3.44.76 \(\int \frac {e^{-2 x} (112+80 x+e^{2 x} (-8-60 x-150 x^2-125 x^3))}{8+60 x+150 x^2+125 x^3} \, dx\)

Optimal. Leaf size=24 \[ -1+e^5-x-\frac {2 e^{-2 x}}{\left (1+\frac {5 x}{2}\right )^2} \]

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Rubi [A]  time = 0.19, antiderivative size = 18, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 2, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6688, 2197} \begin {gather*} -x-\frac {8 e^{-2 x}}{(5 x+2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(112 + 80*x + E^(2*x)*(-8 - 60*x - 150*x^2 - 125*x^3))/(E^(2*x)*(8 + 60*x + 150*x^2 + 125*x^3)),x]

[Out]

-x - 8/(E^(2*x)*(2 + 5*x)^2)

Rule 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+\frac {16 e^{-2 x} (7+5 x)}{(2+5 x)^3}\right ) \, dx\\ &=-x+16 \int \frac {e^{-2 x} (7+5 x)}{(2+5 x)^3} \, dx\\ &=-x-\frac {8 e^{-2 x}}{(2+5 x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 28, normalized size = 1.17 \begin {gather*} -x-\frac {8 e^{\frac {4}{5}-\frac {2}{5} (2+5 x)}}{(2+5 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(112 + 80*x + E^(2*x)*(-8 - 60*x - 150*x^2 - 125*x^3))/(E^(2*x)*(8 + 60*x + 150*x^2 + 125*x^3)),x]

[Out]

-x - (8*E^(4/5 - (2*(2 + 5*x))/5))/(2 + 5*x)^2

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fricas [A]  time = 0.51, size = 39, normalized size = 1.62 \begin {gather*} -\frac {{\left ({\left (25 \, x^{3} + 20 \, x^{2} + 4 \, x\right )} e^{\left (2 \, x\right )} + 8\right )} e^{\left (-2 \, x\right )}}{25 \, x^{2} + 20 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x^3-150*x^2-60*x-8)*exp(x)^2+80*x+112)/(125*x^3+150*x^2+60*x+8)/exp(x)^2,x, algorithm="fricas
")

[Out]

-((25*x^3 + 20*x^2 + 4*x)*e^(2*x) + 8)*e^(-2*x)/(25*x^2 + 20*x + 4)

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giac [A]  time = 0.14, size = 34, normalized size = 1.42 \begin {gather*} -\frac {25 \, x^{3} + 20 \, x^{2} + 4 \, x + 8 \, e^{\left (-2 \, x\right )}}{25 \, x^{2} + 20 \, x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x^3-150*x^2-60*x-8)*exp(x)^2+80*x+112)/(125*x^3+150*x^2+60*x+8)/exp(x)^2,x, algorithm="giac")

[Out]

-(25*x^3 + 20*x^2 + 4*x + 8*e^(-2*x))/(25*x^2 + 20*x + 4)

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maple [A]  time = 0.04, size = 18, normalized size = 0.75

method	result	size
risch	\(-x -\frac {8 \,{\mathrm e}^{-2 x}}{\left (5 x +2\right )^{2}}\)	\(18\)
default	\(-x +\frac {56 \,{\mathrm e}^{-2 x} \left (10 x -1\right )}{25 \left (25 x^{2}+20 x +4\right )}-\frac {16 \,{\mathrm e}^{-2 x} \left (35 x +9\right )}{25 \left (25 x^{2}+20 x +4\right )}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-125*x^3-150*x^2-60*x-8)*exp(x)^2+80*x+112)/(125*x^3+150*x^2+60*x+8)/exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x-8/(5*x+2)^2*exp(-2*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {125 \, x^{4} + 150 \, x^{3} + 60 \, x^{2} + 40 \, x e^{\left (-2 \, x\right )} + 8 \, x}{125 \, x^{3} + 150 \, x^{2} + 60 \, x + 8} - \frac {112 \, e^{\frac {4}{5}} E_{3}\left (2 \, x + \frac {4}{5}\right )}{5 \, {\left (5 \, x + 2\right )}^{2}} - \int \frac {80 \, {\left (5 \, x - 1\right )} e^{\left (-2 \, x\right )}}{625 \, x^{4} + 1000 \, x^{3} + 600 \, x^{2} + 160 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x^3-150*x^2-60*x-8)*exp(x)^2+80*x+112)/(125*x^3+150*x^2+60*x+8)/exp(x)^2,x, algorithm="maxima
")

[Out]

-(125*x^4 + 150*x^3 + 60*x^2 + 40*x*e^(-2*x) + 8*x)/(125*x^3 + 150*x^2 + 60*x + 8) - 112/5*e^(4/5)*exp_integra
l_e(3, 2*x + 4/5)/(5*x + 2)^2 - integrate(80*(5*x - 1)*e^(-2*x)/(625*x^4 + 1000*x^3 + 600*x^2 + 160*x + 16), x
)

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mupad [B]  time = 0.12, size = 17, normalized size = 0.71 \begin {gather*} -x-\frac {8\,{\mathrm {e}}^{-2\,x}}{{\left (5\,x+2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x)*(80*x - exp(2*x)*(60*x + 150*x^2 + 125*x^3 + 8) + 112))/(60*x + 150*x^2 + 125*x^3 + 8),x)

[Out]

- x - (8*exp(-2*x))/(5*x + 2)^2

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sympy [A]  time = 0.13, size = 19, normalized size = 0.79 \begin {gather*} - x - \frac {8 e^{- 2 x}}{25 x^{2} + 20 x + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-125*x**3-150*x**2-60*x-8)*exp(x)**2+80*x+112)/(125*x**3+150*x**2+60*x+8)/exp(x)**2,x)

[Out]

-x - 8*exp(-2*x)/(25*x**2 + 20*x + 4)

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