3.70 \(\int x (1+x^2) (1+2 x^2+x^4)^5 \, dx\)

Optimal. Leaf size=11 \[ \frac {1}{24} \left (x^2+1\right )^{12} \]

[Out]

1/24*(x^2+1)^12

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {28, 261} \[ \frac {1}{24} \left (x^2+1\right )^{12} \]

Antiderivative was successfully verified.

[In]

Int[x*(1 + x^2)*(1 + 2*x^2 + x^4)^5,x]

[Out]

(1 + x^2)^12/24

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps

\begin {align*} \int x \left (1+x^2\right ) \left (1+2 x^2+x^4\right )^5 \, dx &=\int x \left (1+x^2\right )^{11} \, dx\\ &=\frac {1}{24} \left (1+x^2\right )^{12}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 11, normalized size = 1.00 \[ \frac {1}{24} \left (x^2+1\right )^{12} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(1 + x^2)*(1 + 2*x^2 + x^4)^5,x]

[Out]

(1 + x^2)^12/24

________________________________________________________________________________________

fricas [B]  time = 0.58, size = 61, normalized size = 5.55 \[ \frac {1}{24} x^{24} + \frac {1}{2} x^{22} + \frac {11}{4} x^{20} + \frac {55}{6} x^{18} + \frac {165}{8} x^{16} + 33 x^{14} + \frac {77}{2} x^{12} + 33 x^{10} + \frac {165}{8} x^{8} + \frac {55}{6} x^{6} + \frac {11}{4} x^{4} + \frac {1}{2} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)*(x^4+2*x^2+1)^5,x, algorithm="fricas")

[Out]

1/24*x^24 + 1/2*x^22 + 11/4*x^20 + 55/6*x^18 + 165/8*x^16 + 33*x^14 + 77/2*x^12 + 33*x^10 + 165/8*x^8 + 55/6*x
^6 + 11/4*x^4 + 1/2*x^2

________________________________________________________________________________________

giac [B]  time = 0.31, size = 76, normalized size = 6.91 \[ \frac {1}{24} \, {\left (x^{4} + 2 \, x^{2}\right )}^{6} + \frac {1}{4} \, {\left (x^{4} + 2 \, x^{2}\right )}^{5} + \frac {5}{8} \, {\left (x^{4} + 2 \, x^{2}\right )}^{4} + \frac {1}{4} \, x^{4} + \frac {5}{6} \, {\left (x^{4} + 2 \, x^{2}\right )}^{3} + \frac {5}{8} \, {\left (x^{4} + 2 \, x^{2}\right )}^{2} + \frac {1}{2} \, x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)*(x^4+2*x^2+1)^5,x, algorithm="giac")

[Out]

1/24*(x^4 + 2*x^2)^6 + 1/4*(x^4 + 2*x^2)^5 + 5/8*(x^4 + 2*x^2)^4 + 1/4*x^4 + 5/6*(x^4 + 2*x^2)^3 + 5/8*(x^4 +
2*x^2)^2 + 1/2*x^2

________________________________________________________________________________________

maple [B]  time = 0.00, size = 62, normalized size = 5.64 \[ \frac {1}{24} x^{24}+\frac {1}{2} x^{22}+\frac {11}{4} x^{20}+\frac {55}{6} x^{18}+\frac {165}{8} x^{16}+33 x^{14}+\frac {77}{2} x^{12}+33 x^{10}+\frac {165}{8} x^{8}+\frac {55}{6} x^{6}+\frac {11}{4} x^{4}+\frac {1}{2} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2+1)*(x^4+2*x^2+1)^5,x)

[Out]

1/24*x^24+1/2*x^22+11/4*x^20+55/6*x^18+165/8*x^16+33*x^14+77/2*x^12+33*x^10+165/8*x^8+55/6*x^6+11/4*x^4+1/2*x^
2

________________________________________________________________________________________

maxima [B]  time = 0.78, size = 61, normalized size = 5.55 \[ \frac {1}{24} \, x^{24} + \frac {1}{2} \, x^{22} + \frac {11}{4} \, x^{20} + \frac {55}{6} \, x^{18} + \frac {165}{8} \, x^{16} + 33 \, x^{14} + \frac {77}{2} \, x^{12} + 33 \, x^{10} + \frac {165}{8} \, x^{8} + \frac {55}{6} \, x^{6} + \frac {11}{4} \, x^{4} + \frac {1}{2} \, x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x^2+1)*(x^4+2*x^2+1)^5,x, algorithm="maxima")

[Out]

1/24*x^24 + 1/2*x^22 + 11/4*x^20 + 55/6*x^18 + 165/8*x^16 + 33*x^14 + 77/2*x^12 + 33*x^10 + 165/8*x^8 + 55/6*x
^6 + 11/4*x^4 + 1/2*x^2

________________________________________________________________________________________

mupad [B]  time = 0.06, size = 61, normalized size = 5.55 \[ \frac {x^{24}}{24}+\frac {x^{22}}{2}+\frac {11\,x^{20}}{4}+\frac {55\,x^{18}}{6}+\frac {165\,x^{16}}{8}+33\,x^{14}+\frac {77\,x^{12}}{2}+33\,x^{10}+\frac {165\,x^8}{8}+\frac {55\,x^6}{6}+\frac {11\,x^4}{4}+\frac {x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(x^2 + 1)*(2*x^2 + x^4 + 1)^5,x)

[Out]

x^2/2 + (11*x^4)/4 + (55*x^6)/6 + (165*x^8)/8 + 33*x^10 + (77*x^12)/2 + 33*x^14 + (165*x^16)/8 + (55*x^18)/6 +
(11*x^20)/4 + x^22/2 + x^24/24

________________________________________________________________________________________

sympy [B]  time = 0.07, size = 71, normalized size = 6.45 \[ \frac {x^{24}}{24} + \frac {x^{22}}{2} + \frac {11 x^{20}}{4} + \frac {55 x^{18}}{6} + \frac {165 x^{16}}{8} + 33 x^{14} + \frac {77 x^{12}}{2} + 33 x^{10} + \frac {165 x^{8}}{8} + \frac {55 x^{6}}{6} + \frac {11 x^{4}}{4} + \frac {x^{2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(x**2+1)*(x**4+2*x**2+1)**5,x)

[Out]

x**24/24 + x**22/2 + 11*x**20/4 + 55*x**18/6 + 165*x**16/8 + 33*x**14 + 77*x**12/2 + 33*x**10 + 165*x**8/8 + 5
5*x**6/6 + 11*x**4/4 + x**2/2

________________________________________________________________________________________