\(\int (1+4 x+4 x^2+4 x^4)^2 \, dx\) [53]

Optimal result

Integrand size = 17, antiderivative size = 45 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=x+4 x^2+8 x^3+8 x^4+\frac {24 x^5}{5}+\frac {16 x^6}{3}+\frac {32 x^7}{7}+\frac {16 x^9}{9} \]

[Out]

x+4*x^2+8*x^3+8*x^4+24/5*x^5+16/3*x^6+32/7*x^7+16/9*x^9

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2086} \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=\frac {16 x^9}{9}+\frac {32 x^7}{7}+\frac {16 x^6}{3}+\frac {24 x^5}{5}+8 x^4+8 x^3+4 x^2+x \]

[In]

Int[(1 + 4*x + 4*x^2 + 4*x^4)^2,x]

[Out]

x + 4*x^2 + 8*x^3 + 8*x^4 + (24*x^5)/5 + (16*x^6)/3 + (32*x^7)/7 + (16*x^9)/9

Rule 2086

Int[(P_)^(p_), x_Symbol] :> Int[ExpandToSum[P^p, x], x] /; PolyQ[P, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+8 x+24 x^2+32 x^3+24 x^4+32 x^5+32 x^6+16 x^8\right ) \, dx \\ & = x+4 x^2+8 x^3+8 x^4+\frac {24 x^5}{5}+\frac {16 x^6}{3}+\frac {32 x^7}{7}+\frac {16 x^9}{9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=x+4 x^2+8 x^3+8 x^4+\frac {24 x^5}{5}+\frac {16 x^6}{3}+\frac {32 x^7}{7}+\frac {16 x^9}{9} \]

[In]

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

[Out]

x + 4*x^2 + 8*x^3 + 8*x^4 + (24*x^5)/5 + (16*x^6)/3 + (32*x^7)/7 + (16*x^9)/9

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.84

method	result	size
gosper	\(x +4 x^{2}+8 x^{3}+8 x^{4}+\frac {24}{5} x^{5}+\frac {16}{3} x^{6}+\frac {32}{7} x^{7}+\frac {16}{9} x^{9}\)	\(38\)
default	\(x +4 x^{2}+8 x^{3}+8 x^{4}+\frac {24}{5} x^{5}+\frac {16}{3} x^{6}+\frac {32}{7} x^{7}+\frac {16}{9} x^{9}\)	\(38\)
norman	\(x +4 x^{2}+8 x^{3}+8 x^{4}+\frac {24}{5} x^{5}+\frac {16}{3} x^{6}+\frac {32}{7} x^{7}+\frac {16}{9} x^{9}\)	\(38\)
risch	\(x +4 x^{2}+8 x^{3}+8 x^{4}+\frac {24}{5} x^{5}+\frac {16}{3} x^{6}+\frac {32}{7} x^{7}+\frac {16}{9} x^{9}\)	\(38\)
parallelrisch	\(x +4 x^{2}+8 x^{3}+8 x^{4}+\frac {24}{5} x^{5}+\frac {16}{3} x^{6}+\frac {32}{7} x^{7}+\frac {16}{9} x^{9}\)	\(38\)

[In]

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

[Out]

x+4*x^2+8*x^3+8*x^4+24/5*x^5+16/3*x^6+32/7*x^7+16/9*x^9

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=\frac {16}{9} \, x^{9} + \frac {32}{7} \, x^{7} + \frac {16}{3} \, x^{6} + \frac {24}{5} \, x^{5} + 8 \, x^{4} + 8 \, x^{3} + 4 \, x^{2} + x \]

[In]

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

[Out]

16/9*x^9 + 32/7*x^7 + 16/3*x^6 + 24/5*x^5 + 8*x^4 + 8*x^3 + 4*x^2 + x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.93 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=\frac {16 x^{9}}{9} + \frac {32 x^{7}}{7} + \frac {16 x^{6}}{3} + \frac {24 x^{5}}{5} + 8 x^{4} + 8 x^{3} + 4 x^{2} + x \]

[In]

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

[Out]

16*x**9/9 + 32*x**7/7 + 16*x**6/3 + 24*x**5/5 + 8*x**4 + 8*x**3 + 4*x**2 + x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=\frac {16}{9} \, x^{9} + \frac {32}{7} \, x^{7} + \frac {16}{3} \, x^{6} + \frac {24}{5} \, x^{5} + 8 \, x^{4} + 8 \, x^{3} + 4 \, x^{2} + x \]

[In]

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

[Out]

16/9*x^9 + 32/7*x^7 + 16/3*x^6 + 24/5*x^5 + 8*x^4 + 8*x^3 + 4*x^2 + x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=\frac {16}{9} \, x^{9} + \frac {32}{7} \, x^{7} + \frac {16}{3} \, x^{6} + \frac {24}{5} \, x^{5} + 8 \, x^{4} + 8 \, x^{3} + 4 \, x^{2} + x \]

[In]

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

[Out]

16/9*x^9 + 32/7*x^7 + 16/3*x^6 + 24/5*x^5 + 8*x^4 + 8*x^3 + 4*x^2 + x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82 \[ \int \left (1+4 x+4 x^2+4 x^4\right )^2 \, dx=\frac {16\,x^9}{9}+\frac {32\,x^7}{7}+\frac {16\,x^6}{3}+\frac {24\,x^5}{5}+8\,x^4+8\,x^3+4\,x^2+x \]

[In]

int((4*x + 4*x^2 + 4*x^4 + 1)^2,x)

[Out]

x + 4*x^2 + 8*x^3 + 8*x^4 + (24*x^5)/5 + (16*x^6)/3 + (32*x^7)/7 + (16*x^9)/9