3.19.10 \(\int \frac {e^{-4 x} (2048-3072 x+1536 x^2-256 x^3+e^{4 x} (-6+12 x-6 x^2+x^3))}{-8+12 x-6 x^2+x^3} \, dx\)

Optimal. Leaf size=19 \[ 4+64 e^{-4 x}-\frac {1}{(2-x)^2}+x \]

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Rubi [A]  time = 0.35, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 3, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6688, 2194, 1850} \begin {gather*} x+64 e^{-4 x}-\frac {1}{(2-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2048 - 3072*x + 1536*x^2 - 256*x^3 + E^(4*x)*(-6 + 12*x - 6*x^2 + x^3))/(E^(4*x)*(-8 + 12*x - 6*x^2 + x^3
)),x]

[Out]

64/E^(4*x) - (2 - x)^(-2) + x

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])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-256 e^{-4 x}+\frac {-6+12 x-6 x^2+x^3}{(-2+x)^3}\right ) \, dx\\ &=-\left (256 \int e^{-4 x} \, dx\right )+\int \frac {-6+12 x-6 x^2+x^3}{(-2+x)^3} \, dx\\ &=64 e^{-4 x}+\int \left (1+\frac {2}{(-2+x)^3}\right ) \, dx\\ &=64 e^{-4 x}-\frac {1}{(2-x)^2}+x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 0.84 \begin {gather*} 64 e^{-4 x}-\frac {1}{(-2+x)^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2048 - 3072*x + 1536*x^2 - 256*x^3 + E^(4*x)*(-6 + 12*x - 6*x^2 + x^3))/(E^(4*x)*(-8 + 12*x - 6*x^2
+ x^3)),x]

[Out]

64/E^(4*x) - (-2 + x)^(-2) + x

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fricas [B]  time = 0.74, size = 43, normalized size = 2.26 \begin {gather*} \frac {{\left (64 \, x^{2} + {\left (x^{3} - 4 \, x^{2} + 4 \, x - 1\right )} e^{\left (4 \, x\right )} - 256 \, x + 256\right )} e^{\left (-4 \, x\right )}}{x^{2} - 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-6*x^2+12*x-6)*exp(x)^4-256*x^3+1536*x^2-3072*x+2048)/(x^3-6*x^2+12*x-8)/exp(x)^4,x, algorithm=
"fricas")

[Out]

(64*x^2 + (x^3 - 4*x^2 + 4*x - 1)*e^(4*x) - 256*x + 256)*e^(-4*x)/(x^2 - 4*x + 4)

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giac [B]  time = 0.20, size = 46, normalized size = 2.42 \begin {gather*} \frac {x^{3} + 64 \, x^{2} e^{\left (-4 \, x\right )} - 4 \, x^{2} - 256 \, x e^{\left (-4 \, x\right )} + 4 \, x + 256 \, e^{\left (-4 \, x\right )} - 1}{x^{2} - 4 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-6*x^2+12*x-6)*exp(x)^4-256*x^3+1536*x^2-3072*x+2048)/(x^3-6*x^2+12*x-8)/exp(x)^4,x, algorithm=
"giac")

[Out]

(x^3 + 64*x^2*e^(-4*x) - 4*x^2 - 256*x*e^(-4*x) + 4*x + 256*e^(-4*x) - 1)/(x^2 - 4*x + 4)

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maple [A]  time = 0.07, size = 21, normalized size = 1.11

method	result	size
risch	\(x -\frac {1}{x^{2}-4 x +4}+64 \,{\mathrm e}^{-4 x}\)	\(21\)
norman	\(\frac {\left (256+x^{3} {\mathrm e}^{4 x}+15 \,{\mathrm e}^{4 x}-12 x \,{\mathrm e}^{4 x}-256 x +64 x^{2}\right ) {\mathrm e}^{-4 x}}{\left (x -2\right )^{2}}\)	\(42\)
default	\(x -\frac {1}{\left (x -2\right )^{2}}+\frac {1024 \,{\mathrm e}^{-4 x} \left (4 x -9\right )}{x^{2}-4 x +4}-\frac {3072 \,{\mathrm e}^{-4 x} \left (3 x -7\right )}{x^{2}-4 x +4}+\frac {3072 \,{\mathrm e}^{-4 x} \left (2 x -5\right )}{x^{2}-4 x +4}+64 \,{\mathrm e}^{-4 x}-\frac {1024 \,{\mathrm e}^{-4 x} \left (x -3\right )}{x^{2}-4 x +4}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-6*x^2+12*x-6)*exp(x)^4-256*x^3+1536*x^2-3072*x+2048)/(x^3-6*x^2+12*x-8)/exp(x)^4,x,method=_RETURNVER
BOSE)

[Out]

x-1/(x^2-4*x+4)+64*exp(-4*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {x^{4} - 6 \, x^{3} + 12 \, x^{2} + 64 \, {\left (x^{3} - 6 \, x^{2} + 12 \, x\right )} e^{\left (-4 \, x\right )} - 9 \, x + 2}{x^{3} - 6 \, x^{2} + 12 \, x - 8} - \frac {2048 \, e^{\left (-8\right )} E_{3}\left (4 \, x - 8\right )}{{\left (x - 2\right )}^{2}} + 1536 \, \int \frac {e^{\left (-4 \, x\right )}}{x^{4} - 8 \, x^{3} + 24 \, x^{2} - 32 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-6*x^2+12*x-6)*exp(x)^4-256*x^3+1536*x^2-3072*x+2048)/(x^3-6*x^2+12*x-8)/exp(x)^4,x, algorithm=
"maxima")

[Out]

(x^4 - 6*x^3 + 12*x^2 + 64*(x^3 - 6*x^2 + 12*x)*e^(-4*x) - 9*x + 2)/(x^3 - 6*x^2 + 12*x - 8) - 2048*e^(-8)*exp
_integral_e(3, 4*x - 8)/(x - 2)^2 + 1536*integrate(e^(-4*x)/(x^4 - 8*x^3 + 24*x^2 - 32*x + 16), x)

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mupad [B]  time = 0.12, size = 20, normalized size = 1.05 \begin {gather*} x+64\,{\mathrm {e}}^{-4\,x}-\frac {1}{x^2-4\,x+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4*x)*(1536*x^2 - 3072*x - 256*x^3 + exp(4*x)*(12*x - 6*x^2 + x^3 - 6) + 2048))/(12*x - 6*x^2 + x^3 -
8),x)

[Out]

x + 64*exp(-4*x) - 1/(x^2 - 4*x + 4)

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sympy [A]  time = 0.16, size = 17, normalized size = 0.89 \begin {gather*} x + 64 e^{- 4 x} - \frac {1}{x^{2} - 4 x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-6*x**2+12*x-6)*exp(x)**4-256*x**3+1536*x**2-3072*x+2048)/(x**3-6*x**2+12*x-8)/exp(x)**4,x)

[Out]

x + 64*exp(-4*x) - 1/(x**2 - 4*x + 4)

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