3.62.20 \(\int \frac {8-8 e^3-10 x+5 x \log (x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=15 \[ \frac {-4+x+4 \left (e^3+x\right )}{\log ^2(x)} \]

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Rubi [A]  time = 0.25, antiderivative size = 21, normalized size of antiderivative = 1.40, number of steps used = 12, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {6741, 6742, 2353, 2297, 2298, 2302, 30} \begin {gather*} \frac {5 x}{\log ^2(x)}-\frac {4 \left (1-e^3\right )}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(8 - 8*E^3 - 10*x + 5*x*Log[x])/(x*Log[x]^3),x]

[Out]

(-4*(1 - E^3))/Log[x]^2 + (5*x)/Log[x]^2

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} &=\int \frac {8 \left (1-e^3\right )-10 x+5 x \log (x)}{x \log ^3(x)} \, dx\\ &=\int \left (-\frac {2 \left (-4+4 e^3+5 x\right )}{x \log ^3(x)}+\frac {5}{\log ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {-4+4 e^3+5 x}{x \log ^3(x)} \, dx\right )+5 \int \frac {1}{\log ^2(x)} \, dx\\ &=-\frac {5 x}{\log (x)}-2 \int \left (\frac {5}{\log ^3(x)}+\frac {4 \left (-1+e^3\right )}{x \log ^3(x)}\right ) \, dx+5 \int \frac {1}{\log (x)} \, dx\\ &=-\frac {5 x}{\log (x)}+5 \text {li}(x)-10 \int \frac {1}{\log ^3(x)} \, dx+\left (8 \left (1-e^3\right )\right ) \int \frac {1}{x \log ^3(x)} \, dx\\ &=\frac {5 x}{\log ^2(x)}-\frac {5 x}{\log (x)}+5 \text {li}(x)-5 \int \frac {1}{\log ^2(x)} \, dx+\left (8 \left (1-e^3\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right )\\ &=-\frac {4 \left (1-e^3\right )}{\log ^2(x)}+\frac {5 x}{\log ^2(x)}+5 \text {li}(x)-5 \int \frac {1}{\log (x)} \, dx\\ &=-\frac {4 \left (1-e^3\right )}{\log ^2(x)}+\frac {5 x}{\log ^2(x)}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.03, size = 44, normalized size = 2.93 \begin {gather*} -\frac {4}{\log ^2(x)}+\frac {4 e^3}{\log ^2(x)}-\frac {5 x}{\log (x)}-5 \left (\text {Ei}(\log (x))-\frac {x (1+\log (x))}{\log ^2(x)}\right )+5 \text {li}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(8 - 8*E^3 - 10*x + 5*x*Log[x])/(x*Log[x]^3),x]

[Out]

-4/Log[x]^2 + (4*E^3)/Log[x]^2 - (5*x)/Log[x] - 5*(ExpIntegralEi[Log[x]] - (x*(1 + Log[x]))/Log[x]^2) + 5*LogI
ntegral[x]

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fricas [A]  time = 1.31, size = 14, normalized size = 0.93 \begin {gather*} \frac {5 \, x + 4 \, e^{3} - 4}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(x)-8*exp(3)-10*x+8)/x/log(x)^3,x, algorithm="fricas")

[Out]

(5*x + 4*e^3 - 4)/log(x)^2

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giac [A]  time = 0.17, size = 14, normalized size = 0.93 \begin {gather*} \frac {5 \, x + 4 \, e^{3} - 4}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(x)-8*exp(3)-10*x+8)/x/log(x)^3,x, algorithm="giac")

[Out]

(5*x + 4*e^3 - 4)/log(x)^2

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maple [A]  time = 0.04, size = 15, normalized size = 1.00

method	result	size
norman	\(\frac {4 \,{\mathrm e}^{3}+5 x -4}{\ln \relax (x )^{2}}\)	\(15\)
risch	\(\frac {4 \,{\mathrm e}^{3}+5 x -4}{\ln \relax (x )^{2}}\)	\(15\)
default	\(\frac {4 \,{\mathrm e}^{3}}{\ln \relax (x )^{2}}+\frac {5 x}{\ln \relax (x )^{2}}-\frac {4}{\ln \relax (x )^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*ln(x)-8*exp(3)-10*x+8)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

(4*exp(3)+5*x-4)/ln(x)^2

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maxima [C]  time = 0.36, size = 31, normalized size = 2.07 \begin {gather*} \frac {4 \, e^{3}}{\log \relax (x)^{2}} - \frac {4}{\log \relax (x)^{2}} + 5 \, \Gamma \left (-1, -\log \relax (x)\right ) + 10 \, \Gamma \left (-2, -\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(x)-8*exp(3)-10*x+8)/x/log(x)^3,x, algorithm="maxima")

[Out]

4*e^3/log(x)^2 - 4/log(x)^2 + 5*gamma(-1, -log(x)) + 10*gamma(-2, -log(x))

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mupad [B]  time = 4.27, size = 14, normalized size = 0.93 \begin {gather*} \frac {5\,x+4\,{\mathrm {e}}^3-4}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + 8*exp(3) - 5*x*log(x) - 8)/(x*log(x)^3),x)

[Out]

(5*x + 4*exp(3) - 4)/log(x)^2

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sympy [A]  time = 0.09, size = 14, normalized size = 0.93 \begin {gather*} \frac {5 x - 4 + 4 e^{3}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*ln(x)-8*exp(3)-10*x+8)/x/ln(x)**3,x)

[Out]

(5*x - 4 + 4*exp(3))/log(x)**2

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