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

Optimal. Leaf size=45 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {28, 199, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(2*a*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*a^(3/2)*Sqrt[b])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 205

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

Rubi steps

\begin {align*} \int \frac {1}{a^2+2 a b x^2+b^2 x^4} \, dx &=b^2 \int \frac {1}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=\frac {x}{2 a \left (a+b x^2\right )}+\frac {b \int \frac {1}{a b+b^2 x^2} \, dx}{2 a}\\ &=\frac {x}{2 a \left (a+b x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(2*a*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*a^(3/2)*Sqrt[b])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{a^2+2 a b x^2+b^2 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 120, normalized size = 2.67 \begin {gather*} \left [\frac {2 \, a b x - {\left (b x^{2} + a\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{4 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}, \frac {a b x + {\left (b x^{2} + a\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{2 \, {\left (a^{2} b^{2} x^{2} + a^{3} b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 35, normalized size = 0.78 \begin {gather*} \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} + \frac {x}{2 \, {\left (b x^{2} + a\right )} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 36, normalized size = 0.80 \begin {gather*} \frac {x}{2 \left (b \,x^{2}+a \right ) a}+\frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.85, size = 35, normalized size = 0.78 \begin {gather*} \frac {x}{2 \, {\left (a b x^{2} + a^{2}\right )}} + \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 33, normalized size = 0.73 \begin {gather*} \frac {x}{2\,a\,\left (b\,x^2+a\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^4 + 2*a*b*x^2),x)

[Out]

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

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sympy [B]  time = 0.22, size = 78, normalized size = 1.73 \begin {gather*} \frac {x}{2 a^{2} + 2 a b x^{2}} - \frac {\sqrt {- \frac {1}{a^{3} b}} \log {\left (- a^{2} \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{a^{3} b}} \log {\left (a^{2} \sqrt {- \frac {1}{a^{3} b}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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