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

Optimal. Leaf size=95 \[ \frac{35}{48 a^3 x \left (a+b x^2\right )}+\frac{7}{24 a^2 x \left (a+b x^2\right )^2}-\frac{35 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{9/2}}-\frac{35}{16 a^4 x}+\frac{1}{6 a x \left (a+b x^2\right )^3} \]

[Out]

-35/(16*a^4*x) + 1/(6*a*x*(a + b*x^2)^3) + 7/(24*a^2*x*(a + b*x^2)^2) + 35/(48*a^3*x*(a + b*x^2)) - (35*Sqrt[b
]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*a^(9/2))

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Rubi [A]  time = 0.0560155, antiderivative size = 95, 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} \[ \frac{35}{48 a^3 x \left (a+b x^2\right )}+\frac{7}{24 a^2 x \left (a+b x^2\right )^2}-\frac{35 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{9/2}}-\frac{35}{16 a^4 x}+\frac{1}{6 a x \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-35/(16*a^4*x) + 1/(6*a*x*(a + b*x^2)^3) + 7/(24*a^2*x*(a + b*x^2)^2) + 35/(48*a^3*x*(a + b*x^2)) - (35*Sqrt[b
]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*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^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac{1}{x^2 \left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac{1}{6 a x \left (a+b x^2\right )^3}+\frac{\left (7 b^3\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )^3} \, dx}{6 a}\\ &=\frac{1}{6 a x \left (a+b x^2\right )^3}+\frac{7}{24 a^2 x \left (a+b x^2\right )^2}+\frac{\left (35 b^2\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{24 a^2}\\ &=\frac{1}{6 a x \left (a+b x^2\right )^3}+\frac{7}{24 a^2 x \left (a+b x^2\right )^2}+\frac{35}{48 a^3 x \left (a+b x^2\right )}+\frac{(35 b) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{16 a^3}\\ &=-\frac{35}{16 a^4 x}+\frac{1}{6 a x \left (a+b x^2\right )^3}+\frac{7}{24 a^2 x \left (a+b x^2\right )^2}+\frac{35}{48 a^3 x \left (a+b x^2\right )}-\frac{\left (35 b^2\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{16 a^4}\\ &=-\frac{35}{16 a^4 x}+\frac{1}{6 a x \left (a+b x^2\right )^3}+\frac{7}{24 a^2 x \left (a+b x^2\right )^2}+\frac{35}{48 a^3 x \left (a+b x^2\right )}-\frac{35 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0427745, size = 79, normalized size = 0.83 \[ -\frac{231 a^2 b x^2+48 a^3+280 a b^2 x^4+105 b^3 x^6}{48 a^4 x \left (a+b x^2\right )^3}-\frac{35 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(48*a^3 + 231*a^2*b*x^2 + 280*a*b^2*x^4 + 105*b^3*x^6)/(48*a^4*x*(a + b*x^2)^3) - (35*Sqrt[b]*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(16*a^(9/2))

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Maple [A]  time = 0.055, size = 86, normalized size = 0.9 \begin{align*} -{\frac{1}{{a}^{4}x}}-{\frac{19\,{b}^{3}{x}^{5}}{16\,{a}^{4} \left ( b{x}^{2}+a \right ) ^{3}}}-{\frac{17\,{b}^{2}{x}^{3}}{6\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{3}}}-{\frac{29\,bx}{16\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{3}}}-{\frac{35\,b}{16\,{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^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

-1/a^4/x-19/16/a^4*b^3/(b*x^2+a)^3*x^5-17/6/a^3*b^2/(b*x^2+a)^3*x^3-29/16/a^2*b/(b*x^2+a)^3*x-35/16/a^4*b/(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86338, size = 574, normalized size = 6.04 \begin{align*} \left [-\frac{210 \, b^{3} x^{6} + 560 \, a b^{2} x^{4} + 462 \, a^{2} b x^{2} + 96 \, a^{3} - 105 \,{\left (b^{3} x^{7} + 3 \, a b^{2} x^{5} + 3 \, a^{2} b x^{3} + a^{3} 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 )}{96 \,{\left (a^{4} b^{3} x^{7} + 3 \, a^{5} b^{2} x^{5} + 3 \, a^{6} b x^{3} + a^{7} x\right )}}, -\frac{105 \, b^{3} x^{6} + 280 \, a b^{2} x^{4} + 231 \, a^{2} b x^{2} + 48 \, a^{3} + 105 \,{\left (b^{3} x^{7} + 3 \, a b^{2} x^{5} + 3 \, a^{2} b x^{3} + a^{3} x\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{48 \,{\left (a^{4} b^{3} x^{7} + 3 \, a^{5} b^{2} x^{5} + 3 \, a^{6} b x^{3} + a^{7} 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)^2,x, algorithm="fricas")

[Out]

[-1/96*(210*b^3*x^6 + 560*a*b^2*x^4 + 462*a^2*b*x^2 + 96*a^3 - 105*(b^3*x^7 + 3*a*b^2*x^5 + 3*a^2*b*x^3 + a^3*
x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^4*b^3*x^7 + 3*a^5*b^2*x^5 + 3*a^6*b*x^3 + a^
7*x), -1/48*(105*b^3*x^6 + 280*a*b^2*x^4 + 231*a^2*b*x^2 + 48*a^3 + 105*(b^3*x^7 + 3*a*b^2*x^5 + 3*a^2*b*x^3 +
a^3*x)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^4*b^3*x^7 + 3*a^5*b^2*x^5 + 3*a^6*b*x^3 + a^7*x)]

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Sympy [A]  time = 0.976444, size = 138, normalized size = 1.45 \begin{align*} \frac{35 \sqrt{- \frac{b}{a^{9}}} \log{\left (- \frac{a^{5} \sqrt{- \frac{b}{a^{9}}}}{b} + x \right )}}{32} - \frac{35 \sqrt{- \frac{b}{a^{9}}} \log{\left (\frac{a^{5} \sqrt{- \frac{b}{a^{9}}}}{b} + x \right )}}{32} - \frac{48 a^{3} + 231 a^{2} b x^{2} + 280 a b^{2} x^{4} + 105 b^{3} x^{6}}{48 a^{7} x + 144 a^{6} b x^{3} + 144 a^{5} b^{2} x^{5} + 48 a^{4} b^{3} x^{7}} \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)**2,x)

[Out]

35*sqrt(-b/a**9)*log(-a**5*sqrt(-b/a**9)/b + x)/32 - 35*sqrt(-b/a**9)*log(a**5*sqrt(-b/a**9)/b + x)/32 - (48*a
**3 + 231*a**2*b*x**2 + 280*a*b**2*x**4 + 105*b**3*x**6)/(48*a**7*x + 144*a**6*b*x**3 + 144*a**5*b**2*x**5 + 4
8*a**4*b**3*x**7)

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Giac [A]  time = 1.11721, size = 92, normalized size = 0.97 \begin{align*} -\frac{35 \, b \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{16 \, \sqrt{a b} a^{4}} - \frac{1}{a^{4} x} - \frac{57 \, b^{3} x^{5} + 136 \, a b^{2} x^{3} + 87 \, a^{2} b x}{48 \,{\left (b x^{2} + a\right )}^{3} a^{4}} \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)^2,x, algorithm="giac")

[Out]

-35/16*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/(a^4*x) - 1/48*(57*b^3*x^5 + 136*a*b^2*x^3 + 87*a^2*b*x)/((
b*x^2 + a)^3*a^4)