3.22.28 \(\int \frac {459-45 e^5-288 x+84 x^2-8 x^3}{36-24 x+4 x^2} \, dx\)

Optimal. Leaf size=28 \[ \left (9+\frac {45 \left (1-\frac {e^5}{x}\right )}{4 (3-x)}-x\right ) x \]

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Rubi [A]  time = 0.03, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {27, 12, 1850} \begin {gather*} -x^2+9 x+\frac {45 \left (3-e^5\right )}{4 (3-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(459 - 45*E^5 - 288*x + 84*x^2 - 8*x^3)/(36 - 24*x + 4*x^2),x]

[Out]

(45*(3 - E^5))/(4*(3 - x)) + 9*x - 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 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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {459-45 e^5-288 x+84 x^2-8 x^3}{4 (-3+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {459-45 e^5-288 x+84 x^2-8 x^3}{(-3+x)^2} \, dx\\ &=\frac {1}{4} \int \left (36-\frac {45 \left (-3+e^5\right )}{(-3+x)^2}-8 x\right ) \, dx\\ &=\frac {45 \left (3-e^5\right )}{4 (3-x)}+9 x-x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 29, normalized size = 1.04 \begin {gather*} \frac {1}{4} \left (\frac {45 \left (-3+e^5\right )}{-3+x}+12 (-3+x)-4 (-3+x)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(459 - 45*E^5 - 288*x + 84*x^2 - 8*x^3)/(36 - 24*x + 4*x^2),x]

[Out]

((45*(-3 + E^5))/(-3 + x) + 12*(-3 + x) - 4*(-3 + x)^2)/4

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fricas [A]  time = 0.85, size = 26, normalized size = 0.93 \begin {gather*} -\frac {4 \, x^{3} - 48 \, x^{2} + 108 \, x - 45 \, e^{5} + 135}{4 \, {\left (x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-45*exp(5)-8*x^3+84*x^2-288*x+459)/(4*x^2-24*x+36),x, algorithm="fricas")

[Out]

-1/4*(4*x^3 - 48*x^2 + 108*x - 45*e^5 + 135)/(x - 3)

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giac [A]  time = 0.35, size = 20, normalized size = 0.71 \begin {gather*} -x^{2} + 9 \, x + \frac {45 \, {\left (e^{5} - 3\right )}}{4 \, {\left (x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-45*exp(5)-8*x^3+84*x^2-288*x+459)/(4*x^2-24*x+36),x, algorithm="giac")

[Out]

-x^2 + 9*x + 45/4*(e^5 - 3)/(x - 3)

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maple [A]  time = 0.40, size = 23, normalized size = 0.82

method	result	size
default	\(9 x -x^{2}-\frac {135-45 \,{\mathrm e}^{5}}{4 \left (x -3\right )}\)	\(23\)
norman	\(\frac {12 x^{2}-x^{3}-\frac {459}{4}+\frac {45 \,{\mathrm e}^{5}}{4}}{x -3}\)	\(23\)
gosper	\(\frac {-4 x^{3}+48 x^{2}+45 \,{\mathrm e}^{5}-459}{4 x -12}\)	\(24\)
risch	\(-x^{2}+9 x -\frac {135}{4 \left (x -3\right )}+\frac {45 \,{\mathrm e}^{5}}{4 \left (x -3\right )}\)	\(26\)
meijerg	\(-\frac {45 x}{4 \left (1-\frac {x}{3}\right )}-\frac {5 \,{\mathrm e}^{5} x}{4 \left (1-\frac {x}{3}\right )}-\frac {3 x \left (-\frac {2}{9} x^{2}-2 x +12\right )}{2 \left (1-\frac {x}{3}\right )}+\frac {7 x \left (-x +6\right )}{1-\frac {x}{3}}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-45*exp(5)-8*x^3+84*x^2-288*x+459)/(4*x^2-24*x+36),x,method=_RETURNVERBOSE)

[Out]

9*x-x^2-1/4*(135-45*exp(5))/(x-3)

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maxima [A]  time = 0.37, size = 20, normalized size = 0.71 \begin {gather*} -x^{2} + 9 \, x + \frac {45 \, {\left (e^{5} - 3\right )}}{4 \, {\left (x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-45*exp(5)-8*x^3+84*x^2-288*x+459)/(4*x^2-24*x+36),x, algorithm="maxima")

[Out]

-x^2 + 9*x + 45/4*(e^5 - 3)/(x - 3)

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mupad [B]  time = 0.06, size = 23, normalized size = 0.82 \begin {gather*} 9\,x+\frac {45\,{\mathrm {e}}^5-135}{4\,x-12}-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(288*x + 45*exp(5) - 84*x^2 + 8*x^3 - 459)/(4*x^2 - 24*x + 36),x)

[Out]

9*x + (45*exp(5) - 135)/(4*x - 12) - x^2

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sympy [A]  time = 0.12, size = 17, normalized size = 0.61 \begin {gather*} - x^{2} + 9 x - \frac {135 - 45 e^{5}}{4 x - 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-45*exp(5)-8*x**3+84*x**2-288*x+459)/(4*x**2-24*x+36),x)

[Out]

-x**2 + 9*x - (135 - 45*exp(5))/(4*x - 12)

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