\(\int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx\) [38]

Optimal result

Integrand size = 31, antiderivative size = 60 \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {1-2 b x}{\sqrt {1-4 a b}}\right )}{\sqrt {1-4 a b}}-\frac {\text {arctanh}\left (\frac {1+2 b x}{\sqrt {1-4 a b}}\right )}{\sqrt {1-4 a b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1175, 632, 212} \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {1-2 b x}{\sqrt {1-4 a b}}\right )}{\sqrt {1-4 a b}}-\frac {\text {arctanh}\left (\frac {2 b x+1}{\sqrt {1-4 a b}}\right )}{\sqrt {1-4 a b}} \]

[In]

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

[Out]

ArcTanh[(1 - 2*b*x)/Sqrt[1 - 4*a*b]]/Sqrt[1 - 4*a*b] - ArcTanh[(1 + 2*b*x)/Sqrt[1 - 4*a*b]]/Sqrt[1 - 4*a*b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1175

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), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/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[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {1}{\frac {a}{b}-\frac {x}{b}+x^2} \, dx}{2 b}+\frac {\int \frac {1}{\frac {a}{b}+\frac {x}{b}+x^2} \, dx}{2 b} \\ & = -\frac {\text {Subst}\left (\int \frac {1}{\frac {1-4 a b}{b^2}-x^2} \, dx,x,-\frac {1}{b}+2 x\right )}{b}-\frac {\text {Subst}\left (\int \frac {1}{\frac {1-4 a b}{b^2}-x^2} \, dx,x,\frac {1}{b}+2 x\right )}{b} \\ & = \frac {\tanh ^{-1}\left (\frac {1-2 b x}{\sqrt {1-4 a b}}\right )}{\sqrt {1-4 a b}}-\frac {\tanh ^{-1}\left (\frac {1+2 b x}{\sqrt {1-4 a b}}\right )}{\sqrt {1-4 a b}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(138\) vs. \(2(60)=120\).

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.30 \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {\frac {\left (1+\sqrt {1-4 a b}\right ) \arctan \left (\frac {b x}{\sqrt {-\frac {1}{2}+a b-\frac {1}{2} \sqrt {1-4 a b}}}\right )}{\sqrt {-1+2 a b-\sqrt {1-4 a b}}}+\frac {\left (-1+\sqrt {1-4 a b}\right ) \arctan \left (\frac {\sqrt {2} b x}{\sqrt {-1+2 a b+\sqrt {1-4 a b}}}\right )}{\sqrt {-1+2 a b+\sqrt {1-4 a b}}}}{\sqrt {2-8 a b}} \]

[In]

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

[Out]

(((1 + Sqrt[1 - 4*a*b])*ArcTan[(b*x)/Sqrt[-1/2 + a*b - Sqrt[1 - 4*a*b]/2]])/Sqrt[-1 + 2*a*b - Sqrt[1 - 4*a*b]]
+ ((-1 + Sqrt[1 - 4*a*b])*ArcTan[(Sqrt[2]*b*x)/Sqrt[-1 + 2*a*b + Sqrt[1 - 4*a*b]]])/Sqrt[-1 + 2*a*b + Sqrt[1
- 4*a*b]])/Sqrt[2 - 8*a*b]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87

method	result	size
default	\(\frac {\arctan \left (\frac {2 b x +1}{\sqrt {4 a b -1}}\right )}{\sqrt {4 a b -1}}+\frac {\arctan \left (\frac {2 b x -1}{\sqrt {4 a b -1}}\right )}{\sqrt {4 a b -1}}\)	\(52\)
risch	\(-\frac {\ln \left (b \,x^{2} \sqrt {-4 a b +1}+\left (-4 a b +1\right ) x -a \sqrt {-4 a b +1}\right )}{2 \sqrt {-4 a b +1}}+\frac {\ln \left (b \,x^{2} \sqrt {-4 a b +1}+x \left (4 a b -1\right )-a \sqrt {-4 a b +1}\right )}{2 \sqrt {-4 a b +1}}\)	\(90\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.73 \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\left [-\frac {\sqrt {-4 \, a b + 1} \log \left (\frac {b^{2} x^{4} - {\left (6 \, a b - 1\right )} x^{2} + a^{2} - 2 \, {\left (b x^{3} - a x\right )} \sqrt {-4 \, a b + 1}}{b^{2} x^{4} + {\left (2 \, a b - 1\right )} x^{2} + a^{2}}\right )}{2 \, {\left (4 \, a b - 1\right )}}, \frac {\sqrt {4 \, a b - 1} \arctan \left (\frac {b x}{\sqrt {4 \, a b - 1}}\right ) + \sqrt {4 \, a b - 1} \arctan \left (\frac {{\left (b^{2} x^{3} + {\left (3 \, a b - 1\right )} x\right )} \sqrt {4 \, a b - 1}}{4 \, a^{2} b - a}\right )}{4 \, a b - 1}\right ] \]

[In]

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

[Out]

[-1/2*sqrt(-4*a*b + 1)*log((b^2*x^4 - (6*a*b - 1)*x^2 + a^2 - 2*(b*x^3 - a*x)*sqrt(-4*a*b + 1))/(b^2*x^4 + (2*
a*b - 1)*x^2 + a^2))/(4*a*b - 1), (sqrt(4*a*b - 1)*arctan(b*x/sqrt(4*a*b - 1)) + sqrt(4*a*b - 1)*arctan((b^2*x
^3 + (3*a*b - 1)*x)*sqrt(4*a*b - 1)/(4*a^2*b - a)))/(4*a*b - 1)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (56) = 112\).

Time = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.95 \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=- \frac {\sqrt {- \frac {1}{4 a b - 1}} \log {\left (- \frac {a}{b} + x^{2} + \frac {x \left (- 4 a b \sqrt {- \frac {1}{4 a b - 1}} + \sqrt {- \frac {1}{4 a b - 1}}\right )}{b} \right )}}{2} + \frac {\sqrt {- \frac {1}{4 a b - 1}} \log {\left (- \frac {a}{b} + x^{2} + \frac {x \left (4 a b \sqrt {- \frac {1}{4 a b - 1}} - \sqrt {- \frac {1}{4 a b - 1}}\right )}{b} \right )}}{2} \]

[In]

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

[Out]

-sqrt(-1/(4*a*b - 1))*log(-a/b + x**2 + x*(-4*a*b*sqrt(-1/(4*a*b - 1)) + sqrt(-1/(4*a*b - 1)))/b)/2 + sqrt(-1/
(4*a*b - 1))*log(-a/b + x**2 + x*(4*a*b*sqrt(-1/(4*a*b - 1)) - sqrt(-1/(4*a*b - 1)))/b)/2

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*b-0.25>0)', see `assume?` fo
r more detai

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {\arctan \left (\frac {2 \, b x + 1}{\sqrt {4 \, a b - 1}}\right )}{\sqrt {4 \, a b - 1}} + \frac {\arctan \left (\frac {2 \, b x - 1}{\sqrt {4 \, a b - 1}}\right )}{\sqrt {4 \, a b - 1}} \]

[In]

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

[Out]

arctan((2*b*x + 1)/sqrt(4*a*b - 1))/sqrt(4*a*b - 1) + arctan((2*b*x - 1)/sqrt(4*a*b - 1))/sqrt(4*a*b - 1)

Mupad [B] (verification not implemented)

Time = 13.77 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x^2}{a^2+(-1+2 a b) x^2+b^2 x^4} \, dx=\frac {\mathrm {atan}\left (\frac {b\,x}{\sqrt {4\,a\,b-1}}\right )+\mathrm {atan}\left (\frac {\frac {3\,x\,\left (4\,a\,b-1\right )}{4}-\frac {x}{4}+b^2\,x^3}{a\,\sqrt {4\,a\,b-1}}\right )}{\sqrt {4\,a\,b-1}} \]

[In]

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

[Out]

(atan((b*x)/(4*a*b - 1)^(1/2)) + atan(((3*x*(4*a*b - 1))/4 - x/4 + b^2*x^3)/(a*(4*a*b - 1)^(1/2))))/(4*a*b - 1
)^(1/2)