3.57.40 \(\int \frac {-240+64 x-40 x^2+16 x^3+5 x^4}{e^2 x^2 (25 x^2-10 x^3+x^4)} \, dx\)

Optimal. Leaf size=23 \[ \frac {\left (\frac {4}{x}+x\right )^2}{e^2 (5-x) x} \]

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Rubi [A]  time = 0.07, antiderivative size = 45, normalized size of antiderivative = 1.96, number of steps used = 5, number of rules used = 4, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {12, 1585, 27, 1620} \begin {gather*} \frac {16}{5 e^2 x^3}+\frac {16}{25 e^2 x^2}+\frac {216}{125 e^2 x}+\frac {841}{125 e^2 (5-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-240 + 64*x - 40*x^2 + 16*x^3 + 5*x^4)/(E^2*x^2*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

841/(125*E^2*(5 - x)) + 16/(5*E^2*x^3) + 16/(25*E^2*x^2) + 216/(125*E^2*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 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-240+64 x-40 x^2+16 x^3+5 x^4}{x^2 \left (25 x^2-10 x^3+x^4\right )} \, dx}{e^2}\\ &=\frac {\int \frac {-240+64 x-40 x^2+16 x^3+5 x^4}{x^4 \left (25-10 x+x^2\right )} \, dx}{e^2}\\ &=\frac {\int \frac {-240+64 x-40 x^2+16 x^3+5 x^4}{(-5+x)^2 x^4} \, dx}{e^2}\\ &=\frac {\int \left (\frac {841}{125 (-5+x)^2}-\frac {48}{5 x^4}-\frac {32}{25 x^3}-\frac {216}{125 x^2}\right ) \, dx}{e^2}\\ &=\frac {841}{125 e^2 (5-x)}+\frac {16}{5 e^2 x^3}+\frac {16}{25 e^2 x^2}+\frac {216}{125 e^2 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.09 \begin {gather*} -\frac {16+8 x^2+5 x^3}{e^2 (-5+x) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-240 + 64*x - 40*x^2 + 16*x^3 + 5*x^4)/(E^2*x^2*(25*x^2 - 10*x^3 + x^4)),x]

[Out]

-((16 + 8*x^2 + 5*x^3)/(E^2*(-5 + x)*x^3))

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fricas [A]  time = 0.60, size = 27, normalized size = 1.17 \begin {gather*} -\frac {{\left (5 \, x^{3} + 8 \, x^{2} + 16\right )} e^{\left (-2\right )}}{x^{4} - 5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4+16*x^3-40*x^2+64*x-240)*exp(5)/(x^4-10*x^3+25*x^2)/exp(3)/exp(log(x)+2)^2,x, algorithm="frica
s")

[Out]

-(5*x^3 + 8*x^2 + 16)*e^(-2)/(x^4 - 5*x^3)

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giac [A]  time = 0.12, size = 27, normalized size = 1.17 \begin {gather*} -\frac {841 \, e^{\left (-2\right )}}{125 \, {\left (x - 5\right )}} + \frac {8 \, {\left (27 \, x^{2} + 10 \, x + 50\right )} e^{\left (-2\right )}}{125 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4+16*x^3-40*x^2+64*x-240)*exp(5)/(x^4-10*x^3+25*x^2)/exp(3)/exp(log(x)+2)^2,x, algorithm="giac"
)

[Out]

-841/125*e^(-2)/(x - 5) + 8/125*(27*x^2 + 10*x + 50)*e^(-2)/x^3

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maple [A]  time = 0.07, size = 24, normalized size = 1.04

method	result	size
risch	\(\frac {{\mathrm e}^{-2} \left (-5 x^{3}-8 x^{2}-16\right )}{x^{3} \left (x -5\right )}\)	\(24\)
default	\({\mathrm e}^{-3} {\mathrm e}^{5} {\mathrm e}^{-4} \left (\frac {16}{5 x^{3}}+\frac {16}{25 x^{2}}+\frac {216}{125 x}-\frac {841}{125 \left (x -5\right )}\right )\)	\(33\)
gosper	\(-\frac {{\mathrm e}^{5} \left (5 x^{3}+8 x^{2}+16\right ) {\mathrm e}^{-4} {\mathrm e}^{-3}}{x^{3} \left (x -5\right )}\)	\(36\)
norman	\(\frac {\left (-5 \,{\mathrm e}^{-2} {\mathrm e}^{-3} {\mathrm e}^{5} x^{3}-16 \,{\mathrm e}^{-2} {\mathrm e}^{-3} {\mathrm e}^{5}-8 \,{\mathrm e}^{-2} {\mathrm e}^{-3} {\mathrm e}^{5} x^{2}\right ) {\mathrm e}^{-2}}{x^{3} \left (x -5\right )}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^4+16*x^3-40*x^2+64*x-240)*exp(5)/(x^4-10*x^3+25*x^2)/exp(3)/exp(ln(x)+2)^2,x,method=_RETURNVERBOSE)

[Out]

exp(-2)*(-5*x^3-8*x^2-16)/x^3/(x-5)

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maxima [A]  time = 0.37, size = 27, normalized size = 1.17 \begin {gather*} -\frac {{\left (5 \, x^{3} + 8 \, x^{2} + 16\right )} e^{\left (-2\right )}}{x^{4} - 5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^4+16*x^3-40*x^2+64*x-240)*exp(5)/(x^4-10*x^3+25*x^2)/exp(3)/exp(log(x)+2)^2,x, algorithm="maxim
a")

[Out]

-(5*x^3 + 8*x^2 + 16)*e^(-2)/(x^4 - 5*x^3)

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mupad [B]  time = 0.11, size = 19, normalized size = 0.83 \begin {gather*} -\frac {{\mathrm {e}}^{-2}\,{\left (x^2+4\right )}^2}{x^3\,\left (x-5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(- 2*log(x) - 4)*exp(2)*(64*x - 40*x^2 + 16*x^3 + 5*x^4 - 240))/(25*x^2 - 10*x^3 + x^4),x)

[Out]

-(exp(-2)*(x^2 + 4)^2)/(x^3*(x - 5))

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sympy [A]  time = 0.27, size = 27, normalized size = 1.17 \begin {gather*} \frac {- 5 x^{3} - 8 x^{2} - 16}{x^{4} e^{2} - 5 x^{3} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**4+16*x**3-40*x**2+64*x-240)*exp(5)/(x**4-10*x**3+25*x**2)/exp(3)/exp(ln(x)+2)**2,x)

[Out]

(-5*x**3 - 8*x**2 - 16)/(x**4*exp(2) - 5*x**3*exp(2))

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