3.482 \(\int \frac{1}{x^3 (a^2+2 a b x^2+b^2 x^4)} \, dx\)

Optimal. Leaf size=49 \[ -\frac{b}{2 a^2 \left (a+b x^2\right )}+\frac{b \log \left (a+b x^2\right )}{a^3}-\frac{2 b \log (x)}{a^3}-\frac{1}{2 a^2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0489452, antiderivative size = 49, 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} \[ -\frac{b}{2 a^2 \left (a+b x^2\right )}+\frac{b \log \left (a+b x^2\right )}{a^3}-\frac{2 b \log (x)}{a^3}-\frac{1}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

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

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

Mathematica [A]  time = 0.0362084, size = 41, normalized size = 0.84 \[ -\frac{a \left (\frac{b}{a+b x^2}+\frac{1}{x^2}\right )-2 b \log \left (a+b x^2\right )+4 b \log (x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]

[Out]

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

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Maple [A]  time = 0.055, size = 46, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{a}^{2}{x}^{2}}}-{\frac{b}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}-2\,{\frac{b\ln \left ( x \right ) }{{a}^{3}}}+{\frac{b\ln \left ( b{x}^{2}+a \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x)

[Out]

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

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Maxima [A]  time = 0.991222, size = 70, normalized size = 1.43 \begin{align*} -\frac{2 \, b x^{2} + a}{2 \,{\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} + \frac{b \log \left (b x^{2} + a\right )}{a^{3}} - \frac{b \log \left (x^{2}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.7226, size = 157, normalized size = 3.2 \begin{align*} -\frac{2 \, a b x^{2} + a^{2} - 2 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \log \left (x\right )}{2 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.534295, size = 49, normalized size = 1. \begin{align*} - \frac{a + 2 b x^{2}}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} - \frac{2 b \log{\left (x \right )}}{a^{3}} + \frac{b \log{\left (\frac{a}{b} + x^{2} \right )}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b**2*x**4+2*a*b*x**2+a**2),x)

[Out]

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

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Giac [A]  time = 1.16721, size = 69, normalized size = 1.41 \begin{align*} -\frac{b \log \left (x^{2}\right )}{a^{3}} + \frac{b \log \left ({\left | b x^{2} + a \right |}\right )}{a^{3}} - \frac{2 \, b x^{2} + a}{2 \,{\left (b x^{4} + a x^{2}\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")

[Out]

-b*log(x^2)/a^3 + b*log(abs(b*x^2 + a))/a^3 - 1/2*(2*b*x^2 + a)/((b*x^4 + a*x^2)*a^2)