\(\int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx\) [1800]

Optimal result

Integrand size = 38, antiderivative size = 22 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \log \left (2+x+3 \left (-3-x+\frac {x^2}{5}\right )^2\right ) \]

[Out]

2*ln(2+3*(-3-x+1/5*x^2)^2+x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1601} \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \log \left (3 x^4-30 x^3-15 x^2+475 x+725\right ) \]

[In]

Int[(950 - 60*x - 180*x^2 + 24*x^3)/(725 + 475*x - 15*x^2 - 30*x^3 + 3*x^4),x]

[Out]

2*Log[725 + 475*x - 15*x^2 - 30*x^3 + 3*x^4]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps \begin{align*} \text {integral}& = 2 \log \left (725+475 x-15 x^2-30 x^3+3 x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \log \left (725+475 x-15 x^2-30 x^3+3 x^4\right ) \]

[In]

Integrate[(950 - 60*x - 180*x^2 + 24*x^3)/(725 + 475*x - 15*x^2 - 30*x^3 + 3*x^4),x]

[Out]

2*Log[725 + 475*x - 15*x^2 - 30*x^3 + 3*x^4]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(2 \ln \left (x^{4}-10 x^{3}-5 x^{2}+\frac {475}{3} x +\frac {725}{3}\right )\)	\(22\)
default	\(2 \ln \left (3 x^{4}-30 x^{3}-15 x^{2}+475 x +725\right )\)	\(24\)
norman	\(2 \ln \left (3 x^{4}-30 x^{3}-15 x^{2}+475 x +725\right )\)	\(24\)
risch	\(2 \ln \left (3 x^{4}-30 x^{3}-15 x^{2}+475 x +725\right )\)	\(24\)

[In]

int((24*x^3-180*x^2-60*x+950)/(3*x^4-30*x^3-15*x^2+475*x+725),x,method=_RETURNVERBOSE)

[Out]

2*ln(x^4-10*x^3-5*x^2+475/3*x+725/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \, \log \left (3 \, x^{4} - 30 \, x^{3} - 15 \, x^{2} + 475 \, x + 725\right ) \]

[In]

integrate((24*x^3-180*x^2-60*x+950)/(3*x^4-30*x^3-15*x^2+475*x+725),x, algorithm="fricas")

[Out]

2*log(3*x^4 - 30*x^3 - 15*x^2 + 475*x + 725)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \log {\left (3 x^{4} - 30 x^{3} - 15 x^{2} + 475 x + 725 \right )} \]

[In]

integrate((24*x**3-180*x**2-60*x+950)/(3*x**4-30*x**3-15*x**2+475*x+725),x)

[Out]

2*log(3*x**4 - 30*x**3 - 15*x**2 + 475*x + 725)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \, \log \left (3 \, x^{4} - 30 \, x^{3} - 15 \, x^{2} + 475 \, x + 725\right ) \]

[In]

integrate((24*x^3-180*x^2-60*x+950)/(3*x^4-30*x^3-15*x^2+475*x+725),x, algorithm="maxima")

[Out]

2*log(3*x^4 - 30*x^3 - 15*x^2 + 475*x + 725)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2 \, \log \left ({\left | 3 \, x^{4} - 30 \, x^{3} - 15 \, x^{2} + 475 \, x + 725 \right |}\right ) \]

[In]

integrate((24*x^3-180*x^2-60*x+950)/(3*x^4-30*x^3-15*x^2+475*x+725),x, algorithm="giac")

[Out]

2*log(abs(3*x^4 - 30*x^3 - 15*x^2 + 475*x + 725))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {950-60 x-180 x^2+24 x^3}{725+475 x-15 x^2-30 x^3+3 x^4} \, dx=2\,\ln \left (x^4-10\,x^3-5\,x^2+\frac {475\,x}{3}+\frac {725}{3}\right ) \]

[In]

int(-(60*x + 180*x^2 - 24*x^3 - 950)/(475*x - 15*x^2 - 30*x^3 + 3*x^4 + 725),x)

[Out]

2*log((475*x)/3 - 5*x^2 - 10*x^3 + x^4 + 725/3)