3.100.42 \(\int \frac {32-96 x+36 x^2-4 x^3+3 e^{\frac {2}{x}-x} (32-48 x+34 x^2-10 x^3+x^4)}{16-8 x+x^2} \, dx\)

Optimal. Leaf size=34 \[ x \left (2-x-\left (1+3 e^{\frac {2}{x}-x}+\frac {2}{4-x}\right ) x\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 45, normalized size of antiderivative = 1.32, number of steps used = 10, number of rules used = 4, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {27, 6742, 43, 2288} \begin {gather*} -2 x^2-\frac {3 e^{\frac {2}{x}-x} \left (x^2+2\right )}{\frac {2}{x^2}+1}+4 x-\frac {32}{4-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32 - 96*x + 36*x^2 - 4*x^3 + 3*E^(2/x - x)*(32 - 48*x + 34*x^2 - 10*x^3 + x^4))/(16 - 8*x + x^2),x]

[Out]

-32/(4 - x) + 4*x - 2*x^2 - (3*E^(2/x - x)*(2 + x^2))/(1 + 2/x^2)

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32-96 x+36 x^2-4 x^3+3 e^{\frac {2}{x}-x} \left (32-48 x+34 x^2-10 x^3+x^4\right )}{(-4+x)^2} \, dx\\ &=\int \left (\frac {32}{(-4+x)^2}-\frac {96 x}{(-4+x)^2}+\frac {36 x^2}{(-4+x)^2}-\frac {4 x^3}{(-4+x)^2}+3 e^{\frac {2}{x}-x} \left (2-2 x+x^2\right )\right ) \, dx\\ &=\frac {32}{4-x}+3 \int e^{\frac {2}{x}-x} \left (2-2 x+x^2\right ) \, dx-4 \int \frac {x^3}{(-4+x)^2} \, dx+36 \int \frac {x^2}{(-4+x)^2} \, dx-96 \int \frac {x}{(-4+x)^2} \, dx\\ &=\frac {32}{4-x}-\frac {3 e^{\frac {2}{x}-x} \left (2+x^2\right )}{1+\frac {2}{x^2}}-4 \int \left (8+\frac {64}{(-4+x)^2}+\frac {48}{-4+x}+x\right ) \, dx+36 \int \left (1+\frac {16}{(-4+x)^2}+\frac {8}{-4+x}\right ) \, dx-96 \int \left (\frac {4}{(-4+x)^2}+\frac {1}{-4+x}\right ) \, dx\\ &=-\frac {32}{4-x}+4 x-2 x^2-\frac {3 e^{\frac {2}{x}-x} \left (2+x^2\right )}{1+\frac {2}{x^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 1.38 \begin {gather*} \frac {32}{-4+x}-12 (-4+x)-2 (-4+x)^2+\frac {3 e^{\frac {2}{x}-x} \left (2+x^2\right )}{-1-\frac {2}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 96*x + 36*x^2 - 4*x^3 + 3*E^(2/x - x)*(32 - 48*x + 34*x^2 - 10*x^3 + x^4))/(16 - 8*x + x^2),x]

[Out]

32/(-4 + x) - 12*(-4 + x) - 2*(-4 + x)^2 + (3*E^(2/x - x)*(2 + x^2))/(-1 - 2/x^2)

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fricas [A]  time = 1.05, size = 48, normalized size = 1.41 \begin {gather*} -\frac {2 \, x^{3} - 12 \, x^{2} + {\left (x^{3} - 4 \, x^{2}\right )} e^{\left (-\frac {x^{2} - x \log \relax (3) - 2}{x}\right )} + 16 \, x - 32}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4-10*x^3+34*x^2-48*x+32)*exp(log(3*exp(2/x))-x)-4*x^3+36*x^2-96*x+32)/(x^2-8*x+16),x, algorithm=
"fricas")

[Out]

-(2*x^3 - 12*x^2 + (x^3 - 4*x^2)*e^(-(x^2 - x*log(3) - 2)/x) + 16*x - 32)/(x - 4)

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giac [A]  time = 0.33, size = 54, normalized size = 1.59 \begin {gather*} -\frac {3 \, x^{3} e^{\left (-\frac {x^{2} - 2}{x}\right )} + 2 \, x^{3} - 12 \, x^{2} e^{\left (-\frac {x^{2} - 2}{x}\right )} - 12 \, x^{2} + 16 \, x - 32}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4-10*x^3+34*x^2-48*x+32)*exp(log(3*exp(2/x))-x)-4*x^3+36*x^2-96*x+32)/(x^2-8*x+16),x, algorithm=
"giac")

[Out]

-(3*x^3*e^(-(x^2 - 2)/x) + 2*x^3 - 12*x^2*e^(-(x^2 - 2)/x) - 12*x^2 + 16*x - 32)/(x - 4)

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maple [A]  time = 0.22, size = 36, normalized size = 1.06

method	result	size
default	\(-{\mathrm e}^{\ln \left (3 \,{\mathrm e}^{\frac {2}{x}}\right )-x} x^{2}-2 x^{2}+4 x +\frac {32}{x -4}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4-10*x^3+34*x^2-48*x+32)*exp(ln(3*exp(2/x))-x)-4*x^3+36*x^2-96*x+32)/(x^2-8*x+16),x,method=_RETURNVERB
OSE)

[Out]

-exp(ln(3*exp(2/x))-x)*x^2-2*x^2+4*x+32/(x-4)

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maxima [A]  time = 0.40, size = 31, normalized size = 0.91 \begin {gather*} -3 \, x^{2} e^{\left (-x + \frac {2}{x}\right )} - 2 \, x^{2} + 4 \, x + \frac {32}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^4-10*x^3+34*x^2-48*x+32)*exp(log(3*exp(2/x))-x)-4*x^3+36*x^2-96*x+32)/(x^2-8*x+16),x, algorithm=
"maxima")

[Out]

-3*x^2*e^(-x + 2/x) - 2*x^2 + 4*x + 32/(x - 4)

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mupad [B]  time = 7.67, size = 31, normalized size = 0.91 \begin {gather*} 4\,x-3\,x^2\,{\mathrm {e}}^{\frac {2}{x}-x}+\frac {32}{x-4}-2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((36*x^2 - 96*x - 4*x^3 + exp(log(3*exp(2/x)) - x)*(34*x^2 - 48*x - 10*x^3 + x^4 + 32) + 32)/(x^2 - 8*x + 1
6),x)

[Out]

4*x - 3*x^2*exp(2/x - x) + 32/(x - 4) - 2*x^2

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sympy [A]  time = 10.36, size = 26, normalized size = 0.76 \begin {gather*} - 3 x^{2} e^{\frac {2}{x}} e^{- x} - 2 x^{2} + 4 x + \frac {32}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**4-10*x**3+34*x**2-48*x+32)*exp(ln(3*exp(2/x))-x)-4*x**3+36*x**2-96*x+32)/(x**2-8*x+16),x)

[Out]

-3*x**2*exp(2/x)*exp(-x) - 2*x**2 + 4*x + 32/(x - 4)

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