3.2.63 \(\int \frac {e^{-2 x} (-100-240 x-40 x^2)}{25 x^2+10 x^3+x^4} \, dx\)

Optimal. Leaf size=15 \[ \frac {20 e^{-2 x}}{x (5+x)} \]

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Rubi [A]  time = 0.33, antiderivative size = 23, normalized size of antiderivative = 1.53, number of steps used = 10, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1594, 27, 6742, 2177, 2178} \begin {gather*} \frac {4 e^{-2 x}}{x}-\frac {4 e^{-2 x}}{x+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100 - 240*x - 40*x^2)/(E^(2*x)*(25*x^2 + 10*x^3 + x^4)),x]

[Out]

4/(E^(2*x)*x) - 4/(E^(2*x)*(5 + 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^{-2 x} \left (-100-240 x-40 x^2\right )}{x^2 \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{x^2 (5+x)^2} \, dx\\ &=\int \left (-\frac {4 e^{-2 x}}{x^2}-\frac {8 e^{-2 x}}{x}+\frac {4 e^{-2 x}}{(5+x)^2}+\frac {8 e^{-2 x}}{5+x}\right ) \, dx\\ &=-\left (4 \int \frac {e^{-2 x}}{x^2} \, dx\right )+4 \int \frac {e^{-2 x}}{(5+x)^2} \, dx-8 \int \frac {e^{-2 x}}{x} \, dx+8 \int \frac {e^{-2 x}}{5+x} \, dx\\ &=\frac {4 e^{-2 x}}{x}-\frac {4 e^{-2 x}}{5+x}-8 \text {Ei}(-2 x)+8 e^{10} \text {Ei}(-2 (5+x))+8 \int \frac {e^{-2 x}}{x} \, dx-8 \int \frac {e^{-2 x}}{5+x} \, dx\\ &=\frac {4 e^{-2 x}}{x}-\frac {4 e^{-2 x}}{5+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 16, normalized size = 1.07 \begin {gather*} \frac {20 e^{-2 x}}{5 x+x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 - 240*x - 40*x^2)/(E^(2*x)*(25*x^2 + 10*x^3 + x^4)),x]

[Out]

20/(E^(2*x)*(5*x + x^2))

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fricas [A]  time = 0.59, size = 15, normalized size = 1.00 \begin {gather*} \frac {20 \, e^{\left (-2 \, x\right )}}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-40*x^2-240*x-100)*exp(x)/(x^4+10*x^3+25*x^2)/exp(3*x),x, algorithm="fricas")

[Out]

20*e^(-2*x)/(x^2 + 5*x)

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giac [A]  time = 0.41, size = 15, normalized size = 1.00 \begin {gather*} \frac {20 \, e^{\left (-2 \, x\right )}}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-40*x^2-240*x-100)*exp(x)/(x^4+10*x^3+25*x^2)/exp(3*x),x, algorithm="giac")

[Out]

20*e^(-2*x)/(x^2 + 5*x)

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maple [A]  time = 0.06, size = 15, normalized size = 1.00

method	result	size
norman	\(\frac {20 \,{\mathrm e}^{-2 x}}{x \left (5+x \right )}\)	\(15\)
risch	\(\frac {20 \,{\mathrm e}^{-2 x}}{x \left (5+x \right )}\)	\(15\)
gosper	\(\frac {20 \,{\mathrm e}^{x} {\mathrm e}^{-3 x}}{x \left (5+x \right )}\)	\(19\)
default	\(\frac {4 \,{\mathrm e}^{-2 x} \left (5+2 x \right )}{\left (5+x \right ) x}-\frac {8 \,{\mathrm e}^{-2 x}}{5+x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-40*x^2-240*x-100)*exp(x)/(x^4+10*x^3+25*x^2)/exp(3*x),x,method=_RETURNVERBOSE)

[Out]

20/exp(x)^2/x/(5+x)

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maxima [A]  time = 0.60, size = 15, normalized size = 1.00 \begin {gather*} \frac {20 \, e^{\left (-2 \, x\right )}}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-40*x^2-240*x-100)*exp(x)/(x^4+10*x^3+25*x^2)/exp(3*x),x, algorithm="maxima")

[Out]

20*e^(-2*x)/(x^2 + 5*x)

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mupad [B]  time = 0.33, size = 14, normalized size = 0.93 \begin {gather*} \frac {20\,{\mathrm {e}}^{-2\,x}}{x\,\left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-2*x)*(240*x + 40*x^2 + 100))/(25*x^2 + 10*x^3 + x^4),x)

[Out]

(20*exp(-2*x))/(x*(x + 5))

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sympy [A]  time = 0.11, size = 12, normalized size = 0.80 \begin {gather*} \frac {20 e^{- 2 x}}{x^{2} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-40*x**2-240*x-100)*exp(x)/(x**4+10*x**3+25*x**2)/exp(3*x),x)

[Out]

20*exp(-2*x)/(x**2 + 5*x)

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