\(\int \frac {x^3}{(a+c x^4)^3} \, dx\) [676]

   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 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8 c \left (a+c x^4\right )^2} \]

[Out]

-1/8/c/(c*x^4+a)^2

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} \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8 c \left (a+c x^4\right )^2} \]

[In]

Int[x^3/(a + c*x^4)^3,x]

[Out]

-1/8*1/(c*(a + c*x^4)^2)

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}{8 c \left (a+c x^4\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8 c \left (a+c x^4\right )^2} \]

[In]

Integrate[x^3/(a + c*x^4)^3,x]

[Out]

-1/8*1/(c*(a + c*x^4)^2)

Maple [A] (verified)

Time = 3.86 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method result size
gosper \(-\frac {1}{8 c \left (x^{4} c +a \right )^{2}}\) \(15\)
derivativedivides \(-\frac {1}{8 c \left (x^{4} c +a \right )^{2}}\) \(15\)
default \(-\frac {1}{8 c \left (x^{4} c +a \right )^{2}}\) \(15\)
norman \(-\frac {1}{8 c \left (x^{4} c +a \right )^{2}}\) \(15\)
risch \(-\frac {1}{8 c \left (x^{4} c +a \right )^{2}}\) \(15\)
parallelrisch \(-\frac {1}{8 c \left (x^{4} c +a \right )^{2}}\) \(15\)

[In]

int(x^3/(c*x^4+a)^3,x,method=_RETURNVERBOSE)

[Out]

-1/8/c/(c*x^4+a)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8 \, {\left (c^{3} x^{8} + 2 \, a c^{2} x^{4} + a^{2} c\right )}} \]

[In]

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

[Out]

-1/8/(c^3*x^8 + 2*a*c^2*x^4 + a^2*c)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=- \frac {1}{8 a^{2} c + 16 a c^{2} x^{4} + 8 c^{3} x^{8}} \]

[In]

integrate(x**3/(c*x**4+a)**3,x)

[Out]

-1/(8*a**2*c + 16*a*c**2*x**4 + 8*c**3*x**8)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8 \, {\left (c x^{4} + a\right )}^{2} c} \]

[In]

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

[Out]

-1/8/((c*x^4 + a)^2*c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8 \, {\left (c x^{4} + a\right )}^{2} c} \]

[In]

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

[Out]

-1/8/((c*x^4 + a)^2*c)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {x^3}{\left (a+c x^4\right )^3} \, dx=-\frac {1}{8\,c\,\left (a^2+2\,a\,c\,x^4+c^2\,x^8\right )} \]

[In]

int(x^3/(a + c*x^4)^3,x)

[Out]

-1/(8*c*(a^2 + c^2*x^8 + 2*a*c*x^4))