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

Optimal. Leaf size=57 \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{3}{2 a^2 x}+\frac{1}{2 a x \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0274065, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {28, 290, 325, 205} \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{3}{2 a^2 x}+\frac{1}{2 a x \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0362392, size = 54, normalized size = 0.95 \[ -\frac{b x}{2 a^2 \left (a+b x^2\right )}-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{1}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 46, normalized size = 0.8 \begin{align*} -{\frac{1}{{a}^{2}x}}-{\frac{bx}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}-{\frac{3\,b}{2\,{a}^{2}}\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^2/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

-1/a^2/x-1/2*b/a^2*x/(b*x^2+a)-3/2*b/a^2/(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^2/(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 = 1.76015, size = 288, normalized size = 5.05 \begin{align*} \left [-\frac{6 \, b x^{2} - 3 \,{\left (b x^{3} + a x\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}{4 \,{\left (a^{2} b x^{3} + a^{3} x\right )}}, -\frac{3 \, b x^{2} + 3 \,{\left (b x^{3} + a x\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right ) + 2 \, a}{2 \,{\left (a^{2} b x^{3} + a^{3} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*b*x^2 - 3*(b*x^3 + a*x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 4*a)/(a^2*b*x^3
+ a^3*x), -1/2*(3*b*x^2 + 3*(b*x^3 + a*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 2*a)/(a^2*b*x^3 + a^3*x)]

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Sympy [A]  time = 0.452937, size = 90, normalized size = 1.58 \begin{align*} \frac{3 \sqrt{- \frac{b}{a^{5}}} \log{\left (- \frac{a^{3} \sqrt{- \frac{b}{a^{5}}}}{b} + x \right )}}{4} - \frac{3 \sqrt{- \frac{b}{a^{5}}} \log{\left (\frac{a^{3} \sqrt{- \frac{b}{a^{5}}}}{b} + x \right )}}{4} - \frac{2 a + 3 b x^{2}}{2 a^{3} x + 2 a^{2} b x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16988, size = 63, normalized size = 1.11 \begin{align*} -\frac{3 \, b \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{2}} - \frac{3 \, b x^{2} + 2 \, a}{2 \,{\left (b x^{3} + a x\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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