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

Optimal. Leaf size=68 \[ \frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}+\frac{5 b}{2 a^3 x}-\frac{5}{6 a^2 x^3}+\frac{1}{2 a x^3 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0359153, antiderivative size = 68, normalized size of antiderivative = 1., 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, 290, 325, 205} \[ \frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}+\frac{5 b}{2 a^3 x}-\frac{5}{6 a^2 x^3}+\frac{1}{2 a x^3 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.037667, size = 67, normalized size = 0.99 \[ \frac{b^2 x}{2 a^3 \left (a+b x^2\right )}+\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}+\frac{2 b}{a^3 x}-\frac{1}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.054, size = 59, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{a}^{2}{x}^{3}}}+2\,{\frac{b}{x{a}^{3}}}+{\frac{{b}^{2}x}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{5\,{b}^{2}}{2\,{a}^{3}}\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^4/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

-1/3/x^3/a^2+2*b/a^3/x+1/2*b^2/a^3*x/(b*x^2+a)+5/2*b^2/a^3/(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^4/(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 = 1.8174, size = 359, normalized size = 5.28 \begin{align*} \left [\frac{30 \, b^{2} x^{4} + 20 \, a b x^{2} + 15 \,{\left (b^{2} x^{5} + a b x^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right ) - 4 \, a^{2}}{12 \,{\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac{15 \, b^{2} x^{4} + 10 \, a b x^{2} + 15 \,{\left (b^{2} x^{5} + a b x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) - 2 \, a^{2}}{6 \,{\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.550768, size = 114, normalized size = 1.68 \begin{align*} - \frac{5 \sqrt{- \frac{b^{3}}{a^{7}}} \log{\left (- \frac{a^{4} \sqrt{- \frac{b^{3}}{a^{7}}}}{b^{2}} + x \right )}}{4} + \frac{5 \sqrt{- \frac{b^{3}}{a^{7}}} \log{\left (\frac{a^{4} \sqrt{- \frac{b^{3}}{a^{7}}}}{b^{2}} + x \right )}}{4} + \frac{- 2 a^{2} + 10 a b x^{2} + 15 b^{2} x^{4}}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*sqrt(-b**3/a**7)*log(-a**4*sqrt(-b**3/a**7)/b**2 + x)/4 + 5*sqrt(-b**3/a**7)*log(a**4*sqrt(-b**3/a**7)/b**2
+ x)/4 + (-2*a**2 + 10*a*b*x**2 + 15*b**2*x**4)/(6*a**4*x**3 + 6*a**3*b*x**5)

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Giac [A]  time = 1.13932, size = 80, normalized size = 1.18 \begin{align*} \frac{5 \, b^{2} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{3}} + \frac{b^{2} x}{2 \,{\left (b x^{2} + a\right )} a^{3}} + \frac{6 \, b x^{2} - a}{3 \, a^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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