\(\int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx\) [2564]

Optimal result

Integrand size = 27, antiderivative size = 17 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\log \left (-x+\frac {4 x}{-\frac {8}{3}-x}\right ) \]

[Out]

ln(4*x/(-x-8/3)-x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1608, 1642} \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\log (x)-\log (3 x+8)+\log (3 x+20) \]

[In]

Int[(160 + 48*x + 9*x^2)/(160*x + 84*x^2 + 9*x^3),x]

[Out]

Log[x] - Log[8 + 3*x] + Log[20 + 3*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 {160+48 x+9 x^2}{x \left (160+84 x+9 x^2\right )} \, dx \\ & = \int \left (\frac {1}{x}-\frac {3}{8+3 x}+\frac {3}{20+3 x}\right ) \, dx \\ & = \log (x)-\log (8+3 x)+\log (20+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\log (x)-\log (8+3 x)+\log (20+3 x) \]

[In]

Integrate[(160 + 48*x + 9*x^2)/(160*x + 84*x^2 + 9*x^3),x]

[Out]

Log[x] - Log[8 + 3*x] + Log[20 + 3*x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(\ln \left (x \right )-\ln \left (x +\frac {8}{3}\right )+\ln \left (x +\frac {20}{3}\right )\)	\(14\)
default	\(\ln \left (x \right )+\ln \left (3 x +20\right )-\ln \left (8+3 x \right )\)	\(18\)
norman	\(\ln \left (x \right )+\ln \left (3 x +20\right )-\ln \left (8+3 x \right )\)	\(18\)
risch	\(-\ln \left (8+3 x \right )+\ln \left (3 x^{2}+20 x \right )\)	\(20\)

[In]

int((9*x^2+48*x+160)/(9*x^3+84*x^2+160*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)-ln(x+8/3)+ln(x+20/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\log \left (3 \, x^{2} + 20 \, x\right ) - \log \left (3 \, x + 8\right ) \]

[In]

integrate((9*x^2+48*x+160)/(9*x^3+84*x^2+160*x),x, algorithm="fricas")

[Out]

log(3*x^2 + 20*x) - log(3*x + 8)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=- \log {\left (3 x + 8 \right )} + \log {\left (3 x^{2} + 20 x \right )} \]

[In]

integrate((9*x**2+48*x+160)/(9*x**3+84*x**2+160*x),x)

[Out]

-log(3*x + 8) + log(3*x**2 + 20*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\log \left (3 \, x + 20\right ) - \log \left (3 \, x + 8\right ) + \log \left (x\right ) \]

[In]

integrate((9*x^2+48*x+160)/(9*x^3+84*x^2+160*x),x, algorithm="maxima")

[Out]

log(3*x + 20) - log(3*x + 8) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\log \left ({\left | 3 \, x + 20 \right |}\right ) - \log \left ({\left | 3 \, x + 8 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((9*x^2+48*x+160)/(9*x^3+84*x^2+160*x),x, algorithm="giac")

[Out]

log(abs(3*x + 20)) - log(abs(3*x + 8)) + log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {160+48 x+9 x^2}{160 x+84 x^2+9 x^3} \, dx=\ln \left (x\,\left (3\,x+20\right )\right )-\ln \left (x+\frac {8}{3}\right ) \]

[In]

int((48*x + 9*x^2 + 160)/(160*x + 84*x^2 + 9*x^3),x)

[Out]

log(x*(3*x + 20)) - log(x + 8/3)