3.94.79 \(\int \frac {2+4 x}{e^2 (5 x^2+10 x^3+5 x^4)} \, dx\)

Optimal. Leaf size=21 \[ \frac {2 x}{5 e^2 \left (x-x \left (1+x+x^2\right )\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.71, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 1594, 27, 74} \begin {gather*} -\frac {2}{5 e^2 x (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 4*x)/(E^2*(5*x^2 + 10*x^3 + 5*x^4)),x]

[Out]

-2/(5*E^2*x*(1 + 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 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {2+4 x}{5 x^2+10 x^3+5 x^4} \, dx}{e^2}\\ &=\frac {\int \frac {2+4 x}{x^2 \left (5+10 x+5 x^2\right )} \, dx}{e^2}\\ &=\frac {\int \frac {2+4 x}{5 x^2 (1+x)^2} \, dx}{e^2}\\ &=\frac {\int \frac {2+4 x}{x^2 (1+x)^2} \, dx}{5 e^2}\\ &=-\frac {2}{5 e^2 x (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.86 \begin {gather*} \frac {2 \left (-\frac {1}{x}+\frac {1}{1+x}\right )}{5 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 4*x)/(E^2*(5*x^2 + 10*x^3 + 5*x^4)),x]

[Out]

(2*(-x^(-1) + (1 + x)^(-1)))/(5*E^2)

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fricas [A]  time = 0.74, size = 11, normalized size = 0.52 \begin {gather*} -\frac {2 \, e^{\left (-2\right )}}{5 \, {\left (x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)/(5*x^4+10*x^3+5*x^2)/exp(2),x, algorithm="fricas")

[Out]

-2/5*e^(-2)/(x^2 + x)

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giac [A]  time = 0.12, size = 11, normalized size = 0.52 \begin {gather*} -\frac {2 \, e^{\left (-2\right )}}{5 \, {\left (x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)/(5*x^4+10*x^3+5*x^2)/exp(2),x, algorithm="giac")

[Out]

-2/5*e^(-2)/(x^2 + x)

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maple [A]  time = 0.04, size = 13, normalized size = 0.62

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{-2}}{5 x \left (x +1\right )}\)	\(13\)
gosper	\(-\frac {2 \,{\mathrm e}^{-2}}{5 x \left (x +1\right )}\)	\(15\)
norman	\(-\frac {2 \,{\mathrm e}^{-2}}{5 x \left (x +1\right )}\)	\(15\)
default	\(\frac {2 \,{\mathrm e}^{-2} \left (\frac {1}{x +1}-\frac {1}{x}\right )}{5}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x+2)/(5*x^4+10*x^3+5*x^2)/exp(2),x,method=_RETURNVERBOSE)

[Out]

-2/5/x*exp(-2)/(x+1)

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maxima [A]  time = 0.36, size = 11, normalized size = 0.52 \begin {gather*} -\frac {2 \, e^{\left (-2\right )}}{5 \, {\left (x^{2} + x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)/(5*x^4+10*x^3+5*x^2)/exp(2),x, algorithm="maxima")

[Out]

-2/5*e^(-2)/(x^2 + x)

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mupad [B]  time = 0.10, size = 12, normalized size = 0.57 \begin {gather*} -\frac {2\,{\mathrm {e}}^{-2}}{5\,x\,\left (x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2)*(4*x + 2))/(5*x^2 + 10*x^3 + 5*x^4),x)

[Out]

-(2*exp(-2))/(5*x*(x + 1))

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sympy [A]  time = 0.14, size = 17, normalized size = 0.81 \begin {gather*} - \frac {2}{5 x^{2} e^{2} + 5 x e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x+2)/(5*x**4+10*x**3+5*x**2)/exp(2),x)

[Out]

-2/(5*x**2*exp(2) + 5*x*exp(2))

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