\(\int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx\) [9201]

Optimal result

Integrand size = 25, antiderivative size = 20 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=-2 \log \left (x+x \left (1+\frac {5-x}{5+x}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {1608, 1642} \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=-2 \log (x)+2 \log (x+5)-2 \log (x+15) \]

[In]

Int[(-150 - 20*x - 2*x^2)/(75*x + 20*x^2 + x^3),x]

[Out]

-2*Log[x] + 2*Log[5 + x] - 2*Log[15 + x]

Rule 1608

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-150-20 x-2 x^2}{x \left (75+20 x+x^2\right )} \, dx \\ & = \int \left (-\frac {2}{x}+\frac {2}{5+x}-\frac {2}{15+x}\right ) \, dx \\ & = -2 \log (x)+2 \log (5+x)-2 \log (15+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=-2 (\log (x)-\log (5+x)+\log (15+x)) \]

[In]

Integrate[(-150 - 20*x - 2*x^2)/(75*x + 20*x^2 + x^3),x]

[Out]

-2*(Log[x] - Log[5 + x] + Log[15 + x])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
default	\(-2 \ln \left (x \right )+2 \ln \left (5+x \right )-2 \ln \left (x +15\right )\)	\(18\)
norman	\(-2 \ln \left (x \right )+2 \ln \left (5+x \right )-2 \ln \left (x +15\right )\)	\(18\)
risch	\(2 \ln \left (5+x \right )-2 \ln \left (x^{2}+15 x \right )\)	\(18\)
parallelrisch	\(-2 \ln \left (x \right )+2 \ln \left (5+x \right )-2 \ln \left (x +15\right )\)	\(18\)

[In]

int((-2*x^2-20*x-150)/(x^3+20*x^2+75*x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x)+2*ln(5+x)-2*ln(x+15)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=-2 \, \log \left (x^{2} + 15 \, x\right ) + 2 \, \log \left (x + 5\right ) \]

[In]

integrate((-2*x^2-20*x-150)/(x^3+20*x^2+75*x),x, algorithm="fricas")

[Out]

-2*log(x^2 + 15*x) + 2*log(x + 5)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=2 \log {\left (x + 5 \right )} - 2 \log {\left (x^{2} + 15 x \right )} \]

[In]

integrate((-2*x**2-20*x-150)/(x**3+20*x**2+75*x),x)

[Out]

2*log(x + 5) - 2*log(x**2 + 15*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=-2 \, \log \left (x + 15\right ) + 2 \, \log \left (x + 5\right ) - 2 \, \log \left (x\right ) \]

[In]

integrate((-2*x^2-20*x-150)/(x^3+20*x^2+75*x),x, algorithm="maxima")

[Out]

-2*log(x + 15) + 2*log(x + 5) - 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=-2 \, \log \left ({\left | x + 15 \right |}\right ) + 2 \, \log \left ({\left | x + 5 \right |}\right ) - 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

integrate((-2*x^2-20*x-150)/(x^3+20*x^2+75*x),x, algorithm="giac")

[Out]

-2*log(abs(x + 15)) + 2*log(abs(x + 5)) - 2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-150-20 x-2 x^2}{75 x+20 x^2+x^3} \, dx=2\,\ln \left (x+5\right )-2\,\ln \left (x\,\left (x+15\right )\right ) \]

[In]

int(-(20*x + 2*x^2 + 150)/(75*x + 20*x^2 + x^3),x)

[Out]

2*log(x + 5) - 2*log(x*(x + 15))