\(\int (a+b x^2+c x^4)^2 \, dx\) [827]

Optimal result

Integrand size = 14, antiderivative size = 49 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {1}{5} \left (b^2+2 a c\right ) x^5+\frac {2}{7} b c x^7+\frac {c^2 x^9}{9} \]

[Out]

a^2*x+2/3*a*b*x^3+1/5*(2*a*c+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 = 49, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1104} \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=a^2 x+\frac {1}{5} x^5 \left (2 a c+b^2\right )+\frac {2}{3} a b x^3+\frac {2}{7} b c x^7+\frac {c^2 x^9}{9} \]

[In]

Int[(a + b*x^2 + c*x^4)^2,x]

[Out]

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

Rule 1104

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2+2 a b x^2+b^2 \left (1+\frac {2 a c}{b^2}\right ) x^4+2 b c x^6+c^2 x^8\right ) \, dx \\ & = a^2 x+\frac {2}{3} a b x^3+\frac {1}{5} \left (b^2+2 a c\right ) x^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 = 49, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=a^2 x+\frac {2}{3} a b x^3+\frac {1}{5} \left (b^2+2 a c\right ) x^5+\frac {2}{7} b c x^7+\frac {c^2 x^9}{9} \]

[In]

Integrate[(a + b*x^2 + c*x^4)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86

method	result	size
default	\(a^{2} x +\frac {2 a b \,x^{3}}{3}+\frac {\left (2 a c +b^{2}\right ) x^{5}}{5}+\frac {2 b c \,x^{7}}{7}+\frac {c^{2} x^{9}}{9}\)	\(42\)
norman	\(\frac {c^{2} x^{9}}{9}+\frac {2 b c \,x^{7}}{7}+\left (\frac {2 a c}{5}+\frac {b^{2}}{5}\right ) x^{5}+\frac {2 a b \,x^{3}}{3}+a^{2} x\)	\(43\)
gosper	\(\frac {1}{9} c^{2} x^{9}+\frac {2}{7} b c \,x^{7}+\frac {2}{5} x^{5} a c +\frac {1}{5} b^{2} x^{5}+\frac {2}{3} a b \,x^{3}+a^{2} x\)	\(44\)
risch	\(\frac {1}{9} c^{2} x^{9}+\frac {2}{7} b c \,x^{7}+\frac {2}{5} x^{5} a c +\frac {1}{5} b^{2} x^{5}+\frac {2}{3} a b \,x^{3}+a^{2} x\)	\(44\)
parallelrisch	\(\frac {1}{9} c^{2} x^{9}+\frac {2}{7} b c \,x^{7}+\frac {2}{5} x^{5} a c +\frac {1}{5} b^{2} x^{5}+\frac {2}{3} a b \,x^{3}+a^{2} x\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=\frac {1}{9} \, c^{2} x^{9} + \frac {2}{7} \, b c x^{7} + \frac {1}{5} \, {\left (b^{2} + 2 \, a c\right )} x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=a^{2} x + \frac {2 a b x^{3}}{3} + \frac {2 b c x^{7}}{7} + \frac {c^{2} x^{9}}{9} + x^{5} \cdot \left (\frac {2 a c}{5} + \frac {b^{2}}{5}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.92 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=\frac {1}{9} \, c^{2} x^{9} + \frac {2}{7} \, b c x^{7} + \frac {1}{5} \, b^{2} x^{5} + a^{2} x + \frac {2}{15} \, {\left (3 \, c x^{5} + 5 \, b x^{3}\right )} a \]

[In]

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

[Out]

1/9*c^2*x^9 + 2/7*b*c*x^7 + 1/5*b^2*x^5 + a^2*x + 2/15*(3*c*x^5 + 5*b*x^3)*a

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=\frac {1}{9} \, c^{2} x^{9} + \frac {2}{7} \, b c x^{7} + \frac {1}{5} \, b^{2} x^{5} + \frac {2}{5} \, a c x^{5} + \frac {2}{3} \, a b x^{3} + a^{2} x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86 \[ \int \left (a+b x^2+c x^4\right )^2 \, dx=a^2\,x+x^5\,\left (\frac {b^2}{5}+\frac {2\,a\,c}{5}\right )+\frac {c^2\,x^9}{9}+\frac {2\,a\,b\,x^3}{3}+\frac {2\,b\,c\,x^7}{7} \]

[In]

int((a + b*x^2 + c*x^4)^2,x)

[Out]

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