3.38.31 \(\int \frac {3+x-2 x \log (x)}{135 x+135 x^2+45 x^3+5 x^4} \, dx\)

Optimal. Leaf size=22 \[ 3+e^{e^{64 e^4}}+\frac {\log (x)}{5 (3+x)^2} \]

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Rubi [A]  time = 0.17, antiderivative size = 11, normalized size of antiderivative = 0.50, number of steps used = 9, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6688, 12, 6742, 44, 2319} \begin {gather*} \frac {\log (x)}{5 (x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + x - 2*x*Log[x])/(135*x + 135*x^2 + 45*x^3 + 5*x^4),x]

[Out]

Log[x]/(5*(3 + x)^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 6688

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

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 {3+x-2 x \log (x)}{5 x (3+x)^3} \, dx\\ &=\frac {1}{5} \int \frac {3+x-2 x \log (x)}{x (3+x)^3} \, dx\\ &=\frac {1}{5} \int \left (\frac {1}{x (3+x)^2}-\frac {2 \log (x)}{(3+x)^3}\right ) \, dx\\ &=\frac {1}{5} \int \frac {1}{x (3+x)^2} \, dx-\frac {2}{5} \int \frac {\log (x)}{(3+x)^3} \, dx\\ &=\frac {\log (x)}{5 (3+x)^2}-\frac {1}{5} \int \frac {1}{x (3+x)^2} \, dx+\frac {1}{5} \int \left (\frac {1}{9 x}-\frac {1}{3 (3+x)^2}-\frac {1}{9 (3+x)}\right ) \, dx\\ &=\frac {1}{15 (3+x)}+\frac {\log (x)}{45}+\frac {\log (x)}{5 (3+x)^2}-\frac {1}{45} \log (3+x)-\frac {1}{5} \int \left (\frac {1}{9 x}-\frac {1}{3 (3+x)^2}-\frac {1}{9 (3+x)}\right ) \, dx\\ &=\frac {\log (x)}{5 (3+x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 11, normalized size = 0.50 \begin {gather*} \frac {\log (x)}{5 (3+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + x - 2*x*Log[x])/(135*x + 135*x^2 + 45*x^3 + 5*x^4),x]

[Out]

Log[x]/(5*(3 + x)^2)

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fricas [A]  time = 0.50, size = 14, normalized size = 0.64 \begin {gather*} \frac {\log \relax (x)}{5 \, {\left (x^{2} + 6 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)+3+x)/(5*x^4+45*x^3+135*x^2+135*x),x, algorithm="fricas")

[Out]

1/5*log(x)/(x^2 + 6*x + 9)

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giac [A]  time = 0.17, size = 14, normalized size = 0.64 \begin {gather*} \frac {\log \relax (x)}{5 \, {\left (x^{2} + 6 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)+3+x)/(5*x^4+45*x^3+135*x^2+135*x),x, algorithm="giac")

[Out]

1/5*log(x)/(x^2 + 6*x + 9)

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maple [A]  time = 0.04, size = 10, normalized size = 0.45

method	result	size
norman	\(\frac {\ln \relax (x )}{5 \left (3+x \right )^{2}}\)	\(10\)
risch	\(\frac {\ln \relax (x )}{5 x^{2}+30 x +45}\)	\(15\)
default	\(\frac {\ln \relax (x )}{45}-\frac {\ln \relax (x ) x \left (x +6\right )}{45 \left (3+x \right )^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x)+3+x)/(5*x^4+45*x^3+135*x^2+135*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.39, size = 51, normalized size = 2.32 \begin {gather*} \frac {2 \, x + 9}{30 \, {\left (x^{2} + 6 \, x + 9\right )}} + \frac {\log \relax (x)}{5 \, {\left (x^{2} + 6 \, x + 9\right )}} - \frac {1}{10 \, {\left (x^{2} + 6 \, x + 9\right )}} - \frac {1}{15 \, {\left (x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)+3+x)/(5*x^4+45*x^3+135*x^2+135*x),x, algorithm="maxima")

[Out]

1/30*(2*x + 9)/(x^2 + 6*x + 9) + 1/5*log(x)/(x^2 + 6*x + 9) - 1/10/(x^2 + 6*x + 9) - 1/15/(x + 3)

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mupad [B]  time = 2.38, size = 9, normalized size = 0.41 \begin {gather*} \frac {\ln \relax (x)}{5\,{\left (x+3\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 2*x*log(x) + 3)/(135*x + 135*x^2 + 45*x^3 + 5*x^4),x)

[Out]

log(x)/(5*(x + 3)^2)

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sympy [A]  time = 0.13, size = 12, normalized size = 0.55 \begin {gather*} \frac {\log {\relax (x )}}{5 x^{2} + 30 x + 45} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x)+3+x)/(5*x**4+45*x**3+135*x**2+135*x),x)

[Out]

log(x)/(5*x**2 + 30*x + 45)

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