3.62.78 \(\int \frac {e^{-x} (-24-30 x-6 x^2+e^x (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 (-4 x^3-4 x^4-x^5)))}{4 x^3+4 x^4+x^5} \, dx\)

Optimal. Leaf size=31 \[ x \left (-25-e^2-x+\frac {6 e^{-x}}{x^2 (x+x (1+x))}\right ) \]

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Rubi [A]  time = 0.92, antiderivative size = 50, normalized size of antiderivative = 1.61, number of steps used = 15, number of rules used = 5, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1594, 27, 6742, 2177, 2178} \begin {gather*} -x^2+\frac {3 e^{-x}}{x^2}-\left (25+e^2\right ) x+\frac {3 e^{-x}}{2 (x+2)}-\frac {3 e^{-x}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24 - 30*x - 6*x^2 + E^x*(-100*x^3 - 108*x^4 - 33*x^5 - 2*x^6 + E^2*(-4*x^3 - 4*x^4 - x^5)))/(E^x*(4*x^3
+ 4*x^4 + x^5)),x]

[Out]

3/(E^x*x^2) - 3/(2*E^x*x) - (25 + E^2)*x - x^2 + 3/(2*E^x*(2 + 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 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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

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 {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{x^3 \left (4+4 x+x^2\right )} \, dx\\ &=\int \frac {e^{-x} \left (-24-30 x-6 x^2+e^x \left (-100 x^3-108 x^4-33 x^5-2 x^6+e^2 \left (-4 x^3-4 x^4-x^5\right )\right )\right )}{x^3 (2+x)^2} \, dx\\ &=\int \left (-25-e^2-2 x-\frac {6 e^{-x} \left (4+5 x+x^2\right )}{x^3 (2+x)^2}\right ) \, dx\\ &=-\left (\left (25+e^2\right ) x\right )-x^2-6 \int \frac {e^{-x} \left (4+5 x+x^2\right )}{x^3 (2+x)^2} \, dx\\ &=-\left (\left (25+e^2\right ) x\right )-x^2-6 \int \left (\frac {e^{-x}}{x^3}+\frac {e^{-x}}{4 x^2}-\frac {e^{-x}}{4 x}+\frac {e^{-x}}{4 (2+x)^2}+\frac {e^{-x}}{4 (2+x)}\right ) \, dx\\ &=-\left (\left (25+e^2\right ) x\right )-x^2-\frac {3}{2} \int \frac {e^{-x}}{x^2} \, dx+\frac {3}{2} \int \frac {e^{-x}}{x} \, dx-\frac {3}{2} \int \frac {e^{-x}}{(2+x)^2} \, dx-\frac {3}{2} \int \frac {e^{-x}}{2+x} \, dx-6 \int \frac {e^{-x}}{x^3} \, dx\\ &=\frac {3 e^{-x}}{x^2}+\frac {3 e^{-x}}{2 x}-\left (25+e^2\right ) x-x^2+\frac {3 e^{-x}}{2 (2+x)}-\frac {3}{2} e^2 \text {Ei}(-2-x)+\frac {3 \text {Ei}(-x)}{2}+\frac {3}{2} \int \frac {e^{-x}}{x} \, dx+\frac {3}{2} \int \frac {e^{-x}}{2+x} \, dx+3 \int \frac {e^{-x}}{x^2} \, dx\\ &=\frac {3 e^{-x}}{x^2}-\frac {3 e^{-x}}{2 x}-\left (25+e^2\right ) x-x^2+\frac {3 e^{-x}}{2 (2+x)}+3 \text {Ei}(-x)-3 \int \frac {e^{-x}}{x} \, dx\\ &=\frac {3 e^{-x}}{x^2}-\frac {3 e^{-x}}{2 x}-\left (25+e^2\right ) x-x^2+\frac {3 e^{-x}}{2 (2+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.35, size = 28, normalized size = 0.90 \begin {gather*} -e^2 x+\frac {6 e^{-x}}{x^2 (2+x)}-x (25+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 - 30*x - 6*x^2 + E^x*(-100*x^3 - 108*x^4 - 33*x^5 - 2*x^6 + E^2*(-4*x^3 - 4*x^4 - x^5)))/(E^x*(
4*x^3 + 4*x^4 + x^5)),x]

[Out]

-(E^2*x) + 6/(E^x*x^2*(2 + x)) - x*(25 + x)

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fricas [A]  time = 1.38, size = 48, normalized size = 1.55 \begin {gather*} -\frac {{\left ({\left (x^{5} + 27 \, x^{4} + 50 \, x^{3} + {\left (x^{4} + 2 \, x^{3}\right )} e^{2}\right )} e^{x} - 6\right )} e^{\left (-x\right )}}{x^{3} + 2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x)-6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/ex
p(x),x, algorithm="fricas")

[Out]

-((x^5 + 27*x^4 + 50*x^3 + (x^4 + 2*x^3)*e^2)*e^x - 6)*e^(-x)/(x^3 + 2*x^2)

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giac [A]  time = 0.25, size = 46, normalized size = 1.48 \begin {gather*} -\frac {x^{5} + x^{4} e^{2} + 27 \, x^{4} + 2 \, x^{3} e^{2} + 50 \, x^{3} - 6 \, e^{\left (-x\right )}}{x^{3} + 2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x)-6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/ex
p(x),x, algorithm="giac")

[Out]

-(x^5 + x^4*e^2 + 27*x^4 + 2*x^3*e^2 + 50*x^3 - 6*e^(-x))/(x^3 + 2*x^2)

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maple [A]  time = 0.11, size = 29, normalized size = 0.94

method	result	size
risch	\(-{\mathrm e}^{2} x -x^{2}-25 x +\frac {6 \,{\mathrm e}^{-x}}{x^{2} \left (2+x \right )}\)	\(29\)
norman	\(\frac {\left (6+\left (-27-{\mathrm e}^{2}\right ) x^{4} {\mathrm e}^{x}+\left (100+4 \,{\mathrm e}^{2}\right ) x^{2} {\mathrm e}^{x}-x^{5} {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x^{2} \left (2+x \right )}\)	\(47\)
default	\(-x^{2}-25 x -\frac {3 \,{\mathrm e}^{-x} \left (4 x^{2}+5 x -2\right )}{x^{2} \left (2+x \right )}-{\mathrm e}^{2} x +\frac {15 \,{\mathrm e}^{-x} \left (x +1\right )}{\left (2+x \right ) x}-\frac {3 \,{\mathrm e}^{-x}}{2+x}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x)-6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/exp(x),x
,method=_RETURNVERBOSE)

[Out]

-exp(2)*x-x^2-25*x+6/x^2/(2+x)*exp(-x)

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maxima [A]  time = 0.39, size = 40, normalized size = 1.29 \begin {gather*} -\frac {x^{5} + x^{4} {\left (e^{2} + 27\right )} + 2 \, x^{3} {\left (e^{2} + 25\right )} - 6 \, e^{\left (-x\right )}}{x^{3} + 2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^5-4*x^4-4*x^3)*exp(2)-2*x^6-33*x^5-108*x^4-100*x^3)*exp(x)-6*x^2-30*x-24)/(x^5+4*x^4+4*x^3)/ex
p(x),x, algorithm="maxima")

[Out]

-(x^5 + x^4*(e^2 + 27) + 2*x^3*(e^2 + 25) - 6*e^(-x))/(x^3 + 2*x^2)

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mupad [B]  time = 4.25, size = 54, normalized size = 1.74 \begin {gather*} \frac {6}{2\,x^2\,{\mathrm {e}}^x+x^3\,{\mathrm {e}}^x}-\frac {x^5+\left ({\mathrm {e}}^2+27\right )\,x^4+\left (2\,{\mathrm {e}}^2+50\right )\,x^3}{x^3+2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(30*x + 6*x^2 + exp(x)*(exp(2)*(4*x^3 + 4*x^4 + x^5) + 100*x^3 + 108*x^4 + 33*x^5 + 2*x^6) + 24)
)/(4*x^3 + 4*x^4 + x^5),x)

[Out]

6/(2*x^2*exp(x) + x^3*exp(x)) - (x^3*(2*exp(2) + 50) + x^5 + x^4*(exp(2) + 27))/(2*x^2 + x^3)

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sympy [A]  time = 0.15, size = 24, normalized size = 0.77 \begin {gather*} - x^{2} + x \left (-25 - e^{2}\right ) + \frac {6 e^{- x}}{x^{3} + 2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**5-4*x**4-4*x**3)*exp(2)-2*x**6-33*x**5-108*x**4-100*x**3)*exp(x)-6*x**2-30*x-24)/(x**5+4*x**4
+4*x**3)/exp(x),x)

[Out]

-x**2 + x*(-25 - exp(2)) + 6*exp(-x)/(x**3 + 2*x**2)

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