3.642 \(\int \frac{1}{(a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=135 \[ \frac{3 x \left (a+b x^2\right )^2}{8 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{x \left (a+b x^2\right )}{4 a \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{3 \left (a+b x^2\right )^3 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

[Out]

(x*(a + b*x^2))/(4*a*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)) + (3*x*(a + b*x^2)^2)/(8*a^2*(a^2 + 2*a*b*x^2 + b^2*x^
4)^(3/2)) + (3*(a + b*x^2)^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)
)

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Rubi [A]  time = 0.0383879, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {1088, 199, 205} \[ \frac{3 x \left (a+b x^2\right )^2}{8 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{x \left (a+b x^2\right )}{4 a \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{3 \left (a+b x^2\right )^3 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-3/2),x]

[Out]

(x*(a + b*x^2))/(4*a*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)) + (3*x*(a + b*x^2)^2)/(8*a^2*(a^2 + 2*a*b*x^2 + b^2*x^
4)^(3/2)) + (3*(a + b*x^2)^3*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)
)

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 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}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac{\left (2 a b+2 b^2 x^2\right )^3 \int \frac{1}{\left (2 a b+2 b^2 x^2\right )^3} \, dx}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}\\ &=\frac{x \left (a+b x^2\right )}{4 a \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{\left (3 \left (2 a b+2 b^2 x^2\right )^3\right ) \int \frac{1}{\left (2 a b+2 b^2 x^2\right )^2} \, dx}{8 a b \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}\\ &=\frac{x \left (a+b x^2\right )}{4 a \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{3 x \left (a+b x^2\right )^2}{8 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{\left (3 \left (2 a b+2 b^2 x^2\right )^3\right ) \int \frac{1}{2 a b+2 b^2 x^2} \, dx}{32 a^2 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}\\ &=\frac{x \left (a+b x^2\right )}{4 a \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{3 x \left (a+b x^2\right )^2}{8 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}+\frac{3 \left (a+b x^2\right )^3 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0246737, size = 83, normalized size = 0.61 \[ \frac{\sqrt{a} \sqrt{b} x \left (5 a+3 b x^2\right )+3 \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{b} \left (a+b x^2\right ) \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-3/2),x]

[Out]

(Sqrt[a]*Sqrt[b]*x*(5*a + 3*b*x^2) + 3*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2)*Sqrt[b]*(a + b*x^
2)*Sqrt[(a + b*x^2)^2])

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Maple [A]  time = 0.221, size = 97, normalized size = 0.7 \begin{align*}{\frac{b{x}^{2}+a}{8\,{a}^{2}} \left ( 3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{4}{b}^{2}+3\,\sqrt{ab}{x}^{3}b+6\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){x}^{2}ab+5\,\sqrt{ab}xa+3\,{a}^{2}\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

1/8*(3*arctan(b*x/(a*b)^(1/2))*x^4*b^2+3*(a*b)^(1/2)*x^3*b+6*arctan(b*x/(a*b)^(1/2))*x^2*a*b+5*(a*b)^(1/2)*x*a
+3*a^2*arctan(b*x/(a*b)^(1/2)))*(b*x^2+a)/(a*b)^(1/2)/a^2/((b*x^2+a)^2)^(3/2)

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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^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.31085, size = 401, normalized size = 2.97 \begin{align*} \left [\frac{6 \, a b^{2} x^{3} + 10 \, a^{2} b x - 3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{16 \,{\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}, \frac{3 \, a b^{2} x^{3} + 5 \, a^{2} b x + 3 \,{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{8 \,{\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(6*a*b^2*x^3 + 10*a^2*b*x - 3*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b
*x^2 + a)))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b), 1/8*(3*a*b^2*x^3 + 5*a^2*b*x + 3*(b^2*x^4 + 2*a*b*x^2 + a^2
)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x