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

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

[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))

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Rubi [A]  time = 0.0460682, antiderivative size = 81, normalized size of antiderivative = 1., 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 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]

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}{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*}

Mathematica [A]  time = 0.0440794, size = 80, normalized size = 0.99 \[ -\frac{b^3 x}{2 a^4 \left (a+b x^2\right )}-\frac{3 b^2}{a^4 x}-\frac{7 b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{9/2}}+\frac{2 b}{3 a^3 x^3}-\frac{1}{5 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(5*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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Maple [A]  time = 0.055, size = 70, normalized size = 0.9 \begin{align*} -{\frac{1}{5\,{a}^{2}{x}^{5}}}+{\frac{2\,b}{3\,{a}^{3}{x}^{3}}}-3\,{\frac{{b}^{2}}{{a}^{4}x}}-{\frac{{b}^{3}x}{2\,{a}^{4} \left ( b{x}^{2}+a \right ) }}-{\frac{7\,{b}^{3}}{2\,{a}^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.7181, size = 423, normalized size = 5.22 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 0.69883, size = 126, normalized size = 1.56 \begin{align*} \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}} \end{align*}

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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Giac [A]  time = 1.1281, size = 95, normalized size = 1.17 \begin{align*} -\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}} \end{align*}

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)