3.9 \(\int \frac {1}{-1+a^2+2 a x^2+x^4} \, dx\)

Optimal. Leaf size=47 \[ -\frac {\tan ^{-1}\left (\frac {x}{\sqrt {a+1}}\right )}{2 \sqrt {a+1}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {1-a}}\right )}{2 \sqrt {1-a}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1093, 207, 203} \[ -\frac {\tan ^{-1}\left (\frac {x}{\sqrt {a+1}}\right )}{2 \sqrt {a+1}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {1-a}}\right )}{2 \sqrt {1-a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 207

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

Rule 1093

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

Rubi steps

\begin {align*} \int \frac {1}{-1+a^2+2 a x^2+x^4} \, dx &=\frac {1}{2} \int \frac {1}{-1+a+x^2} \, dx-\frac {1}{2} \int \frac {1}{1+a+x^2} \, dx\\ &=-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {1+a}}\right )}{2 \sqrt {1+a}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {1-a}}\right )}{2 \sqrt {1-a}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.91 \[ \frac {\tan ^{-1}\left (\frac {x}{\sqrt {a-1}}\right )}{2 \sqrt {a-1}}-\frac {\tan ^{-1}\left (\frac {x}{\sqrt {a+1}}\right )}{2 \sqrt {a+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 269, normalized size = 5.72 \[ \left [-\frac {{\left (a - 1\right )} \sqrt {-a - 1} \log \left (\frac {x^{2} + 2 \, \sqrt {-a - 1} x - a - 1}{x^{2} + a + 1}\right ) + {\left (a + 1\right )} \sqrt {-a + 1} \log \left (\frac {x^{2} - 2 \, \sqrt {-a + 1} x - a + 1}{x^{2} + a - 1}\right )}{4 \, {\left (a^{2} - 1\right )}}, \frac {2 \, {\left (a + 1\right )} \sqrt {a - 1} \arctan \left (\frac {x}{\sqrt {a - 1}}\right ) - {\left (a - 1\right )} \sqrt {-a - 1} \log \left (\frac {x^{2} + 2 \, \sqrt {-a - 1} x - a - 1}{x^{2} + a + 1}\right )}{4 \, {\left (a^{2} - 1\right )}}, -\frac {2 \, \sqrt {a + 1} {\left (a - 1\right )} \arctan \left (\frac {x}{\sqrt {a + 1}}\right ) + {\left (a + 1\right )} \sqrt {-a + 1} \log \left (\frac {x^{2} - 2 \, \sqrt {-a + 1} x - a + 1}{x^{2} + a - 1}\right )}{4 \, {\left (a^{2} - 1\right )}}, -\frac {\sqrt {a + 1} {\left (a - 1\right )} \arctan \left (\frac {x}{\sqrt {a + 1}}\right ) - {\left (a + 1\right )} \sqrt {a - 1} \arctan \left (\frac {x}{\sqrt {a - 1}}\right )}{2 \, {\left (a^{2} - 1\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 31, normalized size = 0.66 \[ -\frac {\arctan \left (\frac {x}{\sqrt {a + 1}}\right )}{2 \, \sqrt {a + 1}} + \frac {\arctan \left (\frac {x}{\sqrt {a - 1}}\right )}{2 \, \sqrt {a - 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 32, normalized size = 0.68 \[ -\frac {\arctan \left (\frac {x}{\sqrt {a +1}}\right )}{2 \sqrt {a +1}}+\frac {\arctan \left (\frac {x}{\sqrt {a -1}}\right )}{2 \sqrt {a -1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^4+2*a*x^2+a^2-1),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-1.0>0)', see `assume?` for m
ore details)Is a-1.0 positive or negative?

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mupad [B]  time = 0.10, size = 85, normalized size = 1.81 \[ \frac {\mathrm {atanh}\left (\frac {2\,x\,\left (\frac {a}{2}-\frac {1}{2}\right )}{\sqrt {1-a}}+\frac {2\,a\,x\,\left (\frac {a}{2}-\frac {1}{2}\right )}{{\left (1-a\right )}^{3/2}}\right )}{2\,\sqrt {1-a}}+\frac {\mathrm {atanh}\left (\frac {2\,x\,\left (\frac {a}{2}+\frac {1}{2}\right )}{\sqrt {-a-1}}+\frac {2\,a\,x\,\left (\frac {a}{2}+\frac {1}{2}\right )}{{\left (-a-1\right )}^{3/2}}\right )}{2\,\sqrt {-a-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*a*x^2 + a^2 + x^4 - 1),x)

[Out]

atanh((2*x*(a/2 - 1/2))/(1 - a)^(1/2) + (2*a*x*(a/2 - 1/2))/(1 - a)^(3/2))/(2*(1 - a)^(1/2)) + atanh((2*x*(a/2
+ 1/2))/(- a - 1)^(1/2) + (2*a*x*(a/2 + 1/2))/(- a - 1)^(3/2))/(2*(- a - 1)^(1/2))

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sympy [B]  time = 0.63, size = 257, normalized size = 5.47 \[ \frac {\sqrt {- \frac {1}{a - 1}} \log {\left (- a^{3} \left (- \frac {1}{a - 1}\right )^{\frac {3}{2}} - a^{2} \sqrt {- \frac {1}{a - 1}} + a \left (- \frac {1}{a - 1}\right )^{\frac {3}{2}} + x - \sqrt {- \frac {1}{a - 1}} \right )}}{4} - \frac {\sqrt {- \frac {1}{a - 1}} \log {\left (a^{3} \left (- \frac {1}{a - 1}\right )^{\frac {3}{2}} + a^{2} \sqrt {- \frac {1}{a - 1}} - a \left (- \frac {1}{a - 1}\right )^{\frac {3}{2}} + x + \sqrt {- \frac {1}{a - 1}} \right )}}{4} + \frac {\sqrt {- \frac {1}{a + 1}} \log {\left (- a^{3} \left (- \frac {1}{a + 1}\right )^{\frac {3}{2}} - a^{2} \sqrt {- \frac {1}{a + 1}} + a \left (- \frac {1}{a + 1}\right )^{\frac {3}{2}} + x - \sqrt {- \frac {1}{a + 1}} \right )}}{4} - \frac {\sqrt {- \frac {1}{a + 1}} \log {\left (a^{3} \left (- \frac {1}{a + 1}\right )^{\frac {3}{2}} + a^{2} \sqrt {- \frac {1}{a + 1}} - a \left (- \frac {1}{a + 1}\right )^{\frac {3}{2}} + x + \sqrt {- \frac {1}{a + 1}} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-1/(a - 1))*log(-a**3*(-1/(a - 1))**(3/2) - a**2*sqrt(-1/(a - 1)) + a*(-1/(a - 1))**(3/2) + x - sqrt(-1/(
a - 1)))/4 - sqrt(-1/(a - 1))*log(a**3*(-1/(a - 1))**(3/2) + a**2*sqrt(-1/(a - 1)) - a*(-1/(a - 1))**(3/2) + x
+ sqrt(-1/(a - 1)))/4 + sqrt(-1/(a + 1))*log(-a**3*(-1/(a + 1))**(3/2) - a**2*sqrt(-1/(a + 1)) + a*(-1/(a + 1
))**(3/2) + x - sqrt(-1/(a + 1)))/4 - sqrt(-1/(a + 1))*log(a**3*(-1/(a + 1))**(3/2) + a**2*sqrt(-1/(a + 1)) -
a*(-1/(a + 1))**(3/2) + x + sqrt(-1/(a + 1)))/4

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