3.69.59 \(\int \frac {e^{-x} (e^x (-4+2 x+17 x^2)+e (-64 x^2+32 x^3-4 x^4))}{64 x^2-32 x^3+4 x^4} \, dx\)

Optimal. Leaf size=29 \[ e^{1-x}-\frac {-25-\frac {1}{x}+2 x}{4 (4-x)} \]

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Rubi [A]  time = 0.34, antiderivative size = 26, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 6, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1594, 27, 12, 6688, 2194, 893} \begin {gather*} e^{1-x}+\frac {1}{16 x}+\frac {69}{16 (4-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-4 + 2*x + 17*x^2) + E*(-64*x^2 + 32*x^3 - 4*x^4))/(E^x*(64*x^2 - 32*x^3 + 4*x^4)),x]

[Out]

E^(1 - x) + 69/(16*(4 - x)) + 1/(16*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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 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 \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{x^2 \left (64-32 x+4 x^2\right )} \, dx\\ &=\int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{4 (-4+x)^2 x^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{-x} \left (e^x \left (-4+2 x+17 x^2\right )+e \left (-64 x^2+32 x^3-4 x^4\right )\right )}{(-4+x)^2 x^2} \, dx\\ &=\frac {1}{4} \int \left (-4 e^{1-x}+\frac {-4+2 x+17 x^2}{(-4+x)^2 x^2}\right ) \, dx\\ &=\frac {1}{4} \int \frac {-4+2 x+17 x^2}{(-4+x)^2 x^2} \, dx-\int e^{1-x} \, dx\\ &=e^{1-x}+\frac {1}{4} \int \left (\frac {69}{4 (-4+x)^2}-\frac {1}{4 x^2}\right ) \, dx\\ &=e^{1-x}+\frac {69}{16 (4-x)}+\frac {1}{16 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 24, normalized size = 0.83 \begin {gather*} e^{1-x}-\frac {69}{16 (-4+x)}+\frac {1}{16 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-4 + 2*x + 17*x^2) + E*(-64*x^2 + 32*x^3 - 4*x^4))/(E^x*(64*x^2 - 32*x^3 + 4*x^4)),x]

[Out]

E^(1 - x) - 69/(16*(-4 + x)) + 1/(16*x)

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fricas [A]  time = 0.66, size = 36, normalized size = 1.24 \begin {gather*} \frac {{\left (4 \, {\left (x^{2} - 4 \, x\right )} e - {\left (17 \, x + 1\right )} e^{x}\right )} e^{\left (-x\right )}}{4 \, {\left (x^{2} - 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32*x^3+64*x^2)/exp(x),x, algorithm="fri
cas")

[Out]

1/4*(4*(x^2 - 4*x)*e - (17*x + 1)*e^x)*e^(-x)/(x^2 - 4*x)

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giac [A]  time = 0.15, size = 36, normalized size = 1.24 \begin {gather*} \frac {4 \, x^{2} e^{\left (-x + 1\right )} - 16 \, x e^{\left (-x + 1\right )} - 17 \, x - 1}{4 \, {\left (x^{2} - 4 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32*x^3+64*x^2)/exp(x),x, algorithm="gia
c")

[Out]

1/4*(4*x^2*e^(-x + 1) - 16*x*e^(-x + 1) - 17*x - 1)/(x^2 - 4*x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.76

method	result	size
risch	\(\frac {-\frac {17 x}{4}-\frac {1}{4}}{\left (x -4\right ) x}+{\mathrm e}^{1-x}\)	\(22\)
norman	\(\frac {\left (x^{2} {\mathrm e}-\frac {17 \,{\mathrm e}^{x} x}{4}-4 x \,{\mathrm e}-\frac {{\mathrm e}^{x}}{4}\right ) {\mathrm e}^{-x}}{x \left (x -4\right )}\)	\(35\)
default	\(\frac {1}{16 x}-\frac {69}{16 \left (x -4\right )}-16 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-x}}{x -4}+{\mathrm e}^{-4} \expIntegralEi \left (1, x -4\right )\right )+8 \,{\mathrm e} \left (-\frac {4 \,{\mathrm e}^{-x}}{x -4}+3 \,{\mathrm e}^{-4} \expIntegralEi \left (1, x -4\right )\right )-{\mathrm e} \left (-{\mathrm e}^{-x}-\frac {16 \,{\mathrm e}^{-x}}{x -4}+8 \,{\mathrm e}^{-4} \expIntegralEi \left (1, x -4\right )\right )\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32*x^3+64*x^2)/exp(x),x,method=_RETURNVERBOSE
)

[Out]

(-17/4*x-1/4)/(x-4)/x+exp(1-x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {16 \, e^{\left (-3\right )} E_{2}\left (x - 4\right )}{x - 4} - \frac {17 \, x^{2} - 4 \, {\left (x^{3} e - 8 \, x^{2} e\right )} e^{\left (-x\right )} - 67 \, x - 4}{4 \, {\left (x^{3} - 8 \, x^{2} + 16 \, x\right )}} - 32 \, \int \frac {e^{\left (-x + 1\right )}}{x^{3} - 12 \, x^{2} + 48 \, x - 64}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((17*x^2+2*x-4)*exp(x)+(-4*x^4+32*x^3-64*x^2)*exp(1))/(4*x^4-32*x^3+64*x^2)/exp(x),x, algorithm="max
ima")

[Out]

16*e^(-3)*exp_integral_e(2, x - 4)/(x - 4) - 1/4*(17*x^2 - 4*(x^3*e - 8*x^2*e)*e^(-x) - 67*x - 4)/(x^3 - 8*x^2
+ 16*x) - 32*integrate(e^(-x + 1)/(x^3 - 12*x^2 + 48*x - 64), x)

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mupad [B]  time = 0.13, size = 33, normalized size = 1.14 \begin {gather*} {\mathrm {e}}^{-x}\,\mathrm {e}+\frac {17\,x}{4\,\left (4\,x-x^2\right )}+\frac {1}{4\,\left (4\,x-x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(x)*(2*x + 17*x^2 - 4) - exp(1)*(64*x^2 - 32*x^3 + 4*x^4)))/(64*x^2 - 32*x^3 + 4*x^4),x)

[Out]

exp(-x)*exp(1) + (17*x)/(4*(4*x - x^2)) + 1/(4*(4*x - x^2))

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sympy [A]  time = 0.21, size = 20, normalized size = 0.69 \begin {gather*} \frac {- 17 x - 1}{4 x^{2} - 16 x} + e e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((17*x**2+2*x-4)*exp(x)+(-4*x**4+32*x**3-64*x**2)*exp(1))/(4*x**4-32*x**3+64*x**2)/exp(x),x)

[Out]

(-17*x - 1)/(4*x**2 - 16*x) + E*exp(-x)

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