3.58.33 \(\int \frac {-48-123 x-66 x^2-20 x^3-6 x^4-x^5+(18 x+12 x^2+2 x^3) \log (4 x)}{9 x^3+6 x^4+x^5} \, dx\)

Optimal. Leaf size=31 \[ \frac {9-x^2+2 \left (x+\frac {4}{x (3+x)}-\log (4 x)\right )}{x} \]

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Rubi [A]  time = 0.34, antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 15, number of rules used = 6, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {1594, 27, 6742, 44, 43, 2304} \begin {gather*} \frac {8}{3 x^2}-x+\frac {8}{9 (x+3)}+\frac {73}{9 x}-\frac {2 \log (4 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48 - 123*x - 66*x^2 - 20*x^3 - 6*x^4 - x^5 + (18*x + 12*x^2 + 2*x^3)*Log[4*x])/(9*x^3 + 6*x^4 + x^5),x]

[Out]

8/(3*x^2) + 73/(9*x) - x + 8/(9*(3 + x)) - (2*Log[4*x])/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 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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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 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 {-48-123 x-66 x^2-20 x^3-6 x^4-x^5+\left (18 x+12 x^2+2 x^3\right ) \log (4 x)}{x^3 \left (9+6 x+x^2\right )} \, dx\\ &=\int \frac {-48-123 x-66 x^2-20 x^3-6 x^4-x^5+\left (18 x+12 x^2+2 x^3\right ) \log (4 x)}{x^3 (3+x)^2} \, dx\\ &=\int \left (-\frac {20}{(3+x)^2}-\frac {48}{x^3 (3+x)^2}-\frac {123}{x^2 (3+x)^2}-\frac {66}{x (3+x)^2}-\frac {6 x}{(3+x)^2}-\frac {x^2}{(3+x)^2}+\frac {2 \log (4 x)}{x^2}\right ) \, dx\\ &=\frac {20}{3+x}+2 \int \frac {\log (4 x)}{x^2} \, dx-6 \int \frac {x}{(3+x)^2} \, dx-48 \int \frac {1}{x^3 (3+x)^2} \, dx-66 \int \frac {1}{x (3+x)^2} \, dx-123 \int \frac {1}{x^2 (3+x)^2} \, dx-\int \frac {x^2}{(3+x)^2} \, dx\\ &=-\frac {2}{x}+\frac {20}{3+x}-\frac {2 \log (4 x)}{x}-6 \int \left (-\frac {3}{(3+x)^2}+\frac {1}{3+x}\right ) \, dx-48 \int \left (\frac {1}{9 x^3}-\frac {2}{27 x^2}+\frac {1}{27 x}-\frac {1}{27 (3+x)^2}-\frac {1}{27 (3+x)}\right ) \, dx-66 \int \left (\frac {1}{9 x}-\frac {1}{3 (3+x)^2}-\frac {1}{9 (3+x)}\right ) \, dx-123 \int \left (\frac {1}{9 x^2}-\frac {2}{27 x}+\frac {1}{9 (3+x)^2}+\frac {2}{27 (3+x)}\right ) \, dx-\int \left (1+\frac {9}{(3+x)^2}-\frac {6}{3+x}\right ) \, dx\\ &=\frac {8}{3 x^2}+\frac {73}{9 x}-x+\frac {8}{9 (3+x)}-\frac {2 \log (4 x)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 36, normalized size = 1.16 \begin {gather*} \frac {8}{3 x^2}+\frac {73}{9 x}-x+\frac {8}{9 (3+x)}-\frac {2 \log (4 x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48 - 123*x - 66*x^2 - 20*x^3 - 6*x^4 - x^5 + (18*x + 12*x^2 + 2*x^3)*Log[4*x])/(9*x^3 + 6*x^4 + x^
5),x]

[Out]

8/(3*x^2) + 73/(9*x) - x + 8/(9*(3 + x)) - (2*Log[4*x])/x

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fricas [A]  time = 0.71, size = 44, normalized size = 1.42 \begin {gather*} -\frac {x^{4} + 3 \, x^{3} - 9 \, x^{2} + 2 \, {\left (x^{2} + 3 \, x\right )} \log \left (4 \, x\right ) - 27 \, x - 8}{x^{3} + 3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+12*x^2+18*x)*log(4*x)-x^5-6*x^4-20*x^3-66*x^2-123*x-48)/(x^5+6*x^4+9*x^3),x, algorithm="fric
as")

[Out]

-(x^4 + 3*x^3 - 9*x^2 + 2*(x^2 + 3*x)*log(4*x) - 27*x - 8)/(x^3 + 3*x^2)

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giac [A]  time = 0.12, size = 30, normalized size = 0.97 \begin {gather*} -x - \frac {2 \, \log \left (4 \, x\right )}{x} + \frac {8}{9 \, {\left (x + 3\right )}} + \frac {73 \, x + 24}{9 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+12*x^2+18*x)*log(4*x)-x^5-6*x^4-20*x^3-66*x^2-123*x-48)/(x^5+6*x^4+9*x^3),x, algorithm="giac
")

[Out]

-x - 2*log(4*x)/x + 8/9/(x + 3) + 1/9*(73*x + 24)/x^2

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maple [A]  time = 0.06, size = 33, normalized size = 1.06

method	result	size
derivativedivides	\(-\frac {2 \ln \left (4 x \right )}{x}+\frac {73}{9 x}-x +\frac {8}{3 x^{2}}+\frac {32}{9 \left (4 x +12\right )}\)	\(33\)
default	\(-\frac {2 \ln \left (4 x \right )}{x}+\frac {73}{9 x}-x +\frac {8}{3 x^{2}}+\frac {32}{9 \left (4 x +12\right )}\)	\(33\)
risch	\(-\frac {2 \ln \left (4 x \right )}{x}-\frac {x^{4}+3 x^{3}-9 x^{2}-27 x -8}{x^{2} \left (3+x \right )}\)	\(39\)
norman	\(\frac {8+18 x^{2}-2 x^{2} \ln \left (4 x \right )+27 x -x^{4}-6 x \ln \left (4 x \right )}{x^{2} \left (3+x \right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+12*x^2+18*x)*ln(4*x)-x^5-6*x^4-20*x^3-66*x^2-123*x-48)/(x^5+6*x^4+9*x^3),x,method=_RETURNVERBOSE)

[Out]

-2*ln(4*x)/x+73/9/x-x+8/3/x^2+32/9/(4*x+12)

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maxima [B]  time = 0.68, size = 63, normalized size = 2.03 \begin {gather*} -x - \frac {8 \, {\left (2 \, x^{2} + 3 \, x - 3\right )}}{3 \, {\left (x^{3} + 3 \, x^{2}\right )}} + \frac {41 \, {\left (2 \, x + 3\right )}}{3 \, {\left (x^{2} + 3 \, x\right )}} - \frac {2 \, {\left (2 \, \log \relax (2) + \log \relax (x) + 1\right )}}{x} - \frac {11}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+12*x^2+18*x)*log(4*x)-x^5-6*x^4-20*x^3-66*x^2-123*x-48)/(x^5+6*x^4+9*x^3),x, algorithm="maxi
ma")

[Out]

-x - 8/3*(2*x^2 + 3*x - 3)/(x^3 + 3*x^2) + 41/3*(2*x + 3)/(x^2 + 3*x) - 2*(2*log(2) + log(x) + 1)/x - 11/(x +
3)

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mupad [B]  time = 3.63, size = 38, normalized size = 1.23 \begin {gather*} -x-\frac {x^2\,\left (2\,\ln \left (4\,x\right )-9\right )+x\,\left (6\,\ln \left (4\,x\right )-27\right )-8}{x^2\,\left (x+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(123*x - log(4*x)*(18*x + 12*x^2 + 2*x^3) + 66*x^2 + 20*x^3 + 6*x^4 + x^5 + 48)/(9*x^3 + 6*x^4 + x^5),x)

[Out]

- x - (x^2*(2*log(4*x) - 9) + x*(6*log(4*x) - 27) - 8)/(x^2*(x + 3))

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sympy [A]  time = 0.17, size = 31, normalized size = 1.00 \begin {gather*} - x - \frac {- 9 x^{2} - 27 x - 8}{x^{3} + 3 x^{2}} - \frac {2 \log {\left (4 x \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+12*x**2+18*x)*ln(4*x)-x**5-6*x**4-20*x**3-66*x**2-123*x-48)/(x**5+6*x**4+9*x**3),x)

[Out]

-x - (-9*x**2 - 27*x - 8)/(x**3 + 3*x**2) - 2*log(4*x)/x

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