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

Optimal. Leaf size=92 \[ -\frac{a+b x^2}{a x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

-((a + b*x^2)/(a*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])) - (Sqrt[b]*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3
/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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Rubi [A]  time = 0.0333592, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1112, 325, 205} \[ -\frac{a+b x^2}{a x \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{a^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*x^2)/(a*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])) - (Sqrt[b]*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(a^(3
/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A]  time = 0.0135005, size = 56, normalized size = 0.61 \[ -\frac{\left (a+b x^2\right ) \left (\sqrt{b} x \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )+\sqrt{a}\right )}{a^{3/2} x \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.171, size = 50, normalized size = 0.5 \begin{align*} -{\frac{b{x}^{2}+a}{ax} \left ( b\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ) x+\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/x^2/((b*x^2+a)^2)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35675, size = 173, normalized size = 1.88 \begin{align*} \left [\frac{x \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} - 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right ) - 2}{2 \, a x}, -\frac{x \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) + 1}{a x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.363345, size = 65, normalized size = 0.71 \begin{align*} \frac{\sqrt{- \frac{b}{a^{3}}} \log{\left (- \frac{a^{2} \sqrt{- \frac{b}{a^{3}}}}{b} + x \right )}}{2} - \frac{\sqrt{- \frac{b}{a^{3}}} \log{\left (\frac{a^{2} \sqrt{- \frac{b}{a^{3}}}}{b} + x \right )}}{2} - \frac{1}{a x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13449, size = 50, normalized size = 0.54 \begin{align*} -{\left (\frac{b \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{\sqrt{a b} a} + \frac{1}{a x}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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