3.33.37 $\int \frac{-72-24x-5x^2-x^3+(6+2x)\log (2)+(72+12x+(-6-x)\log (2))\log (6+x)}{6x^2+x^3}dx$

Optimal. Leaf size=20 $$3-x+\frac{(-12+x+\log (2))(-1+\log (6+x))}{x}$$

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Rubi [B]  time = 0.37, antiderivative size = 84, normalized size of antiderivative = 4.20, number of steps used = 9, number of rules used = 7, integrand size = 53, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.132, Rules used = {1593, 6742, 1620, 2395, 36, 29, 31} $$\begin{array}{c}-x-\frac{1}{36}(72-\log (64))\log (x)+\frac{1}{6}(12-\log (2))\log (x)+\frac{1}{36}(108-\log (64))\log (x+6)-\frac{1}{6}(12-\log (2))\log (x+6)-\frac{(12-\log (2))\log (x+6)}{x}+\frac{72-\log (64)}{6x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(-72 - 24*x - 5*x^2 - x^3 + (6 + 2*x)*Log[2] + (72 + 12*x + (-6 - x)*Log[2])*Log[6 + x])/(6*x^2 + x^3),x]

[Out]

-x + (72 - Log[64])/(6*x) + ((12 - Log[2])*Log[x])/6 - ((72 - Log[64])*Log[x])/36 - ((12 - Log[2])*Log[6 + x])
/6 - ((12 - Log[2])*Log[6 + x])/x + ((108 - Log[64])*Log[6 + x])/36

Rule 29

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

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1593

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 2395

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{-72-24x-5x^2-x^3+(6+2x)\log (2)+(72+12x+(-6-x)\log (2))\log (6+x)}{x^2(6+x)}dx\\ & =\int (\frac{-72-5x^2-x^3-x(24-\log (4))+\log (64)}{x^2(6+x)}-\frac{(-12+\log (2))\log (6+x)}{x^2})dx\\ & =(12-\log (2))\int \frac{\log (6+x)}{x^2}dx+\int \frac{-72-5x^2-x^3-x(24-\log (4))+\log (64)}{x^2(6+x)}dx\\ & =-\frac{(12-\log (2))\log (6+x)}{x}+(12-\log (2))\int \frac{1}{x(6+x)}dx+\int (-1+\frac{108-\log (64)}{36(6+x)}+\frac{-72+\log (64)}{6x^2}+\frac{-72+\log (64)}{36x})dx\\ & =-x+\frac{72-\log (64)}{6x}-\frac{1}{36}(72-\log (64))\log (x)-\frac{(12-\log (2))\log (6+x)}{x}+\frac{1}{36}(108-\log (64))\log (6+x)+\frac{1}{6}(12-\log (2))\int \frac{1}{x}dx+\frac{1}{6}(-12+\log (2))\int \frac{1}{6+x}dx\\ & =-x+\frac{72-\log (64)}{6x}+\frac{1}{6}(12-\log (2))\log (x)-\frac{1}{36}(72-\log (64))\log (x)-\frac{1}{6}(12-\log (2))\log (6+x)-\frac{(12-\log (2))\log (6+x)}{x}+\frac{1}{36}(108-\log (64))\log (6+x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.11, size = 39, normalized size = 1.95 $$\begin{array}{c}\frac{12}{x}-x-\frac{\log (2)}{x}+\log (6+x)-\frac{12\log (6+x)}{x}+\frac{\log (2)\log (6+x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-72 - 24*x - 5*x^2 - x^3 + (6 + 2*x)*Log[2] + (72 + 12*x + (-6 - x)*Log[2])*Log[6 + x])/(6*x^2 + x^
3),x]

[Out]

12/x - x - Log[2]/x + Log[6 + x] - (12*Log[6 + x])/x + (Log[2]*Log[6 + x])/x

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fricas [A]  time = 0.52, size = 23, normalized size = 1.15 $$\begin{array}{c}-\frac{x^2-(x+\log (2)-12)\log (x+6)+\log (2)-12}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-6)*log(2)+12*x+72)*log(x+6)+(2*x+6)*log(2)-x^3-5*x^2-24*x-72)/(x^3+6*x^2),x, algorithm="fricas
")

[Out]

-(x^2 - (x + log(2) - 12)*log(x + 6) + log(2) - 12)/x

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giac [A]  time = 0.21, size = 29, normalized size = 1.45 $$\begin{array}{c}-x+\frac{(\log (2)-12)\log (x+6)}{x}-\frac{\log (2)-12}{x}+\log (x+6)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-6)*log(2)+12*x+72)*log(x+6)+(2*x+6)*log(2)-x^3-5*x^2-24*x-72)/(x^3+6*x^2),x, algorithm="giac")

[Out]

-x + (log(2) - 12)*log(x + 6)/x - (log(2) - 12)/x + log(x + 6)

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maple [A]  time = 0.62, size = 31, normalized size = 1.55

method	result	size
norman	$\frac{(\ln (2)-12)\ln (x+6)+x\ln (x+6)-x^2-\ln (2)+12}{x}$	$31$
risch	$\frac{(\ln (2)-12)\ln (x+6)}{x}+\frac{x\ln (x+6)-x^2-\ln (2)+12}{x}$	$35$
derivativedivides	$\frac{\ln (2)\ln (x+6)(x+6)}{6x}-\frac{\ln (2)}{x}-\frac{\ln (x+6)\ln (2)}{6}-\frac{2\ln (x+6)(x+6)}{x}-x-6+\frac{12}{x}+3\ln (x+6)$	$58$
default	$\frac{\ln (2)\ln (x+6)(x+6)}{6x}-\frac{\ln (2)}{x}-\frac{\ln (x+6)\ln (2)}{6}-\frac{2\ln (x+6)(x+6)}{x}-x-6+\frac{12}{x}+3\ln (x+6)$	$58$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x-6)*ln(2)+12*x+72)*ln(x+6)+(2*x+6)*ln(2)-x^3-5*x^2-24*x-72)/(x^3+6*x^2),x,method=_RETURNVERBOSE)

[Out]

((ln(2)-12)*ln(x+6)+x*ln(x+6)-x^2-ln(2)+12)/x

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maxima [B]  time = 0.53, size = 79, normalized size = 3.95 $$\begin{array}{c}-\frac{1}{6}(\frac{6}{x}-\log (x+6)+\log (x))\log (2)-\frac{1}{3}(\log (x+6)-\log (x))\log (2)-\frac{1}{6}(\log (2)-12)\log (x)-x+\frac{(x(\log (2)-12)+6\log (2)-72)\log (x+6)}{6x}+\frac{12}{x}+3\log (x+6)-2\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-6)*log(2)+12*x+72)*log(x+6)+(2*x+6)*log(2)-x^3-5*x^2-24*x-72)/(x^3+6*x^2),x, algorithm="maxima
")

[Out]

-1/6*(6/x - log(x + 6) + log(x))*log(2) - 1/3*(log(x + 6) - log(x))*log(2) - 1/6*(log(2) - 12)*log(x) - x + 1/
6*(x*(log(2) - 12) + 6*log(2) - 72)*log(x + 6)/x + 12/x + 3*log(x + 6) - 2*log(x)

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mupad [B]  time = 1.99, size = 22, normalized size = 1.10 $$\begin{array}{c}\ln (x+6)-x+\frac{(\ln (x+6)-1)(\ln (2)-12)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*x - log(2)*(2*x + 6) - log(x + 6)*(12*x - log(2)*(x + 6) + 72) + 5*x^2 + x^3 + 72)/(6*x^2 + x^3),x)

[Out]

log(x + 6) - x + ((log(x + 6) - 1)*(log(2) - 12))/x

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sympy [A]  time = 0.34, size = 24, normalized size = 1.20 $$\begin{array}{c}-x+\log (x+6)+\frac{(-12+\log (2))\log (x+6)}{x}-\frac{-12+\log (2)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x-6)*ln(2)+12*x+72)*ln(x+6)+(2*x+6)*ln(2)-x**3-5*x**2-24*x-72)/(x**3+6*x**2),x)

[Out]

-x + log(x + 6) + (-12 + log(2))*log(x + 6)/x - (-12 + log(2))/x

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