3.86.3 \(\int \frac {e}{(-8+x) (-16+2 x)} \, dx\)

Optimal. Leaf size=12 \[ -4-\frac {e}{-16+2 x} \]

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Rubi [A]  time = 0.00, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 21, 32} \begin {gather*} \frac {e}{2 (8-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E/((-8 + x)*(-16 + 2*x)),x]

[Out]

E/(2*(8 - 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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 10, normalized size = 0.83 \begin {gather*} -\frac {e}{2 (-8+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E/((-8 + x)*(-16 + 2*x)),x]

[Out]

-1/2*E/(-8 + x)

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fricas [A]  time = 0.98, size = 9, normalized size = 0.75 \begin {gather*} -\frac {e}{2 \, {\left (x - 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-log(2*x-16)+1)/(-8+x),x, algorithm="fricas")

[Out]

-1/2*e/(x - 8)

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giac [A]  time = 0.15, size = 9, normalized size = 0.75 \begin {gather*} -\frac {e}{2 \, {\left (x - 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-log(2*x-16)+1)/(-8+x),x, algorithm="giac")

[Out]

-1/2*e/(x - 8)

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maple [A]  time = 0.36, size = 10, normalized size = 0.83

method	result	size
default	\(-\frac {{\mathrm e}}{2 \left (-8+x \right )}\)	\(10\)
norman	\(-\frac {{\mathrm e}}{2 \left (-8+x \right )}\)	\(10\)
risch	\(-\frac {{\mathrm e}}{2 \left (-8+x \right )}\)	\(10\)
gosper	\(-{\mathrm e}^{-\ln \left (2 x -16\right )+1}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-ln(2*x-16)+1)/(-8+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.37, size = 9, normalized size = 0.75 \begin {gather*} -\frac {e}{2 \, {\left (x - 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-log(2*x-16)+1)/(-8+x),x, algorithm="maxima")

[Out]

-1/2*e/(x - 8)

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mupad [B]  time = 0.06, size = 11, normalized size = 0.92 \begin {gather*} -\frac {\mathrm {e}}{2\,\left (x-8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(1 - log(2*x - 16))/(x - 8),x)

[Out]

-exp(1)/(2*(x - 8))

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sympy [A]  time = 0.09, size = 8, normalized size = 0.67 \begin {gather*} - \frac {e}{2 x - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(-ln(2*x-16)+1)/(-8+x),x)

[Out]

-E/(2*x - 16)

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