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

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 {gather*} -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)}{6 x} \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \frac {-72-24 x-5 x^2-x^3+(6+2 x) \log (2)+(72+12 x+(-6-x) \log (2)) \log (6+x)}{x^2 (6+x)} \, dx\\ &=\int \left (\frac {-72-5 x^2-x^3-x (24-\log (4))+\log (64)}{x^2 (6+x)}-\frac {(-12+\log (2)) \log (6+x)}{x^2}\right ) \, dx\\ &=(12-\log (2)) \int \frac {\log (6+x)}{x^2} \, dx+\int \frac {-72-5 x^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 \left (-1+\frac {108-\log (64)}{36 (6+x)}+\frac {-72+\log (64)}{6 x^2}+\frac {-72+\log (64)}{36 x}\right ) \, dx\\ &=-x+\frac {72-\log (64)}{6 x}-\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)}{6 x}+\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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 39, normalized size = 1.95 \begin {gather*} \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 {gather*}

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 {gather*} -\frac {x^{2} - {\left (x + \log \relax (2) - 12\right )} \log \left (x + 6\right ) + \log \relax (2) - 12}{x} \end {gather*}

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 {gather*} -x + \frac {{\left (\log \relax (2) - 12\right )} \log \left (x + 6\right )}{x} - \frac {\log \relax (2) - 12}{x} + \log \left (x + 6\right ) \end {gather*}

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	\(\frac {\left (\ln \relax (2)-12\right ) \ln \left (x +6\right )+x \ln \left (x +6\right )-x^{2}-\ln \relax (2)+12}{x}\)	\(31\)
risch	\(\frac {\left (\ln \relax (2)-12\right ) \ln \left (x +6\right )}{x}+\frac {x \ln \left (x +6\right )-x^{2}-\ln \relax (2)+12}{x}\)	\(35\)
derivativedivides	\(\frac {\ln \relax (2) \ln \left (x +6\right ) \left (x +6\right )}{6 x}-\frac {\ln \relax (2)}{x}-\frac {\ln \left (x +6\right ) \ln \relax (2)}{6}-\frac {2 \ln \left (x +6\right ) \left (x +6\right )}{x}-x -6+\frac {12}{x}+3 \ln \left (x +6\right )\)	\(58\)
default	\(\frac {\ln \relax (2) \ln \left (x +6\right ) \left (x +6\right )}{6 x}-\frac {\ln \relax (2)}{x}-\frac {\ln \left (x +6\right ) \ln \relax (2)}{6}-\frac {2 \ln \left (x +6\right ) \left (x +6\right )}{x}-x -6+\frac {12}{x}+3 \ln \left (x +6\right )\)	\(58\)

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 {gather*} -\frac {1}{6} \, {\left (\frac {6}{x} - \log \left (x + 6\right ) + \log \relax (x)\right )} \log \relax (2) - \frac {1}{3} \, {\left (\log \left (x + 6\right ) - \log \relax (x)\right )} \log \relax (2) - \frac {1}{6} \, {\left (\log \relax (2) - 12\right )} \log \relax (x) - x + \frac {{\left (x {\left (\log \relax (2) - 12\right )} + 6 \, \log \relax (2) - 72\right )} \log \left (x + 6\right )}{6 \, x} + \frac {12}{x} + 3 \, \log \left (x + 6\right ) - 2 \, \log \relax (x) \end {gather*}

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 {gather*} \ln \left (x+6\right )-x+\frac {\left (\ln \left (x+6\right )-1\right )\,\left (\ln \relax (2)-12\right )}{x} \end {gather*}

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 {gather*} - x + \log {\left (x + 6 \right )} + \frac {\left (-12 + \log {\relax (2 )}\right ) \log {\left (x + 6 \right )}}{x} - \frac {-12 + \log {\relax (2 )}}{x} \end {gather*}

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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