\(\int \frac {e^{24} (-16+(-8 x-8 \log (3)) \log (x)+(5-x-x^2-2 x \log (3)-\log ^2(3)) \log ^2(x)) (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+(x+2 x^2+2 x \log (3)) \log ^3(x))}{\log ^2(x) (16 x \log (x)+(8 x^2+8 x \log (3)) \log ^2(x)+(-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)) \log ^3(x))} \, dx\) [2628]

   Optimal result
   Rubi [B] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 136, antiderivative size = 23 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=e^{24} \left (5-x-\left (x+\log (3)+\frac {4}{\log (x)}\right )^2\right ) \]

[Out]

exp(ln(-x+5-(4/ln(x)+ln(3)+x)^2)+24)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(23)=46\).

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6820, 6874, 2339, 30, 2395, 2334, 2335} \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-e^{24} x^2-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-e^{24} x (1+\log (9))-\frac {8 e^{24} \log (3)}{\log (x)} \]

[In]

Int[(E^24*(-16 + (-8*x - 8*Log[3])*Log[x] + (5 - x - x^2 - 2*x*Log[3] - Log[3]^2)*Log[x]^2)*(-32 + (-8*x - 8*L
og[3])*Log[x] + 8*x*Log[x]^2 + (x + 2*x^2 + 2*x*Log[3])*Log[x]^3))/(Log[x]^2*(16*x*Log[x] + (8*x^2 + 8*x*Log[3
])*Log[x]^2 + (-5*x + x^2 + x^3 + 2*x^2*Log[3] + x*Log[3]^2)*Log[x]^3)),x]

[Out]

-(E^24*x^2) - E^24*x*(1 + Log[9]) - (16*E^24)/Log[x]^2 - (8*E^24*x)/Log[x] - (8*E^24*Log[3])/Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2334

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 2335

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

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 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 6820

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = e^{24} \int \frac {\left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx \\ & = e^{24} \int \frac {32+8 (x+\log (3)) \log (x)-8 x \log ^2(x)-x (1+2 x+\log (9)) \log ^3(x)}{x \log ^3(x)} \, dx \\ & = e^{24} \int \left (-1-2 x-\log (9)+\frac {32}{x \log ^3(x)}+\frac {8 (x+\log (3))}{x \log ^2(x)}-\frac {8}{\log (x)}\right ) \, dx \\ & = -e^{24} x^2-e^{24} x (1+\log (9))+\left (8 e^{24}\right ) \int \frac {x+\log (3)}{x \log ^2(x)} \, dx-\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (32 e^{24}\right ) \int \frac {1}{x \log ^3(x)} \, dx \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-8 e^{24} \operatorname {LogIntegral}(x)+\left (8 e^{24}\right ) \int \left (\frac {1}{\log ^2(x)}+\frac {\log (3)}{x \log ^2(x)}\right ) \, dx+\left (32 e^{24}\right ) \text {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (x)\right ) \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-8 e^{24} \operatorname {LogIntegral}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log ^2(x)} \, dx+\left (8 e^{24} \log (3)\right ) \int \frac {1}{x \log ^2(x)} \, dx \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-8 e^{24} \operatorname {LogIntegral}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (8 e^{24} \log (3)\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right ) \\ & = -e^{24} x^2-e^{24} x (1+\log (9))-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(53\) vs. \(2(23)=46\).

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-e^{24} x-e^{24} x^2-e^{24} x \log (9)-\frac {16 e^{24}}{\log ^2(x)}-\frac {8 e^{24} x}{\log (x)}-\frac {8 e^{24} \log (3)}{\log (x)} \]

[In]

Integrate[(E^24*(-16 + (-8*x - 8*Log[3])*Log[x] + (5 - x - x^2 - 2*x*Log[3] - Log[3]^2)*Log[x]^2)*(-32 + (-8*x
 - 8*Log[3])*Log[x] + 8*x*Log[x]^2 + (x + 2*x^2 + 2*x*Log[3])*Log[x]^3))/(Log[x]^2*(16*x*Log[x] + (8*x^2 + 8*x
*Log[3])*Log[x]^2 + (-5*x + x^2 + x^3 + 2*x^2*Log[3] + x*Log[3]^2)*Log[x]^3)),x]

[Out]

-(E^24*x) - E^24*x^2 - E^24*x*Log[9] - (16*E^24)/Log[x]^2 - (8*E^24*x)/Log[x] - (8*E^24*Log[3])/Log[x]

Maple [A] (verified)

Time = 142.32 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70

method result size
default \({\mathrm e}^{24} \left (-2 x \ln \left (3\right )-x^{2}-x -\frac {8 \ln \left (3\right )}{\ln \left (x \right )}-\frac {8 x}{\ln \left (x \right )}-\frac {16}{\ln \left (x \right )^{2}}\right )\) \(39\)
parallelrisch \(\text {Expression too large to display}\) \(1635\)

[In]

int(((2*x*ln(3)+2*x^2+x)*ln(x)^3+8*x*ln(x)^2+(-8*ln(3)-8*x)*ln(x)-32)*exp(ln(((-ln(3)^2-2*x*ln(3)-x^2-x+5)*ln(
x)^2+(-8*ln(3)-8*x)*ln(x)-16)/ln(x)^2)+24)/((x*ln(3)^2+2*x^2*ln(3)+x^3+x^2-5*x)*ln(x)^3+(8*x*ln(3)+8*x^2)*ln(x
)^2+16*x*ln(x)),x,method=_RETURNVERBOSE)

[Out]

exp(24)*(-2*x*ln(3)-x^2-x-8*ln(3)/ln(x)-8*x/ln(x)-16/ln(x)^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.00 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {{\left (2 \, x e^{24} \log \left (3\right ) + {\left (x^{2} + x\right )} e^{24}\right )} \log \left (x\right )^{2} + 8 \, {\left (x e^{24} + e^{24} \log \left (3\right )\right )} \log \left (x\right ) + 16 \, e^{24}}{\log \left (x\right )^{2}} \]

[In]

integrate(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log(x)-32)*exp(log(((-log(3)^2-2*x*log(3
)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*log(x)-16)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(
8*x*log(3)+8*x^2)*log(x)^2+16*x*log(x)),x, algorithm="fricas")

[Out]

-((2*x*e^24*log(3) + (x^2 + x)*e^24)*log(x)^2 + 8*(x*e^24 + e^24*log(3))*log(x) + 16*e^24)/log(x)^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (17) = 34\).

Time = 0.16 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.22 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=- x^{2} e^{24} + x \left (- 2 e^{24} \log {\left (3 \right )} - e^{24}\right ) + \frac {\left (- 8 x e^{24} - 8 e^{24} \log {\left (3 \right )}\right ) \log {\left (x \right )} - 16 e^{24}}{\log {\left (x \right )}^{2}} \]

[In]

integrate(((2*x*ln(3)+2*x**2+x)*ln(x)**3+8*x*ln(x)**2+(-8*ln(3)-8*x)*ln(x)-32)*exp(ln(((-ln(3)**2-2*x*ln(3)-x*
*2-x+5)*ln(x)**2+(-8*ln(3)-8*x)*ln(x)-16)/ln(x)**2)+24)/((x*ln(3)**2+2*x**2*ln(3)+x**3+x**2-5*x)*ln(x)**3+(8*x
*ln(3)+8*x**2)*ln(x)**2+16*x*ln(x)),x)

[Out]

-x**2*exp(24) + x*(-2*exp(24)*log(3) - exp(24)) + ((-8*x*exp(24) - 8*exp(24)*log(3))*log(x) - 16*exp(24))/log(
x)**2

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {{\left ({\left (x^{2} + x {\left (2 \, \log \left (3\right ) + 1\right )}\right )} \log \left (x\right )^{2} + 8 \, {\left (x + \log \left (3\right )\right )} \log \left (x\right ) + 16\right )} e^{24}}{\log \left (x\right )^{2}} \]

[In]

integrate(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log(x)-32)*exp(log(((-log(3)^2-2*x*log(3
)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*log(x)-16)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(
8*x*log(3)+8*x^2)*log(x)^2+16*x*log(x)),x, algorithm="maxima")

[Out]

-((x^2 + x*(2*log(3) + 1))*log(x)^2 + 8*(x + log(3))*log(x) + 16)*e^24/log(x)^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.39 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-\frac {x^{2} e^{24} \log \left (x\right )^{2} + 2 \, x e^{24} \log \left (3\right ) \log \left (x\right )^{2} + x e^{24} \log \left (x\right )^{2} + 8 \, x e^{24} \log \left (x\right ) + 8 \, e^{24} \log \left (3\right ) \log \left (x\right ) + 16 \, e^{24}}{\log \left (x\right )^{2}} \]

[In]

integrate(((2*x*log(3)+2*x^2+x)*log(x)^3+8*x*log(x)^2+(-8*log(3)-8*x)*log(x)-32)*exp(log(((-log(3)^2-2*x*log(3
)-x^2-x+5)*log(x)^2+(-8*log(3)-8*x)*log(x)-16)/log(x)^2)+24)/((x*log(3)^2+2*x^2*log(3)+x^3+x^2-5*x)*log(x)^3+(
8*x*log(3)+8*x^2)*log(x)^2+16*x*log(x)),x, algorithm="giac")

[Out]

-(x^2*e^24*log(x)^2 + 2*x*e^24*log(3)*log(x)^2 + x*e^24*log(x)^2 + 8*x*e^24*log(x) + 8*e^24*log(3)*log(x) + 16
*e^24)/log(x)^2

Mupad [B] (verification not implemented)

Time = 8.68 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \frac {e^{24} \left (-16+(-8 x-8 \log (3)) \log (x)+\left (5-x-x^2-2 x \log (3)-\log ^2(3)\right ) \log ^2(x)\right ) \left (-32+(-8 x-8 \log (3)) \log (x)+8 x \log ^2(x)+\left (x+2 x^2+2 x \log (3)\right ) \log ^3(x)\right )}{\log ^2(x) \left (16 x \log (x)+\left (8 x^2+8 x \log (3)\right ) \log ^2(x)+\left (-5 x+x^2+x^3+2 x^2 \log (3)+x \log ^2(3)\right ) \log ^3(x)\right )} \, dx=-x^2\,{\mathrm {e}}^{24}-\frac {16\,{\mathrm {e}}^{24}+8\,{\mathrm {e}}^{24}\,\ln \left (3\right )\,\ln \left (x\right )}{{\ln \left (x\right )}^2}-\frac {x\,\left ({\mathrm {e}}^{24}\,\left (2\,\ln \left (3\right )+1\right )\,{\ln \left (x\right )}^2+8\,{\mathrm {e}}^{24}\,\ln \left (x\right )\right )}{{\ln \left (x\right )}^2} \]

[In]

int((exp(log(-(log(x)*(8*x + 8*log(3)) + log(x)^2*(x + 2*x*log(3) + log(3)^2 + x^2 - 5) + 16)/log(x)^2) + 24)*
(8*x*log(x)^2 - log(x)*(8*x + 8*log(3)) + log(x)^3*(x + 2*x*log(3) + 2*x^2) - 32))/(log(x)^2*(8*x*log(3) + 8*x
^2) + 16*x*log(x) + log(x)^3*(x*log(3)^2 - 5*x + 2*x^2*log(3) + x^2 + x^3)),x)

[Out]

- x^2*exp(24) - (16*exp(24) + 8*exp(24)*log(3)*log(x))/log(x)^2 - (x*(8*exp(24)*log(x) + exp(24)*log(x)^2*(2*l
og(3) + 1)))/log(x)^2