\(\int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx\) [6056]

Optimal result

Integrand size = 34, antiderivative size = 23 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=\log \left (\frac {1-(6+x) \left (4+x-4 x^2\right )}{4 x}\right ) \]

[Out]

ln(1/4*(1-(6+x)*(-4*x^2+x+4))/x)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2099, 1601} \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=\log \left (-4 x^3-23 x^2+10 x+23\right )-\log (x) \]

[In]

Int[(23 + 23*x^2 + 8*x^3)/(-23*x - 10*x^2 + 23*x^3 + 4*x^4),x]

[Out]

-Log[x] + Log[23 + 10*x - 23*x^2 - 4*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]

Rule 2099

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{x}+\frac {2 \left (-5+23 x+6 x^2\right )}{-23-10 x+23 x^2+4 x^3}\right ) \, dx \\ & = -\log (x)+2 \int \frac {-5+23 x+6 x^2}{-23-10 x+23 x^2+4 x^3} \, dx \\ & = -\log (x)+\log \left (23+10 x-23 x^2-4 x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=-\log (x)+\log \left (23+10 x-23 x^2-4 x^3\right ) \]

[In]

Integrate[(23 + 23*x^2 + 8*x^3)/(-23*x - 10*x^2 + 23*x^3 + 4*x^4),x]

[Out]

-Log[x] + Log[23 + 10*x - 23*x^2 - 4*x^3]

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
parallelrisch	\(-\ln \left (x \right )+\ln \left (x^{3}+\frac {23}{4} x^{2}-\frac {5}{2} x -\frac {23}{4}\right )\)	\(20\)
default	\(\ln \left (4 x^{3}+23 x^{2}-10 x -23\right )-\ln \left (x \right )\)	\(22\)
norman	\(\ln \left (4 x^{3}+23 x^{2}-10 x -23\right )-\ln \left (x \right )\)	\(22\)
risch	\(\ln \left (4 x^{3}+23 x^{2}-10 x -23\right )-\ln \left (x \right )\)	\(22\)

[In]

int((8*x^3+23*x^2+23)/(4*x^4+23*x^3-10*x^2-23*x),x,method=_RETURNVERBOSE)

[Out]

-ln(x)+ln(x^3+23/4*x^2-5/2*x-23/4)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=\log \left (4 \, x^{3} + 23 \, x^{2} - 10 \, x - 23\right ) - \log \left (x\right ) \]

[In]

integrate((8*x^3+23*x^2+23)/(4*x^4+23*x^3-10*x^2-23*x),x, algorithm="fricas")

[Out]

log(4*x^3 + 23*x^2 - 10*x - 23) - log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=- \log {\left (x \right )} + \log {\left (4 x^{3} + 23 x^{2} - 10 x - 23 \right )} \]

[In]

integrate((8*x**3+23*x**2+23)/(4*x**4+23*x**3-10*x**2-23*x),x)

[Out]

-log(x) + log(4*x**3 + 23*x**2 - 10*x - 23)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=\log \left (4 \, x^{3} + 23 \, x^{2} - 10 \, x - 23\right ) - \log \left (x\right ) \]

[In]

integrate((8*x^3+23*x^2+23)/(4*x^4+23*x^3-10*x^2-23*x),x, algorithm="maxima")

[Out]

log(4*x^3 + 23*x^2 - 10*x - 23) - log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=\log \left ({\left | 4 \, x^{3} + 23 \, x^{2} - 10 \, x - 23 \right |}\right ) - \log \left ({\left | x \right |}\right ) \]

[In]

integrate((8*x^3+23*x^2+23)/(4*x^4+23*x^3-10*x^2-23*x),x, algorithm="giac")

[Out]

log(abs(4*x^3 + 23*x^2 - 10*x - 23)) - log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {23+23 x^2+8 x^3}{-23 x-10 x^2+23 x^3+4 x^4} \, dx=\ln \left (4\,x^3+23\,x^2-10\,x-23\right )-\ln \left (x\right ) \]

[In]

int(-(23*x^2 + 8*x^3 + 23)/(23*x + 10*x^2 - 23*x^3 - 4*x^4),x)

[Out]

log(23*x^2 - 10*x + 4*x^3 - 23) - log(x)