\(\int \frac {(b x^2+c x^4)^2}{x^6} \, dx\) [151]

Optimal result

Integrand size = 17, antiderivative size = 24 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=-\frac {b^2}{x}+2 b c x+\frac {c^2 x^3}{3} \]

[Out]

-b^2/x+2*b*c*x+1/3*c^2*x^3

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {1598, 276} \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=-\frac {b^2}{x}+2 b c x+\frac {c^2 x^3}{3} \]

[In]

Int[(b*x^2 + c*x^4)^2/x^6,x]

[Out]

-(b^2/x) + 2*b*c*x + (c^2*x^3)/3

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=-\frac {b^2}{x}+2 b c x+\frac {c^2 x^3}{3} \]

[In]

Integrate[(b*x^2 + c*x^4)^2/x^6,x]

[Out]

-(b^2/x) + 2*b*c*x + (c^2*x^3)/3

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {b^{2}}{x}+2 b c x +\frac {c^{2} x^{3}}{3}\)	\(23\)
risch	\(-\frac {b^{2}}{x}+2 b c x +\frac {c^{2} x^{3}}{3}\)	\(23\)
parallelrisch	\(\frac {c^{2} x^{4}+6 b c \,x^{2}-3 b^{2}}{3 x}\)	\(26\)
gosper	\(-\frac {-c^{2} x^{4}-6 b c \,x^{2}+3 b^{2}}{3 x}\)	\(27\)
norman	\(\frac {-b^{2} x^{4}+\frac {1}{3} c^{2} x^{8}+2 b c \,x^{6}}{x^{5}}\)	\(29\)

[In]

int((c*x^4+b*x^2)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

-b^2/x+2*b*c*x+1/3*c^2*x^3

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=\frac {c^{2} x^{4} + 6 \, b c x^{2} - 3 \, b^{2}}{3 \, x} \]

[In]

integrate((c*x^4+b*x^2)^2/x^6,x, algorithm="fricas")

[Out]

1/3*(c^2*x^4 + 6*b*c*x^2 - 3*b^2)/x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=- \frac {b^{2}}{x} + 2 b c x + \frac {c^{2} x^{3}}{3} \]

[In]

integrate((c*x**4+b*x**2)**2/x**6,x)

[Out]

-b**2/x + 2*b*c*x + c**2*x**3/3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=\frac {1}{3} \, c^{2} x^{3} + 2 \, b c x - \frac {b^{2}}{x} \]

[In]

integrate((c*x^4+b*x^2)^2/x^6,x, algorithm="maxima")

[Out]

1/3*c^2*x^3 + 2*b*c*x - b^2/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=\frac {1}{3} \, c^{2} x^{3} + 2 \, b c x - \frac {b^{2}}{x} \]

[In]

integrate((c*x^4+b*x^2)^2/x^6,x, algorithm="giac")

[Out]

1/3*c^2*x^3 + 2*b*c*x - b^2/x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (b x^2+c x^4\right )^2}{x^6} \, dx=\frac {c^2\,x^3}{3}-\frac {b^2}{x}+2\,b\,c\,x \]

[In]

int((b*x^2 + c*x^4)^2/x^6,x)

[Out]

(c^2*x^3)/3 - b^2/x + 2*b*c*x