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

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

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Rubi [A]  time = 0.04, antiderivative size = 135, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 83, normalized size = 0.61 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 6.01, size = 76, normalized size = 0.56 \begin {gather*} \frac {\left (a+b x^2\right ) \left (\frac {3 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b}}+\frac {5 a x+3 b x^3}{8 a^2 \left (a+b x^2\right )^2}\right )}{\sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.16, size = 188, normalized size = 1.39 \begin {gather*} \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 {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)^(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \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)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.01, size = 97, normalized size = 0.72 \begin {gather*} \frac {\left (3 b^{2} x^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+6 a b \,x^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+3 \sqrt {a b}\, b \,x^{3}+3 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+5 \sqrt {a b}\, a x \right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a^{2}} \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)^(3/2),x)

[Out]

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

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maxima [A]  time = 2.99, size = 58, normalized size = 0.43 \begin {gather*} \frac {3 \, b x^{3} + 5 \, a x}{8 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2}} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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