\(\int \frac {-4+10 e-5 x}{8 e-4 x} \, dx\) [9465]

Optimal result

Integrand size = 18, antiderivative size = 15 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=-\frac {3}{10}+\frac {5 x}{4}+\log (-2 e+x) \]

[Out]

5/4*x+ln(-2*exp(1)+x)-3/10

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {45} \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5 x}{4}+\log (2 e-x) \]

[In]

Int[(-4 + 10*E - 5*x)/(8*E - 4*x),x]

[Out]

(5*x)/4 + Log[2*E - x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5}{4}+\frac {1}{-2 e+x}\right ) \, dx \\ & = \frac {5 x}{4}+\log (2 e-x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=-\frac {5}{16} (8 e-4 x)+\log (8 e-4 x) \]

[In]

Integrate[(-4 + 10*E - 5*x)/(8*E - 4*x),x]

[Out]

(-5*(8*E - 4*x))/16 + Log[8*E - 4*x]

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method	result	size
default	\(\frac {5 x}{4}+\ln \left (-2 \,{\mathrm e}+x \right )\)	\(12\)
risch	\(\frac {5 x}{4}+\ln \left (-2 \,{\mathrm e}+x \right )\)	\(12\)
parallelrisch	\(\frac {5 x}{4}+\ln \left (-2 \,{\mathrm e}+x \right )\)	\(12\)
norman	\(\frac {5 x}{4}+\ln \left (8 \,{\mathrm e}-4 x \right )\)	\(14\)
meijerg	\(-\frac {5 \,{\mathrm e} \ln \left (1-\frac {{\mathrm e}^{-1} x}{2}\right )}{2}+\ln \left (1-\frac {{\mathrm e}^{-1} x}{2}\right )-\frac {5 \,{\mathrm e} \left (-\frac {{\mathrm e}^{-1} x}{2}-\ln \left (1-\frac {{\mathrm e}^{-1} x}{2}\right )\right )}{2}\)	\(42\)

[In]

int((10*exp(1)-5*x-4)/(8*exp(1)-4*x),x,method=_RETURNVERBOSE)

[Out]

5/4*x+ln(-2*exp(1)+x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5}{4} \, x + \log \left (x - 2 \, e\right ) \]

[In]

integrate((10*exp(1)-5*x-4)/(8*exp(1)-4*x),x, algorithm="fricas")

[Out]

5/4*x + log(x - 2*e)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5 x}{4} + \log {\left (x - 2 e \right )} \]

[In]

integrate((10*exp(1)-5*x-4)/(8*exp(1)-4*x),x)

[Out]

5*x/4 + log(x - 2*E)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5}{4} \, x + \log \left (x - 2 \, e\right ) \]

[In]

integrate((10*exp(1)-5*x-4)/(8*exp(1)-4*x),x, algorithm="maxima")

[Out]

5/4*x + log(x - 2*e)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5}{4} \, x + \log \left ({\left | x - 2 \, e \right |}\right ) \]

[In]

integrate((10*exp(1)-5*x-4)/(8*exp(1)-4*x),x, algorithm="giac")

[Out]

5/4*x + log(abs(x - 2*e))

Mupad [B] (verification not implemented)

Time = 14.83 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 e-5 x}{8 e-4 x} \, dx=\frac {5\,x}{4}+\ln \left (x-2\,\mathrm {e}\right ) \]

[In]

int((5*x - 10*exp(1) + 4)/(4*x - 8*exp(1)),x)

[Out]

(5*x)/4 + log(x - 2*exp(1))