3.100.59 \(\int \frac {32 x+160 \log (3)-32 \log ^2(3)}{-x^4-9 x^3 \log (3)-27 x^2 \log ^2(3)-27 x \log ^3(3)+(-15 x^3 \log (3)+(-90 x^2+3 x^3) \log ^2(3)+(-135 x+18 x^2) \log ^3(3)+27 x \log ^4(3)) \log (x)+(-75 x^2 \log ^2(3)+(-225 x+30 x^2) \log ^3(3)+(90 x-3 x^2) \log ^4(3)-9 x \log ^5(3)) \log ^2(x)+(-125 x \log ^3(3)+75 x \log ^4(3)-15 x \log ^5(3)+x \log ^6(3)) \log ^3(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {16}{(x+\log (3) (3+(5-\log (3)) \log (x)))^2} \]

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Rubi [A]  time = 0.31, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 3, integrand size = 177, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 12, 6686} \begin {gather*} \frac {16}{(x+(5-\log (3)) \log (3) \log (x)+\log (27))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*x + 160*Log[3] - 32*Log[3]^2)/(-x^4 - 9*x^3*Log[3] - 27*x^2*Log[3]^2 - 27*x*Log[3]^3 + (-15*x^3*Log[3]
+ (-90*x^2 + 3*x^3)*Log[3]^2 + (-135*x + 18*x^2)*Log[3]^3 + 27*x*Log[3]^4)*Log[x] + (-75*x^2*Log[3]^2 + (-225
*x + 30*x^2)*Log[3]^3 + (90*x - 3*x^2)*Log[3]^4 - 9*x*Log[3]^5)*Log[x]^2 + (-125*x*Log[3]^3 + 75*x*Log[3]^4 -
15*x*Log[3]^5 + x*Log[3]^6)*Log[x]^3),x]

[Out]

16/(x + Log[27] + (5 - Log[3])*Log[3]*Log[x])^2

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 (-x+(-5+\log (3)) \log (3))}{x (x+\log (27)-(-5+\log (3)) \log (3) \log (x))^3} \, dx\\ &=32 \int \frac {-x+(-5+\log (3)) \log (3)}{x (x+\log (27)-(-5+\log (3)) \log (3) \log (x))^3} \, dx\\ &=\frac {16}{(x+\log (27)+(5-\log (3)) \log (3) \log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 0.90 \begin {gather*} \frac {16}{(x+\log (27)-(-5+\log (3)) \log (3) \log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x + 160*Log[3] - 32*Log[3]^2)/(-x^4 - 9*x^3*Log[3] - 27*x^2*Log[3]^2 - 27*x*Log[3]^3 + (-15*x^3*
Log[3] + (-90*x^2 + 3*x^3)*Log[3]^2 + (-135*x + 18*x^2)*Log[3]^3 + 27*x*Log[3]^4)*Log[x] + (-75*x^2*Log[3]^2 +
(-225*x + 30*x^2)*Log[3]^3 + (90*x - 3*x^2)*Log[3]^4 - 9*x*Log[3]^5)*Log[x]^2 + (-125*x*Log[3]^3 + 75*x*Log[3
]^4 - 15*x*Log[3]^5 + x*Log[3]^6)*Log[x]^3),x]

[Out]

16/(x + Log[27] - (-5 + Log[3])*Log[3]*Log[x])^2

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fricas [B]  time = 1.09, size = 65, normalized size = 3.25 \begin {gather*} \frac {16}{{\left (\log \relax (3)^{4} - 10 \, \log \relax (3)^{3} + 25 \, \log \relax (3)^{2}\right )} \log \relax (x)^{2} + x^{2} + 6 \, x \log \relax (3) + 9 \, \log \relax (3)^{2} - 2 \, {\left ({\left (x - 15\right )} \log \relax (3)^{2} + 3 \, \log \relax (3)^{3} - 5 \, x \log \relax (3)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(3)^2+160*log(3)+32*x)/((x*log(3)^6-15*x*log(3)^5+75*x*log(3)^4-125*x*log(3)^3)*log(x)^3+(-9
*x*log(3)^5+(-3*x^2+90*x)*log(3)^4+(30*x^2-225*x)*log(3)^3-75*x^2*log(3)^2)*log(x)^2+(27*x*log(3)^4+(18*x^2-13
5*x)*log(3)^3+(3*x^3-90*x^2)*log(3)^2-15*x^3*log(3))*log(x)-27*x*log(3)^3-27*x^2*log(3)^2-9*x^3*log(3)-x^4),x,
algorithm="fricas")

[Out]

16/((log(3)^4 - 10*log(3)^3 + 25*log(3)^2)*log(x)^2 + x^2 + 6*x*log(3) + 9*log(3)^2 - 2*((x - 15)*log(3)^2 + 3
*log(3)^3 - 5*x*log(3))*log(x))

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giac [B]  time = 0.22, size = 206, normalized size = 10.30 \begin {gather*} \frac {16 \, {\left (\log \relax (3)^{2} - x - 5 \, \log \relax (3)\right )}}{\log \relax (3)^{6} \log \relax (x)^{2} - x \log \relax (3)^{4} \log \relax (x)^{2} - 15 \, \log \relax (3)^{5} \log \relax (x)^{2} - 2 \, x \log \relax (3)^{4} \log \relax (x) - 6 \, \log \relax (3)^{5} \log \relax (x) + 10 \, x \log \relax (3)^{3} \log \relax (x)^{2} + 75 \, \log \relax (3)^{4} \log \relax (x)^{2} + 2 \, x^{2} \log \relax (3)^{2} \log \relax (x) + 26 \, x \log \relax (3)^{3} \log \relax (x) + 60 \, \log \relax (3)^{4} \log \relax (x) - 25 \, x \log \relax (3)^{2} \log \relax (x)^{2} - 125 \, \log \relax (3)^{3} \log \relax (x)^{2} + x^{2} \log \relax (3)^{2} + 6 \, x \log \relax (3)^{3} + 9 \, \log \relax (3)^{4} - 10 \, x^{2} \log \relax (3) \log \relax (x) - 80 \, x \log \relax (3)^{2} \log \relax (x) - 150 \, \log \relax (3)^{3} \log \relax (x) - x^{3} - 11 \, x^{2} \log \relax (3) - 39 \, x \log \relax (3)^{2} - 45 \, \log \relax (3)^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(3)^2+160*log(3)+32*x)/((x*log(3)^6-15*x*log(3)^5+75*x*log(3)^4-125*x*log(3)^3)*log(x)^3+(-9
*x*log(3)^5+(-3*x^2+90*x)*log(3)^4+(30*x^2-225*x)*log(3)^3-75*x^2*log(3)^2)*log(x)^2+(27*x*log(3)^4+(18*x^2-13
5*x)*log(3)^3+(3*x^3-90*x^2)*log(3)^2-15*x^3*log(3))*log(x)-27*x*log(3)^3-27*x^2*log(3)^2-9*x^3*log(3)-x^4),x,
algorithm="giac")

[Out]

16*(log(3)^2 - x - 5*log(3))/(log(3)^6*log(x)^2 - x*log(3)^4*log(x)^2 - 15*log(3)^5*log(x)^2 - 2*x*log(3)^4*lo
g(x) - 6*log(3)^5*log(x) + 10*x*log(3)^3*log(x)^2 + 75*log(3)^4*log(x)^2 + 2*x^2*log(3)^2*log(x) + 26*x*log(3)
^3*log(x) + 60*log(3)^4*log(x) - 25*x*log(3)^2*log(x)^2 - 125*log(3)^3*log(x)^2 + x^2*log(3)^2 + 6*x*log(3)^3
+ 9*log(3)^4 - 10*x^2*log(3)*log(x) - 80*x*log(3)^2*log(x) - 150*log(3)^3*log(x) - x^3 - 11*x^2*log(3) - 39*x*
log(3)^2 - 45*log(3)^3)

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maple [A]  time = 0.46, size = 26, normalized size = 1.30

method	result	size
norman	\(\frac {16}{\left (\ln \relax (3)^{2} \ln \relax (x )-5 \ln \relax (3) \ln \relax (x )-3 \ln \relax (3)-x \right )^{2}}\)	\(26\)
risch	\(\frac {16}{\left (\ln \relax (3)^{2} \ln \relax (x )-5 \ln \relax (3) \ln \relax (x )-3 \ln \relax (3)-x \right )^{2}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*ln(3)^2+160*ln(3)+32*x)/((x*ln(3)^6-15*x*ln(3)^5+75*x*ln(3)^4-125*x*ln(3)^3)*ln(x)^3+(-9*x*ln(3)^5+(-
3*x^2+90*x)*ln(3)^4+(30*x^2-225*x)*ln(3)^3-75*x^2*ln(3)^2)*ln(x)^2+(27*x*ln(3)^4+(18*x^2-135*x)*ln(3)^3+(3*x^3
-90*x^2)*ln(3)^2-15*x^3*ln(3))*ln(x)-27*x*ln(3)^3-27*x^2*ln(3)^2-9*x^3*ln(3)-x^4),x,method=_RETURNVERBOSE)

[Out]

16/(ln(3)^2*ln(x)-5*ln(3)*ln(x)-3*ln(3)-x)^2

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maxima [B]  time = 0.48, size = 69, normalized size = 3.45 \begin {gather*} \frac {16}{{\left (\log \relax (3)^{4} - 10 \, \log \relax (3)^{3} + 25 \, \log \relax (3)^{2}\right )} \log \relax (x)^{2} + x^{2} + 6 \, x \log \relax (3) + 9 \, \log \relax (3)^{2} - 2 \, {\left (3 \, \log \relax (3)^{3} + {\left (\log \relax (3)^{2} - 5 \, \log \relax (3)\right )} x - 15 \, \log \relax (3)^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(3)^2+160*log(3)+32*x)/((x*log(3)^6-15*x*log(3)^5+75*x*log(3)^4-125*x*log(3)^3)*log(x)^3+(-9
*x*log(3)^5+(-3*x^2+90*x)*log(3)^4+(30*x^2-225*x)*log(3)^3-75*x^2*log(3)^2)*log(x)^2+(27*x*log(3)^4+(18*x^2-13
5*x)*log(3)^3+(3*x^3-90*x^2)*log(3)^2-15*x^3*log(3))*log(x)-27*x*log(3)^3-27*x^2*log(3)^2-9*x^3*log(3)-x^4),x,
algorithm="maxima")

[Out]

16/((log(3)^4 - 10*log(3)^3 + 25*log(3)^2)*log(x)^2 + x^2 + 6*x*log(3) + 9*log(3)^2 - 2*(3*log(3)^3 + (log(3)^
2 - 5*log(3))*x - 15*log(3)^2)*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {32\,x+160\,\ln \relax (3)-32\,{\ln \relax (3)}^2}{27\,x^2\,{\ln \relax (3)}^2+\ln \relax (x)\,\left ({\ln \relax (3)}^3\,\left (135\,x-18\,x^2\right )+15\,x^3\,\ln \relax (3)-27\,x\,{\ln \relax (3)}^4+{\ln \relax (3)}^2\,\left (90\,x^2-3\,x^3\right )\right )+{\ln \relax (x)}^3\,\left (125\,x\,{\ln \relax (3)}^3-75\,x\,{\ln \relax (3)}^4+15\,x\,{\ln \relax (3)}^5-x\,{\ln \relax (3)}^6\right )+27\,x\,{\ln \relax (3)}^3+9\,x^3\,\ln \relax (3)+x^4+{\ln \relax (x)}^2\,\left (75\,x^2\,{\ln \relax (3)}^2-{\ln \relax (3)}^4\,\left (90\,x-3\,x^2\right )+{\ln \relax (3)}^3\,\left (225\,x-30\,x^2\right )+9\,x\,{\ln \relax (3)}^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(32*x + 160*log(3) - 32*log(3)^2)/(27*x^2*log(3)^2 + log(x)*(log(3)^3*(135*x - 18*x^2) + 15*x^3*log(3) -
27*x*log(3)^4 + log(3)^2*(90*x^2 - 3*x^3)) + log(x)^3*(125*x*log(3)^3 - 75*x*log(3)^4 + 15*x*log(3)^5 - x*log(
3)^6) + 27*x*log(3)^3 + 9*x^3*log(3) + x^4 + log(x)^2*(75*x^2*log(3)^2 - log(3)^4*(90*x - 3*x^2) + log(3)^3*(2
25*x - 30*x^2) + 9*x*log(3)^5)),x)

[Out]

int(-(32*x + 160*log(3) - 32*log(3)^2)/(27*x^2*log(3)^2 + log(x)*(log(3)^3*(135*x - 18*x^2) + 15*x^3*log(3) -
27*x*log(3)^4 + log(3)^2*(90*x^2 - 3*x^3)) + log(x)^3*(125*x*log(3)^3 - 75*x*log(3)^4 + 15*x*log(3)^5 - x*log(
3)^6) + 27*x*log(3)^3 + 9*x^3*log(3) + x^4 + log(x)^2*(75*x^2*log(3)^2 - log(3)^4*(90*x - 3*x^2) + log(3)^3*(2
25*x - 30*x^2) + 9*x*log(3)^5)), x)

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sympy [B]  time = 0.40, size = 73, normalized size = 3.65 \begin {gather*} \frac {16}{x^{2} + 6 x \log {\relax (3 )} + \left (- 2 x \log {\relax (3 )}^{2} + 10 x \log {\relax (3 )} - 6 \log {\relax (3 )}^{3} + 30 \log {\relax (3 )}^{2}\right ) \log {\relax (x )} + \left (- 10 \log {\relax (3 )}^{3} + \log {\relax (3 )}^{4} + 25 \log {\relax (3 )}^{2}\right ) \log {\relax (x )}^{2} + 9 \log {\relax (3 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*ln(3)**2+160*ln(3)+32*x)/((x*ln(3)**6-15*x*ln(3)**5+75*x*ln(3)**4-125*x*ln(3)**3)*ln(x)**3+(-9*
x*ln(3)**5+(-3*x**2+90*x)*ln(3)**4+(30*x**2-225*x)*ln(3)**3-75*x**2*ln(3)**2)*ln(x)**2+(27*x*ln(3)**4+(18*x**2
-135*x)*ln(3)**3+(3*x**3-90*x**2)*ln(3)**2-15*x**3*ln(3))*ln(x)-27*x*ln(3)**3-27*x**2*ln(3)**2-9*x**3*ln(3)-x*
*4),x)

[Out]

16/(x**2 + 6*x*log(3) + (-2*x*log(3)**2 + 10*x*log(3) - 6*log(3)**3 + 30*log(3)**2)*log(x) + (-10*log(3)**3 +
log(3)**4 + 25*log(3)**2)*log(x)**2 + 9*log(3)**2)

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