\(\int \frac {x}{(-3+x) (5+x)^2} \, dx\) [126]

Optimal result

Integrand size = 12, antiderivative size = 22 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {5}{8 (5+x)}-\frac {3}{32} \text {arctanh}\left (\frac {1+x}{4}\right ) \]

[Out]

-5/(8*x+40)-3/32*arctanh(1/4+1/4*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {78} \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {5}{8 (x+5)}+\frac {3}{64} \log (3-x)-\frac {3}{64} \log (x+5) \]

[In]

Int[x/((-3 + x)*(5 + x)^2),x]

[Out]

-5/(8*(5 + x)) + (3*Log[3 - x])/64 - (3*Log[5 + x])/64

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 {3}{64 (-3+x)}+\frac {5}{8 (5+x)^2}-\frac {3}{64 (5+x)}\right ) \, dx \\ & = -\frac {5}{8 (5+x)}+\frac {3}{64} \log (3-x)-\frac {3}{64} \log (5+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {5}{8 (5+x)}+\frac {3}{64} \log (3-x)-\frac {3}{64} \log (5+x) \]

[In]

Integrate[x/((-3 + x)*(5 + x)^2),x]

[Out]

-5/(8*(5 + x)) + (3*Log[3 - x])/64 - (3*Log[5 + x])/64

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {5}{8 \left (5+x \right )}-\frac {3 \ln \left (5+x \right )}{64}+\frac {3 \ln \left (-3+x \right )}{64}\)	\(21\)
norman	\(-\frac {5}{8 \left (5+x \right )}-\frac {3 \ln \left (5+x \right )}{64}+\frac {3 \ln \left (-3+x \right )}{64}\)	\(21\)
risch	\(-\frac {5}{8 \left (5+x \right )}-\frac {3 \ln \left (5+x \right )}{64}+\frac {3 \ln \left (-3+x \right )}{64}\)	\(21\)
parallelrisch	\(-\frac {3 \ln \left (5+x \right ) x -3 \ln \left (-3+x \right ) x +40+15 \ln \left (5+x \right )-15 \ln \left (-3+x \right )}{64 \left (5+x \right )}\)	\(36\)

[In]

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

[Out]

-5/8/(5+x)-3/64*ln(5+x)+3/64*ln(-3+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {3 \, {\left (x + 5\right )} \log \left (x + 5\right ) - 3 \, {\left (x + 5\right )} \log \left (x - 3\right ) + 40}{64 \, {\left (x + 5\right )}} \]

[In]

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

[Out]

-1/64*(3*(x + 5)*log(x + 5) - 3*(x + 5)*log(x - 3) + 40)/(x + 5)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=\frac {3 \log {\left (x - 3 \right )}}{64} - \frac {3 \log {\left (x + 5 \right )}}{64} - \frac {5}{8 x + 40} \]

[In]

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

[Out]

3*log(x - 3)/64 - 3*log(x + 5)/64 - 5/(8*x + 40)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {5}{8 \, {\left (x + 5\right )}} - \frac {3}{64} \, \log \left (x + 5\right ) + \frac {3}{64} \, \log \left (x - 3\right ) \]

[In]

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

[Out]

-5/8/(x + 5) - 3/64*log(x + 5) + 3/64*log(x - 3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {5}{8 \, {\left (x + 5\right )}} + \frac {3}{64} \, \log \left ({\left | -\frac {8}{x + 5} + 1 \right |}\right ) \]

[In]

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

[Out]

-5/8/(x + 5) + 3/64*log(abs(-8/(x + 5) + 1))

Mupad [B] (verification not implemented)

Time = 16.61 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {x}{(-3+x) (5+x)^2} \, dx=-\frac {3\,\ln \left (\frac {x+5}{x-3}\right )}{64}-\frac {5}{8\,\left (x+5\right )} \]

[In]

int(x/((x - 3)*(x + 5)^2),x)

[Out]

- (3*log((x + 5)/(x - 3)))/64 - 5/(8*(x + 5))