\(\int x (a x+b x^3+c x^5)^2 \, dx\) [74]

Optimal result

Integrand size = 18, antiderivative size = 54 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {a^2 x^4}{4}+\frac {1}{3} a b x^6+\frac {1}{8} \left (b^2+2 a c\right ) x^8+\frac {1}{5} b c x^{10}+\frac {c^2 x^{12}}{12} \]

[Out]

1/4*a^2*x^4+1/3*a*b*x^6+1/8*(2*a*c+b^2)*x^8+1/5*b*c*x^10+1/12*c^2*x^12

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1599, 1128, 645} \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {a^2 x^4}{4}+\frac {1}{8} x^8 \left (2 a c+b^2\right )+\frac {1}{3} a b x^6+\frac {1}{5} b c x^{10}+\frac {c^2 x^{12}}{12} \]

[In]

Int[x*(a*x + b*x^3 + c*x^5)^2,x]

[Out]

(a^2*x^4)/4 + (a*b*x^6)/3 + ((b^2 + 2*a*c)*x^8)/8 + (b*c*x^10)/5 + (c^2*x^12)/12

Rule 645

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)
*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0]
|| EqQ[a, 0])

Rule 1128

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rule 1599

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps \begin{align*} \text {integral}& = \int x^3 \left (a+b x^2+c x^4\right )^2 \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int x \left (a+b x+c x^2\right )^2 \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (a^2 x+2 a b x^2+\left (b^2+2 a c\right ) x^3+2 b c x^4+c^2 x^5\right ) \, dx,x,x^2\right ) \\ & = \frac {a^2 x^4}{4}+\frac {1}{3} a b x^6+\frac {1}{8} \left (b^2+2 a c\right ) x^8+\frac {1}{5} b c x^{10}+\frac {c^2 x^{12}}{12} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {1}{120} x^4 \left (30 a^2+40 a b x^2+15 \left (b^2+2 a c\right ) x^4+24 b c x^6+10 c^2 x^8\right ) \]

[In]

Integrate[x*(a*x + b*x^3 + c*x^5)^2,x]

[Out]

(x^4*(30*a^2 + 40*a*b*x^2 + 15*(b^2 + 2*a*c)*x^4 + 24*b*c*x^6 + 10*c^2*x^8))/120

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {a^{2} x^{4}}{4}+\frac {a b \,x^{6}}{3}+\frac {\left (2 a c +b^{2}\right ) x^{8}}{8}+\frac {b c \,x^{10}}{5}+\frac {c^{2} x^{12}}{12}\)	\(45\)
norman	\(\frac {c^{2} x^{12}}{12}+\frac {b c \,x^{10}}{5}+\left (\frac {a c}{4}+\frac {b^{2}}{8}\right ) x^{8}+\frac {a b \,x^{6}}{3}+\frac {a^{2} x^{4}}{4}\)	\(46\)
risch	\(\frac {1}{4} a^{2} x^{4}+\frac {1}{3} a b \,x^{6}+\frac {1}{4} x^{8} a c +\frac {1}{8} b^{2} x^{8}+\frac {1}{5} b c \,x^{10}+\frac {1}{12} c^{2} x^{12}\)	\(47\)
parallelrisch	\(\frac {1}{4} a^{2} x^{4}+\frac {1}{3} a b \,x^{6}+\frac {1}{4} x^{8} a c +\frac {1}{8} b^{2} x^{8}+\frac {1}{5} b c \,x^{10}+\frac {1}{12} c^{2} x^{12}\)	\(47\)
gosper	\(\frac {x^{4} \left (10 c^{2} x^{8}+24 b c \,x^{6}+30 a c \,x^{4}+15 b^{2} x^{4}+40 a b \,x^{2}+30 a^{2}\right )}{120}\)	\(49\)

[In]

int(x*(c*x^5+b*x^3+a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*a^2*x^4+1/3*a*b*x^6+1/8*(2*a*c+b^2)*x^8+1/5*b*c*x^10+1/12*c^2*x^12

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {1}{12} \, c^{2} x^{12} + \frac {1}{5} \, b c x^{10} + \frac {1}{8} \, {\left (b^{2} + 2 \, a c\right )} x^{8} + \frac {1}{3} \, a b x^{6} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

integrate(x*(c*x^5+b*x^3+a*x)^2,x, algorithm="fricas")

[Out]

1/12*c^2*x^12 + 1/5*b*c*x^10 + 1/8*(b^2 + 2*a*c)*x^8 + 1/3*a*b*x^6 + 1/4*a^2*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {a^{2} x^{4}}{4} + \frac {a b x^{6}}{3} + \frac {b c x^{10}}{5} + \frac {c^{2} x^{12}}{12} + x^{8} \left (\frac {a c}{4} + \frac {b^{2}}{8}\right ) \]

[In]

integrate(x*(c*x**5+b*x**3+a*x)**2,x)

[Out]

a**2*x**4/4 + a*b*x**6/3 + b*c*x**10/5 + c**2*x**12/12 + x**8*(a*c/4 + b**2/8)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {1}{12} \, c^{2} x^{12} + \frac {1}{5} \, b c x^{10} + \frac {1}{8} \, {\left (b^{2} + 2 \, a c\right )} x^{8} + \frac {1}{3} \, a b x^{6} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

integrate(x*(c*x^5+b*x^3+a*x)^2,x, algorithm="maxima")

[Out]

1/12*c^2*x^12 + 1/5*b*c*x^10 + 1/8*(b^2 + 2*a*c)*x^8 + 1/3*a*b*x^6 + 1/4*a^2*x^4

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=\frac {1}{12} \, c^{2} x^{12} + \frac {1}{5} \, b c x^{10} + \frac {1}{8} \, b^{2} x^{8} + \frac {1}{4} \, a c x^{8} + \frac {1}{3} \, a b x^{6} + \frac {1}{4} \, a^{2} x^{4} \]

[In]

integrate(x*(c*x^5+b*x^3+a*x)^2,x, algorithm="giac")

[Out]

1/12*c^2*x^12 + 1/5*b*c*x^10 + 1/8*b^2*x^8 + 1/4*a*c*x^8 + 1/3*a*b*x^6 + 1/4*a^2*x^4

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int x \left (a x+b x^3+c x^5\right )^2 \, dx=x^8\,\left (\frac {b^2}{8}+\frac {a\,c}{4}\right )+\frac {a^2\,x^4}{4}+\frac {c^2\,x^{12}}{12}+\frac {a\,b\,x^6}{3}+\frac {b\,c\,x^{10}}{5} \]

[In]

int(x*(a*x + b*x^3 + c*x^5)^2,x)

[Out]

x^8*((a*c)/4 + b^2/8) + (a^2*x^4)/4 + (c^2*x^12)/12 + (a*b*x^6)/3 + (b*c*x^10)/5