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

Optimal. Leaf size=79 \[ \frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}}+\frac {5 x}{16 a^3 \left (a+b x^2\right )}+\frac {5 x}{24 a^2 \left (a+b x^2\right )^2}+\frac {x}{6 a \left (a+b x^2\right )^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {28, 199, 205} \begin {gather*} \frac {5 x}{16 a^3 \left (a+b x^2\right )}+\frac {5 x}{24 a^2 \left (a+b x^2\right )^2}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}}+\frac {x}{6 a \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/(6*a*(a + b*x^2)^3) + (5*x)/(24*a^2*(a + b*x^2)^2) + (5*x)/(16*a^3*(a + b*x^2)) + (5*ArcTan[(Sqrt[b]*x)/Sqrt
[a]])/(16*a^(7/2)*Sqrt[b])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 )^2} \, dx &=b^4 \int \frac {1}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac {x}{6 a \left (a+b x^2\right )^3}+\frac {\left (5 b^3\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^3} \, dx}{6 a}\\ &=\frac {x}{6 a \left (a+b x^2\right )^3}+\frac {5 x}{24 a^2 \left (a+b x^2\right )^2}+\frac {\left (5 b^2\right ) \int \frac {1}{\left (a b+b^2 x^2\right )^2} \, dx}{8 a^2}\\ &=\frac {x}{6 a \left (a+b x^2\right )^3}+\frac {5 x}{24 a^2 \left (a+b x^2\right )^2}+\frac {5 x}{16 a^3 \left (a+b x^2\right )}+\frac {(5 b) \int \frac {1}{a b+b^2 x^2} \, dx}{16 a^3}\\ &=\frac {x}{6 a \left (a+b x^2\right )^3}+\frac {5 x}{24 a^2 \left (a+b x^2\right )^2}+\frac {5 x}{16 a^3 \left (a+b x^2\right )}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 66, normalized size = 0.84 \begin {gather*} \frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}}+\frac {33 a^2 x+40 a b x^3+15 b^2 x^5}{48 a^3 \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(33*a^2*x + 40*a*b*x^3 + 15*b^2*x^5)/(48*a^3*(a + b*x^2)^3) + (5*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*a^(7/2)*Sqrt
[b])

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IntegrateAlgebraic [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 )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 254, normalized size = 3.22 \begin {gather*} \left [\frac {30 \, a b^{3} x^{5} + 80 \, a^{2} b^{2} x^{3} + 66 \, a^{3} b x - 15 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{96 \, {\left (a^{4} b^{4} x^{6} + 3 \, a^{5} b^{3} x^{4} + 3 \, a^{6} b^{2} x^{2} + a^{7} b\right )}}, \frac {15 \, a b^{3} x^{5} + 40 \, a^{2} b^{2} x^{3} + 33 \, a^{3} b x + 15 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{48 \, {\left (a^{4} b^{4} x^{6} + 3 \, a^{5} b^{3} x^{4} + 3 \, a^{6} b^{2} x^{2} + a^{7} 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)^2,x, algorithm="fricas")

[Out]

[1/96*(30*a*b^3*x^5 + 80*a^2*b^2*x^3 + 66*a^3*b*x - 15*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(-a*b)*
log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^4*b^4*x^6 + 3*a^5*b^3*x^4 + 3*a^6*b^2*x^2 + a^7*b), 1/48*(15
*a*b^3*x^5 + 40*a^2*b^2*x^3 + 33*a^3*b*x + 15*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(a*b)*arctan(sqr
t(a*b)*x/a))/(a^4*b^4*x^6 + 3*a^5*b^3*x^4 + 3*a^6*b^2*x^2 + a^7*b)]

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giac [A]  time = 0.15, size = 56, normalized size = 0.71 \begin {gather*} \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} + \frac {15 \, b^{2} x^{5} + 40 \, a b x^{3} + 33 \, a^{2} x}{48 \, {\left (b x^{2} + a\right )}^{3} a^{3}} \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)^2,x, algorithm="giac")

[Out]

5/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/48*(15*b^2*x^5 + 40*a*b*x^3 + 33*a^2*x)/((b*x^2 + a)^3*a^3)

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maple [A]  time = 0.00, size = 66, normalized size = 0.84 \begin {gather*} \frac {x}{6 \left (b \,x^{2}+a \right )^{3} a}+\frac {5 x}{24 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {5 x}{16 \left (b \,x^{2}+a \right ) a^{3}}+\frac {5 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}\, a^{3}} \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)^2,x)

[Out]

1/6*x/a/(b*x^2+a)^3+5/24*x/a^2/(b*x^2+a)^2+5/16*x/a^3/(b*x^2+a)+5/16/a^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 3.01, size = 80, normalized size = 1.01 \begin {gather*} \frac {15 \, b^{2} x^{5} + 40 \, a b x^{3} + 33 \, a^{2} x}{48 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} + \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} \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)^2,x, algorithm="maxima")

[Out]

1/48*(15*b^2*x^5 + 40*a*b*x^3 + 33*a^2*x)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6) + 5/16*arctan(b*x/
sqrt(a*b))/(sqrt(a*b)*a^3)

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mupad [B]  time = 4.36, size = 77, normalized size = 0.97 \begin {gather*} \frac {\frac {11\,x}{16\,a}+\frac {5\,b\,x^3}{6\,a^2}+\frac {5\,b^2\,x^5}{16\,a^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}+\frac {5\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{7/2}\,\sqrt {b}} \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)^2,x)

[Out]

((11*x)/(16*a) + (5*b*x^3)/(6*a^2) + (5*b^2*x^5)/(16*a^3))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) + (5*at
an((b^(1/2)*x)/a^(1/2)))/(16*a^(7/2)*b^(1/2))

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sympy [A]  time = 0.44, size = 129, normalized size = 1.63 \begin {gather*} - \frac {5 \sqrt {- \frac {1}{a^{7} b}} \log {\left (- a^{4} \sqrt {- \frac {1}{a^{7} b}} + x \right )}}{32} + \frac {5 \sqrt {- \frac {1}{a^{7} b}} \log {\left (a^{4} \sqrt {- \frac {1}{a^{7} b}} + x \right )}}{32} + \frac {33 a^{2} x + 40 a b x^{3} + 15 b^{2} x^{5}}{48 a^{6} + 144 a^{5} b x^{2} + 144 a^{4} b^{2} x^{4} + 48 a^{3} b^{3} x^{6}} \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)**2,x)

[Out]

-5*sqrt(-1/(a**7*b))*log(-a**4*sqrt(-1/(a**7*b)) + x)/32 + 5*sqrt(-1/(a**7*b))*log(a**4*sqrt(-1/(a**7*b)) + x)
/32 + (33*a**2*x + 40*a*b*x**3 + 15*b**2*x**5)/(48*a**6 + 144*a**5*b*x**2 + 144*a**4*b**2*x**4 + 48*a**3*b**3*
x**6)

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