3.19.54 \(\int \frac {-e^4+e^{4+x} (-3+x)}{32-32 x+8 x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {e^4 \left (1+e^x\right )}{4 (-4+2 x)} \]

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Rubi [A]  time = 0.17, antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 12, 6741, 6742, 2197} \begin {gather*} -\frac {e^{x+4}}{8 (2-x)}-\frac {e^4}{8 (2-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^4 + E^(4 + x)*(-3 + x))/(32 - 32*x + 8*x^2),x]

[Out]

-1/8*E^4/(2 - x) - E^(4 + x)/(8*(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 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 6741

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

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^4+e^{4+x} (-3+x)}{8 (-2+x)^2} \, dx\\ &=\frac {1}{8} \int \frac {-e^4+e^{4+x} (-3+x)}{(-2+x)^2} \, dx\\ &=\frac {1}{8} \int \frac {e^4 \left (-1-3 e^x+e^x x\right )}{(2-x)^2} \, dx\\ &=\frac {1}{8} e^4 \int \frac {-1-3 e^x+e^x x}{(2-x)^2} \, dx\\ &=\frac {1}{8} e^4 \int \left (-\frac {1}{(-2+x)^2}+\frac {e^x (-3+x)}{(-2+x)^2}\right ) \, dx\\ &=-\frac {e^4}{8 (2-x)}+\frac {1}{8} e^4 \int \frac {e^x (-3+x)}{(-2+x)^2} \, dx\\ &=-\frac {e^4}{8 (2-x)}-\frac {e^{4+x}}{8 (2-x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 0.89 \begin {gather*} \frac {e^4 \left (1+e^x\right )}{8 (-2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^4 + E^(4 + x)*(-3 + x))/(32 - 32*x + 8*x^2),x]

[Out]

(E^4*(1 + E^x))/(8*(-2 + x))

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fricas [A]  time = 0.94, size = 14, normalized size = 0.74 \begin {gather*} \frac {e^{4} + e^{\left (x + 4\right )}}{8 \, {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*exp(2)^2*exp(x)-exp(2)^2)/(8*x^2-32*x+32),x, algorithm="fricas")

[Out]

1/8*(e^4 + e^(x + 4))/(x - 2)

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giac [A]  time = 0.23, size = 14, normalized size = 0.74 \begin {gather*} \frac {e^{4} + e^{\left (x + 4\right )}}{8 \, {\left (x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*exp(2)^2*exp(x)-exp(2)^2)/(8*x^2-32*x+32),x, algorithm="giac")

[Out]

1/8*(e^4 + e^(x + 4))/(x - 2)

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maple [A]  time = 0.37, size = 22, normalized size = 1.16

method	result	size
norman	\(\frac {\frac {{\mathrm e}^{4} {\mathrm e}^{x}}{8}+\frac {{\mathrm e}^{4}}{8}}{x -2}\)	\(22\)
risch	\(\frac {{\mathrm e}^{4}}{8 x -16}+\frac {{\mathrm e}^{4+x}}{8 x -16}\)	\(22\)
default	\(\frac {{\mathrm e}^{4} \left (-\frac {2 \,{\mathrm e}^{x}}{x -2}-3 \,{\mathrm e}^{2} \expIntegralEi \left (1, 2-x \right )\right )}{8}+\frac {{\mathrm e}^{4}}{8 x -16}-\frac {3 \,{\mathrm e}^{4} \left (-\frac {{\mathrm e}^{x}}{x -2}-{\mathrm e}^{2} \expIntegralEi \left (1, 2-x \right )\right )}{8}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-3)*exp(2)^2*exp(x)-exp(2)^2)/(8*x^2-32*x+32),x,method=_RETURNVERBOSE)

[Out]

(1/8*exp(2)^2*exp(x)+1/8*exp(2)^2)/(x-2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*exp(2)^2*exp(x)-exp(2)^2)/(8*x^2-32*x+32),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.16, size = 17, normalized size = 0.89 \begin {gather*} \frac {\frac {{\mathrm {e}}^{x+4}}{8}+\frac {{\mathrm {e}}^4}{8}}{x-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4) - exp(4)*exp(x)*(x - 3))/(8*x^2 - 32*x + 32),x)

[Out]

(exp(x + 4)/8 + exp(4)/8)/(x - 2)

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sympy [A]  time = 0.16, size = 19, normalized size = 1.00 \begin {gather*} \frac {e^{4} e^{x}}{8 x - 16} + \frac {e^{4}}{8 x - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-3)*exp(2)**2*exp(x)-exp(2)**2)/(8*x**2-32*x+32),x)

[Out]

exp(4)*exp(x)/(8*x - 16) + exp(4)/(8*x - 16)

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