\(\int \frac {x^5}{(b x^2+c x^4)^2} \, dx\) [197]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2 c \left (b+c x^2\right )} \]

[Out]

-1/2/c/(c*x^2+b)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 267} \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2 c \left (b+c x^2\right )} \]

[In]

Int[x^5/(b*x^2 + c*x^4)^2,x]

[Out]

-1/2*1/(c*(b + c*x^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]

Rule 1598

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x}{\left (b+c x^2\right )^2} \, dx \\ & = -\frac {1}{2 c \left (b+c x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2 c \left (b+c x^2\right )} \]

[In]

Integrate[x^5/(b*x^2 + c*x^4)^2,x]

[Out]

-1/2*1/(c*(b + c*x^2))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

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

[In]

int(x^5/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/c/(c*x^2+b)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2 \, {\left (c^{2} x^{2} + b c\right )}} \]

[In]

integrate(x^5/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

-1/2/(c^2*x^2 + b*c)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=- \frac {1}{2 b c + 2 c^{2} x^{2}} \]

[In]

integrate(x**5/(c*x**4+b*x**2)**2,x)

[Out]

-1/(2*b*c + 2*c**2*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2 \, {\left (c^{2} x^{2} + b c\right )}} \]

[In]

integrate(x^5/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

-1/2/(c^2*x^2 + b*c)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2 \, {\left (c x^{2} + b\right )} c} \]

[In]

integrate(x^5/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

-1/2/((c*x^2 + b)*c)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {x^5}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {1}{2\,c\,\left (c\,x^2+b\right )} \]

[In]

int(x^5/(b*x^2 + c*x^4)^2,x)

[Out]

-1/(2*c*(b + c*x^2))