3.19.48 \(\int \frac {-2-80 x+8 x \log (3)+(-120 x+12 x \log (3)) \log (x)+(-60 x+6 x \log (3)) \log ^2(x)+(-10 x+x \log (3)) \log ^3(x)}{8 x+12 x \log (x)+6 x \log ^2(x)+x \log ^3(x)} \, dx\)

Optimal. Leaf size=17 \[ 4-x+x (-9+\log (3))+\frac {1}{(2+\log (x))^2} \]

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Rubi [A]  time = 0.39, antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6, 6688, 6742, 2302, 30} \begin {gather*} \frac {1}{(\log (x)+2)^2}-x (10-\log (3)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 80*x + 8*x*Log[3] + (-120*x + 12*x*Log[3])*Log[x] + (-60*x + 6*x*Log[3])*Log[x]^2 + (-10*x + x*Log[3
])*Log[x]^3)/(8*x + 12*x*Log[x] + 6*x*Log[x]^2 + x*Log[x]^3),x]

[Out]

-(x*(10 - Log[3])) + (2 + Log[x])^(-2)

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 30

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

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 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} &=\int \frac {-2+x (-80+8 \log (3))+(-120 x+12 x \log (3)) \log (x)+(-60 x+6 x \log (3)) \log ^2(x)+(-10 x+x \log (3)) \log ^3(x)}{8 x+12 x \log (x)+6 x \log ^2(x)+x \log ^3(x)} \, dx\\ &=\int \frac {-2+8 x (-10+\log (3))+12 x (-10+\log (3)) \log (x)+6 x (-10+\log (3)) \log ^2(x)+x (-10+\log (3)) \log ^3(x)}{x (2+\log (x))^3} \, dx\\ &=\int \left (-10 \left (1-\frac {\log (3)}{10}\right )-\frac {2}{x (2+\log (x))^3}\right ) \, dx\\ &=-x (10-\log (3))-2 \int \frac {1}{x (2+\log (x))^3} \, dx\\ &=-x (10-\log (3))-2 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,2+\log (x)\right )\\ &=-x (10-\log (3))+\frac {1}{(2+\log (x))^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 14, normalized size = 0.82 \begin {gather*} -10 x+x \log (3)+\frac {1}{(2+\log (x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 80*x + 8*x*Log[3] + (-120*x + 12*x*Log[3])*Log[x] + (-60*x + 6*x*Log[3])*Log[x]^2 + (-10*x + x
*Log[3])*Log[x]^3)/(8*x + 12*x*Log[x] + 6*x*Log[x]^2 + x*Log[x]^3),x]

[Out]

-10*x + x*Log[3] + (2 + Log[x])^(-2)

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fricas [B]  time = 0.60, size = 48, normalized size = 2.82 \begin {gather*} \frac {{\left (x \log \relax (3) - 10 \, x\right )} \log \relax (x)^{2} + 4 \, x \log \relax (3) + 4 \, {\left (x \log \relax (3) - 10 \, x\right )} \log \relax (x) - 40 \, x + 1}{\log \relax (x)^{2} + 4 \, \log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(3)-10*x)*log(x)^3+(6*x*log(3)-60*x)*log(x)^2+(12*x*log(3)-120*x)*log(x)+8*x*log(3)-80*x-2)/(
x*log(x)^3+6*x*log(x)^2+12*x*log(x)+8*x),x, algorithm="fricas")

[Out]

((x*log(3) - 10*x)*log(x)^2 + 4*x*log(3) + 4*(x*log(3) - 10*x)*log(x) - 40*x + 1)/(log(x)^2 + 4*log(x) + 4)

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giac [A]  time = 0.19, size = 19, normalized size = 1.12 \begin {gather*} x {\left (\log \relax (3) - 10\right )} + \frac {1}{\log \relax (x)^{2} + 4 \, \log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(3)-10*x)*log(x)^3+(6*x*log(3)-60*x)*log(x)^2+(12*x*log(3)-120*x)*log(x)+8*x*log(3)-80*x-2)/(
x*log(x)^3+6*x*log(x)^2+12*x*log(x)+8*x),x, algorithm="giac")

[Out]

x*(log(3) - 10) + 1/(log(x)^2 + 4*log(x) + 4)

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maple [A]  time = 0.05, size = 15, normalized size = 0.88

method	result	size
risch	\(x \ln \relax (3)-10 x +\frac {1}{\left (\ln \relax (x )+2\right )^{2}}\)	\(15\)
default	\(\frac {1-40 x -40 x \ln \relax (x )-10 x \ln \relax (x )^{2}}{\left (\ln \relax (x )+2\right )^{2}}+x \ln \relax (3)\)	\(30\)
norman	\(\frac {\left (-40+4 \ln \relax (3)\right ) x +1+\left (-40+4 \ln \relax (3)\right ) x \ln \relax (x )+\left (\ln \relax (3)-10\right ) x \ln \relax (x )^{2}}{\left (\ln \relax (x )+2\right )^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(3)-10*x)*ln(x)^3+(6*x*ln(3)-60*x)*ln(x)^2+(12*x*ln(3)-120*x)*ln(x)+8*x*ln(3)-80*x-2)/(x*ln(x)^3+6*x
*ln(x)^2+12*x*ln(x)+8*x),x,method=_RETURNVERBOSE)

[Out]

x*ln(3)-10*x+1/(ln(x)+2)^2

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maxima [A]  time = 0.60, size = 19, normalized size = 1.12 \begin {gather*} x {\left (\log \relax (3) - 10\right )} + \frac {1}{\log \relax (x)^{2} + 4 \, \log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(3)-10*x)*log(x)^3+(6*x*log(3)-60*x)*log(x)^2+(12*x*log(3)-120*x)*log(x)+8*x*log(3)-80*x-2)/(
x*log(x)^3+6*x*log(x)^2+12*x*log(x)+8*x),x, algorithm="maxima")

[Out]

x*(log(3) - 10) + 1/(log(x)^2 + 4*log(x) + 4)

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mupad [B]  time = 1.28, size = 35, normalized size = 2.06 \begin {gather*} x\,\ln \relax (3)-10\,x+\frac {x^2}{x^2\,{\ln \relax (x)}^2+4\,x^2\,\ln \relax (x)+4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(80*x + log(x)^3*(10*x - x*log(3)) + log(x)^2*(60*x - 6*x*log(3)) - 8*x*log(3) + log(x)*(120*x - 12*x*log
(3)) + 2)/(8*x + 6*x*log(x)^2 + x*log(x)^3 + 12*x*log(x)),x)

[Out]

x*log(3) - 10*x + x^2/(4*x^2*log(x) + x^2*log(x)^2 + 4*x^2)

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sympy [A]  time = 0.12, size = 19, normalized size = 1.12 \begin {gather*} x \left (-10 + \log {\relax (3 )}\right ) + \frac {1}{\log {\relax (x )}^{2} + 4 \log {\relax (x )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(3)-10*x)*ln(x)**3+(6*x*ln(3)-60*x)*ln(x)**2+(12*x*ln(3)-120*x)*ln(x)+8*x*ln(3)-80*x-2)/(x*ln(
x)**3+6*x*ln(x)**2+12*x*ln(x)+8*x),x)

[Out]

x*(-10 + log(3)) + 1/(log(x)**2 + 4*log(x) + 4)

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