\(\int \frac {e^{-2 x} (-100-240 x-40 x^2)}{25 x^2+10 x^3+x^4} \, dx\) [163]

Optimal result

Integrand size = 32, antiderivative size = 15 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20 e^{-2 x}}{x (5+x)} \]

[Out]

20*exp(x)/x/exp(3*x)/(5+x)

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1608, 27, 6874, 2208, 2209} \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {4 e^{-2 x}}{x}-\frac {4 e^{-2 x}}{x+5} \]

[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 1608

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 2208

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] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \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 \operatorname {ExpIntegralEi}(-2 x)+8 e^{10} \operatorname {ExpIntegralEi}(-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{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20 e^{-2 x}}{5 x+x^2} \]

[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))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 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	\(\frac {20 \,{\mathrm e}^{x} {\mathrm e}^{-3 x}}{x \left (5+x \right )}\)	\(19\)
parallelrisch	\(\frac {20 \,{\mathrm e}^{x} {\mathrm e}^{-3 x}}{x \left (5+x \right )}\)	\(19\)
default	\(\frac {4 \,{\mathrm e}^{-2 x} \left (5+2 x \right )}{\left (5+x \right ) x}-\frac {8 \,{\mathrm e}^{-2 x}}{5+x}\)	\(32\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20 \, e^{\left (-2 \, x\right )}}{x^{2} + 5 \, x} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20 e^{- 2 x}}{x^{2} + 5 x} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20 \, e^{\left (-2 \, x\right )}}{x^{2} + 5 \, x} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20 \, e^{\left (-2 \, x\right )}}{x^{2} + 5 \, x} \]

[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)

Mupad [B] (verification not implemented)

Time = 8.43 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-2 x} \left (-100-240 x-40 x^2\right )}{25 x^2+10 x^3+x^4} \, dx=\frac {20\,{\mathrm {e}}^{-2\,x}}{x\,\left (x+5\right )} \]

[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))