3.34.27 \(\int \frac {-2 e^{-\frac {x}{10+5 x}}+e^{-6-2 x} (40+40 x+10 x^2)}{20+20 x+5 x^2} \, dx\)

Optimal. Leaf size=24 \[ -e^{-6-2 x}+e^{\frac {x}{5 (-2-x)}} \]

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Rubi [A]  time = 0.27, antiderivative size = 25, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {27, 12, 6742, 2194, 2230, 2209} \begin {gather*} e^{\frac {2}{5 (x+2)}-\frac {1}{5}}-e^{-2 x-6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2/E^(x/(10 + 5*x)) + E^(-6 - 2*x)*(40 + 40*x + 10*x^2))/(20 + 20*x + 5*x^2),x]

[Out]

-E^(-6 - 2*x) + E^(-1/5 + 2/(5*(2 + 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 2194

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

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 {-2 e^{-\frac {x}{10+5 x}}+e^{-6-2 x} \left (40+40 x+10 x^2\right )}{5 (2+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-2 e^{-\frac {x}{10+5 x}}+e^{-6-2 x} \left (40+40 x+10 x^2\right )}{(2+x)^2} \, dx\\ &=\frac {1}{5} \int \left (10 e^{-6-2 x}-\frac {2 e^{-\frac {x}{5 (2+x)}}}{(2+x)^2}\right ) \, dx\\ &=-\left (\frac {2}{5} \int \frac {e^{-\frac {x}{5 (2+x)}}}{(2+x)^2} \, dx\right )+2 \int e^{-6-2 x} \, dx\\ &=-e^{-6-2 x}-\frac {2}{5} \int \frac {e^{-\frac {1}{5}+\frac {2}{5 (2+x)}}}{(2+x)^2} \, dx\\ &=-e^{-6-2 x}+e^{-\frac {1}{5}+\frac {2}{5 (2+x)}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 28, normalized size = 1.17 \begin {gather*} \frac {1}{5} \left (5 e^{-\frac {x}{5 (2+x)}}-5 e^{-2 (3+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2/E^(x/(10 + 5*x)) + E^(-6 - 2*x)*(40 + 40*x + 10*x^2))/(20 + 20*x + 5*x^2),x]

[Out]

(5/E^(x/(5*(2 + x))) - 5/E^(2*(3 + x)))/5

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fricas [A]  time = 0.69, size = 18, normalized size = 0.75 \begin {gather*} -e^{\left (-2 \, x - 6\right )} + e^{\left (-\frac {x}{5 \, {\left (x + 2\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(-x/(5*x+10))+(10*x^2+40*x+40)*exp(-2*x-6))/(5*x^2+20*x+20),x, algorithm="fricas")

[Out]

-e^(-2*x - 6) + e^(-1/5*x/(x + 2))

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giac [A]  time = 0.32, size = 18, normalized size = 0.75 \begin {gather*} -e^{\left (-2 \, x - 6\right )} + e^{\left (-\frac {x}{5 \, {\left (x + 2\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(-x/(5*x+10))+(10*x^2+40*x+40)*exp(-2*x-6))/(5*x^2+20*x+20),x, algorithm="giac")

[Out]

-e^(-2*x - 6) + e^(-1/5*x/(x + 2))

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maple [A]  time = 0.46, size = 19, normalized size = 0.79

method	result	size
risch	\(-{\mathrm e}^{-2 x -6}+{\mathrm e}^{-\frac {x}{5 \left (2+x \right )}}\)	\(19\)
norman	\(\frac {x \,{\mathrm e}^{-\frac {x}{5 x +10}}-{\mathrm e}^{-2 x -6} x +2 \,{\mathrm e}^{-\frac {x}{5 x +10}}-2 \,{\mathrm e}^{-2 x -6}}{2+x}\)	\(51\)
default	\({\mathrm e}^{-\frac {1}{5}+\frac {2}{5 \left (2+x \right )}}+8 \,{\mathrm e}^{-6} \left (-\frac {{\mathrm e}^{-2 x}}{2+x}+2 \,{\mathrm e}^{4} \expIntegralEi \left (1, 2 x +4\right )\right )+8 \,{\mathrm e}^{-6} \left (\frac {2 \,{\mathrm e}^{-2 x}}{2+x}-5 \,{\mathrm e}^{4} \expIntegralEi \left (1, 2 x +4\right )\right )+2 \,{\mathrm e}^{-6} \left (-\frac {{\mathrm e}^{-2 x}}{2}-\frac {4 \,{\mathrm e}^{-2 x}}{2+x}+12 \,{\mathrm e}^{4} \expIntegralEi \left (1, 2 x +4\right )\right )\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(-x/(5*x+10))+(10*x^2+40*x+40)*exp(-2*x-6))/(5*x^2+20*x+20),x,method=_RETURNVERBOSE)

[Out]

-exp(-2*x-6)+exp(-1/5*x/(2+x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {8 \, e^{\left (-2\right )} E_{2}\left (2 \, x + 4\right )}{x + 2} - \frac {{\left (x^{2} e^{\frac {1}{5}} + 4 \, x e^{\frac {1}{5}}\right )} e^{\left (-2 \, x\right )} - {\left (x^{2} e^{6} + 4 \, x e^{6} + 4 \, e^{6}\right )} e^{\left (\frac {2}{5 \, {\left (x + 2\right )}}\right )}}{x^{2} e^{\frac {31}{5}} + 4 \, x e^{\frac {31}{5}} + 4 \, e^{\frac {31}{5}}} + 8 \, \int \frac {e^{\left (-2 \, x\right )}}{x^{3} e^{6} + 6 \, x^{2} e^{6} + 12 \, x e^{6} + 8 \, e^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(-x/(5*x+10))+(10*x^2+40*x+40)*exp(-2*x-6))/(5*x^2+20*x+20),x, algorithm="maxima")

[Out]

-8*e^(-2)*exp_integral_e(2, 2*x + 4)/(x + 2) - ((x^2*e^(1/5) + 4*x*e^(1/5))*e^(-2*x) - (x^2*e^6 + 4*x*e^6 + 4*
e^6)*e^(2/5/(x + 2)))/(x^2*e^(31/5) + 4*x*e^(31/5) + 4*e^(31/5)) + 8*integrate(e^(-2*x)/(x^3*e^6 + 6*x^2*e^6 +
12*x*e^6 + 8*e^6), x)

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mupad [B]  time = 2.08, size = 20, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{-\frac {x}{5\,x+10}}-{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^{-6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*exp(-x/(5*x + 10)) - exp(- 2*x - 6)*(40*x + 10*x^2 + 40))/(20*x + 5*x^2 + 20),x)

[Out]

exp(-x/(5*x + 10)) - exp(-2*x)*exp(-6)

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sympy [A]  time = 0.36, size = 17, normalized size = 0.71 \begin {gather*} - e^{- 2 x - 6} + e^{- \frac {x}{5 x + 10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(-x/(5*x+10))+(10*x**2+40*x+40)*exp(-2*x-6))/(5*x**2+20*x+20),x)

[Out]

-exp(-2*x - 6) + exp(-x/(5*x + 10))

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