3.1878 \(\int \frac {x}{(a+\frac {b}{x^2})^3} \, dx\)

Optimal. Leaf size=65 \[ \frac {b^3}{4 a^4 \left (a x^2+b\right )^2}-\frac {3 b^2}{2 a^4 \left (a x^2+b\right )}-\frac {3 b \log \left (a x^2+b\right )}{2 a^4}+\frac {x^2}{2 a^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {263, 266, 43} \[ \frac {b^3}{4 a^4 \left (a x^2+b\right )^2}-\frac {3 b^2}{2 a^4 \left (a x^2+b\right )}-\frac {3 b \log \left (a x^2+b\right )}{2 a^4}+\frac {x^2}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x^2)^3,x]

[Out]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 263

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

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 {x}{\left (a+\frac {b}{x^2}\right )^3} \, dx &=\int \frac {x^7}{\left (b+a x^2\right )^3} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{(b+a x)^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{a^3}-\frac {b^3}{a^3 (b+a x)^3}+\frac {3 b^2}{a^3 (b+a x)^2}-\frac {3 b}{a^3 (b+a x)}\right ) \, dx,x,x^2\right )\\ &=\frac {x^2}{2 a^3}+\frac {b^3}{4 a^4 \left (b+a x^2\right )^2}-\frac {3 b^2}{2 a^4 \left (b+a x^2\right )}-\frac {3 b \log \left (b+a x^2\right )}{2 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 48, normalized size = 0.74 \[ -\frac {\frac {b^2 \left (6 a x^2+5 b\right )}{\left (a x^2+b\right )^2}+6 b \log \left (a x^2+b\right )-2 a x^2}{4 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x^2)^3,x]

[Out]

-1/4*(-2*a*x^2 + (b^2*(5*b + 6*a*x^2))/(b + a*x^2)^2 + 6*b*Log[b + a*x^2])/a^4

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fricas [A]  time = 0.99, size = 91, normalized size = 1.40 \[ \frac {2 \, a^{3} x^{6} + 4 \, a^{2} b x^{4} - 4 \, a b^{2} x^{2} - 5 \, b^{3} - 6 \, {\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \log \left (a x^{2} + b\right )}{4 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 53, normalized size = 0.82 \[ \frac {x^{2}}{2 \, a^{3}} - \frac {3 \, b \log \left ({\left | a x^{2} + b \right |}\right )}{2 \, a^{4}} - \frac {6 \, a b^{2} x^{2} + 5 \, b^{3}}{4 \, {\left (a x^{2} + b\right )}^{2} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 58, normalized size = 0.89 \[ \frac {x^{2}}{2 a^{3}}+\frac {b^{3}}{4 \left (a \,x^{2}+b \right )^{2} a^{4}}-\frac {3 b^{2}}{2 \left (a \,x^{2}+b \right ) a^{4}}-\frac {3 b \ln \left (a \,x^{2}+b \right )}{2 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 66, normalized size = 1.02 \[ -\frac {6 \, a b^{2} x^{2} + 5 \, b^{3}}{4 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}} + \frac {x^{2}}{2 \, a^{3}} - \frac {3 \, b \log \left (a x^{2} + b\right )}{2 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.11, size = 68, normalized size = 1.05 \[ \frac {x^2}{2\,a^3}-\frac {\frac {5\,b^3}{4\,a}+\frac {3\,b^2\,x^2}{2}}{a^5\,x^4+2\,a^4\,b\,x^2+a^3\,b^2}-\frac {3\,b\,\ln \left (a\,x^2+b\right )}{2\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b/x^2)^3,x)

[Out]

x^2/(2*a^3) - ((5*b^3)/(4*a) + (3*b^2*x^2)/2)/(a^3*b^2 + a^5*x^4 + 2*a^4*b*x^2) - (3*b*log(b + a*x^2))/(2*a^4)

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sympy [A]  time = 0.40, size = 68, normalized size = 1.05 \[ \frac {- 6 a b^{2} x^{2} - 5 b^{3}}{4 a^{6} x^{4} + 8 a^{5} b x^{2} + 4 a^{4} b^{2}} + \frac {x^{2}}{2 a^{3}} - \frac {3 b \log {\left (a x^{2} + b \right )}}{2 a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-6*a*b**2*x**2 - 5*b**3)/(4*a**6*x**4 + 8*a**5*b*x**2 + 4*a**4*b**2) + x**2/(2*a**3) - 3*b*log(a*x**2 + b)/(2
*a**4)

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