3.1.44 \(\int \frac {-5+7 x+8 x^2-4 x^3+(-6-2 x+8 x^2-8 x^3) \log (x)}{1-4 x+4 x^2} \, dx\)

Optimal. Leaf size=26 \[ 2+x+x \left (-x+\frac {2}{\frac {1}{3} (-1-x)+x}\right ) \log (x) \]

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Rubi [A]  time = 0.19, antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 14, number of rules used = 7, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 6742, 43, 2357, 2304, 2314, 31} \begin {gather*} x^2 (-\log (x))+x-\frac {6 x \log (x)}{1-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 + 7*x + 8*x^2 - 4*x^3 + (-6 - 2*x + 8*x^2 - 8*x^3)*Log[x])/(1 - 4*x + 4*x^2),x]

[Out]

x - (6*x*Log[x])/(1 - 2*x) - x^2*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

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 {-5+7 x+8 x^2-4 x^3+\left (-6-2 x+8 x^2-8 x^3\right ) \log (x)}{(-1+2 x)^2} \, dx\\ &=\int \left (-\frac {5}{(-1+2 x)^2}+\frac {7 x}{(-1+2 x)^2}+\frac {8 x^2}{(-1+2 x)^2}-\frac {4 x^3}{(-1+2 x)^2}-\frac {2 \left (3+x-4 x^2+4 x^3\right ) \log (x)}{(-1+2 x)^2}\right ) \, dx\\ &=-\frac {5}{2 (1-2 x)}-2 \int \frac {\left (3+x-4 x^2+4 x^3\right ) \log (x)}{(-1+2 x)^2} \, dx-4 \int \frac {x^3}{(-1+2 x)^2} \, dx+7 \int \frac {x}{(-1+2 x)^2} \, dx+8 \int \frac {x^2}{(-1+2 x)^2} \, dx\\ &=-\frac {5}{2 (1-2 x)}-2 \int \left (x \log (x)+\frac {3 \log (x)}{(-1+2 x)^2}\right ) \, dx-4 \int \left (\frac {1}{4}+\frac {x}{4}+\frac {1}{8 (-1+2 x)^2}+\frac {3}{8 (-1+2 x)}\right ) \, dx+7 \int \left (\frac {1}{2 (-1+2 x)^2}+\frac {1}{2 (-1+2 x)}\right ) \, dx+8 \int \left (\frac {1}{4}+\frac {1}{4 (-1+2 x)^2}+\frac {1}{2 (-1+2 x)}\right ) \, dx\\ &=x-\frac {x^2}{2}+3 \log (1-2 x)-2 \int x \log (x) \, dx-6 \int \frac {\log (x)}{(-1+2 x)^2} \, dx\\ &=x+3 \log (1-2 x)-\frac {6 x \log (x)}{1-2 x}-x^2 \log (x)-6 \int \frac {1}{-1+2 x} \, dx\\ &=x-\frac {6 x \log (x)}{1-2 x}-x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 19, normalized size = 0.73 \begin {gather*} x+x \left (-x+\frac {6}{-1+2 x}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 7*x + 8*x^2 - 4*x^3 + (-6 - 2*x + 8*x^2 - 8*x^3)*Log[x])/(1 - 4*x + 4*x^2),x]

[Out]

x + x*(-x + 6/(-1 + 2*x))*Log[x]

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fricas [A]  time = 0.64, size = 35, normalized size = 1.35 \begin {gather*} \frac {2 \, x^{2} - {\left (2 \, x^{3} - x^{2} - 6 \, x\right )} \log \relax (x) - x}{2 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+8*x^2-2*x-6)*log(x)-4*x^3+8*x^2+7*x-5)/(4*x^2-4*x+1),x, algorithm="fricas")

[Out]

(2*x^2 - (2*x^3 - x^2 - 6*x)*log(x) - x)/(2*x - 1)

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giac [A]  time = 0.29, size = 23, normalized size = 0.88 \begin {gather*} -{\left (x^{2} - \frac {3}{2 \, x - 1}\right )} \log \relax (x) + x + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+8*x^2-2*x-6)*log(x)-4*x^3+8*x^2+7*x-5)/(4*x^2-4*x+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 22, normalized size = 0.85

method	result	size
default	\(x -x^{2} \ln \relax (x )+\frac {6 \ln \relax (x ) x}{2 x -1}\)	\(22\)
risch	\(-\frac {\left (2 x^{3}-x^{2}-3\right ) \ln \relax (x )}{2 x -1}+x +3 \ln \relax (x )\)	\(30\)
norman	\(\frac {x^{2} \ln \relax (x )-x +2 x^{2}+6 x \ln \relax (x )-2 x^{3} \ln \relax (x )}{2 x -1}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^3+8*x^2-2*x-6)*ln(x)-4*x^3+8*x^2+7*x-5)/(4*x^2-4*x+1),x,method=_RETURNVERBOSE)

[Out]

x-x^2*ln(x)+6*ln(x)*x/(2*x-1)

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maxima [A]  time = 0.49, size = 24, normalized size = 0.92 \begin {gather*} -x^{2} \log \relax (x) + x + \frac {3 \, \log \relax (x)}{2 \, x - 1} + 3 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+8*x^2-2*x-6)*log(x)-4*x^3+8*x^2+7*x-5)/(4*x^2-4*x+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.31, size = 24, normalized size = 0.92 \begin {gather*} x+3\,\ln \relax (x)-x^2\,\ln \relax (x)+\frac {3\,\ln \relax (x)}{2\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x^3 - 8*x^2 - 7*x + log(x)*(2*x - 8*x^2 + 8*x^3 + 6) + 5)/(4*x^2 - 4*x + 1),x)

[Out]

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

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sympy [A]  time = 0.18, size = 24, normalized size = 0.92 \begin {gather*} x + 3 \log {\relax (x )} + \frac {\left (- 2 x^{3} + x^{2} + 3\right ) \log {\relax (x )}}{2 x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**3+8*x**2-2*x-6)*ln(x)-4*x**3+8*x**2+7*x-5)/(4*x**2-4*x+1),x)

[Out]

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

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