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

Optimal. Leaf size=89 \[ \frac{x \left (a+b x^2\right )}{b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{a} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.031588, antiderivative size = 89, 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, 321, 205} \[ \frac{x \left (a+b x^2\right )}{b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{a} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^2))/(b*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (Sqrt[a]*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(b^(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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{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{x^2}{a b+b^2 x^2} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{x \left (a+b x^2\right )}{b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (a \left (a b+b^2 x^2\right )\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{b \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{x \left (a+b x^2\right )}{b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{a} \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{3/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0136017, size = 54, normalized size = 0.61 \[ \frac{\left (a+b x^2\right ) \left (\sqrt{b} x-\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right )}{b^{3/2} \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.213, size = 48, normalized size = 0.5 \begin{align*}{\frac{b{x}^{2}+a}{b} \left ( x\sqrt{ab}-a\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ) \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(x^2/((b*x^2+a)^2)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.284, size = 165, normalized size = 1.85 \begin{align*} \left [\frac{\sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} - 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right ) + 2 \, x}{2 \, b}, -\frac{\sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right ) - x}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.33345, size = 56, normalized size = 0.63 \begin{align*} \frac{\sqrt{- \frac{a}{b^{3}}} \log{\left (- b \sqrt{- \frac{a}{b^{3}}} + x \right )}}{2} - \frac{\sqrt{- \frac{a}{b^{3}}} \log{\left (b \sqrt{- \frac{a}{b^{3}}} + x \right )}}{2} + \frac{x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13674, size = 57, normalized size = 0.64 \begin{align*} -\frac{a \arctan \left (\frac{b x}{\sqrt{a b}}\right ) \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{a b} b} + \frac{x \mathrm{sgn}\left (b x^{2} + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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