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

Optimal. Leaf size=84 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{3/2} b^{5/2}}+\frac {x}{16 a b^2 \left (a+b x^2\right )}-\frac {x}{8 b^2 \left (a+b x^2\right )^2}-\frac {x^3}{6 b \left (a+b x^2\right )^3} \]

[Out]

-1/6*x^3/b/(b*x^2+a)^3-1/8*x/b^2/(b*x^2+a)^2+1/16*x/a/b^2/(b*x^2+a)+1/16*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/b^(
5/2)

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Rubi [A]  time = 0.04, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 288, 199, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{3/2} b^{5/2}}+\frac {x}{16 a b^2 \left (a+b x^2\right )}-\frac {x}{8 b^2 \left (a+b x^2\right )^2}-\frac {x^3}{6 b \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x^3/(6*b*(a + b*x^2)^3) - x/(8*b^2*(a + b*x^2)^2) + x/(16*a*b^2*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(1
6*a^(3/2)*b^(5/2))

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]

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.82 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{3/2} b^{5/2}}+\frac {-3 a^2 x-8 a b x^3+3 b^2 x^5}{48 a b^2 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^2*x - 8*a*b*x^3 + 3*b^2*x^5)/(48*a*b^2*(a + b*x^2)^3) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(16*a^(3/2)*b^(5/2))

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fricas [A]  time = 1.02, size = 258, normalized size = 3.07 \[ \left [\frac {6 \, a b^{3} x^{5} - 16 \, a^{2} b^{2} x^{3} - 6 \, a^{3} b x - 3 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{96 \, {\left (a^{2} b^{6} x^{6} + 3 \, a^{3} b^{5} x^{4} + 3 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}}, \frac {3 \, a b^{3} x^{5} - 8 \, a^{2} b^{2} x^{3} - 3 \, a^{3} b x + 3 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{48 \, {\left (a^{2} b^{6} x^{6} + 3 \, a^{3} b^{5} x^{4} + 3 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(6*a*b^3*x^5 - 16*a^2*b^2*x^3 - 6*a^3*b*x - 3*(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^2*b^6*x^6 + 3*a^3*b^5*x^4 + 3*a^4*b^4*x^2 + a^5*b^3), 1/48*(3*a
*b^3*x^5 - 8*a^2*b^2*x^3 - 3*a^3*b*x + 3*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(a*b)*arctan(sqrt(a*b
)*x/a))/(a^2*b^6*x^6 + 3*a^3*b^5*x^4 + 3*a^4*b^4*x^2 + a^5*b^3)]

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giac [A]  time = 0.16, size = 62, normalized size = 0.74 \[ \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a b^{2}} + \frac {3 \, b^{2} x^{5} - 8 \, a b x^{3} - 3 \, a^{2} x}{48 \, {\left (b x^{2} + a\right )}^{3} a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^2) + 1/48*(3*b^2*x^5 - 8*a*b*x^3 - 3*a^2*x)/((b*x^2 + a)^3*a*b^2)

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maple [A]  time = 0.01, size = 58, normalized size = 0.69 \[ \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}\, a \,b^{2}}+\frac {\frac {x^{5}}{16 a}-\frac {x^{3}}{6 b}-\frac {a x}{16 b^{2}}}{\left (b \,x^{2}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.93, size = 87, normalized size = 1.04 \[ \frac {3 \, b^{2} x^{5} - 8 \, a b x^{3} - 3 \, a^{2} x}{48 \, {\left (a b^{5} x^{6} + 3 \, a^{2} b^{4} x^{4} + 3 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} + \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*b^2*x^5 - 8*a*b*x^3 - 3*a^2*x)/(a*b^5*x^6 + 3*a^2*b^4*x^4 + 3*a^3*b^3*x^2 + a^4*b^2) + 1/16*arctan(b*x
/sqrt(a*b))/(sqrt(a*b)*a*b^2)

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mupad [B]  time = 4.35, size = 75, normalized size = 0.89 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{3/2}\,b^{5/2}}-\frac {\frac {x^3}{6\,b}-\frac {x^5}{16\,a}+\frac {a\,x}{16\,b^2}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan((b^(1/2)*x)/a^(1/2))/(16*a^(3/2)*b^(5/2)) - (x^3/(6*b) - x^5/(16*a) + (a*x)/(16*b^2))/(a^3 + b^3*x^6 + 3*
a^2*b*x^2 + 3*a*b^2*x^4)

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sympy [B]  time = 0.45, size = 143, normalized size = 1.70 \[ - \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \log {\left (- a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{3} b^{5}}} \log {\left (a^{2} b^{2} \sqrt {- \frac {1}{a^{3} b^{5}}} + x \right )}}{32} + \frac {- 3 a^{2} x - 8 a b x^{3} + 3 b^{2} x^{5}}{48 a^{4} b^{2} + 144 a^{3} b^{3} x^{2} + 144 a^{2} b^{4} x^{4} + 48 a b^{5} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**3*b**5))*log(-a**2*b**2*sqrt(-1/(a**3*b**5)) + x)/32 + sqrt(-1/(a**3*b**5))*log(a**2*b**2*sqrt(-1
/(a**3*b**5)) + x)/32 + (-3*a**2*x - 8*a*b*x**3 + 3*b**2*x**5)/(48*a**4*b**2 + 144*a**3*b**3*x**2 + 144*a**2*b
**4*x**4 + 48*a*b**5*x**6)

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