3.43.29 \(\int \frac {-3-x+2 x^2+(-3+4 x+2 x^2) \log (x)}{(-3 x-x^2+2 x^3) \log (x)} \, dx\)

Optimal. Leaf size=23 \[ \log (x)-\log \left (\frac {2}{\left (-2+\frac {5}{1+x}\right ) \log (x)}\right ) \]

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Rubi [A]  time = 0.29, antiderivative size = 18, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 5, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {1594, 6728, 1628, 2302, 29} \begin {gather*} \log (3-2 x)+\log (x)-\log (x+1)+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-3 - x + 2*x^2 + (-3 + 4*x + 2*x^2)*Log[x])/((-3*x - x^2 + 2*x^3)*Log[x]),x]

[Out]

Log[3 - 2*x] + Log[x] - Log[1 + x] + Log[Log[x]]

Rule 29

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

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 {-3-x+2 x^2+\left (-3+4 x+2 x^2\right ) \log (x)}{x \left (-3-x+2 x^2\right ) \log (x)} \, dx\\ &=\int \left (\frac {-3+4 x+2 x^2}{x \left (-3-x+2 x^2\right )}+\frac {1}{x \log (x)}\right ) \, dx\\ &=\int \frac {-3+4 x+2 x^2}{x \left (-3-x+2 x^2\right )} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=\int \left (\frac {1}{-1-x}+\frac {1}{x}+\frac {2}{-3+2 x}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\log (3-2 x)+\log (x)-\log (1+x)+\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 18, normalized size = 0.78 \begin {gather*} \log (3-2 x)+\log (x)-\log (1+x)+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 - x + 2*x^2 + (-3 + 4*x + 2*x^2)*Log[x])/((-3*x - x^2 + 2*x^3)*Log[x]),x]

[Out]

Log[3 - 2*x] + Log[x] - Log[1 + x] + Log[Log[x]]

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fricas [A]  time = 0.76, size = 20, normalized size = 0.87 \begin {gather*} \log \left (2 \, x^{2} - 3 \, x\right ) - \log \left (x + 1\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2+4*x-3)*log(x)+2*x^2-x-3)/(2*x^3-x^2-3*x)/log(x),x, algorithm="fricas")

[Out]

log(2*x^2 - 3*x) - log(x + 1) + log(log(x))

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giac [A]  time = 0.24, size = 18, normalized size = 0.78 \begin {gather*} \log \left (2 \, x - 3\right ) - \log \left (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(((2*x^2+4*x-3)*log(x)+2*x^2-x-3)/(2*x^3-x^2-3*x)/log(x),x, algorithm="giac")

[Out]

log(2*x - 3) - log(x + 1) + log(x) + log(log(x))

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maple [A]  time = 0.03, size = 19, normalized size = 0.83

method	result	size
default	\(\ln \left (\ln \relax (x )\right )-\ln \left (x +1\right )+\ln \left (2 x -3\right )+\ln \relax (x )\)	\(19\)
norman	\(\ln \left (\ln \relax (x )\right )-\ln \left (x +1\right )+\ln \left (2 x -3\right )+\ln \relax (x )\)	\(19\)
risch	\(-\ln \left (x +1\right )+\ln \left (2 x^{2}-3 x \right )+\ln \left (\ln \relax (x )\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2+4*x-3)*ln(x)+2*x^2-x-3)/(2*x^3-x^2-3*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))-ln(x+1)+ln(2*x-3)+ln(x)

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maxima [A]  time = 0.38, size = 18, normalized size = 0.78 \begin {gather*} \log \left (2 \, x - 3\right ) - \log \left (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(((2*x^2+4*x-3)*log(x)+2*x^2-x-3)/(2*x^3-x^2-3*x)/log(x),x, algorithm="maxima")

[Out]

log(2*x - 3) - log(x + 1) + log(x) + log(log(x))

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mupad [B]  time = 3.04, size = 16, normalized size = 0.70 \begin {gather*} \ln \left (x-\frac {3}{2}\right )-\ln \left (x+1\right )+\ln \left (\ln \relax (x)\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x - 3/2) - log(x + 1) + log(log(x)) + log(x)

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sympy [A]  time = 0.15, size = 19, normalized size = 0.83 \begin {gather*} - \log {\left (x + 1 \right )} + \log {\left (2 x^{2} - 3 x \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2+4*x-3)*ln(x)+2*x**2-x-3)/(2*x**3-x**2-3*x)/ln(x),x)

[Out]

-log(x + 1) + log(2*x**2 - 3*x) + log(log(x))

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