\(\int \frac {(a x+b x^3+c x^5)^2}{x} \, dx\) [76]

Optimal result

Integrand size = 20, antiderivative size = 54 \[ \int \frac {\left (a x+b x^3+c x^5\right )^2}{x} \, dx=\frac {a^2 x^2}{2}+\frac {1}{2} a b x^4+\frac {1}{6} \left (b^2+2 a c\right ) x^6+\frac {1}{4} b c x^8+\frac {c^2 x^{10}}{10} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {1599, 1121, 625} \[ \int \frac {\left (a x+b x^3+c x^5\right )^2}{x} \, dx=\frac {a^2 x^2}{2}+\frac {1}{6} x^6 \left (2 a c+b^2\right )+\frac {1}{2} a b x^4+\frac {1}{4} b c x^8+\frac {c^2 x^{10}}{10} \]

[In]

Int[(a*x + b*x^3 + c*x^5)^2/x,x]

[Out]

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

Rule 625

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && (EqQ[a, 0] ||  !PerfectSquareQ[b^2 - 4*a*c])

Rule 1121

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a x+b x^3+c x^5\right )^2}{x} \, dx=\frac {1}{60} x^2 \left (30 a^2+30 a b x^2+10 \left (b^2+2 a c\right ) x^4+15 b c x^6+6 c^2 x^8\right ) \]

[In]

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

[Out]

(x^2*(30*a^2 + 30*a*b*x^2 + 10*(b^2 + 2*a*c)*x^4 + 15*b*c*x^6 + 6*c^2*x^8))/60

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {a^{2} x^{2}}{2}+\frac {a b \,x^{4}}{2}+\frac {\left (2 a c +b^{2}\right ) x^{6}}{6}+\frac {b c \,x^{8}}{4}+\frac {c^{2} x^{10}}{10}\)	\(45\)
norman	\(\frac {c^{2} x^{10}}{10}+\frac {b c \,x^{8}}{4}+\left (\frac {a c}{3}+\frac {b^{2}}{6}\right ) x^{6}+\frac {a b \,x^{4}}{2}+\frac {a^{2} x^{2}}{2}\)	\(46\)
risch	\(\frac {1}{2} a^{2} x^{2}+\frac {1}{2} a b \,x^{4}+\frac {1}{3} x^{6} a c +\frac {1}{6} b^{2} x^{6}+\frac {1}{4} b c \,x^{8}+\frac {1}{10} c^{2} x^{10}\)	\(47\)
parallelrisch	\(\frac {1}{2} a^{2} x^{2}+\frac {1}{2} a b \,x^{4}+\frac {1}{3} x^{6} a c +\frac {1}{6} b^{2} x^{6}+\frac {1}{4} b c \,x^{8}+\frac {1}{10} c^{2} x^{10}\)	\(47\)
gosper	\(\frac {x^{2} \left (6 c^{2} x^{8}+15 b c \,x^{6}+20 a c \,x^{4}+10 b^{2} x^{4}+30 a b \,x^{2}+30 a^{2}\right )}{60}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a x+b x^3+c x^5\right )^2}{x} \, dx=\frac {1}{10} \, c^{2} x^{10} + \frac {1}{4} \, b c x^{8} + \frac {1}{6} \, {\left (b^{2} + 2 \, a c\right )} x^{6} + \frac {1}{2} \, a b x^{4} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a x+b x^3+c x^5\right )^2}{x} \, dx=\frac {1}{10} \, c^{2} x^{10} + \frac {1}{4} \, b c x^{8} + \frac {1}{6} \, b^{2} x^{6} + \frac {1}{3} \, a c x^{6} + \frac {1}{2} \, a b x^{4} + \frac {1}{2} \, a^{2} x^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

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

[In]

int((a*x + b*x^3 + c*x^5)^2/x,x)

[Out]

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