3.1981 \(\int \frac {x^2}{(a+\frac {b}{x^3})^2} \, dx\)

Optimal. Leaf size=46 \[ -\frac {b^2}{3 a^3 \left (a x^3+b\right )}-\frac {2 b \log \left (a x^3+b\right )}{3 a^3}+\frac {x^3}{3 a^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {263, 266, 43} \[ -\frac {b^2}{3 a^3 \left (a x^3+b\right )}-\frac {2 b \log \left (a x^3+b\right )}{3 a^3}+\frac {x^3}{3 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/x^3)^2,x]

[Out]

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

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^2}{\left (a+\frac {b}{x^3}\right )^2} \, dx &=\int \frac {x^8}{\left (b+a x^3\right )^2} \, dx\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(b+a x)^2} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {1}{a^2}+\frac {b^2}{a^2 (b+a x)^2}-\frac {2 b}{a^2 (b+a x)}\right ) \, dx,x,x^3\right )\\ &=\frac {x^3}{3 a^2}-\frac {b^2}{3 a^3 \left (b+a x^3\right )}-\frac {2 b \log \left (b+a x^3\right )}{3 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.83 \[ \frac {-\frac {b^2}{a x^3+b}-2 b \log \left (a x^3+b\right )+a x^3}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/x^3)^2,x]

[Out]

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

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fricas [A]  time = 0.97, size = 56, normalized size = 1.22 \[ \frac {a^{2} x^{6} + a b x^{3} - b^{2} - 2 \, {\left (a b x^{3} + b^{2}\right )} \log \left (a x^{3} + b\right )}{3 \, {\left (a^{4} x^{3} + a^{3} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 41, normalized size = 0.89 \[ \frac {x^{3}}{3 \, a^{2}} - \frac {2 \, b \log \left ({\left | a x^{3} + b \right |}\right )}{3 \, a^{3}} - \frac {b^{2}}{3 \, {\left (a x^{3} + b\right )} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 41, normalized size = 0.89 \[ \frac {x^{3}}{3 a^{2}}-\frac {b^{2}}{3 \left (a \,x^{3}+b \right ) a^{3}}-\frac {2 b \ln \left (a \,x^{3}+b \right )}{3 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.90, size = 43, normalized size = 0.93 \[ -\frac {b^{2}}{3 \, {\left (a^{4} x^{3} + a^{3} b\right )}} + \frac {x^{3}}{3 \, a^{2}} - \frac {2 \, b \log \left (a x^{3} + b\right )}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 45, normalized size = 0.98 \[ \frac {x^3}{3\,a^2}-\frac {b^2}{3\,\left (a^4\,x^3+b\,a^3\right )}-\frac {2\,b\,\ln \left (a\,x^3+b\right )}{3\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/x^3)^2,x)

[Out]

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

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sympy [A]  time = 0.37, size = 42, normalized size = 0.91 \[ - \frac {b^{2}}{3 a^{4} x^{3} + 3 a^{3} b} + \frac {x^{3}}{3 a^{2}} - \frac {2 b \log {\left (a x^{3} + b \right )}}{3 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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