\(\int \frac {(b x^2+c x^4)^3}{x^2} \, dx\) [158]

Optimal result

Integrand size = 17, antiderivative size = 43 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {b^3 x^5}{5}+\frac {3}{7} b^2 c x^7+\frac {1}{3} b c^2 x^9+\frac {c^3 x^{11}}{11} \]

[Out]

1/5*b^3*x^5+3/7*b^2*c*x^7+1/3*b*c^2*x^9+1/11*c^3*x^11

Rubi [A] (verified)

Time = 0.02 (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} \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {b^3 x^5}{5}+\frac {3}{7} b^2 c x^7+\frac {1}{3} b c^2 x^9+\frac {c^3 x^{11}}{11} \]

[In]

Int[(b*x^2 + c*x^4)^3/x^2,x]

[Out]

(b^3*x^5)/5 + (3*b^2*c*x^7)/7 + (b*c^2*x^9)/3 + (c^3*x^11)/11

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^4 \left (b+c x^2\right )^3 \, dx \\ & = \int \left (b^3 x^4+3 b^2 c x^6+3 b c^2 x^8+c^3 x^{10}\right ) \, dx \\ & = \frac {b^3 x^5}{5}+\frac {3}{7} b^2 c x^7+\frac {1}{3} b c^2 x^9+\frac {c^3 x^{11}}{11} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {b^3 x^5}{5}+\frac {3}{7} b^2 c x^7+\frac {1}{3} b c^2 x^9+\frac {c^3 x^{11}}{11} \]

[In]

Integrate[(b*x^2 + c*x^4)^3/x^2,x]

[Out]

(b^3*x^5)/5 + (3*b^2*c*x^7)/7 + (b*c^2*x^9)/3 + (c^3*x^11)/11

Maple [A] (verified)

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

method	result	size
default	\(\frac {1}{5} b^{3} x^{5}+\frac {3}{7} b^{2} c \,x^{7}+\frac {1}{3} b \,c^{2} x^{9}+\frac {1}{11} c^{3} x^{11}\)	\(36\)
risch	\(\frac {1}{5} b^{3} x^{5}+\frac {3}{7} b^{2} c \,x^{7}+\frac {1}{3} b \,c^{2} x^{9}+\frac {1}{11} c^{3} x^{11}\)	\(36\)
parallelrisch	\(\frac {1}{5} b^{3} x^{5}+\frac {3}{7} b^{2} c \,x^{7}+\frac {1}{3} b \,c^{2} x^{9}+\frac {1}{11} c^{3} x^{11}\)	\(36\)
gosper	\(\frac {x^{5} \left (105 c^{3} x^{6}+385 b \,c^{2} x^{4}+495 b^{2} c \,x^{2}+231 b^{3}\right )}{1155}\)	\(38\)
norman	\(\frac {\frac {1}{5} b^{3} x^{6}+\frac {1}{11} c^{3} x^{12}+\frac {1}{3} b \,c^{2} x^{10}+\frac {3}{7} b^{2} c \,x^{8}}{x}\)	\(40\)

[In]

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

[Out]

1/5*b^3*x^5+3/7*b^2*c*x^7+1/3*b*c^2*x^9+1/11*c^3*x^11

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{11} \, c^{3} x^{11} + \frac {1}{3} \, b c^{2} x^{9} + \frac {3}{7} \, b^{2} c x^{7} + \frac {1}{5} \, b^{3} x^{5} \]

[In]

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

[Out]

1/11*c^3*x^11 + 1/3*b*c^2*x^9 + 3/7*b^2*c*x^7 + 1/5*b^3*x^5

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {b^{3} x^{5}}{5} + \frac {3 b^{2} c x^{7}}{7} + \frac {b c^{2} x^{9}}{3} + \frac {c^{3} x^{11}}{11} \]

[In]

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

[Out]

b**3*x**5/5 + 3*b**2*c*x**7/7 + b*c**2*x**9/3 + c**3*x**11/11

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{11} \, c^{3} x^{11} + \frac {1}{3} \, b c^{2} x^{9} + \frac {3}{7} \, b^{2} c x^{7} + \frac {1}{5} \, b^{3} x^{5} \]

[In]

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

[Out]

1/11*c^3*x^11 + 1/3*b*c^2*x^9 + 3/7*b^2*c*x^7 + 1/5*b^3*x^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{11} \, c^{3} x^{11} + \frac {1}{3} \, b c^{2} x^{9} + \frac {3}{7} \, b^{2} c x^{7} + \frac {1}{5} \, b^{3} x^{5} \]

[In]

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

[Out]

1/11*c^3*x^11 + 1/3*b*c^2*x^9 + 3/7*b^2*c*x^7 + 1/5*b^3*x^5

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {b^3\,x^5}{5}+\frac {3\,b^2\,c\,x^7}{7}+\frac {b\,c^2\,x^9}{3}+\frac {c^3\,x^{11}}{11} \]

[In]

int((b*x^2 + c*x^4)^3/x^2,x)

[Out]

(b^3*x^5)/5 + (c^3*x^11)/11 + (3*b^2*c*x^7)/7 + (b*c^2*x^9)/3