\(\int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx\) [37]

Optimal result

Integrand size = 32, antiderivative size = 29 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=-\frac {1}{2} \log \left (a-x+b x^2\right )+\frac {1}{2} \log \left (a+x+b x^2\right ) \]

[Out]

-1/2*ln(b*x^2+a-x)+1/2*ln(b*x^2+a+x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1178, 642} \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {1}{2} \log \left (a+b x^2+x\right )-\frac {1}{2} \log \left (a+b x^2-x\right ) \]

[In]

Int[(a - b*x^2)/(a^2 + (-1 + 2*a*b)*x^2 + b^2*x^4),x]

[Out]

-1/2*Log[a - x + b*x^2] + Log[a + x + b*x^2]/2

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e) - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{2} \int \frac {\frac {1}{b}+2 x}{-\frac {a}{b}-\frac {x}{b}-x^2} \, dx\right )-\frac {1}{2} \int \frac {\frac {1}{b}-2 x}{-\frac {a}{b}+\frac {x}{b}-x^2} \, dx \\ & = -\frac {1}{2} \log \left (a-x+b x^2\right )+\frac {1}{2} \log \left (a+x+b x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=-\frac {1}{2} \log \left (a-x+b x^2\right )+\frac {1}{2} \log \left (a+x+b x^2\right ) \]

[In]

Integrate[(a - b*x^2)/(a^2 + (-1 + 2*a*b)*x^2 + b^2*x^4),x]

[Out]

-1/2*Log[a - x + b*x^2] + Log[a + x + b*x^2]/2

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {\ln \left (b \,x^{2}+a -x \right )}{2}+\frac {\ln \left (b \,x^{2}+a +x \right )}{2}\)	\(26\)
norman	\(-\frac {\ln \left (b \,x^{2}+a -x \right )}{2}+\frac {\ln \left (b \,x^{2}+a +x \right )}{2}\)	\(26\)
risch	\(-\frac {\ln \left (b \,x^{2}+a -x \right )}{2}+\frac {\ln \left (b \,x^{2}+a +x \right )}{2}\)	\(26\)
parallelrisch	\(-\frac {\ln \left (b \,x^{2}+a -x \right )}{2}+\frac {\ln \left (b \,x^{2}+a +x \right )}{2}\)	\(26\)

[In]

int((-b*x^2+a)/(a^2+(2*a*b-1)*x^2+b^2*x^4),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(b*x^2+a-x)+1/2*ln(b*x^2+a+x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {1}{2} \, \log \left (b x^{2} + a + x\right ) - \frac {1}{2} \, \log \left (b x^{2} + a - x\right ) \]

[In]

integrate((-b*x^2+a)/(a^2+(2*a*b-1)*x^2+b^2*x^4),x, algorithm="fricas")

[Out]

1/2*log(b*x^2 + a + x) - 1/2*log(b*x^2 + a - x)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=- \frac {\log {\left (\frac {a}{b} + x^{2} - \frac {x}{b} \right )}}{2} + \frac {\log {\left (\frac {a}{b} + x^{2} + \frac {x}{b} \right )}}{2} \]

[In]

integrate((-b*x**2+a)/(a**2+(2*a*b-1)*x**2+b**2*x**4),x)

[Out]

-log(a/b + x**2 - x/b)/2 + log(a/b + x**2 + x/b)/2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {1}{2} \, \log \left (b x^{2} + a + x\right ) - \frac {1}{2} \, \log \left (b x^{2} + a - x\right ) \]

[In]

integrate((-b*x^2+a)/(a^2+(2*a*b-1)*x^2+b^2*x^4),x, algorithm="maxima")

[Out]

1/2*log(b*x^2 + a + x) - 1/2*log(b*x^2 + a - x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {1}{2} \, \log \left (b x^{2} + a + x\right ) - \frac {1}{2} \, \log \left (b x^{2} + a - x\right ) \]

[In]

integrate((-b*x^2+a)/(a^2+(2*a*b-1)*x^2+b^2*x^4),x, algorithm="giac")

[Out]

1/2*log(b*x^2 + a + x) - 1/2*log(b*x^2 + a - x)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.41 \[ \int \frac {a-b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\mathrm {atanh}\left (\frac {x}{b\,x^2+a}\right ) \]

[In]

int((a - b*x^2)/(x^2*(2*a*b - 1) + a^2 + b^2*x^4),x)

[Out]

atanh(x/(a + b*x^2))