\(\int \frac {(b x^2+c x^4)^3}{x^4} \, dx\) [160]

Optimal result

Integrand size = 17, antiderivative size = 43 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {b^3 x^3}{3}+\frac {3}{5} b^2 c x^5+\frac {3}{7} b c^2 x^7+\frac {c^3 x^9}{9} \]

[Out]

1/3*b^3*x^3+3/5*b^2*c*x^5+3/7*b*c^2*x^7+1/9*c^3*x^9

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, 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} \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {b^3 x^3}{3}+\frac {3}{5} b^2 c x^5+\frac {3}{7} b c^2 x^7+\frac {c^3 x^9}{9} \]

[In]

Int[(b*x^2 + c*x^4)^3/x^4,x]

[Out]

(b^3*x^3)/3 + (3*b^2*c*x^5)/5 + (3*b*c^2*x^7)/7 + (c^3*x^9)/9

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 x^2 \left (b+c x^2\right )^3 \, dx \\ & = \int \left (b^3 x^2+3 b^2 c x^4+3 b c^2 x^6+c^3 x^8\right ) \, dx \\ & = \frac {b^3 x^3}{3}+\frac {3}{5} b^2 c x^5+\frac {3}{7} b c^2 x^7+\frac {c^3 x^9}{9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {b^3 x^3}{3}+\frac {3}{5} b^2 c x^5+\frac {3}{7} b c^2 x^7+\frac {c^3 x^9}{9} \]

[In]

Integrate[(b*x^2 + c*x^4)^3/x^4,x]

[Out]

(b^3*x^3)/3 + (3*b^2*c*x^5)/5 + (3*b*c^2*x^7)/7 + (c^3*x^9)/9

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
default	\(\frac {1}{3} b^{3} x^{3}+\frac {3}{5} b^{2} c \,x^{5}+\frac {3}{7} b \,c^{2} x^{7}+\frac {1}{9} c^{3} x^{9}\)	\(36\)
risch	\(\frac {1}{3} b^{3} x^{3}+\frac {3}{5} b^{2} c \,x^{5}+\frac {3}{7} b \,c^{2} x^{7}+\frac {1}{9} c^{3} x^{9}\)	\(36\)
parallelrisch	\(\frac {1}{3} b^{3} x^{3}+\frac {3}{5} b^{2} c \,x^{5}+\frac {3}{7} b \,c^{2} x^{7}+\frac {1}{9} c^{3} x^{9}\)	\(36\)
gosper	\(\frac {x^{3} \left (35 c^{3} x^{6}+135 b \,c^{2} x^{4}+189 b^{2} c \,x^{2}+105 b^{3}\right )}{315}\)	\(38\)
norman	\(\frac {\frac {1}{3} b^{3} x^{6}+\frac {1}{9} c^{3} x^{12}+\frac {3}{7} b \,c^{2} x^{10}+\frac {3}{5} b^{2} c \,x^{8}}{x^{3}}\)	\(40\)

[In]

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

[Out]

1/3*b^3*x^3+3/5*b^2*c*x^5+3/7*b*c^2*x^7+1/9*c^3*x^9

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{9} \, c^{3} x^{9} + \frac {3}{7} \, b c^{2} x^{7} + \frac {3}{5} \, b^{2} c x^{5} + \frac {1}{3} \, b^{3} x^{3} \]

[In]

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

[Out]

1/9*c^3*x^9 + 3/7*b*c^2*x^7 + 3/5*b^2*c*x^5 + 1/3*b^3*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {b^{3} x^{3}}{3} + \frac {3 b^{2} c x^{5}}{5} + \frac {3 b c^{2} x^{7}}{7} + \frac {c^{3} x^{9}}{9} \]

[In]

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

[Out]

b**3*x**3/3 + 3*b**2*c*x**5/5 + 3*b*c**2*x**7/7 + c**3*x**9/9

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{9} \, c^{3} x^{9} + \frac {3}{7} \, b c^{2} x^{7} + \frac {3}{5} \, b^{2} c x^{5} + \frac {1}{3} \, b^{3} x^{3} \]

[In]

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

[Out]

1/9*c^3*x^9 + 3/7*b*c^2*x^7 + 3/5*b^2*c*x^5 + 1/3*b^3*x^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{9} \, c^{3} x^{9} + \frac {3}{7} \, b c^{2} x^{7} + \frac {3}{5} \, b^{2} c x^{5} + \frac {1}{3} \, b^{3} x^{3} \]

[In]

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

[Out]

1/9*c^3*x^9 + 3/7*b*c^2*x^7 + 3/5*b^2*c*x^5 + 1/3*b^3*x^3

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {b^3\,x^3}{3}+\frac {3\,b^2\,c\,x^5}{5}+\frac {3\,b\,c^2\,x^7}{7}+\frac {c^3\,x^9}{9} \]

[In]

int((b*x^2 + c*x^4)^3/x^4,x)

[Out]

(b^3*x^3)/3 + (c^3*x^9)/9 + (3*b^2*c*x^5)/5 + (3*b*c^2*x^7)/7