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

Optimal. Leaf size=70 \[ \frac{1}{2 a^3 \left (a+b x^2\right )}+\frac{1}{4 a^2 \left (a+b x^2\right )^2}-\frac{\log \left (a+b x^2\right )}{2 a^4}+\frac{\log (x)}{a^4}+\frac{1}{6 a \left (a+b x^2\right )^3} \]

[Out]

1/(6*a*(a + b*x^2)^3) + 1/(4*a^2*(a + b*x^2)^2) + 1/(2*a^3*(a + b*x^2)) + Log[x]/a^4 - Log[a + b*x^2]/(2*a^4)

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Rubi [A]  time = 0.0758043, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {28, 266, 44} \[ \frac{1}{2 a^3 \left (a+b x^2\right )}+\frac{1}{4 a^2 \left (a+b x^2\right )^2}-\frac{\log \left (a+b x^2\right )}{2 a^4}+\frac{\log (x)}{a^4}+\frac{1}{6 a \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(6*a*(a + b*x^2)^3) + 1/(4*a^2*(a + b*x^2)^2) + 1/(2*a^3*(a + b*x^2)) + Log[x]/a^4 - Log[a + b*x^2]/(2*a^4)

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac{1}{x \left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac{1}{2} b^4 \operatorname{Subst}\left (\int \frac{1}{x \left (a b+b^2 x\right )^4} \, dx,x,x^2\right )\\ &=\frac{1}{2} b^4 \operatorname{Subst}\left (\int \left (\frac{1}{a^4 b^4 x}-\frac{1}{a b^3 (a+b x)^4}-\frac{1}{a^2 b^3 (a+b x)^3}-\frac{1}{a^3 b^3 (a+b x)^2}-\frac{1}{a^4 b^3 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac{1}{6 a \left (a+b x^2\right )^3}+\frac{1}{4 a^2 \left (a+b x^2\right )^2}+\frac{1}{2 a^3 \left (a+b x^2\right )}+\frac{\log (x)}{a^4}-\frac{\log \left (a+b x^2\right )}{2 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0355674, size = 54, normalized size = 0.77 \[ \frac{\frac{a \left (11 a^2+15 a b x^2+6 b^2 x^4\right )}{\left (a+b x^2\right )^3}-6 \log \left (a+b x^2\right )+12 \log (x)}{12 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(11*a^2 + 15*a*b*x^2 + 6*b^2*x^4))/(a + b*x^2)^3 + 12*Log[x] - 6*Log[a + b*x^2])/(12*a^4)

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Maple [A]  time = 0.054, size = 63, normalized size = 0.9 \begin{align*}{\frac{1}{6\,a \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{1}{4\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{1}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{\ln \left ( x \right ) }{{a}^{4}}}-{\frac{\ln \left ( b{x}^{2}+a \right ) }{2\,{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a/(b*x^2+a)^3+1/4/a^2/(b*x^2+a)^2+1/2/a^3/(b*x^2+a)+ln(x)/a^4-1/2*ln(b*x^2+a)/a^4

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Maxima [A]  time = 0.997209, size = 111, normalized size = 1.59 \begin{align*} \frac{6 \, b^{2} x^{4} + 15 \, a b x^{2} + 11 \, a^{2}}{12 \,{\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} - \frac{\log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac{\log \left (x^{2}\right )}{2 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*b^2*x^4 + 15*a*b*x^2 + 11*a^2)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6) - 1/2*log(b*x^2 + a)/
a^4 + 1/2*log(x^2)/a^4

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Fricas [B]  time = 1.92034, size = 288, normalized size = 4.11 \begin{align*} \frac{6 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 11 \, a^{3} - 6 \,{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right ) + 12 \,{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (x\right )}{12 \,{\left (a^{4} b^{3} x^{6} + 3 \, a^{5} b^{2} x^{4} + 3 \, a^{6} b x^{2} + a^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a*b^2*x^4 + 15*a^2*b*x^2 + 11*a^3 - 6*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*log(b*x^2 + a) + 12*
(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*log(x))/(a^4*b^3*x^6 + 3*a^5*b^2*x^4 + 3*a^6*b*x^2 + a^7)

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Sympy [A]  time = 0.811277, size = 80, normalized size = 1.14 \begin{align*} \frac{11 a^{2} + 15 a b x^{2} + 6 b^{2} x^{4}}{12 a^{6} + 36 a^{5} b x^{2} + 36 a^{4} b^{2} x^{4} + 12 a^{3} b^{3} x^{6}} + \frac{\log{\left (x \right )}}{a^{4}} - \frac{\log{\left (\frac{a}{b} + x^{2} \right )}}{2 a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(11*a**2 + 15*a*b*x**2 + 6*b**2*x**4)/(12*a**6 + 36*a**5*b*x**2 + 36*a**4*b**2*x**4 + 12*a**3*b**3*x**6) + log
(x)/a**4 - log(a/b + x**2)/(2*a**4)

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Giac [A]  time = 1.14283, size = 95, normalized size = 1.36 \begin{align*} \frac{\log \left (x^{2}\right )}{2 \, a^{4}} - \frac{\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4}} + \frac{11 \, b^{3} x^{6} + 39 \, a b^{2} x^{4} + 48 \, a^{2} b x^{2} + 22 \, a^{3}}{12 \,{\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/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/2*log(x^2)/a^4 - 1/2*log(abs(b*x^2 + a))/a^4 + 1/12*(11*b^3*x^6 + 39*a*b^2*x^4 + 48*a^2*b*x^2 + 22*a^3)/((b*
x^2 + a)^3*a^4)