3.13.64 \(\int \frac {-20+e^2 (-16-56 x+31 x^2-4 x^3)}{e^2 (16-8 x+x^2)} \, dx\)

Optimal. Leaf size=24 \[ \left (-1+\frac {5}{e^2 (-4+x)}-\frac {27}{4 x}-2 x\right ) x \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {12, 27, 1850} \begin {gather*} -2 x^2-x-\frac {20}{e^2 (4-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 + E^2*(-16 - 56*x + 31*x^2 - 4*x^3))/(E^2*(16 - 8*x + x^2)),x]

[Out]

-20/(E^2*(4 - x)) - x - 2*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} &=\frac {\int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{16-8 x+x^2} \, dx}{e^2}\\ &=\frac {\int \frac {-20+e^2 \left (-16-56 x+31 x^2-4 x^3\right )}{(-4+x)^2} \, dx}{e^2}\\ &=\frac {\int \left (-e^2-\frac {20}{(-4+x)^2}-4 e^2 x\right ) \, dx}{e^2}\\ &=-\frac {20}{e^2 (4-x)}-x-2 x^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.79 \begin {gather*} \frac {20}{e^2 (-4+x)}-x-2 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + E^2*(-16 - 56*x + 31*x^2 - 4*x^3))/(E^2*(16 - 8*x + x^2)),x]

[Out]

20/(E^2*(-4 + x)) - x - 2*x^2

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fricas [A]  time = 0.66, size = 28, normalized size = 1.17 \begin {gather*} -\frac {{\left ({\left (2 \, x^{3} - 7 \, x^{2} - 4 \, x\right )} e^{2} - 20\right )} e^{\left (-2\right )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3+31*x^2-56*x-16)*exp(2)-20)/(x^2-8*x+16)/exp(2),x, algorithm="fricas")

[Out]

-((2*x^3 - 7*x^2 - 4*x)*e^2 - 20)*e^(-2)/(x - 4)

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giac [A]  time = 0.25, size = 23, normalized size = 0.96 \begin {gather*} -{\left (2 \, x^{2} e^{2} + x e^{2} - \frac {20}{x - 4}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3+31*x^2-56*x-16)*exp(2)-20)/(x^2-8*x+16)/exp(2),x, algorithm="giac")

[Out]

-(2*x^2*e^2 + x*e^2 - 20/(x - 4))*e^(-2)

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maple [A]  time = 0.24, size = 19, normalized size = 0.79

method	result	size
risch	\(-2 x^{2}-x +\frac {20 \,{\mathrm e}^{-2}}{x -4}\)	\(19\)
default	\({\mathrm e}^{-2} \left (-2 x^{2} {\mathrm e}^{2}-{\mathrm e}^{2} x +\frac {20}{x -4}\right )\)	\(26\)
norman	\(\frac {7 x^{2}-2 x^{3}+4 \left (4 \,{\mathrm e}^{2}+5\right ) {\mathrm e}^{-2}}{x -4}\)	\(30\)
gosper	\(-\frac {\left (2 x^{3} {\mathrm e}^{2}-7 x^{2} {\mathrm e}^{2}-20-16 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-2}}{x -4}\)	\(32\)
meijerg	\(-\frac {5 \,{\mathrm e}^{-2} x}{4 \left (-\frac {x}{4}+1\right )}-\frac {4 x \left (-\frac {1}{8} x^{2}-\frac {3}{2} x +12\right )}{-\frac {x}{4}+1}+\frac {31 x \left (-\frac {3 x}{4}+6\right )}{3 \left (-\frac {x}{4}+1\right )}-\frac {15 x}{-\frac {x}{4}+1}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^3+31*x^2-56*x-16)*exp(2)-20)/(x^2-8*x+16)/exp(2),x,method=_RETURNVERBOSE)

[Out]

-2*x^2-x+20/(x-4)*exp(-2)

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maxima [A]  time = 0.37, size = 23, normalized size = 0.96 \begin {gather*} -{\left (2 \, x^{2} e^{2} + x e^{2} - \frac {20}{x - 4}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3+31*x^2-56*x-16)*exp(2)-20)/(x^2-8*x+16)/exp(2),x, algorithm="maxima")

[Out]

-(2*x^2*e^2 + x*e^2 - 20/(x - 4))*e^(-2)

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mupad [B]  time = 0.89, size = 23, normalized size = 0.96 \begin {gather*} -x-\frac {20}{4\,{\mathrm {e}}^2-x\,{\mathrm {e}}^2}-2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2)*(exp(2)*(56*x - 31*x^2 + 4*x^3 + 16) + 20))/(x^2 - 8*x + 16),x)

[Out]

- x - 20/(4*exp(2) - x*exp(2)) - 2*x^2

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sympy [A]  time = 0.15, size = 17, normalized size = 0.71 \begin {gather*} - 2 x^{2} - x + \frac {20}{x e^{2} - 4 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**3+31*x**2-56*x-16)*exp(2)-20)/(x**2-8*x+16)/exp(2),x)

[Out]

-2*x**2 - x + 20/(x*exp(2) - 4*exp(2))

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