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

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

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Rubi [A]  time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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])

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]]

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*}

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Mathematica [A]  time = 0.04, size = 41, normalized size = 0.84 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.01, size = 73, normalized size = 1.49 \begin {gather*} -\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 \relax (x)}{2 \, {\left (a^{3} b x^{4} + a^{4} x^{2}\right )}} \end {gather*}

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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giac [A]  time = 0.16, size = 51, normalized size = 1.04 \begin {gather*} -\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 {gather*}

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)

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maple [A]  time = 0.01, size = 46, normalized size = 0.94 \begin {gather*} -\frac {b}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {2 b \ln \relax (x )}{a^{3}}+\frac {b \ln \left (b \,x^{2}+a \right )}{a^{3}}-\frac {1}{2 a^{2} x^{2}} \end {gather*}

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 = 1.36, size = 52, normalized size = 1.06 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.08, size = 51, normalized size = 1.04 \begin {gather*} \frac {b\,\ln \left (b\,x^2+a\right )}{a^3}-\frac {\frac {1}{2\,a}+\frac {b\,x^2}{a^2}}{b\,x^4+a\,x^2}-\frac {2\,b\,\ln \relax (x)}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.39, size = 51, normalized size = 1.04 \begin {gather*} \frac {- a - 2 b x^{2}}{2 a^{3} x^{2} + 2 a^{2} b x^{4}} - \frac {2 b \log {\relax (x )}}{a^{3}} + \frac {b \log {\left (\frac {a}{b} + x^{2} \right )}}{a^{3}} \end {gather*}

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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