3.30.27 \(\int \frac {-5-15 x-2 x \log (x)}{50 x-40 x^2+8 x^3} \, dx\)

Optimal. Leaf size=16 \[ \frac {6+x+\log (x)}{4 \left (-\frac {5}{2}+x\right )} \]

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Rubi [A]  time = 0.17, antiderivative size = 32, normalized size of antiderivative = 2.00, number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1594, 27, 12, 6742, 77, 2314, 31} \begin {gather*} -\frac {17}{4 (5-2 x)}-\frac {x \log (x)}{5 (5-2 x)}-\frac {\log (x)}{10} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5 - 15*x - 2*x*Log[x])/(50*x - 40*x^2 + 8*x^3),x]

[Out]

-17/(4*(5 - 2*x)) - Log[x]/10 - (x*Log[x])/(5*(5 - 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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 {-5-15 x-2 x \log (x)}{x \left (50-40 x+8 x^2\right )} \, dx\\ &=\int \frac {-5-15 x-2 x \log (x)}{2 x (-5+2 x)^2} \, dx\\ &=\frac {1}{2} \int \frac {-5-15 x-2 x \log (x)}{x (-5+2 x)^2} \, dx\\ &=\frac {1}{2} \int \left (-\frac {5 (1+3 x)}{x (-5+2 x)^2}-\frac {2 \log (x)}{(-5+2 x)^2}\right ) \, dx\\ &=-\left (\frac {5}{2} \int \frac {1+3 x}{x (-5+2 x)^2} \, dx\right )-\int \frac {\log (x)}{(-5+2 x)^2} \, dx\\ &=-\frac {x \log (x)}{5 (5-2 x)}-\frac {1}{5} \int \frac {1}{-5+2 x} \, dx-\frac {5}{2} \int \left (\frac {1}{25 x}+\frac {17}{5 (-5+2 x)^2}-\frac {2}{25 (-5+2 x)}\right ) \, dx\\ &=-\frac {17}{4 (5-2 x)}-\frac {\log (x)}{10}-\frac {x \log (x)}{5 (5-2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 1.06 \begin {gather*} -\frac {17+2 \log (x)}{2 (10-4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 15*x - 2*x*Log[x])/(50*x - 40*x^2 + 8*x^3),x]

[Out]

-1/2*(17 + 2*Log[x])/(10 - 4*x)

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fricas [A]  time = 0.68, size = 15, normalized size = 0.94 \begin {gather*} \frac {2 \, \log \relax (x) + 17}{4 \, {\left (2 \, x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)-15*x-5)/(8*x^3-40*x^2+50*x),x, algorithm="fricas")

[Out]

1/4*(2*log(x) + 17)/(2*x - 5)

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giac [A]  time = 0.14, size = 21, normalized size = 1.31 \begin {gather*} \frac {\log \relax (x)}{2 \, {\left (2 \, x - 5\right )}} + \frac {17}{4 \, {\left (2 \, x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)-15*x-5)/(8*x^3-40*x^2+50*x),x, algorithm="giac")

[Out]

1/2*log(x)/(2*x - 5) + 17/4/(2*x - 5)

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maple [A]  time = 0.04, size = 15, normalized size = 0.94

method	result	size
norman	\(\frac {\frac {\ln \relax (x )}{2}+\frac {17}{4}}{2 x -5}\)	\(15\)
risch	\(\frac {\ln \relax (x )}{4 x -10}+\frac {17}{4 \left (2 x -5\right )}\)	\(22\)
default	\(\frac {x \ln \relax (x )}{10 x -25}-\frac {\ln \relax (x )}{10}+\frac {17}{4 \left (2 x -5\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x)-15*x-5)/(8*x^3-40*x^2+50*x),x,method=_RETURNVERBOSE)

[Out]

(1/2*ln(x)+17/4)/(2*x-5)

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maxima [A]  time = 0.53, size = 21, normalized size = 1.31 \begin {gather*} \frac {\log \relax (x)}{2 \, {\left (2 \, x - 5\right )}} + \frac {17}{4 \, {\left (2 \, x - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)-15*x-5)/(8*x^3-40*x^2+50*x),x, algorithm="maxima")

[Out]

1/2*log(x)/(2*x - 5) + 17/4/(2*x - 5)

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mupad [B]  time = 1.74, size = 14, normalized size = 0.88 \begin {gather*} \frac {\frac {\ln \relax (x)}{2}+\frac {17}{4}}{2\,x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x + 2*x*log(x) + 5)/(50*x - 40*x^2 + 8*x^3),x)

[Out]

(log(x)/2 + 17/4)/(2*x - 5)

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sympy [A]  time = 0.14, size = 14, normalized size = 0.88 \begin {gather*} \frac {17}{8 x - 20} + \frac {\log {\relax (x )}}{4 x - 10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x)-15*x-5)/(8*x**3-40*x**2+50*x),x)

[Out]

17/(8*x - 20) + log(x)/(4*x - 10)

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