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

Optimal. Leaf size=53 \[ \frac{\left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

((a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0150325, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1088, 205} \[ \frac{\left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

((a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 1088

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

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

Mathematica [A]  time = 0.0117462, size = 44, normalized size = 0.83 \[ \frac{\left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{a} \sqrt{b} \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.167, size = 34, normalized size = 0.6 \begin{align*}{(b{x}^{2}+a)\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.255, size = 151, normalized size = 2.85 \begin{align*} \left [-\frac{\sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{2 \, a b}, \frac{\sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.156914, size = 53, normalized size = 1. \begin{align*} - \frac{\sqrt{- \frac{1}{a b}} \log{\left (- a \sqrt{- \frac{1}{a b}} + x \right )}}{2} + \frac{\sqrt{- \frac{1}{a b}} \log{\left (a \sqrt{- \frac{1}{a b}} + x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12336, size = 31, normalized size = 0.58 \begin{align*} \frac{\arctan \left (\frac{b x}{\sqrt{a b}}\right ) \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{a b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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