\(\int \frac {e}{(-8+x) (-16+2 x)} \, dx\) [8503]

Optimal result

Integrand size = 14, antiderivative size = 12 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=-4-\frac {e}{-16+2 x} \]

[Out]

-exp(-ln(2*x-16)+1)-4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {12, 21, 32} \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=\frac {e}{2 (8-x)} \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=-\frac {e}{2 (-8+x)} \]

[In]

Integrate[E/((-8 + x)*(-16 + 2*x)),x]

[Out]

-1/2*E/(-8 + x)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=-\frac {e}{2 \, {\left (x - 8\right )}} \]

[In]

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

[Out]

-1/2*e/(x - 8)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.67 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=- \frac {e}{2 x - 16} \]

[In]

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

[Out]

-E/(2*x - 16)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=-\frac {e}{2 \, {\left (x - 8\right )}} \]

[In]

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

[Out]

-1/2*e/(x - 8)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.75 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=-\frac {e}{2 \, {\left (x - 8\right )}} \]

[In]

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

[Out]

-1/2*e/(x - 8)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92 \[ \int \frac {e}{(-8+x) (-16+2 x)} \, dx=-\frac {\mathrm {e}}{2\,\left (x-8\right )} \]

[In]

int(exp(1 - log(2*x - 16))/(x - 8),x)

[Out]

-exp(1)/(2*(x - 8))