\(\int (b x^2+c x^4)^2 \, dx\) [145]

Optimal result

Integrand size = 13, antiderivative size = 30 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {b^2 x^5}{5}+\frac {2}{7} b c x^7+\frac {c^2 x^9}{9} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1607, 276} \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {b^2 x^5}{5}+\frac {2}{7} b c x^7+\frac {c^2 x^9}{9} \]

[In]

Int[(b*x^2 + c*x^4)^2,x]

[Out]

(b^2*x^5)/5 + (2*b*c*x^7)/7 + (c^2*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 1607

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {b^2 x^5}{5}+\frac {2}{7} b c x^7+\frac {c^2 x^9}{9} \]

[In]

Integrate[(b*x^2 + c*x^4)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {1}{5} b^{2} x^{5}+\frac {2}{7} b c \,x^{7}+\frac {1}{9} c^{2} x^{9}\)	\(25\)
norman	\(\frac {1}{5} b^{2} x^{5}+\frac {2}{7} b c \,x^{7}+\frac {1}{9} c^{2} x^{9}\)	\(25\)
risch	\(\frac {1}{5} b^{2} x^{5}+\frac {2}{7} b c \,x^{7}+\frac {1}{9} c^{2} x^{9}\)	\(25\)
parallelrisch	\(\frac {1}{5} b^{2} x^{5}+\frac {2}{7} b c \,x^{7}+\frac {1}{9} c^{2} x^{9}\)	\(25\)
gosper	\(\frac {x^{5} \left (35 c^{2} x^{4}+90 b c \,x^{2}+63 b^{2}\right )}{315}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {1}{9} \, c^{2} x^{9} + \frac {2}{7} \, b c x^{7} + \frac {1}{5} \, b^{2} x^{5} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {b^{2} x^{5}}{5} + \frac {2 b c x^{7}}{7} + \frac {c^{2} x^{9}}{9} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {1}{9} \, c^{2} x^{9} + \frac {2}{7} \, b c x^{7} + \frac {1}{5} \, b^{2} x^{5} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {1}{9} \, c^{2} x^{9} + \frac {2}{7} \, b c x^{7} + \frac {1}{5} \, b^{2} x^{5} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \left (b x^2+c x^4\right )^2 \, dx=\frac {b^2\,x^5}{5}+\frac {2\,b\,c\,x^7}{7}+\frac {c^2\,x^9}{9} \]

[In]

int((b*x^2 + c*x^4)^2,x)

[Out]

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