3.73.19 $\int \frac{5+x+x^2\log (-\frac{3x}{4})-5\log (-\frac{3x}{4})\log (\log (-\frac{3x}{4}))}{x^2\log (-\frac{3x}{4})}dx$

Optimal. Leaf size=17 $$(5+x)(1+\frac{\log (\log (-\frac{3x}{4}))}{x})$$

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Rubi [A]  time = 0.36, antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 7, integrand size = 40, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.175, Rules used = {6742, 2353, 2309, 2178, 2302, 29, 2522} $$\begin{array}{c}x+\log (\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(5 + x + x^2*Log[(-3*x)/4] - 5*Log[(-3*x)/4]*Log[Log[(-3*x)/4]])/(x^2*Log[(-3*x)/4]),x]

[Out]

x + Log[Log[(-3*x)/4]] + (5*Log[Log[(-3*x)/4]])/x

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 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 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 2522

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{5+x+x^2\log (-\frac{3x}{4})}{x^2\log (-\frac{3x}{4})}-\frac{5\log (\log (-\frac{3x}{4}))}{x^2})dx\\ & =-(5\int \frac{\log (\log (-\frac{3x}{4}))}{x^2}dx)+\int \frac{5+x+x^2\log (-\frac{3x}{4})}{x^2\log (-\frac{3x}{4})}dx\\ & =\frac{5\log (\log (-\frac{3x}{4}))}{x}-5\int \frac{1}{x^2\log (-\frac{3x}{4})}dx+\int (1+\frac{5+x}{x^2\log (-\frac{3x}{4})})dx\\ & =x+\frac{5\log (\log (-\frac{3x}{4}))}{x}+\frac{15}{4}\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (-\frac{3x}{4}))+\int \frac{5+x}{x^2\log (-\frac{3x}{4})}dx\\ & =x+\frac{15}{4}\text{Ei}(-\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}+\int (\frac{5}{x^2\log (-\frac{3x}{4})}+\frac{1}{x\log (-\frac{3x}{4})})dx\\ & =x+\frac{15}{4}\text{Ei}(-\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}+5\int \frac{1}{x^2\log (-\frac{3x}{4})}dx+\int \frac{1}{x\log (-\frac{3x}{4})}dx\\ & =x+\frac{15}{4}\text{Ei}(-\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}-\frac{15}{4}\mathrm{Subst}(\int \frac{e^{-x}}{x}dx,x,\log (-\frac{3x}{4}))+\mathrm{Subst}(\int \frac{1}{x}dx,x,\log (-\frac{3x}{4}))\\ & =x+\log (\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.04, size = 21, normalized size = 1.24 $$\begin{array}{c}x+\log (\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(5 + x + x^2*Log[(-3*x)/4] - 5*Log[(-3*x)/4]*Log[Log[(-3*x)/4]])/(x^2*Log[(-3*x)/4]),x]

[Out]

x + Log[Log[(-3*x)/4]] + (5*Log[Log[(-3*x)/4]])/x

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fricas [A]  time = 0.58, size = 17, normalized size = 1.00 $$\begin{array}{c}\frac{x^2+(x+5)\log (\log (-\frac{3}{4}x))}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(-3/4*x)*log(log(-3/4*x))+x^2*log(-3/4*x)+5+x)/x^2/log(-3/4*x),x, algorithm="fricas")

[Out]

(x^2 + (x + 5)*log(log(-3/4*x)))/x

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giac [A]  time = 0.15, size = 22, normalized size = 1.29 $$\begin{array}{c}x+\frac{5\log (\log (-\frac{3}{4}x))}{x}+\log (-2\log (2)+\log (-3x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(-3/4*x)*log(log(-3/4*x))+x^2*log(-3/4*x)+5+x)/x^2/log(-3/4*x),x, algorithm="giac")

[Out]

x + 5*log(log(-3/4*x))/x + log(-2*log(2) + log(-3*x))

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maple [A]  time = 0.04, size = 18, normalized size = 1.06

method	result	size
risch	$\frac{5\ln (\ln (-\frac{3x}{4}))}{x}+x+\ln (\ln (-\frac{3x}{4}))$	$18$
norman	$\frac{x^2+5\ln (\ln (-\frac{3x}{4}))}{x}+\ln (\ln (-\frac{3x}{4}))$	$22$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(-3/4*x)*ln(ln(-3/4*x))+x^2*ln(-3/4*x)+5+x)/x^2/ln(-3/4*x),x,method=_RETURNVERBOSE)

[Out]

5*ln(ln(-3/4*x))/x+x+ln(ln(-3/4*x))

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maxima [A]  time = 0.37, size = 17, normalized size = 1.00 $$\begin{array}{c}x+\frac{5\log (\log (-\frac{3}{4}x))}{x}+\log (\log (-\frac{3}{4}x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(-3/4*x)*log(log(-3/4*x))+x^2*log(-3/4*x)+5+x)/x^2/log(-3/4*x),x, algorithm="maxima")

[Out]

x + 5*log(log(-3/4*x))/x + log(log(-3/4*x))

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mupad [B]  time = 4.44, size = 17, normalized size = 1.00 $$\begin{array}{c}x+\ln (\ln (-\frac{3x}{4}))+\frac{5\ln (\ln (-\frac{3x}{4}))}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 5*log(-(3*x)/4)*log(log(-(3*x)/4)) + x^2*log(-(3*x)/4) + 5)/(x^2*log(-(3*x)/4)),x)

[Out]

x + log(log(-(3*x)/4)) + (5*log(log(-(3*x)/4)))/x

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sympy [A]  time = 0.37, size = 24, normalized size = 1.41 $$\begin{array}{c}x+\log (\log (-\frac{3x}{4}))+\frac{5\log (\log (-\frac{3x}{4}))}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(-3/4*x)*ln(ln(-3/4*x))+x**2*ln(-3/4*x)+5+x)/x**2/ln(-3/4*x),x)

[Out]

x + log(log(-3*x/4)) + 5*log(log(-3*x/4))/x

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