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

Optimal. Leaf size=66 \[ \frac {5 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {5 a x}{2 b^3}-\frac {x^5}{2 b \left (a+b x^2\right )}+\frac {5 x^3}{6 b^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 66, 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, 302, 205} \begin {gather*} \frac {5 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}-\frac {5 a x}{2 b^3}-\frac {x^5}{2 b \left (a+b x^2\right )}+\frac {5 x^3}{6 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 60, normalized size = 0.91 \begin {gather*} \frac {5 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{7/2}}+\frac {x \left (-\frac {3 a^2}{a+b x^2}-12 a+2 b x^2\right )}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-12*a + 2*b*x^2 - (3*a^2)/(a + b*x^2)))/(6*b^3) + (5*a^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(7/2))

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

[Out]

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

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fricas [A]  time = 0.66, size = 164, normalized size = 2.48 \begin {gather*} \left [\frac {4 \, b^{2} x^{5} - 20 \, a b x^{3} - 30 \, a^{2} x + 15 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{12 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {2 \, b^{2} x^{5} - 10 \, a b x^{3} - 15 \, a^{2} x + 15 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{6 \, {\left (b^{4} x^{2} + a b^{3}\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),x, algorithm="fricas")

[Out]

[1/12*(4*b^2*x^5 - 20*a*b*x^3 - 30*a^2*x + 15*(a*b*x^2 + a^2)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b
*x^2 + a)))/(b^4*x^2 + a*b^3), 1/6*(2*b^2*x^5 - 10*a*b*x^3 - 15*a^2*x + 15*(a*b*x^2 + a^2)*sqrt(a/b)*arctan(b*
x*sqrt(a/b)/a))/(b^4*x^2 + a*b^3)]

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giac [A]  time = 0.16, size = 61, normalized size = 0.92 \begin {gather*} \frac {5 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} - \frac {a^{2} x}{2 \, {\left (b x^{2} + a\right )} b^{3}} + \frac {b^{4} x^{3} - 6 \, a b^{3} x}{3 \, b^{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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 57, normalized size = 0.86 \begin {gather*} \frac {x^{3}}{3 b^{2}}-\frac {a^{2} x}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {5 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{3}}-\frac {2 a x}{b^{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),x)

[Out]

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

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maxima [A]  time = 3.13, size = 59, normalized size = 0.89 \begin {gather*} -\frac {a^{2} x}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}} + \frac {5 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {b x^{3} - 6 \, a x}{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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 56, normalized size = 0.85 \begin {gather*} \frac {x^3}{3\,b^2}+\frac {5\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,b^{7/2}}-\frac {a^2\,x}{2\,\left (b^4\,x^2+a\,b^3\right )}-\frac {2\,a\,x}{b^3} \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),x)

[Out]

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

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sympy [A]  time = 0.30, size = 107, normalized size = 1.62 \begin {gather*} - \frac {a^{2} x}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a x}{b^{3}} - \frac {5 \sqrt {- \frac {a^{3}}{b^{7}}} \log {\left (x - \frac {b^{3} \sqrt {- \frac {a^{3}}{b^{7}}}}{a} \right )}}{4} + \frac {5 \sqrt {- \frac {a^{3}}{b^{7}}} \log {\left (x + \frac {b^{3} \sqrt {- \frac {a^{3}}{b^{7}}}}{a} \right )}}{4} + \frac {x^{3}}{3 b^{2}} \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),x)

[Out]

-a**2*x/(2*a*b**3 + 2*b**4*x**2) - 2*a*x/b**3 - 5*sqrt(-a**3/b**7)*log(x - b**3*sqrt(-a**3/b**7)/a)/4 + 5*sqrt
(-a**3/b**7)*log(x + b**3*sqrt(-a**3/b**7)/a)/4 + x**3/(3*b**2)

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