3.492 \(\int \frac {1}{x^6 (a^2+2 a b x^2+b^2 x^4)} \, dx\)

Optimal. Leaf size=81 \[ -\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{9/2}}-\frac {7 b^2}{2 a^4 x}+\frac {7 b}{6 a^3 x^3}-\frac {7}{10 a^2 x^5}+\frac {1}{2 a x^5 \left (a+b x^2\right )} \]

[Out]

-7/10/a^2/x^5+7/6*b/a^3/x^3-7/2*b^2/a^4/x+1/2/a/x^5/(b*x^2+a)-7/2*b^(5/2)*arctan(x*b^(1/2)/a^(1/2))/a^(9/2)

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Rubi [A]  time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 290, 325, 205} \[ -\frac {7 b^2}{2 a^4 x}-\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{9/2}}+\frac {7 b}{6 a^3 x^3}-\frac {7}{10 a^2 x^5}+\frac {1}{2 a x^5 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

-7/(10*a^2*x^5) + (7*b)/(6*a^3*x^3) - (7*b^2)/(2*a^4*x) + 1/(2*a*x^5*(a + b*x^2)) - (7*b^(5/2)*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(2*a^(9/2))

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx &=b^2 \int \frac {1}{x^6 \left (a b+b^2 x^2\right )^2} \, dx\\ &=\frac {1}{2 a x^5 \left (a+b x^2\right )}+\frac {(7 b) \int \frac {1}{x^6 \left (a b+b^2 x^2\right )} \, dx}{2 a}\\ &=-\frac {7}{10 a^2 x^5}+\frac {1}{2 a x^5 \left (a+b x^2\right )}-\frac {\left (7 b^2\right ) \int \frac {1}{x^4 \left (a b+b^2 x^2\right )} \, dx}{2 a^2}\\ &=-\frac {7}{10 a^2 x^5}+\frac {7 b}{6 a^3 x^3}+\frac {1}{2 a x^5 \left (a+b x^2\right )}+\frac {\left (7 b^3\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{2 a^3}\\ &=-\frac {7}{10 a^2 x^5}+\frac {7 b}{6 a^3 x^3}-\frac {7 b^2}{2 a^4 x}+\frac {1}{2 a x^5 \left (a+b x^2\right )}-\frac {\left (7 b^4\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{2 a^4}\\ &=-\frac {7}{10 a^2 x^5}+\frac {7 b}{6 a^3 x^3}-\frac {7 b^2}{2 a^4 x}+\frac {1}{2 a x^5 \left (a+b x^2\right )}-\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 80, normalized size = 0.99 \[ -\frac {7 b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{9/2}}-\frac {b^3 x}{2 a^4 \left (a+b x^2\right )}-\frac {3 b^2}{a^4 x}+\frac {2 b}{3 a^3 x^3}-\frac {1}{5 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

-1/5*1/(a^2*x^5) + (2*b)/(3*a^3*x^3) - (3*b^2)/(a^4*x) - (b^3*x)/(2*a^4*(a + b*x^2)) - (7*b^(5/2)*ArcTan[(Sqrt
[b]*x)/Sqrt[a]])/(2*a^(9/2))

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fricas [A]  time = 1.03, size = 198, normalized size = 2.44 \[ \left [-\frac {210 \, b^{3} x^{6} + 140 \, a b^{2} x^{4} - 28 \, a^{2} b x^{2} + 12 \, a^{3} - 105 \, {\left (b^{3} x^{7} + a b^{2} x^{5}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{60 \, {\left (a^{4} b x^{7} + a^{5} x^{5}\right )}}, -\frac {105 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} - 14 \, a^{2} b x^{2} + 6 \, a^{3} + 105 \, {\left (b^{3} x^{7} + a b^{2} x^{5}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{30 \, {\left (a^{4} b x^{7} + a^{5} x^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

[-1/60*(210*b^3*x^6 + 140*a*b^2*x^4 - 28*a^2*b*x^2 + 12*a^3 - 105*(b^3*x^7 + a*b^2*x^5)*sqrt(-b/a)*log((b*x^2
- 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^4*b*x^7 + a^5*x^5), -1/30*(105*b^3*x^6 + 70*a*b^2*x^4 - 14*a^2*b*x^2
+ 6*a^3 + 105*(b^3*x^7 + a*b^2*x^5)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^4*b*x^7 + a^5*x^5)]

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giac [A]  time = 0.15, size = 70, normalized size = 0.86 \[ -\frac {7 \, b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{4}} - \frac {b^{3} x}{2 \, {\left (b x^{2} + a\right )} a^{4}} - \frac {45 \, b^{2} x^{4} - 10 \, a b x^{2} + 3 \, a^{2}}{15 \, a^{4} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

-7/2*b^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/2*b^3*x/((b*x^2 + a)*a^4) - 1/15*(45*b^2*x^4 - 10*a*b*x^2 +
3*a^2)/(a^4*x^5)

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maple [A]  time = 0.01, size = 70, normalized size = 0.86 \[ -\frac {b^{3} x}{2 \left (b \,x^{2}+a \right ) a^{4}}-\frac {7 b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a^{4}}-\frac {3 b^{2}}{a^{4} x}+\frac {2 b}{3 a^{3} x^{3}}-\frac {1}{5 a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

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

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maxima [A]  time = 3.08, size = 75, normalized size = 0.93 \[ -\frac {105 \, b^{3} x^{6} + 70 \, a b^{2} x^{4} - 14 \, a^{2} b x^{2} + 6 \, a^{3}}{30 \, {\left (a^{4} b x^{7} + a^{5} x^{5}\right )}} - \frac {7 \, b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

-1/30*(105*b^3*x^6 + 70*a*b^2*x^4 - 14*a^2*b*x^2 + 6*a^3)/(a^4*b*x^7 + a^5*x^5) - 7/2*b^3*arctan(b*x/sqrt(a*b)
)/(sqrt(a*b)*a^4)

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mupad [B]  time = 4.71, size = 70, normalized size = 0.86 \[ -\frac {\frac {1}{5\,a}-\frac {7\,b\,x^2}{15\,a^2}+\frac {7\,b^2\,x^4}{3\,a^3}+\frac {7\,b^3\,x^6}{2\,a^4}}{b\,x^7+a\,x^5}-\frac {7\,b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)

[Out]

- (1/(5*a) - (7*b*x^2)/(15*a^2) + (7*b^2*x^4)/(3*a^3) + (7*b^3*x^6)/(2*a^4))/(a*x^5 + b*x^7) - (7*b^(5/2)*atan
((b^(1/2)*x)/a^(1/2)))/(2*a^(9/2))

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sympy [A]  time = 0.43, size = 126, normalized size = 1.56 \[ \frac {7 \sqrt {- \frac {b^{5}}{a^{9}}} \log {\left (- \frac {a^{5} \sqrt {- \frac {b^{5}}{a^{9}}}}{b^{3}} + x \right )}}{4} - \frac {7 \sqrt {- \frac {b^{5}}{a^{9}}} \log {\left (\frac {a^{5} \sqrt {- \frac {b^{5}}{a^{9}}}}{b^{3}} + x \right )}}{4} + \frac {- 6 a^{3} + 14 a^{2} b x^{2} - 70 a b^{2} x^{4} - 105 b^{3} x^{6}}{30 a^{5} x^{5} + 30 a^{4} b x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

7*sqrt(-b**5/a**9)*log(-a**5*sqrt(-b**5/a**9)/b**3 + x)/4 - 7*sqrt(-b**5/a**9)*log(a**5*sqrt(-b**5/a**9)/b**3
+ x)/4 + (-6*a**3 + 14*a**2*b*x**2 - 70*a*b**2*x**4 - 105*b**3*x**6)/(30*a**5*x**5 + 30*a**4*b*x**7)

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