\(\int \frac {8-8 x+(-32 x+432 x^2-512 x^3) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx\) [9644]

Optimal result

Integrand size = 38, antiderivative size = 22 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=4 x (-2+2 x) \left (2 x-16 x^2-\log (\log (x))\right ) \]

[Out]

4*x*(2*x-16*x^2-ln(ln(x)))*(-2+2*x)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 18, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {6874, 14, 2367, 2335, 2346, 2209, 2600, 2602} \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=-128 x^4+144 x^3-16 x^2-8 x^2 \log (\log (x))+8 x \log (\log (x)) \]

[In]

Int[(8 - 8*x + (-32*x + 432*x^2 - 512*x^3)*Log[x] + (8 - 16*x)*Log[x]*Log[Log[x]])/Log[x],x]

[Out]

-16*x^2 + 144*x^3 - 128*x^4 + 8*x*Log[Log[x]] - 8*x^2*Log[Log[x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 2367

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 2600

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

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 {8 \left (-1+x+4 x \log (x)-54 x^2 \log (x)+64 x^3 \log (x)\right )}{\log (x)}-8 (-1+2 x) \log (\log (x))\right ) \, dx \\ & = -\left (8 \int \frac {-1+x+4 x \log (x)-54 x^2 \log (x)+64 x^3 \log (x)}{\log (x)} \, dx\right )-8 \int (-1+2 x) \log (\log (x)) \, dx \\ & = -\left (8 \int \left (2 x \left (2-27 x+32 x^2\right )+\frac {-1+x}{\log (x)}\right ) \, dx\right )-8 \int (-\log (\log (x))+2 x \log (\log (x))) \, dx \\ & = -\left (8 \int \frac {-1+x}{\log (x)} \, dx\right )+8 \int \log (\log (x)) \, dx-16 \int x \left (2-27 x+32 x^2\right ) \, dx-16 \int x \log (\log (x)) \, dx \\ & = 8 x \log (\log (x))-8 x^2 \log (\log (x))-8 \int \left (-\frac {1}{\log (x)}+\frac {x}{\log (x)}\right ) \, dx-8 \int \frac {1}{\log (x)} \, dx+8 \int \frac {x}{\log (x)} \, dx-16 \int \left (2 x-27 x^2+32 x^3\right ) \, dx \\ & = -16 x^2+144 x^3-128 x^4+8 x \log (\log (x))-8 x^2 \log (\log (x))-8 \operatorname {LogIntegral}(x)+8 \int \frac {1}{\log (x)} \, dx-8 \int \frac {x}{\log (x)} \, dx+8 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = -16 x^2+144 x^3-128 x^4+8 \operatorname {ExpIntegralEi}(2 \log (x))+8 x \log (\log (x))-8 x^2 \log (\log (x))-8 \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = -16 x^2+144 x^3-128 x^4+8 x \log (\log (x))-8 x^2 \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=-16 x^2+144 x^3-128 x^4+8 x \log (\log (x))-8 x^2 \log (\log (x)) \]

[In]

Integrate[(8 - 8*x + (-32*x + 432*x^2 - 512*x^3)*Log[x] + (8 - 16*x)*Log[x]*Log[Log[x]])/Log[x],x]

[Out]

-16*x^2 + 144*x^3 - 128*x^4 + 8*x*Log[Log[x]] - 8*x^2*Log[Log[x]]

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36

method	result	size
risch	\(\left (-8 x^{2}+8 x \right ) \ln \left (\ln \left (x \right )\right )-128 x^{4}+144 x^{3}-16 x^{2}\)	\(30\)
default	\(-8 x^{2} \ln \left (\ln \left (x \right )\right )+8 x \ln \left (\ln \left (x \right )\right )-16 x^{2}+144 x^{3}-128 x^{4}\)	\(31\)
parallelrisch	\(-8 x^{2} \ln \left (\ln \left (x \right )\right )+8 x \ln \left (\ln \left (x \right )\right )-16 x^{2}+144 x^{3}-128 x^{4}\)	\(31\)
parts	\(-8 x^{2} \ln \left (\ln \left (x \right )\right )+8 x \ln \left (\ln \left (x \right )\right )-16 x^{2}+144 x^{3}-128 x^{4}\)	\(31\)

[In]

int(((-16*x+8)*ln(x)*ln(ln(x))+(-512*x^3+432*x^2-32*x)*ln(x)-8*x+8)/ln(x),x,method=_RETURNVERBOSE)

[Out]

(-8*x^2+8*x)*ln(ln(x))-128*x^4+144*x^3-16*x^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=-128 \, x^{4} + 144 \, x^{3} - 16 \, x^{2} - 8 \, {\left (x^{2} - x\right )} \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-16*x+8)*log(x)*log(log(x))+(-512*x^3+432*x^2-32*x)*log(x)-8*x+8)/log(x),x, algorithm="fricas")

[Out]

-128*x^4 + 144*x^3 - 16*x^2 - 8*(x^2 - x)*log(log(x))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=- 128 x^{4} + 144 x^{3} - 16 x^{2} + \left (- 8 x^{2} + 8 x\right ) \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((-16*x+8)*ln(x)*ln(ln(x))+(-512*x**3+432*x**2-32*x)*ln(x)-8*x+8)/ln(x),x)

[Out]

-128*x**4 + 144*x**3 - 16*x**2 + (-8*x**2 + 8*x)*log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=-128 \, x^{4} + 144 \, x^{3} - 8 \, x^{2} \log \left (\log \left (x\right )\right ) - 16 \, x^{2} + 8 \, x \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-16*x+8)*log(x)*log(log(x))+(-512*x^3+432*x^2-32*x)*log(x)-8*x+8)/log(x),x, algorithm="maxima")

[Out]

-128*x^4 + 144*x^3 - 8*x^2*log(log(x)) - 16*x^2 + 8*x*log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=-128 \, x^{4} + 144 \, x^{3} - 8 \, x^{2} \log \left (\log \left (x\right )\right ) - 16 \, x^{2} + 8 \, x \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-16*x+8)*log(x)*log(log(x))+(-512*x^3+432*x^2-32*x)*log(x)-8*x+8)/log(x),x, algorithm="giac")

[Out]

-128*x^4 + 144*x^3 - 8*x^2*log(log(x)) - 16*x^2 + 8*x*log(log(x))

Mupad [B] (verification not implemented)

Time = 13.84 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {8-8 x+\left (-32 x+432 x^2-512 x^3\right ) \log (x)+(8-16 x) \log (x) \log (\log (x))}{\log (x)} \, dx=-8\,x\,\left (x-1\right )\,\left (\ln \left (\ln \left (x\right )\right )-2\,x+16\,x^2\right ) \]

[In]

int(-(8*x + log(x)*(32*x - 432*x^2 + 512*x^3) + log(log(x))*log(x)*(16*x - 8) - 8)/log(x),x)

[Out]

-8*x*(x - 1)*(log(log(x)) - 2*x + 16*x^2)