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

Optimal. Leaf size=106 \[ \frac{105 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{11/2}}+\frac{21}{16 a^3 x^3 \left (a+b x^2\right )}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{105 b}{16 a^5 x}-\frac{35}{16 a^4 x^3}+\frac{1}{6 a x^3 \left (a+b x^2\right )^3} \]

[Out]

-35/(16*a^4*x^3) + (105*b)/(16*a^5*x) + 1/(6*a*x^3*(a + b*x^2)^3) + 3/(8*a^2*x^3*(a + b*x^2)^2) + 21/(16*a^3*x
^3*(a + b*x^2)) + (105*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*a^(11/2))

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Rubi [A]  time = 0.0661874, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {28, 290, 325, 205} \[ \frac{105 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{11/2}}+\frac{21}{16 a^3 x^3 \left (a+b x^2\right )}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{105 b}{16 a^5 x}-\frac{35}{16 a^4 x^3}+\frac{1}{6 a x^3 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-35/(16*a^4*x^3) + (105*b)/(16*a^5*x) + 1/(6*a*x^3*(a + b*x^2)^3) + 3/(8*a^2*x^3*(a + b*x^2)^2) + 21/(16*a^3*x
^3*(a + b*x^2)) + (105*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*a^(11/2))

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 )^2} \, dx &=b^4 \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac{1}{6 a x^3 \left (a+b x^2\right )^3}+\frac{\left (3 b^3\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^3} \, dx}{2 a}\\ &=\frac{1}{6 a x^3 \left (a+b x^2\right )^3}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{\left (21 b^2\right ) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )^2} \, dx}{8 a^2}\\ &=\frac{1}{6 a x^3 \left (a+b x^2\right )^3}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{21}{16 a^3 x^3 \left (a+b x^2\right )}+\frac{(105 b) \int \frac{1}{x^4 \left (a b+b^2 x^2\right )} \, dx}{16 a^3}\\ &=-\frac{35}{16 a^4 x^3}+\frac{1}{6 a x^3 \left (a+b x^2\right )^3}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{21}{16 a^3 x^3 \left (a+b x^2\right )}-\frac{\left (105 b^2\right ) \int \frac{1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{16 a^4}\\ &=-\frac{35}{16 a^4 x^3}+\frac{105 b}{16 a^5 x}+\frac{1}{6 a x^3 \left (a+b x^2\right )^3}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{21}{16 a^3 x^3 \left (a+b x^2\right )}+\frac{\left (105 b^3\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{16 a^5}\\ &=-\frac{35}{16 a^4 x^3}+\frac{105 b}{16 a^5 x}+\frac{1}{6 a x^3 \left (a+b x^2\right )^3}+\frac{3}{8 a^2 x^3 \left (a+b x^2\right )^2}+\frac{21}{16 a^3 x^3 \left (a+b x^2\right )}+\frac{105 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0474373, size = 91, normalized size = 0.86 \[ \frac{\frac{\sqrt{a} \left (693 a^2 b^2 x^4+144 a^3 b x^2-16 a^4+840 a b^3 x^6+315 b^4 x^8\right )}{x^3 \left (a+b x^2\right )^3}+315 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{48 a^{11/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*(-16*a^4 + 144*a^3*b*x^2 + 693*a^2*b^2*x^4 + 840*a*b^3*x^6 + 315*b^4*x^8))/(x^3*(a + b*x^2)^3) + 315
*b^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(48*a^(11/2))

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Maple [A]  time = 0.056, size = 99, normalized size = 0.9 \begin{align*} -{\frac{1}{3\,{a}^{4}{x}^{3}}}+4\,{\frac{b}{{a}^{5}x}}+{\frac{41\,{b}^{4}{x}^{5}}{16\,{a}^{5} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{35\,{b}^{3}{x}^{3}}{6\,{a}^{4} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{55\,{b}^{2}x}{16\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{105\,{b}^{2}}{16\,{a}^{5}}\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)^2,x)

[Out]

-1/3/a^4/x^3+4*b/a^5/x+41/16/a^5*b^4/(b*x^2+a)^3*x^5+35/6/a^4*b^3/(b*x^2+a)^3*x^3+55/16/a^3*b^2/(b*x^2+a)^3*x+
105/16/a^5*b^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^4/(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 = 1.76567, size = 644, normalized size = 6.08 \begin{align*} \left [\frac{630 \, b^{4} x^{8} + 1680 \, a b^{3} x^{6} + 1386 \, a^{2} b^{2} x^{4} + 288 \, a^{3} b x^{2} - 32 \, a^{4} + 315 \,{\left (b^{4} x^{9} + 3 \, a b^{3} x^{7} + 3 \, a^{2} b^{2} x^{5} + a^{3} 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 )}{96 \,{\left (a^{5} b^{3} x^{9} + 3 \, a^{6} b^{2} x^{7} + 3 \, a^{7} b x^{5} + a^{8} x^{3}\right )}}, \frac{315 \, b^{4} x^{8} + 840 \, a b^{3} x^{6} + 693 \, a^{2} b^{2} x^{4} + 144 \, a^{3} b x^{2} - 16 \, a^{4} + 315 \,{\left (b^{4} x^{9} + 3 \, a b^{3} x^{7} + 3 \, a^{2} b^{2} x^{5} + a^{3} b x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{48 \,{\left (a^{5} b^{3} x^{9} + 3 \, a^{6} b^{2} x^{7} + 3 \, a^{7} b x^{5} + a^{8} 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)^2,x, algorithm="fricas")

[Out]

[1/96*(630*b^4*x^8 + 1680*a*b^3*x^6 + 1386*a^2*b^2*x^4 + 288*a^3*b*x^2 - 32*a^4 + 315*(b^4*x^9 + 3*a*b^3*x^7 +
3*a^2*b^2*x^5 + a^3*b*x^3)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^5*b^3*x^9 + 3*a^6*b
^2*x^7 + 3*a^7*b*x^5 + a^8*x^3), 1/48*(315*b^4*x^8 + 840*a*b^3*x^6 + 693*a^2*b^2*x^4 + 144*a^3*b*x^2 - 16*a^4
+ 315*(b^4*x^9 + 3*a*b^3*x^7 + 3*a^2*b^2*x^5 + a^3*b*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^5*b^3*x^9 + 3*a^6*
b^2*x^7 + 3*a^7*b*x^5 + a^8*x^3)]

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Sympy [A]  time = 1.62734, size = 162, normalized size = 1.53 \begin{align*} - \frac{105 \sqrt{- \frac{b^{3}}{a^{11}}} \log{\left (- \frac{a^{6} \sqrt{- \frac{b^{3}}{a^{11}}}}{b^{2}} + x \right )}}{32} + \frac{105 \sqrt{- \frac{b^{3}}{a^{11}}} \log{\left (\frac{a^{6} \sqrt{- \frac{b^{3}}{a^{11}}}}{b^{2}} + x \right )}}{32} + \frac{- 16 a^{4} + 144 a^{3} b x^{2} + 693 a^{2} b^{2} x^{4} + 840 a b^{3} x^{6} + 315 b^{4} x^{8}}{48 a^{8} x^{3} + 144 a^{7} b x^{5} + 144 a^{6} b^{2} x^{7} + 48 a^{5} b^{3} x^{9}} \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)**2,x)

[Out]

-105*sqrt(-b**3/a**11)*log(-a**6*sqrt(-b**3/a**11)/b**2 + x)/32 + 105*sqrt(-b**3/a**11)*log(a**6*sqrt(-b**3/a*
*11)/b**2 + x)/32 + (-16*a**4 + 144*a**3*b*x**2 + 693*a**2*b**2*x**4 + 840*a*b**3*x**6 + 315*b**4*x**8)/(48*a*
*8*x**3 + 144*a**7*b*x**5 + 144*a**6*b**2*x**7 + 48*a**5*b**3*x**9)

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Giac [A]  time = 1.15613, size = 111, normalized size = 1.05 \begin{align*} \frac{105 \, b^{2} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{16 \, \sqrt{a b} a^{5}} + \frac{315 \, b^{4} x^{8} + 840 \, a b^{3} x^{6} + 693 \, a^{2} b^{2} x^{4} + 144 \, a^{3} b x^{2} - 16 \, a^{4}}{48 \,{\left (b x^{3} + a x\right )}^{3} a^{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)^2,x, algorithm="giac")

[Out]

105/16*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/48*(315*b^4*x^8 + 840*a*b^3*x^6 + 693*a^2*b^2*x^4 + 144*a
^3*b*x^2 - 16*a^4)/((b*x^3 + a*x)^3*a^5)