 1.25

method	result	size
norman	\(\frac {-1-4 x -4 x^{2}+4 x \ln \relax (x )+2 \ln \relax (x )}{\ln \relax (x )^{2}}\)	\(25\)
risch	\(-\frac {4 x^{2}-4 x \ln \relax (x )+4 x -2 \ln \relax (x )+1}{\ln \relax (x )^{2}}\)	\(26\)
default	\(\frac {4 x}{\ln \relax (x )}-\frac {4 x^{2}}{\ln \relax (x )^{2}}+\frac {2}{\ln \relax (x )}-\frac {4 x}{\ln \relax (x )^{2}}-\frac {1}{\ln \relax (x )^{2}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(x)^2+(-8*x^2-8*x-2)*ln(x)+8*x^2+8*x+2)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

(-1-4*x-4*x^2+4*x*ln(x)+2*ln(x))/ln(x)^2

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maxima [C]  time = 0.41, size = 50, normalized size = 2.50 \begin {gather*} \frac {2}{\log \relax (x)} - \frac {1}{\log \relax (x)^{2}} + 4 \, {\rm Ei}\left (\log \relax (x)\right ) - 8 \, \Gamma \left (-1, -\log \relax (x)\right ) - 16 \, \Gamma \left (-1, -2 \, \log \relax (x)\right ) - 8 \, \Gamma \left (-2, -\log \relax (x)\right ) - 32 \, \Gamma \left (-2, -2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+(-8*x^2-8*x-2)*log(x)+8*x^2+8*x+2)/x/log(x)^3,x, algorithm="maxima")

[Out]

2/log(x) - 1/log(x)^2 + 4*Ei(log(x)) - 8*gamma(-1, -log(x)) - 16*gamma(-1, -2*log(x)) - 8*gamma(-2, -log(x)) -
32*gamma(-2, -2*log(x))

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mupad [B]  time = 5.12, size = 24, normalized size = 1.20 \begin {gather*} \frac {4\,x+2}{\ln \relax (x)}-\frac {{\left (2\,x+1\right )}^2}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 4*x*log(x)^2 - log(x)*(8*x + 8*x^2 + 2) + 8*x^2 + 2)/(x*log(x)^3),x)

[Out]

(4*x + 2)/log(x) - (2*x + 1)^2/log(x)^2

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sympy [A]  time = 0.10, size = 22, normalized size = 1.10 \begin {gather*} \frac {- 4 x^{2} - 4 x + \left (4 x + 2\right ) \log {\relax (x )} - 1}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(x)**2+(-8*x**2-8*x-2)*ln(x)+8*x**2+8*x+2)/x/ln(x)**3,x)

[Out]

(-4*x**2 - 4*x + (4*x + 2)*log(x) - 1)/log(x)**2

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