\(\int \frac {(b x^2+c x^4)^2}{x^3} \, dx\) [148]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {\left (b+c x^2\right )^3}{6 c} \]

[Out]

1/6*(c*x^2+b)^3/c

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} \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {\left (b+c x^2\right )^3}{6 c} \]

[In]

Int[(b*x^2 + c*x^4)^2/x^3,x]

[Out]

(b + c*x^2)^3/(6*c)

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 x \left (b+c x^2\right )^2 \, dx \\ & = \frac {\left (b+c x^2\right )^3}{6 c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {\left (b+c x^2\right )^3}{6 c} \]

[In]

Integrate[(b*x^2 + c*x^4)^2/x^3,x]

[Out]

(b + c*x^2)^3/(6*c)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (c \,x^{2}+b \right )^{3}}{6 c}\)	\(15\)
parallelrisch	\(\frac {1}{6} c^{2} x^{6}+\frac {1}{2} b c \,x^{4}+\frac {1}{2} b^{2} x^{2}\)	\(25\)
gosper	\(\frac {x^{2} \left (c^{2} x^{4}+3 b c \,x^{2}+3 b^{2}\right )}{6}\)	\(26\)
norman	\(\frac {\frac {1}{2} b^{2} x^{4}+\frac {1}{6} c^{2} x^{8}+\frac {1}{2} b c \,x^{6}}{x^{2}}\)	\(29\)
risch	\(\frac {c^{2} x^{6}}{6}+\frac {b c \,x^{4}}{2}+\frac {b^{2} x^{2}}{2}+\frac {b^{3}}{6 c}\)	\(33\)

[In]

int((c*x^4+b*x^2)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

1/6*(c*x^2+b)^3/c

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {1}{6} \, c^{2} x^{6} + \frac {1}{2} \, b c x^{4} + \frac {1}{2} \, b^{2} x^{2} \]

[In]

integrate((c*x^4+b*x^2)^2/x^3,x, algorithm="fricas")

[Out]

1/6*c^2*x^6 + 1/2*b*c*x^4 + 1/2*b^2*x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 24 vs. \(2 (10) = 20\).

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {b^{2} x^{2}}{2} + \frac {b c x^{4}}{2} + \frac {c^{2} x^{6}}{6} \]

[In]

integrate((c*x**4+b*x**2)**2/x**3,x)

[Out]

b**2*x**2/2 + b*c*x**4/2 + c**2*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {1}{6} \, c^{2} x^{6} + \frac {1}{2} \, b c x^{4} + \frac {1}{2} \, b^{2} x^{2} \]

[In]

integrate((c*x^4+b*x^2)^2/x^3,x, algorithm="maxima")

[Out]

1/6*c^2*x^6 + 1/2*b*c*x^4 + 1/2*b^2*x^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {1}{6} \, c^{2} x^{6} + \frac {1}{2} \, b c x^{4} + \frac {1}{2} \, b^{2} x^{2} \]

[In]

integrate((c*x^4+b*x^2)^2/x^3,x, algorithm="giac")

[Out]

1/6*c^2*x^6 + 1/2*b*c*x^4 + 1/2*b^2*x^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^3} \, dx=\frac {b^2\,x^2}{2}+\frac {b\,c\,x^4}{2}+\frac {c^2\,x^6}{6} \]

[In]

int((b*x^2 + c*x^4)^2/x^3,x)

[Out]

(b^2*x^2)/2 + (c^2*x^6)/6 + (b*c*x^4)/2