3.75.13 \(\int \frac {-24+20 x-9 x^2+5 x^3-x^4+(6 x-5 x^2+x^3) \log (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x})}{6 x-5 x^2+x^3} \, dx\)

Optimal. Leaf size=27 \[ x \log \left (\frac {5 e^{-\frac {-4+x^2}{x}} (-3+x)}{6-3 x}\right ) \]

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Rubi [A]  time = 0.40, antiderivative size = 30, normalized size of antiderivative = 1.11, number of steps used = 26, number of rules used = 11, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.149, Rules used = {1594, 6728, 616, 31, 705, 29, 632, 703, 701, 2548, 1628} \begin {gather*} x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 + 20*x - 9*x^2 + 5*x^3 - x^4 + (6*x - 5*x^2 + x^3)*Log[(15 - 5*x)/(E^((-4 + x^2)/x)*(-6 + 3*x))])/(6*
x - 5*x^2 + x^3),x]

[Out]

x*Log[(-5*E^(4/x - x)*(3 - x))/(3*(2 - x))]

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 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 701

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)
^m, a + b*x + c*x^2, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2,
0] && NeQ[2*c*d - b*e, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 703

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1))/(c*
(m - 1)), x] + Dist[1/c, Int[((d + e*x)^(m - 2)*Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x])/(a + b*x + c*x^2),
x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*
e, 0] && GtQ[m, 1]

Rule 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 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 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-24+20 x-9 x^2+5 x^3-x^4+\left (6 x-5 x^2+x^3\right ) \log \left (\frac {e^{-\frac {-4+x^2}{x}} (15-5 x)}{-6+3 x}\right )}{x \left (6-5 x+x^2\right )} \, dx\\ &=\int \left (\frac {20}{6-5 x+x^2}-\frac {24}{x \left (6-5 x+x^2\right )}-\frac {9 x}{6-5 x+x^2}+\frac {5 x^2}{6-5 x+x^2}-\frac {x^3}{6-5 x+x^2}+\log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right )\right ) \, dx\\ &=5 \int \frac {x^2}{6-5 x+x^2} \, dx-9 \int \frac {x}{6-5 x+x^2} \, dx+20 \int \frac {1}{6-5 x+x^2} \, dx-24 \int \frac {1}{x \left (6-5 x+x^2\right )} \, dx-\int \frac {x^3}{6-5 x+x^2} \, dx+\int \log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right ) \, dx\\ &=5 x+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )-4 \int \frac {1}{x} \, dx-4 \int \frac {5-x}{6-5 x+x^2} \, dx+5 \int \frac {-6+5 x}{6-5 x+x^2} \, dx+18 \int \frac {1}{-2+x} \, dx+20 \int \frac {1}{-3+x} \, dx-20 \int \frac {1}{-2+x} \, dx-27 \int \frac {1}{-3+x} \, dx-\int \frac {-24+20 x-9 x^2+5 x^3-x^4}{x \left (6-5 x+x^2\right )} \, dx-\int \left (5+x-\frac {30-19 x}{6-5 x+x^2}\right ) \, dx\\ &=-\frac {x^2}{2}-2 \log (2-x)-7 \log (3-x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )-4 \log (x)-8 \int \frac {1}{-3+x} \, dx+12 \int \frac {1}{-2+x} \, dx-20 \int \frac {1}{-2+x} \, dx+45 \int \frac {1}{-3+x} \, dx-\int \left (\frac {3}{-3+x}-\frac {2}{-2+x}-\frac {4}{x}-x\right ) \, dx+\int \frac {30-19 x}{6-5 x+x^2} \, dx\\ &=-8 \log (2-x)+27 \log (3-x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )+8 \int \frac {1}{-2+x} \, dx-27 \int \frac {1}{-3+x} \, dx\\ &=x \log \left (-\frac {5 e^{\frac {4}{x}-x} (3-x)}{3 (2-x)}\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.16, size = 56, normalized size = 2.07 \begin {gather*} -4-2 \log (2-x)+3 \log (3-x)-3 \log (-3+x)+x \log \left (-\frac {5 e^{\frac {4}{x}-x} (-3+x)}{3 (-2+x)}\right )+2 \log (-2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 + 20*x - 9*x^2 + 5*x^3 - x^4 + (6*x - 5*x^2 + x^3)*Log[(15 - 5*x)/(E^((-4 + x^2)/x)*(-6 + 3*x))
])/(6*x - 5*x^2 + x^3),x]

[Out]

-4 - 2*Log[2 - x] + 3*Log[3 - x] - 3*Log[-3 + x] + x*Log[(-5*E^(4/x - x)*(-3 + x))/(3*(-2 + x))] + 2*Log[-2 +
x]

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fricas [A]  time = 0.54, size = 24, normalized size = 0.89 \begin {gather*} x \log \left (-\frac {5 \, {\left (x - 3\right )} e^{\left (-\frac {x^{2} - 4}{x}\right )}}{3 \, {\left (x - 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-5*x^2+6*x)*log((15-5*x)/(3*x-6)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2+20*x-24)/(x^3-5*x^2+6*x),x, al
gorithm="fricas")

[Out]

x*log(-5/3*(x - 3)*e^(-(x^2 - 4)/x)/(x - 2))

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giac [A]  time = 0.31, size = 19, normalized size = 0.70 \begin {gather*} -x^{2} + x \log \left (-\frac {5 \, {\left (x - 3\right )}}{3 \, {\left (x - 2\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-5*x^2+6*x)*log((15-5*x)/(3*x-6)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2+20*x-24)/(x^3-5*x^2+6*x),x, al
gorithm="giac")

[Out]

-x^2 + x*log(-5/3*(x - 3)/(x - 2))

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maple [A]  time = 0.23, size = 29, normalized size = 1.07

method	result	size
default	\(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{3 x -6}\right )\)	\(29\)
norman	\(x \ln \left (\frac {\left (15-5 x \right ) {\mathrm e}^{-\frac {x^{2}-4}{x}}}{3 x -6}\right )\)	\(29\)
risch	\(-x \ln \left ({\mathrm e}^{\frac {\left (x -2\right ) \left (2+x \right )}{x}}\right )-x \ln \left (x -2\right )+\ln \left (x -3\right ) x +\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x -2}\right ) \mathrm {csgn}\left (\frac {i \left (x -3\right )}{x -2}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (x -3\right )}{x -2}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i \left (x -3\right )}{x -2}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}} \left (x -3\right )}{x -2}\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x -2}\right ) \mathrm {csgn}\left (i \left (x -3\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -3\right )}{x -2}\right )}{2}+\frac {i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}} \left (x -3\right )}{x -2}\right )^{3}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (x -3\right )\right ) \mathrm {csgn}\left (\frac {i \left (x -3\right )}{x -2}\right )^{2}}{2}-i \pi x \mathrm {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}} \left (x -3\right )}{x -2}\right )^{2}+\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}} \left (x -3\right )}{x -2}\right )^{2}}{2}+i \pi x -\frac {i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}}\right ) \mathrm {csgn}\left (\frac {i \left (x -3\right )}{x -2}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-\frac {\left (x -2\right ) \left (2+x \right )}{x}} \left (x -3\right )}{x -2}\right )}{2}+x \ln \relax (5)-x \ln \relax (3)\)	\(359\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-5*x^2+6*x)*ln((15-5*x)/(3*x-6)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2+20*x-24)/(x^3-5*x^2+6*x),x,method=_RE
TURNVERBOSE)

[Out]

x*ln((15-5*x)/(3*x-6)/exp((x^2-4)/x))

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maxima [C]  time = 0.54, size = 200, normalized size = 7.41 \begin {gather*} {\left (i \, \pi + \log \relax (5) - \log \relax (3)\right )} x - x^{2} + {\left (6 i \, \pi - x + 6 \, \log \relax (5) - 6 \, \log \relax (3) + 6 \, \log \left (x - 3\right ) + 2\right )} \log \left (x - 2\right ) - 3 \, \log \left (x - 2\right )^{2} + {\left (-6 i \, \pi + x - 6 \, \log \relax (5) + 6 \, \log \relax (3) + 7\right )} \log \left (x - 3\right ) - 3 \, \log \left (x - 3\right )^{2} - 6 \, {\left (\log \left (x - 2\right ) - \log \left (x - 3\right )\right )} \log \left (-\frac {5 \, x e^{\left (-x + \frac {4}{x}\right )}}{3 \, {\left (x - 2\right )}} + \frac {5 \, e^{\left (-x + \frac {4}{x}\right )}}{x - 2}\right ) - \frac {3 \, x \log \left (x - 2\right )^{2} + 3 \, x \log \left (x - 3\right )^{2} + 6 \, {\left (x^{2} - x \log \left (x - 3\right ) - 4\right )} \log \left (x - 2\right ) - 2 \, {\left (3 \, x^{2} - 5 \, x - 12\right )} \log \left (x - 3\right )}{x} - 2 \, \log \left (x - 2\right ) + 3 \, \log \left (x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-5*x^2+6*x)*log((15-5*x)/(3*x-6)/exp((x^2-4)/x))-x^4+5*x^3-9*x^2+20*x-24)/(x^3-5*x^2+6*x),x, al
gorithm="maxima")

[Out]

(I*pi + log(5) - log(3))*x - x^2 + (6*I*pi - x + 6*log(5) - 6*log(3) + 6*log(x - 3) + 2)*log(x - 2) - 3*log(x
- 2)^2 + (-6*I*pi + x - 6*log(5) + 6*log(3) + 7)*log(x - 3) - 3*log(x - 3)^2 - 6*(log(x - 2) - log(x - 3))*log
(-5/3*x*e^(-x + 4/x)/(x - 2) + 5*e^(-x + 4/x)/(x - 2)) - (3*x*log(x - 2)^2 + 3*x*log(x - 3)^2 + 6*(x^2 - x*log
(x - 3) - 4)*log(x - 2) - 2*(3*x^2 - 5*x - 12)*log(x - 3))/x - 2*log(x - 2) + 3*log(x - 3)

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mupad [B]  time = 4.91, size = 23, normalized size = 0.85 \begin {gather*} x\,\ln \left (-\frac {5\,x-15}{3\,x-6}\right )-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*x + log(-(exp(-(x^2 - 4)/x)*(5*x - 15))/(3*x - 6))*(6*x - 5*x^2 + x^3) - 9*x^2 + 5*x^3 - x^4 - 24)/(6*
x - 5*x^2 + x^3),x)

[Out]

x*log(-(5*x - 15)/(3*x - 6)) - x^2

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sympy [A]  time = 0.50, size = 20, normalized size = 0.74 \begin {gather*} x \log {\left (\frac {\left (15 - 5 x\right ) e^{- \frac {x^{2} - 4}{x}}}{3 x - 6} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-5*x**2+6*x)*ln((15-5*x)/(3*x-6)/exp((x**2-4)/x))-x**4+5*x**3-9*x**2+20*x-24)/(x**3-5*x**2+6*x
),x)

[Out]

x*log((15 - 5*x)*exp(-(x**2 - 4)/x)/(3*x - 6))

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