\(\int \frac {5+x+x^2 \log (-\frac {3 x}{4})-5 \log (-\frac {3 x}{4}) \log (\log (-\frac {3 x}{4}))}{x^2 \log (-\frac {3 x}{4})} \, dx\) [7219]

Optimal result

Integrand size = 40, antiderivative size = 17 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=(5+x) \left (1+\frac {\log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x}\right ) \]

[Out]

(ln(ln(-3/4*x))/x+1)*(5+x)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6874, 2395, 2346, 2209, 2339, 29, 2602} \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=x+\log \left (\log \left (-\frac {3 x}{4}\right )\right )+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x} \]

[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 2209

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

Rule 2339

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 2346

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 2395

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 2602

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 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )}{x^2 \log \left (-\frac {3 x}{4}\right )}-\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2}\right ) \, dx \\ & = -\left (5 \int \frac {\log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2} \, dx\right )+\int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx \\ & = \frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x}-5 \int \frac {1}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx+\int \left (1+\frac {5+x}{x^2 \log \left (-\frac {3 x}{4}\right )}\right ) \, dx \\ & = x+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x}+\frac {15}{4} \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (-\frac {3 x}{4}\right )\right )+\int \frac {5+x}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx \\ & = x+\frac {15}{4} \operatorname {ExpIntegralEi}\left (-\log \left (-\frac {3 x}{4}\right )\right )+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x}+\int \left (\frac {5}{x^2 \log \left (-\frac {3 x}{4}\right )}+\frac {1}{x \log \left (-\frac {3 x}{4}\right )}\right ) \, dx \\ & = x+\frac {15}{4} \operatorname {ExpIntegralEi}\left (-\log \left (-\frac {3 x}{4}\right )\right )+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x}+5 \int \frac {1}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx+\int \frac {1}{x \log \left (-\frac {3 x}{4}\right )} \, dx \\ & = x+\frac {15}{4} \operatorname {ExpIntegralEi}\left (-\log \left (-\frac {3 x}{4}\right )\right )+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x}-\frac {15}{4} \text {Subst}\left (\int \frac {e^{-x}}{x} \, dx,x,\log \left (-\frac {3 x}{4}\right )\right )+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (-\frac {3 x}{4}\right )\right ) \\ & = x+\log \left (\log \left (-\frac {3 x}{4}\right )\right )+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=x+\log \left (\log \left (-\frac {3 x}{4}\right )\right )+\frac {5 \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x} \]

[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

Maple [A] (verified)

Time = 1.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06

method	result	size
risch	\(\frac {5 \ln \left (\ln \left (-\frac {3 x}{4}\right )\right )}{x}+x +\ln \left (\ln \left (-\frac {3 x}{4}\right )\right )\)	\(18\)
norman	\(\frac {x^{2}+x \ln \left (\ln \left (-\frac {3 x}{4}\right )\right )+5 \ln \left (\ln \left (-\frac {3 x}{4}\right )\right )}{x}\)	\(23\)
parallelrisch	\(\frac {x^{2}+x \ln \left (\ln \left (-\frac {3 x}{4}\right )\right )+5 \ln \left (\ln \left (-\frac {3 x}{4}\right )\right )}{x}\)	\(23\)
parts	\(x +\frac {15 \,\operatorname {Ei}_{1}\left (\ln \left (-\frac {3 x}{4}\right )\right )}{4}+\ln \left (\ln \left (-\frac {3 x}{4}\right )\right )+\frac {5 \ln \left (\ln \left (3\right )-2 \ln \left (2\right )+\ln \left (-x \right )\right )}{x}-\frac {15 \,\operatorname {Ei}_{1}\left (\ln \left (3\right )-2 \ln \left (2\right )+\ln \left (-x \right )\right )}{4}\)	\(48\)
default	\(x +\ln \left (\ln \left (3\right )-2 \ln \left (2\right )+\ln \left (-x \right )\right )+5 \,{\mathrm e}^{\ln \left (3\right )-2 \ln \left (2\right )} \operatorname {Ei}_{1}\left (\ln \left (3\right )-2 \ln \left (2\right )+\ln \left (-x \right )\right )+\frac {5 \ln \left (\ln \left (3\right )-2 \ln \left (2\right )+\ln \left (-x \right )\right )}{x}-\frac {15 \,\operatorname {Ei}_{1}\left (\ln \left (3\right )-2 \ln \left (2\right )+\ln \left (-x \right )\right )}{4}\)	\(70\)

[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))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=\frac {x^{2} + {\left (x + 5\right )} \log \left (\log \left (-\frac {3}{4} \, x\right )\right )}{x} \]

[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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=x + \log {\left (\log {\left (- \frac {3 x}{4} \right )} \right )} + \frac {5 \log {\left (\log {\left (- \frac {3 x}{4} \right )} \right )}}{x} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=x + \frac {5 \, \log \left (\log \left (-\frac {3}{4} \, x\right )\right )}{x} + \log \left (\log \left (-\frac {3}{4} \, x\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=x + \frac {5 \, \log \left (\log \left (-\frac {3}{4} \, x\right )\right )}{x} + \log \left (-2 \, \log \left (2\right ) + \log \left (-3 \, x\right )\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 13.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {5+x+x^2 \log \left (-\frac {3 x}{4}\right )-5 \log \left (-\frac {3 x}{4}\right ) \log \left (\log \left (-\frac {3 x}{4}\right )\right )}{x^2 \log \left (-\frac {3 x}{4}\right )} \, dx=x+\ln \left (\ln \left (-\frac {3\,x}{4}\right )\right )+\frac {5\,\ln \left (\ln \left (-\frac {3\,x}{4}\right )\right )}{x} \]

[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