3.72.85 \(\int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{120 x+120 x^2+30 x^3} \, dx\)

Optimal. Leaf size=18 \[ 12+\frac {5 x}{3}+\frac {\log (x)}{10 (2+x)} \]

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Rubi [A]  time = 0.23, antiderivative size = 24, normalized size of antiderivative = 1.33, number of steps used = 13, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1594, 27, 12, 6742, 44, 43, 2314, 31} \begin {gather*} \frac {5 x}{3}-\frac {x \log (x)}{20 (x+2)}+\frac {\log (x)}{20} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + 203*x + 200*x^2 + 50*x^3 - 3*x*Log[x])/(120*x + 120*x^2 + 30*x^3),x]

[Out]

(5*x)/3 + Log[x]/20 - (x*Log[x])/(20*(2 + x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 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 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 {6+203 x+200 x^2+50 x^3-3 x \log (x)}{x \left (120+120 x+30 x^2\right )} \, dx\\ &=\int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{30 x (2+x)^2} \, dx\\ &=\frac {1}{30} \int \frac {6+203 x+200 x^2+50 x^3-3 x \log (x)}{x (2+x)^2} \, dx\\ &=\frac {1}{30} \int \left (\frac {203}{(2+x)^2}+\frac {6}{x (2+x)^2}+\frac {200 x}{(2+x)^2}+\frac {50 x^2}{(2+x)^2}-\frac {3 \log (x)}{(2+x)^2}\right ) \, dx\\ &=-\frac {203}{30 (2+x)}-\frac {1}{10} \int \frac {\log (x)}{(2+x)^2} \, dx+\frac {1}{5} \int \frac {1}{x (2+x)^2} \, dx+\frac {5}{3} \int \frac {x^2}{(2+x)^2} \, dx+\frac {20}{3} \int \frac {x}{(2+x)^2} \, dx\\ &=-\frac {203}{30 (2+x)}-\frac {x \log (x)}{20 (2+x)}+\frac {1}{20} \int \frac {1}{2+x} \, dx+\frac {1}{5} \int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx+\frac {5}{3} \int \left (1+\frac {4}{(2+x)^2}-\frac {4}{2+x}\right ) \, dx+\frac {20}{3} \int \left (-\frac {2}{(2+x)^2}+\frac {1}{2+x}\right ) \, dx\\ &=\frac {5 x}{3}+\frac {\log (x)}{20}-\frac {x \log (x)}{20 (2+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 0.94 \begin {gather*} \frac {1}{30} \left (50 x+\frac {3 \log (x)}{2+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + 203*x + 200*x^2 + 50*x^3 - 3*x*Log[x])/(120*x + 120*x^2 + 30*x^3),x]

[Out]

(50*x + (3*Log[x])/(2 + x))/30

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fricas [A]  time = 1.13, size = 20, normalized size = 1.11 \begin {gather*} \frac {50 \, x^{2} + 100 \, x + 3 \, \log \relax (x)}{30 \, {\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*log(x)+50*x^3+200*x^2+203*x+6)/(30*x^3+120*x^2+120*x),x, algorithm="fricas")

[Out]

1/30*(50*x^2 + 100*x + 3*log(x))/(x + 2)

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giac [A]  time = 0.14, size = 13, normalized size = 0.72 \begin {gather*} \frac {5}{3} \, x + \frac {\log \relax (x)}{10 \, {\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*log(x)+50*x^3+200*x^2+203*x+6)/(30*x^3+120*x^2+120*x),x, algorithm="giac")

[Out]

5/3*x + 1/10*log(x)/(x + 2)

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maple [A]  time = 0.03, size = 14, normalized size = 0.78

method	result	size
risch	\(\frac {\ln \relax (x )}{10 x +20}+\frac {5 x}{3}\)	\(14\)
norman	\(\frac {\frac {\ln \relax (x )}{10}+\frac {5 x^{2}}{3}-\frac {20}{3}}{2+x}\)	\(18\)
default	\(\frac {5 x}{3}+\frac {\ln \relax (x )}{20}-\frac {\ln \relax (x ) x}{20 \left (2+x \right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x*ln(x)+50*x^3+200*x^2+203*x+6)/(30*x^3+120*x^2+120*x),x,method=_RETURNVERBOSE)

[Out]

1/10*ln(x)/(2+x)+5/3*x

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maxima [A]  time = 0.36, size = 13, normalized size = 0.72 \begin {gather*} \frac {5}{3} \, x + \frac {\log \relax (x)}{10 \, {\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*log(x)+50*x^3+200*x^2+203*x+6)/(30*x^3+120*x^2+120*x),x, algorithm="maxima")

[Out]

5/3*x + 1/10*log(x)/(x + 2)

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mupad [B]  time = 4.38, size = 14, normalized size = 0.78 \begin {gather*} \frac {5\,x}{3}+\frac {\ln \relax (x)}{10\,\left (x+2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((203*x - 3*x*log(x) + 200*x^2 + 50*x^3 + 6)/(120*x + 120*x^2 + 30*x^3),x)

[Out]

(5*x)/3 + log(x)/(10*(x + 2))

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sympy [A]  time = 0.13, size = 12, normalized size = 0.67 \begin {gather*} \frac {5 x}{3} + \frac {\log {\relax (x )}}{10 x + 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*ln(x)+50*x**3+200*x**2+203*x+6)/(30*x**3+120*x**2+120*x),x)

[Out]

5*x/3 + log(x)/(10*x + 20)

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