\(\int (2 x+x^3) (1+4 x^2+x^4) \, dx\) [220]

Optimal result

Integrand size = 18, antiderivative size = 16 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {1}{8} \left (1+4 x^2+x^4\right )^2 \]

[Out]

1/8*(x^4+4*x^2+1)^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1602} \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {1}{8} \left (x^4+4 x^2+1\right )^2 \]

[In]

Int[(2*x + x^3)*(1 + 4*x^2 + x^4),x]

[Out]

(1 + 4*x^2 + x^4)^2/8

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \left (1+4 x^2+x^4\right )^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=x^2+\frac {9 x^4}{4}+x^6+\frac {x^8}{8} \]

[In]

Integrate[(2*x + x^3)*(1 + 4*x^2 + x^4),x]

[Out]

x^2 + (9*x^4)/4 + x^6 + x^8/8

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (x^{4}+4 x^{2}+1\right )^{2}}{8}\)	\(15\)
norman	\(\frac {1}{8} x^{8}+x^{6}+\frac {9}{4} x^{4}+x^{2}\)	\(18\)
parallelrisch	\(\frac {1}{8} x^{8}+x^{6}+\frac {9}{4} x^{4}+x^{2}\)	\(18\)
risch	\(\frac {1}{8} x^{8}+x^{6}+\frac {9}{4} x^{4}+x^{2}+\frac {1}{8}\)	\(19\)
gosper	\(\frac {x^{2} \left (x^{6}+8 x^{4}+18 x^{2}+8\right )}{8}\)	\(21\)

[In]

int((x^3+2*x)*(x^4+4*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/8*(x^4+4*x^2+1)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {1}{8} \, x^{8} + x^{6} + \frac {9}{4} \, x^{4} + x^{2} \]

[In]

integrate((x^3+2*x)*(x^4+4*x^2+1),x, algorithm="fricas")

[Out]

1/8*x^8 + x^6 + 9/4*x^4 + x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {x^{8}}{8} + x^{6} + \frac {9 x^{4}}{4} + x^{2} \]

[In]

integrate((x**3+2*x)*(x**4+4*x**2+1),x)

[Out]

x**8/8 + x**6 + 9*x**4/4 + x**2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {1}{8} \, {\left (x^{4} + 4 \, x^{2} + 1\right )}^{2} \]

[In]

integrate((x^3+2*x)*(x^4+4*x^2+1),x, algorithm="maxima")

[Out]

1/8*(x^4 + 4*x^2 + 1)^2

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {1}{4} \, x^{4} + \frac {1}{8} \, {\left (x^{4} + 4 \, x^{2}\right )}^{2} + x^{2} \]

[In]

integrate((x^3+2*x)*(x^4+4*x^2+1),x, algorithm="giac")

[Out]

1/4*x^4 + 1/8*(x^4 + 4*x^2)^2 + x^2

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \left (2 x+x^3\right ) \left (1+4 x^2+x^4\right ) \, dx=\frac {x^8}{8}+x^6+\frac {9\,x^4}{4}+x^2 \]

[In]

int((2*x + x^3)*(4*x^2 + x^4 + 1),x)

[Out]

x^2 + (9*x^4)/4 + x^6 + x^8/8