3.31.26 \(\int \frac {-16 x-x^2}{e^{11} (-512-128 x-8 x^2)+e^{16} (64+16 x+x^2)+e^6 (1024+256 x+16 x^2)} \, dx\)

Optimal. Leaf size=25 \[ e^3+\frac {x^2}{e^6 \left (-4+e^5\right )^2 (-8-x)} \]

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Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1593, 1983, 27, 12, 74} \begin {gather*} -\frac {x^2}{e^6 \left (4-e^5\right )^2 (x+8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x - x^2)/(E^11*(-512 - 128*x - 8*x^2) + E^16*(64 + 16*x + x^2) + E^6*(1024 + 256*x + 16*x^2)),x]

[Out]

-(x^2/(E^6*(4 - E^5)^2*(8 + 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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1983

Int[(u_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n*ExpandToSum[w,
x]^p, x] /; FreeQ[{m, n, p}, x] && LinearQ[{u, v}, x] && QuadraticQ[w, x] &&  !(LinearMatchQ[{u, v}, x] && Qua
draticMatchQ[w, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(-16-x) x}{e^{11} \left (-512-128 x-8 x^2\right )+e^{16} \left (64+16 x+x^2\right )+e^6 \left (1024+256 x+16 x^2\right )} \, dx\\ &=\int \frac {(-16-x) x}{64 e^6 \left (4-e^5\right )^2+16 e^6 \left (4-e^5\right )^2 x+e^6 \left (4-e^5\right )^2 x^2} \, dx\\ &=\int \frac {(-16-x) x}{e^6 \left (-4+e^5\right )^2 (8+x)^2} \, dx\\ &=\frac {\int \frac {(-16-x) x}{(8+x)^2} \, dx}{e^6 \left (4-e^5\right )^2}\\ &=-\frac {x^2}{e^6 \left (4-e^5\right )^2 (8+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 0.84 \begin {gather*} -\frac {x+\frac {64}{8+x}}{e^6 \left (-4+e^5\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x - x^2)/(E^11*(-512 - 128*x - 8*x^2) + E^16*(64 + 16*x + x^2) + E^6*(1024 + 256*x + 16*x^2)),x
]

[Out]

-((x + 64/(8 + x))/(E^6*(-4 + E^5)^2))

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fricas [A]  time = 0.45, size = 33, normalized size = 1.32 \begin {gather*} -\frac {x^{2} + 8 \, x + 64}{{\left (x + 8\right )} e^{16} - 8 \, {\left (x + 8\right )} e^{11} + 16 \, {\left (x + 8\right )} e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-16*x)/((x^2+16*x+64)*exp(3)^2*exp(5)^2+(-8*x^2-128*x-512)*exp(3)^2*exp(5)+(16*x^2+256*x+1024)*
exp(3)^2),x, algorithm="fricas")

[Out]

-(x^2 + 8*x + 64)/((x + 8)*e^16 - 8*(x + 8)*e^11 + 16*(x + 8)*e^6)

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giac [A]  time = 0.14, size = 37, normalized size = 1.48 \begin {gather*} -\frac {x}{e^{16} - 8 \, e^{11} + 16 \, e^{6}} - \frac {64}{{\left (x + 8\right )} {\left (e^{16} - 8 \, e^{11} + 16 \, e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-16*x)/((x^2+16*x+64)*exp(3)^2*exp(5)^2+(-8*x^2-128*x-512)*exp(3)^2*exp(5)+(16*x^2+256*x+1024)*
exp(3)^2),x, algorithm="giac")

[Out]

-x/(e^16 - 8*e^11 + 16*e^6) - 64/((x + 8)*(e^16 - 8*e^11 + 16*e^6))

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maple [A]  time = 0.52, size = 21, normalized size = 0.84

method	result	size
norman	\(-\frac {{\mathrm e}^{-6} x^{2}}{\left ({\mathrm e}^{5}-4\right )^{2} \left (x +8\right )}\)	\(21\)
default	\(\frac {{\mathrm e}^{-6} \left (-x -\frac {64}{x +8}\right )}{-8 \,{\mathrm e}^{5}+{\mathrm e}^{10}+16}\)	\(29\)
gosper	\(-\frac {x^{2} {\mathrm e}^{-6}}{x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128}\)	\(38\)
meijerg	\(-\frac {8 \left (\frac {x \left (\frac {3 x}{8}+6\right )}{24+3 x}-2 \ln \left (1+\frac {x}{8}\right )\right )}{{\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}}-\frac {16 \left (-\frac {x}{8 \left (1+\frac {x}{8}\right )}+\ln \left (1+\frac {x}{8}\right )\right )}{{\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}}\)	\(73\)
risch	\(-\frac {x}{{\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}}-\frac {64 \,{\mathrm e}^{-6} {\mathrm e}^{16}}{\left ({\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}\right ) \left (x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128\right )}+\frac {512 \,{\mathrm e}^{-6} {\mathrm e}^{11}}{\left ({\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}\right ) \left (x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128\right )}-\frac {1024 \,{\mathrm e}^{-6} {\mathrm e}^{6}}{\left ({\mathrm e}^{16}-8 \,{\mathrm e}^{11}+16 \,{\mathrm e}^{6}\right ) \left (x \,{\mathrm e}^{10}+8 \,{\mathrm e}^{10}-8 x \,{\mathrm e}^{5}-64 \,{\mathrm e}^{5}+16 x +128\right )}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2-16*x)/((x^2+16*x+64)*exp(3)^2*exp(5)^2+(-8*x^2-128*x-512)*exp(3)^2*exp(5)+(16*x^2+256*x+1024)*exp(3)
^2),x,method=_RETURNVERBOSE)

[Out]

-1/exp(3)^2/(exp(5)-4)^2*x^2/(x+8)

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maxima [B]  time = 0.36, size = 47, normalized size = 1.88 \begin {gather*} -\frac {x}{e^{16} - 8 \, e^{11} + 16 \, e^{6}} - \frac {64}{x {\left (e^{16} - 8 \, e^{11} + 16 \, e^{6}\right )} + 8 \, e^{16} - 64 \, e^{11} + 128 \, e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2-16*x)/((x^2+16*x+64)*exp(3)^2*exp(5)^2+(-8*x^2-128*x-512)*exp(3)^2*exp(5)+(16*x^2+256*x+1024)*
exp(3)^2),x, algorithm="maxima")

[Out]

-x/(e^16 - 8*e^11 + 16*e^6) - 64/(x*(e^16 - 8*e^11 + 16*e^6) + 8*e^16 - 64*e^11 + 128*e^6)

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mupad [B]  time = 1.80, size = 23, normalized size = 0.92 \begin {gather*} -\frac {{\mathrm {e}}^{-6}\,\left (x^2+8\,x+64\right )}{{\left ({\mathrm {e}}^5-4\right )}^2\,\left (x+8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x + x^2)/(exp(6)*(256*x + 16*x^2 + 1024) - exp(11)*(128*x + 8*x^2 + 512) + exp(16)*(16*x + x^2 + 64))
,x)

[Out]

-(exp(-6)*(8*x + x^2 + 64))/((exp(5) - 4)^2*(x + 8))

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sympy [B]  time = 0.29, size = 48, normalized size = 1.92 \begin {gather*} - \frac {x}{- 8 e^{11} + 16 e^{6} + e^{16}} - \frac {64}{x \left (- 8 e^{11} + 16 e^{6} + e^{16}\right ) - 64 e^{11} + 128 e^{6} + 8 e^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2-16*x)/((x**2+16*x+64)*exp(3)**2*exp(5)**2+(-8*x**2-128*x-512)*exp(3)**2*exp(5)+(16*x**2+256*x
+1024)*exp(3)**2),x)

[Out]

-x/(-8*exp(11) + 16*exp(6) + exp(16)) - 64/(x*(-8*exp(11) + 16*exp(6) + exp(16)) - 64*exp(11) + 128*exp(6) + 8
*exp(16))

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