\(\int \frac {e^{-x} (-132-836 x+648 x^2-96 x^3+4 x^4)}{121-22 x+x^2} \, dx\) [5223]

Optimal result

Integrand size = 36, antiderivative size = 19 \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=-4 e^{-x} x \left (\frac {3}{11-x}+x\right ) \]

[Out]

-4*x*(3/(11-x)+x)*exp(-x)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68, number of steps used = 12, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {27, 2230, 2225, 2208, 2209, 2207} \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=-4 e^{-x} x^2+12 e^{-x}-\frac {132 e^{-x}}{11-x} \]

[In]

Int[(-132 - 836*x + 648*x^2 - 96*x^3 + 4*x^4)/(E^x*(121 - 22*x + x^2)),x]

[Out]

12/E^x - 132/(E^x*(11 - x)) - (4*x^2)/E^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 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), 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] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2230

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !TrueQ[$UseGamma]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{(-11+x)^2} \, dx \\ & = \int \left (-12 e^{-x}-\frac {132 e^{-x}}{(-11+x)^2}-\frac {132 e^{-x}}{-11+x}-8 e^{-x} x+4 e^{-x} x^2\right ) \, dx \\ & = 4 \int e^{-x} x^2 \, dx-8 \int e^{-x} x \, dx-12 \int e^{-x} \, dx-132 \int \frac {e^{-x}}{(-11+x)^2} \, dx-132 \int \frac {e^{-x}}{-11+x} \, dx \\ & = 12 e^{-x}-\frac {132 e^{-x}}{11-x}+8 e^{-x} x-4 e^{-x} x^2-\frac {132 \text {Ei}(11-x)}{e^{11}}-8 \int e^{-x} \, dx+8 \int e^{-x} x \, dx+132 \int \frac {e^{-x}}{-11+x} \, dx \\ & = 20 e^{-x}-\frac {132 e^{-x}}{11-x}-4 e^{-x} x^2+8 \int e^{-x} \, dx \\ & = 12 e^{-x}-\frac {132 e^{-x}}{11-x}-4 e^{-x} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=-\frac {4 e^{-x} x \left (-3-11 x+x^2\right )}{-11+x} \]

[In]

Integrate[(-132 - 836*x + 648*x^2 - 96*x^3 + 4*x^4)/(E^x*(121 - 22*x + x^2)),x]

[Out]

(-4*x*(-3 - 11*x + x^2))/(E^x*(-11 + x))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11

method	result	size
gosper	\(-\frac {4 x \left (x^{2}-11 x -3\right ) {\mathrm e}^{-x}}{x -11}\)	\(21\)
risch	\(-\frac {4 x \left (x^{2}-11 x -3\right ) {\mathrm e}^{-x}}{x -11}\)	\(21\)
derivativedivides	\(-4 x^{2} {\mathrm e}^{-x}+12 \,{\mathrm e}^{-x}-\frac {132 \,{\mathrm e}^{-x}}{11-x}\)	\(30\)
default	\(-4 x^{2} {\mathrm e}^{-x}+12 \,{\mathrm e}^{-x}-\frac {132 \,{\mathrm e}^{-x}}{11-x}\)	\(30\)
norman	\(\frac {12 x \,{\mathrm e}^{-x}+44 x^{2} {\mathrm e}^{-x}-4 \,{\mathrm e}^{-x} x^{3}}{x -11}\)	\(33\)
parallelrisch	\(-\frac {4 \,{\mathrm e}^{-x} x^{3}-44 x^{2} {\mathrm e}^{-x}-12 x \,{\mathrm e}^{-x}}{x -11}\)	\(34\)

[In]

int((4*x^4-96*x^3+648*x^2-836*x-132)*exp(-x)/(x^2-22*x+121),x,method=_RETURNVERBOSE)

[Out]

-4*x*(x^2-11*x-3)*exp(-x)/(x-11)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=-\frac {4 \, {\left (x^{3} - 11 \, x^{2} - 3 \, x\right )} e^{\left (-x\right )}}{x - 11} \]

[In]

integrate((4*x^4-96*x^3+648*x^2-836*x-132)*exp(-x)/(x^2-22*x+121),x, algorithm="fricas")

[Out]

-4*(x^3 - 11*x^2 - 3*x)*e^(-x)/(x - 11)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=\frac {\left (- 4 x^{3} + 44 x^{2} + 12 x\right ) e^{- x}}{x - 11} \]

[In]

integrate((4*x**4-96*x**3+648*x**2-836*x-132)*exp(-x)/(x**2-22*x+121),x)

[Out]

(-4*x**3 + 44*x**2 + 12*x)*exp(-x)/(x - 11)

Maxima [F]

\[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=\int { \frac {4 \, {\left (x^{4} - 24 \, x^{3} + 162 \, x^{2} - 209 \, x - 33\right )} e^{\left (-x\right )}}{x^{2} - 22 \, x + 121} \,d x } \]

[In]

integrate((4*x^4-96*x^3+648*x^2-836*x-132)*exp(-x)/(x^2-22*x+121),x, algorithm="maxima")

[Out]

-4*(x^3 - 11*x^2 - 3*x)*e^(-x)/(x - 11) + 132*e^(-11)*exp_integral_e(2, x - 11)/(x - 11) + 132*integrate(e^(-x
)/(x^2 - 22*x + 121), x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=-\frac {4 \, {\left (x^{3} e^{\left (-x\right )} - 11 \, x^{2} e^{\left (-x\right )} - 3 \, x e^{\left (-x\right )}\right )}}{x - 11} \]

[In]

integrate((4*x^4-96*x^3+648*x^2-836*x-132)*exp(-x)/(x^2-22*x+121),x, algorithm="giac")

[Out]

-4*(x^3*e^(-x) - 11*x^2*e^(-x) - 3*x*e^(-x))/(x - 11)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x} \left (-132-836 x+648 x^2-96 x^3+4 x^4\right )}{121-22 x+x^2} \, dx=\frac {4\,x\,{\mathrm {e}}^{-x}\,\left (-x^2+11\,x+3\right )}{x-11} \]

[In]

int(-(exp(-x)*(836*x - 648*x^2 + 96*x^3 - 4*x^4 + 132))/(x^2 - 22*x + 121),x)

[Out]

(4*x*exp(-x)*(11*x - x^2 + 3))/(x - 11)