\(\int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx\) [9974]

Optimal result

Integrand size = 28, antiderivative size = 22 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log \left (2 \left (3-x+x \left (2+x^2-\frac {5 (4+x)}{3}\right )\right )\right ) \]

[Out]

ln(2*(-5/3*x-14/3+x^2)*x-2*x+6)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1601} \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log \left (3 x^3-5 x^2-17 x+9\right ) \]

[In]

Int[(-17 - 10*x + 9*x^2)/(9 - 17*x - 5*x^2 + 3*x^3),x]

[Out]

Log[9 - 17*x - 5*x^2 + 3*x^3]

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}& = \log \left (9-17 x-5 x^2+3 x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log \left (9-17 x-5 x^2+3 x^3\right ) \]

[In]

Integrate[(-17 - 10*x + 9*x^2)/(9 - 17*x - 5*x^2 + 3*x^3),x]

[Out]

Log[9 - 17*x - 5*x^2 + 3*x^3]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(\ln \left (x^{3}-\frac {5}{3} x^{2}-\frac {17}{3} x +3\right )\)	\(15\)
derivativedivides	\(\ln \left (3 x^{3}-5 x^{2}-17 x +9\right )\)	\(17\)
default	\(\ln \left (3 x^{3}-5 x^{2}-17 x +9\right )\)	\(17\)
norman	\(\ln \left (3 x^{3}-5 x^{2}-17 x +9\right )\)	\(17\)
risch	\(\ln \left (3 x^{3}-5 x^{2}-17 x +9\right )\)	\(17\)

[In]

int((9*x^2-10*x-17)/(3*x^3-5*x^2-17*x+9),x,method=_RETURNVERBOSE)

[Out]

ln(x^3-5/3*x^2-17/3*x+3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log \left (3 \, x^{3} - 5 \, x^{2} - 17 \, x + 9\right ) \]

[In]

integrate((9*x^2-10*x-17)/(3*x^3-5*x^2-17*x+9),x, algorithm="fricas")

[Out]

log(3*x^3 - 5*x^2 - 17*x + 9)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log {\left (3 x^{3} - 5 x^{2} - 17 x + 9 \right )} \]

[In]

integrate((9*x**2-10*x-17)/(3*x**3-5*x**2-17*x+9),x)

[Out]

log(3*x**3 - 5*x**2 - 17*x + 9)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log \left (3 \, x^{3} - 5 \, x^{2} - 17 \, x + 9\right ) \]

[In]

integrate((9*x^2-10*x-17)/(3*x^3-5*x^2-17*x+9),x, algorithm="maxima")

[Out]

log(3*x^3 - 5*x^2 - 17*x + 9)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\log \left ({\left | 3 \, x^{3} - 5 \, x^{2} - 17 \, x + 9 \right |}\right ) \]

[In]

integrate((9*x^2-10*x-17)/(3*x^3-5*x^2-17*x+9),x, algorithm="giac")

[Out]

log(abs(3*x^3 - 5*x^2 - 17*x + 9))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-17-10 x+9 x^2}{9-17 x-5 x^2+3 x^3} \, dx=\ln \left (3\,x^3-5\,x^2-17\,x+9\right ) \]

[In]

int((10*x - 9*x^2 + 17)/(17*x + 5*x^2 - 3*x^3 - 9),x)

[Out]

log(3*x^3 - 5*x^2 - 17*x + 9)