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

Optimal. Leaf size=23 \[ e^{24} \left (5-x-\left (x+\log (3)+\frac {4}{\log (x)}\right )^2\right ) \]

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Rubi [B]  time = 0.45, antiderivative size = 49, normalized size of antiderivative = 2.13, number of steps used = 13, number of rules used = 8, integrand size = 136, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 6688, 6742, 2302, 30, 2353, 2297, 2298} \begin {gather*} -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)} \end {gather*}

Antiderivative was successfully verified.

[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 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 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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} \text {li}(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 ) \operatorname {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} \text {li}(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} \text {li}(x)+\left (8 e^{24}\right ) \int \frac {1}{\log (x)} \, dx+\left (8 e^{24} \log (3)\right ) \operatorname {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 {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 53, normalized size = 2.30 \begin {gather*} -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)} \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.56, size = 46, normalized size = 2.00 \begin {gather*} -\frac {{\left (2 \, x e^{24} \log \relax (3) + {\left (x^{2} + x\right )} e^{24}\right )} \log \relax (x)^{2} + 8 \, {\left (x e^{24} + e^{24} \log \relax (3)\right )} \log \relax (x) + 16 \, e^{24}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.31, size = 55, normalized size = 2.39 \begin {gather*} -\frac {x^{2} e^{24} \log \relax (x)^{2} + 2 \, x e^{24} \log \relax (3) \log \relax (x)^{2} + x e^{24} \log \relax (x)^{2} + 8 \, x e^{24} \log \relax (x) + 8 \, e^{24} \log \relax (3) \log \relax (x) + 16 \, e^{24}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.05, size = 39, normalized size = 1.70

method	result	size
default	\({\mathrm e}^{24} \left (-2 x \ln \relax (3)-x^{2}-x -\frac {8 \ln \relax (3)}{\ln \relax (x )}-\frac {8 x}{\ln \relax (x )}-\frac {16}{\ln \relax (x )^{2}}\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.56, size = 35, normalized size = 1.52 \begin {gather*} -\frac {{\left ({\left (x^{2} + x {\left (2 \, \log \relax (3) + 1\right )}\right )} \log \relax (x)^{2} + 8 \, {\left (x + \log \relax (3)\right )} \log \relax (x) + 16\right )} e^{24}}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.66, size = 54, normalized size = 2.35 \begin {gather*} -x^2\,{\mathrm {e}}^{24}-\frac {16\,{\mathrm {e}}^{24}+8\,{\mathrm {e}}^{24}\,\ln \relax (3)\,\ln \relax (x)}{{\ln \relax (x)}^2}-\frac {x\,\left ({\mathrm {e}}^{24}\,\left (2\,\ln \relax (3)+1\right )\,{\ln \relax (x)}^2+8\,{\mathrm {e}}^{24}\,\ln \relax (x)\right )}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [B]  time = 0.30, size = 51, normalized size = 2.22 \begin {gather*} - x^{2} e^{24} + x \left (- 2 e^{24} \log {\relax (3 )} - e^{24}\right ) + \frac {\left (- 8 x e^{24} - 8 e^{24} \log {\relax (3 )}\right ) \log {\relax (x )} - 16 e^{24}}{\log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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