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

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

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.04, size = 79, normalized size = 0.83 \begin {gather*} -\frac {35 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{9/2}}-\frac {48 a^3+231 a^2 b x^2+280 a b^2 x^4+105 b^3 x^6}{48 a^4 x \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/48*(48*a^3 + 231*a^2*b*x^2 + 280*a*b^2*x^4 + 105*b^3*x^6)/(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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 268, normalized size = 2.82 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.16, size = 68, normalized size = 0.72 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.01, size = 86, normalized size = 0.91 \begin {gather*} -\frac {19 b^{3} x^{5}}{16 \left (b \,x^{2}+a \right )^{3} a^{4}}-\frac {17 b^{2} x^{3}}{6 \left (b \,x^{2}+a \right )^{3} a^{3}}-\frac {29 b x}{16 \left (b \,x^{2}+a \right )^{3} a^{2}}-\frac {35 b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}\, a^{4}}-\frac {1}{a^{4} x} \end {gather*}

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*b^3/a^4/(b*x^2+a)^3*x^5-17/6*b^2/a^3/(b*x^2+a)^3*x^3-29/16*b/a^2/(b*x^2+a)^3*x-35/16*b/a^4/(a*b
)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 3.03, size = 93, normalized size = 0.98 \begin {gather*} -\frac {105 \, b^{3} x^{6} + 280 \, a b^{2} x^{4} + 231 \, a^{2} b x^{2} + 48 \, a^{3}}{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 )}} - \frac {35 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{4}} \end {gather*}

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]

-1/48*(105*b^3*x^6 + 280*a*b^2*x^4 + 231*a^2*b*x^2 + 48*a^3)/(a^4*b^3*x^7 + 3*a^5*b^2*x^5 + 3*a^6*b*x^3 + a^7*
x) - 35/16*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4)

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mupad [B]  time = 4.44, size = 88, normalized size = 0.93 \begin {gather*} -\frac {\frac {1}{a}+\frac {77\,b\,x^2}{16\,a^2}+\frac {35\,b^2\,x^4}{6\,a^3}+\frac {35\,b^3\,x^6}{16\,a^4}}{a^3\,x+3\,a^2\,b\,x^3+3\,a\,b^2\,x^5+b^3\,x^7}-\frac {35\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (1/a + (77*b*x^2)/(16*a^2) + (35*b^2*x^4)/(6*a^3) + (35*b^3*x^6)/(16*a^4))/(a^3*x + b^3*x^7 + 3*a^2*b*x^3 +
3*a*b^2*x^5) - (35*b^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(16*a^(9/2))

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sympy [A]  time = 0.58, size = 139, normalized size = 1.46 \begin {gather*} \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 {gather*}

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 +
48*a**4*b**3*x**7)

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