\(\int \frac {x^2}{(a+b x^3)^2} \, dx\) [69]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3 b \left (a+b x^3\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3 b \left (a+b x^3\right )} \]

[In]

Int[x^2/(a + b*x^3)^2,x]

[Out]

-1/3*1/(b*(a + b*x^3))

Rule 267

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

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{3 b \left (a+b x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3 b \left (a+b x^3\right )} \]

[In]

Integrate[x^2/(a + b*x^3)^2,x]

[Out]

-1/3*1/(b*(a + b*x^3))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method result size
gosper \(-\frac {1}{3 b \left (b \,x^{3}+a \right )}\) \(15\)
derivativedivides \(-\frac {1}{3 b \left (b \,x^{3}+a \right )}\) \(15\)
default \(-\frac {1}{3 b \left (b \,x^{3}+a \right )}\) \(15\)
norman \(-\frac {1}{3 b \left (b \,x^{3}+a \right )}\) \(15\)
risch \(-\frac {1}{3 b \left (b \,x^{3}+a \right )}\) \(15\)
parallelrisch \(-\frac {1}{3 b \left (b \,x^{3}+a \right )}\) \(15\)

[In]

int(x^2/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3 \, {\left (b^{2} x^{3} + a b\right )}} \]

[In]

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

[Out]

-1/3/(b^2*x^3 + a*b)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=- \frac {1}{3 a b + 3 b^{2} x^{3}} \]

[In]

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

[Out]

-1/(3*a*b + 3*b**2*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3 \, {\left (b x^{3} + a\right )} b} \]

[In]

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

[Out]

-1/3/((b*x^3 + a)*b)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3 \, {\left (b x^{3} + a\right )} b} \]

[In]

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

[Out]

-1/3/((b*x^3 + a)*b)

Mupad [B] (verification not implemented)

Time = 14.42 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^2}{\left (a+b x^3\right )^2} \, dx=-\frac {1}{3\,b\,\left (b\,x^3+a\right )} \]

[In]

int(x^2/(a + b*x^3)^2,x)

[Out]

-1/(3*b*(a + b*x^3))