3.30.95 \(\int \frac {-1-x-x^2+(-1-2 x-3 x^2) \log (x)}{(x+x^2+x^3) \log (x)} \, dx\)

Optimal. Leaf size=20 \[ 4+\log \left (\frac {4}{x \left (1+x+x^2\right ) \log (x)}\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1594, 6728, 1628, 628, 2302, 29} \begin {gather*} -\log \left (x^2+x+1\right )-\log (x)-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - x - x^2 + (-1 - 2*x - 3*x^2)*Log[x])/((x + x^2 + x^3)*Log[x]),x]

[Out]

-Log[x] - Log[1 + x + x^2] - Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1-x-x^2+\left (-1-2 x-3 x^2\right ) \log (x)}{x \left (1+x+x^2\right ) \log (x)} \, dx\\ &=\int \left (\frac {-1-2 x-3 x^2}{x \left (1+x+x^2\right )}-\frac {1}{x \log (x)}\right ) \, dx\\ &=\int \frac {-1-2 x-3 x^2}{x \left (1+x+x^2\right )} \, dx-\int \frac {1}{x \log (x)} \, dx\\ &=\int \left (-\frac {1}{x}+\frac {-1-2 x}{1+x+x^2}\right ) \, dx-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-\log (x)-\log (\log (x))+\int \frac {-1-2 x}{1+x+x^2} \, dx\\ &=-\log (x)-\log \left (1+x+x^2\right )-\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 19, normalized size = 0.95 \begin {gather*} -\log (x)-\log \left (1+x+x^2\right )-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - x - x^2 + (-1 - 2*x - 3*x^2)*Log[x])/((x + x^2 + x^3)*Log[x]),x]

[Out]

-Log[x] - Log[1 + x + x^2] - Log[Log[x]]

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fricas [A]  time = 0.78, size = 17, normalized size = 0.85 \begin {gather*} -\log \left (x^{3} + x^{2} + x\right ) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-2*x-1)*log(x)-x^2-x-1)/(x^3+x^2+x)/log(x),x, algorithm="fricas")

[Out]

-log(x^3 + x^2 + x) - log(log(x))

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giac [A]  time = 1.14, size = 19, normalized size = 0.95 \begin {gather*} -\log \left (x^{2} + x + 1\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-2*x-1)*log(x)-x^2-x-1)/(x^3+x^2+x)/log(x),x, algorithm="giac")

[Out]

-log(x^2 + x + 1) - log(x) - log(log(x))

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maple [A]  time = 0.03, size = 18, normalized size = 0.90

method	result	size
default	\(-\ln \left (\ln \relax (x )\right )-\ln \left (x \left (x^{2}+x +1\right )\right )\)	\(18\)
risch	\(-\ln \left (x^{3}+x^{2}+x \right )-\ln \left (\ln \relax (x )\right )\)	\(18\)
norman	\(-\ln \relax (x )-\ln \left (\ln \relax (x )\right )-\ln \left (x^{2}+x +1\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2-2*x-1)*ln(x)-x^2-x-1)/(x^3+x^2+x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(x))-ln(x*(x^2+x+1))

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maxima [A]  time = 0.52, size = 19, normalized size = 0.95 \begin {gather*} -\log \left (x^{2} + x + 1\right ) - \log \relax (x) - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2-2*x-1)*log(x)-x^2-x-1)/(x^3+x^2+x)/log(x),x, algorithm="maxima")

[Out]

-log(x^2 + x + 1) - log(x) - log(log(x))

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mupad [B]  time = 1.85, size = 17, normalized size = 0.85 \begin {gather*} -\ln \left (\ln \relax (x)\,\left (x^2+x+1\right )\right )-\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + log(x)*(2*x + 3*x^2 + 1) + x^2 + 1)/(log(x)*(x + x^2 + x^3)),x)

[Out]

- log(log(x)*(x + x^2 + 1)) - log(x)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.75 \begin {gather*} - \log {\left (x^{3} + x^{2} + x \right )} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2-2*x-1)*ln(x)-x**2-x-1)/(x**3+x**2+x)/ln(x),x)

[Out]

-log(x**3 + x**2 + x) - log(log(x))

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