\(\int \frac {3+2 x}{(-2+x) (5+x)} \, dx\) [115]

Optimal result

Integrand size = 16, antiderivative size = 11 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log (2-x)+\log (5+x) \]

[Out]

ln(2-x)+ln(5+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {78} \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log (2-x)+\log (x+5) \]

[In]

Int[(3 + 2*x)/((-2 + x)*(5 + x)),x]

[Out]

Log[2 - x] + Log[5 + x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log (-2+x)+\log (5+x) \]

[In]

Integrate[(3 + 2*x)/((-2 + x)*(5 + x)),x]

[Out]

Log[-2 + x] + Log[5 + x]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82

method	result	size
default	\(\ln \left (\left (-2+x \right ) \left (5+x \right )\right )\)	\(9\)
norman	\(\ln \left (-2+x \right )+\ln \left (5+x \right )\)	\(10\)
risch	\(\ln \left (x^{2}+3 x -10\right )\)	\(10\)
parallelrisch	\(\ln \left (-2+x \right )+\ln \left (5+x \right )\)	\(10\)

[In]

int((3+2*x)/(-2+x)/(5+x),x,method=_RETURNVERBOSE)

[Out]

ln((-2+x)*(5+x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log \left (x^{2} + 3 \, x - 10\right ) \]

[In]

integrate((3+2*x)/(-2+x)/(5+x),x, algorithm="fricas")

[Out]

log(x^2 + 3*x - 10)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log {\left (x^{2} + 3 x - 10 \right )} \]

[In]

integrate((3+2*x)/(-2+x)/(5+x),x)

[Out]

log(x**2 + 3*x - 10)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log \left (x + 5\right ) + \log \left (x - 2\right ) \]

[In]

integrate((3+2*x)/(-2+x)/(5+x),x, algorithm="maxima")

[Out]

log(x + 5) + log(x - 2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\log \left ({\left | x + 5 \right |}\right ) + \log \left ({\left | x - 2 \right |}\right ) \]

[In]

integrate((3+2*x)/(-2+x)/(5+x),x, algorithm="giac")

[Out]

log(abs(x + 5)) + log(abs(x - 2))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82 \[ \int \frac {3+2 x}{(-2+x) (5+x)} \, dx=\ln \left (x^2+3\,x-10\right ) \]

[In]

int((2*x + 3)/((x - 2)*(x + 5)),x)

[Out]

log(3*x + x^2 - 10)