3.81.13 \(\int \frac {-57+2 x+19 x^2+(-6+2 x^2) \log (x)}{9 x^3-6 x^4+x^5} \, dx\)

Optimal. Leaf size=19 \[ 5-\frac {(1+x) (10+\log (x))}{(-3+x) x^2} \]

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Rubi [B]  time = 0.28, antiderivative size = 64, normalized size of antiderivative = 3.37, number of steps used = 12, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1594, 27, 6742, 893, 2357, 2314, 31, 2304} \begin {gather*} \frac {10}{3 x^2}+\frac {\log (x)}{3 x^2}+\frac {40}{9 (3-x)}+\frac {40}{9 x}+\frac {4 x \log (x)}{27 (3-x)}+\frac {4 \log (x)}{9 x}+\frac {4 \log (x)}{27} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-57 + 2*x + 19*x^2 + (-6 + 2*x^2)*Log[x])/(9*x^3 - 6*x^4 + x^5),x]

[Out]

40/(9*(3 - x)) + 10/(3*x^2) + 40/(9*x) + (4*Log[x])/27 + Log[x]/(3*x^2) + (4*Log[x])/(9*x) + (4*x*Log[x])/(27*
(3 - 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 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 {-57+2 x+19 x^2+\left (-6+2 x^2\right ) \log (x)}{x^3 \left (9-6 x+x^2\right )} \, dx\\ &=\int \frac {-57+2 x+19 x^2+\left (-6+2 x^2\right ) \log (x)}{(-3+x)^2 x^3} \, dx\\ &=\int \left (\frac {-57+2 x+19 x^2}{(-3+x)^2 x^3}+\frac {2 \left (-3+x^2\right ) \log (x)}{(-3+x)^2 x^3}\right ) \, dx\\ &=2 \int \frac {\left (-3+x^2\right ) \log (x)}{(-3+x)^2 x^3} \, dx+\int \frac {-57+2 x+19 x^2}{(-3+x)^2 x^3} \, dx\\ &=2 \int \left (\frac {2 \log (x)}{9 (-3+x)^2}-\frac {\log (x)}{3 x^3}-\frac {2 \log (x)}{9 x^2}\right ) \, dx+\int \left (\frac {40}{9 (-3+x)^2}-\frac {4}{27 (-3+x)}-\frac {19}{3 x^3}-\frac {4}{x^2}+\frac {4}{27 x}\right ) \, dx\\ &=\frac {40}{9 (3-x)}+\frac {19}{6 x^2}+\frac {4}{x}-\frac {4}{27} \log (3-x)+\frac {4 \log (x)}{27}+\frac {4}{9} \int \frac {\log (x)}{(-3+x)^2} \, dx-\frac {4}{9} \int \frac {\log (x)}{x^2} \, dx-\frac {2}{3} \int \frac {\log (x)}{x^3} \, dx\\ &=\frac {40}{9 (3-x)}+\frac {10}{3 x^2}+\frac {40}{9 x}-\frac {4}{27} \log (3-x)+\frac {4 \log (x)}{27}+\frac {\log (x)}{3 x^2}+\frac {4 \log (x)}{9 x}+\frac {4 x \log (x)}{27 (3-x)}+\frac {4}{27} \int \frac {1}{-3+x} \, dx\\ &=\frac {40}{9 (3-x)}+\frac {10}{3 x^2}+\frac {40}{9 x}+\frac {4 \log (x)}{27}+\frac {\log (x)}{3 x^2}+\frac {4 \log (x)}{9 x}+\frac {4 x \log (x)}{27 (3-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 17, normalized size = 0.89 \begin {gather*} -\frac {(1+x) (10+\log (x))}{(-3+x) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-57 + 2*x + 19*x^2 + (-6 + 2*x^2)*Log[x])/(9*x^3 - 6*x^4 + x^5),x]

[Out]

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

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fricas [A]  time = 1.20, size = 24, normalized size = 1.26 \begin {gather*} -\frac {{\left (x + 1\right )} \log \relax (x) + 10 \, x + 10}{x^{3} - 3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-6)*log(x)+19*x^2+2*x-57)/(x^5-6*x^4+9*x^3),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.13, size = 40, normalized size = 2.11 \begin {gather*} -\frac {1}{9} \, {\left (\frac {4}{x - 3} - \frac {4 \, x + 3}{x^{2}}\right )} \log \relax (x) - \frac {40}{9 \, {\left (x - 3\right )}} + \frac {10 \, {\left (4 \, x + 3\right )}}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-6)*log(x)+19*x^2+2*x-57)/(x^5-6*x^4+9*x^3),x, algorithm="giac")

[Out]

-1/9*(4/(x - 3) - (4*x + 3)/x^2)*log(x) - 40/9/(x - 3) + 10/9*(4*x + 3)/x^2

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maple [A]  time = 0.05, size = 24, normalized size = 1.26

method	result	size
norman	\(\frac {-10-10 x -x \ln \relax (x )-\ln \relax (x )}{x^{2} \left (x -3\right )}\)	\(24\)
risch	\(-\frac {\left (x +1\right ) \ln \relax (x )}{x^{2} \left (x -3\right )}-\frac {10 \left (x +1\right )}{x^{2} \left (x -3\right )}\)	\(30\)
default	\(\frac {10}{3 x^{2}}+\frac {40}{9 x}+\frac {4 \ln \relax (x )}{27}-\frac {40}{9 \left (x -3\right )}+\frac {4 \ln \relax (x )}{9 x}+\frac {\ln \relax (x )}{3 x^{2}}-\frac {4 \ln \relax (x ) x}{27 \left (x -3\right )}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-6)*ln(x)+19*x^2+2*x-57)/(x^5-6*x^4+9*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.41, size = 93, normalized size = 4.89 \begin {gather*} \frac {24 \, x^{2} - 2 \, {\left (4 \, x^{3} - 12 \, x^{2} + 27 \, x + 27\right )} \log \relax (x) - 63 \, x - 27}{54 \, {\left (x^{3} - 3 \, x^{2}\right )}} + \frac {19 \, {\left (2 \, x^{2} - 3 \, x - 3\right )}}{6 \, {\left (x^{3} - 3 \, x^{2}\right )}} - \frac {2 \, {\left (2 \, x - 3\right )}}{9 \, {\left (x^{2} - 3 \, x\right )}} - \frac {19}{3 \, {\left (x - 3\right )}} + \frac {4}{27} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-6)*log(x)+19*x^2+2*x-57)/(x^5-6*x^4+9*x^3),x, algorithm="maxima")

[Out]

1/54*(24*x^2 - 2*(4*x^3 - 12*x^2 + 27*x + 27)*log(x) - 63*x - 27)/(x^3 - 3*x^2) + 19/6*(2*x^2 - 3*x - 3)/(x^3
- 3*x^2) - 2/9*(2*x - 3)/(x^2 - 3*x) - 19/3/(x - 3) + 4/27*log(x)

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mupad [B]  time = 5.51, size = 17, normalized size = 0.89 \begin {gather*} -\frac {\left (\ln \relax (x)+10\right )\,\left (x+1\right )}{x^2\,\left (x-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 19*x^2 + log(x)*(2*x^2 - 6) - 57)/(9*x^3 - 6*x^4 + x^5),x)

[Out]

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

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sympy [B]  time = 0.18, size = 31, normalized size = 1.63 \begin {gather*} \frac {- 10 x - 10}{x^{3} - 3 x^{2}} + \frac {\left (- x - 1\right ) \log {\relax (x )}}{x^{3} - 3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-6)*ln(x)+19*x**2+2*x-57)/(x**5-6*x**4+9*x**3),x)

[Out]

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

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