3.4.33 \(\int \frac {x^6}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=83 \[ \frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \sqrt {a} b^{7/2}}-\frac {5 x}{16 b^3 \left (a+b x^2\right )}-\frac {5 x^3}{24 b^2 \left (a+b x^2\right )^2}-\frac {x^5}{6 b \left (a+b x^2\right )^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 288, 205} \begin {gather*} -\frac {5 x^3}{24 b^2 \left (a+b x^2\right )^2}-\frac {5 x}{16 b^3 \left (a+b x^2\right )}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \sqrt {a} b^{7/2}}-\frac {x^5}{6 b \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 66, normalized size = 0.80 \begin {gather*} \frac {5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \sqrt {a} b^{7/2}}-\frac {x \left (15 a^2+40 a b x^2+33 b^2 x^4\right )}{48 b^3 \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-1/48*(x*(15*a^2 + 40*a*b*x^2 + 33*b^2*x^4))/(b^3*(a + b*x^2)^3) + (5*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*Sqrt[a]
*b^(7/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^6}{\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[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

IntegrateAlgebraic[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^2, x]

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fricas [A]  time = 0.85, size = 254, normalized size = 3.06 \begin {gather*} \left [-\frac {66 \, a b^{3} x^{5} + 80 \, a^{2} b^{2} x^{3} + 30 \, 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 b^{7} x^{6} + 3 \, a^{2} b^{6} x^{4} + 3 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}, -\frac {33 \, a b^{3} x^{5} + 40 \, a^{2} b^{2} x^{3} + 15 \, 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 b^{7} x^{6} + 3 \, a^{2} b^{6} x^{4} + 3 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 56, normalized size = 0.67 \begin {gather*} \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{3}} - \frac {33 \, b^{2} x^{5} + 40 \, a b x^{3} + 15 \, a^{2} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(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)*b^3) - 1/48*(33*b^2*x^5 + 40*a*b*x^3 + 15*a^2*x)/((b*x^2 + a)^3*b^3)

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maple [A]  time = 0.01, size = 58, normalized size = 0.70 \begin {gather*} \frac {5 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}\, b^{3}}+\frac {-\frac {11 x^{5}}{16 b}-\frac {5 a \,x^{3}}{6 b^{2}}-\frac {5 a^{2} x}{16 b^{3}}}{\left (b \,x^{2}+a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

(-11/16/b*x^5-5/6*a/b^2*x^3-5/16*a^2/b^3*x)/(b*x^2+a)^3+5/16/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 3.07, size = 81, normalized size = 0.98 \begin {gather*} -\frac {33 \, b^{2} x^{5} + 40 \, a b x^{3} + 15 \, a^{2} x}{48 \, {\left (b^{6} x^{6} + 3 \, a b^{5} x^{4} + 3 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}} + \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.39, size = 78, normalized size = 0.94 \begin {gather*} \frac {5\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,\sqrt {a}\,b^{7/2}}-\frac {\frac {11\,x^5}{16\,b}+\frac {5\,a\,x^3}{6\,b^2}+\frac {5\,a^2\,x}{16\,b^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

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

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sympy [A]  time = 0.49, size = 134, normalized size = 1.61 \begin {gather*} - \frac {5 \sqrt {- \frac {1}{a b^{7}}} \log {\left (- a b^{3} \sqrt {- \frac {1}{a b^{7}}} + x \right )}}{32} + \frac {5 \sqrt {- \frac {1}{a b^{7}}} \log {\left (a b^{3} \sqrt {- \frac {1}{a b^{7}}} + x \right )}}{32} + \frac {- 15 a^{2} x - 40 a b x^{3} - 33 b^{2} x^{5}}{48 a^{3} b^{3} + 144 a^{2} b^{4} x^{2} + 144 a b^{5} x^{4} + 48 b^{6} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

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

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