3.72 \(\int \frac {(1+x^2) (1+2 x^2+x^4)^5}{x} \, dx\)

Optimal. Leaf size=80 \[ \frac {x^{22}}{22}+\frac {11 x^{20}}{20}+\frac {55 x^{18}}{18}+\frac {165 x^{16}}{16}+\frac {165 x^{14}}{7}+\frac {77 x^{12}}{2}+\frac {231 x^{10}}{5}+\frac {165 x^8}{4}+\frac {55 x^6}{2}+\frac {55 x^4}{4}+\frac {11 x^2}{2}+\log (x) \]

[Out]

11/2*x^2+55/4*x^4+55/2*x^6+165/4*x^8+231/5*x^10+77/2*x^12+165/7*x^14+165/16*x^16+55/18*x^18+11/20*x^20+1/22*x^
22+ln(x)

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Rubi [A]  time = 0.03, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {28, 266, 43} \[ \frac {x^{22}}{22}+\frac {11 x^{20}}{20}+\frac {55 x^{18}}{18}+\frac {165 x^{16}}{16}+\frac {165 x^{14}}{7}+\frac {77 x^{12}}{2}+\frac {231 x^{10}}{5}+\frac {165 x^8}{4}+\frac {55 x^6}{2}+\frac {55 x^4}{4}+\frac {11 x^2}{2}+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*x^2)/2 + (55*x^4)/4 + (55*x^6)/2 + (165*x^8)/4 + (231*x^10)/5 + (77*x^12)/2 + (165*x^14)/7 + (165*x^16)/16
+ (55*x^18)/18 + (11*x^20)/20 + x^22/22 + Log[x]

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right ) \left (1+2 x^2+x^4\right )^5}{x} \, dx &=\int \frac {\left (1+x^2\right )^{11}}{x} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(1+x)^{11}}{x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (11+\frac {1}{x}+55 x+165 x^2+330 x^3+462 x^4+462 x^5+330 x^6+165 x^7+55 x^8+11 x^9+x^{10}\right ) \, dx,x,x^2\right )\\ &=\frac {11 x^2}{2}+\frac {55 x^4}{4}+\frac {55 x^6}{2}+\frac {165 x^8}{4}+\frac {231 x^{10}}{5}+\frac {77 x^{12}}{2}+\frac {165 x^{14}}{7}+\frac {165 x^{16}}{16}+\frac {55 x^{18}}{18}+\frac {11 x^{20}}{20}+\frac {x^{22}}{22}+\log (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 80, normalized size = 1.00 \[ \frac {x^{22}}{22}+\frac {11 x^{20}}{20}+\frac {55 x^{18}}{18}+\frac {165 x^{16}}{16}+\frac {165 x^{14}}{7}+\frac {77 x^{12}}{2}+\frac {231 x^{10}}{5}+\frac {165 x^8}{4}+\frac {55 x^6}{2}+\frac {55 x^4}{4}+\frac {11 x^2}{2}+\log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(11*x^2)/2 + (55*x^4)/4 + (55*x^6)/2 + (165*x^8)/4 + (231*x^10)/5 + (77*x^12)/2 + (165*x^14)/7 + (165*x^16)/16
+ (55*x^18)/18 + (11*x^20)/20 + x^22/22 + Log[x]

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fricas [A]  time = 0.49, size = 58, normalized size = 0.72 \[ \frac {1}{22} \, x^{22} + \frac {11}{20} \, x^{20} + \frac {55}{18} \, x^{18} + \frac {165}{16} \, x^{16} + \frac {165}{7} \, x^{14} + \frac {77}{2} \, x^{12} + \frac {231}{5} \, x^{10} + \frac {165}{4} \, x^{8} + \frac {55}{2} \, x^{6} + \frac {55}{4} \, x^{4} + \frac {11}{2} \, x^{2} + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/22*x^22 + 11/20*x^20 + 55/18*x^18 + 165/16*x^16 + 165/7*x^14 + 77/2*x^12 + 231/5*x^10 + 165/4*x^8 + 55/2*x^6
+ 55/4*x^4 + 11/2*x^2 + log(x)

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giac [A]  time = 0.23, size = 62, normalized size = 0.78 \[ \frac {1}{22} \, x^{22} + \frac {11}{20} \, x^{20} + \frac {55}{18} \, x^{18} + \frac {165}{16} \, x^{16} + \frac {165}{7} \, x^{14} + \frac {77}{2} \, x^{12} + \frac {231}{5} \, x^{10} + \frac {165}{4} \, x^{8} + \frac {55}{2} \, x^{6} + \frac {55}{4} \, x^{4} + \frac {11}{2} \, x^{2} + \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/22*x^22 + 11/20*x^20 + 55/18*x^18 + 165/16*x^16 + 165/7*x^14 + 77/2*x^12 + 231/5*x^10 + 165/4*x^8 + 55/2*x^6
+ 55/4*x^4 + 11/2*x^2 + 1/2*log(x^2)

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maple [A]  time = 0.00, size = 59, normalized size = 0.74 \[ \frac {x^{22}}{22}+\frac {11 x^{20}}{20}+\frac {55 x^{18}}{18}+\frac {165 x^{16}}{16}+\frac {165 x^{14}}{7}+\frac {77 x^{12}}{2}+\frac {231 x^{10}}{5}+\frac {165 x^{8}}{4}+\frac {55 x^{6}}{2}+\frac {55 x^{4}}{4}+\frac {11 x^{2}}{2}+\ln \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/2*x^2+55/4*x^4+55/2*x^6+165/4*x^8+231/5*x^10+77/2*x^12+165/7*x^14+165/16*x^16+55/18*x^18+11/20*x^20+1/22*x^
22+ln(x)

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maxima [A]  time = 0.82, size = 62, normalized size = 0.78 \[ \frac {1}{22} \, x^{22} + \frac {11}{20} \, x^{20} + \frac {55}{18} \, x^{18} + \frac {165}{16} \, x^{16} + \frac {165}{7} \, x^{14} + \frac {77}{2} \, x^{12} + \frac {231}{5} \, x^{10} + \frac {165}{4} \, x^{8} + \frac {55}{2} \, x^{6} + \frac {55}{4} \, x^{4} + \frac {11}{2} \, x^{2} + \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/22*x^22 + 11/20*x^20 + 55/18*x^18 + 165/16*x^16 + 165/7*x^14 + 77/2*x^12 + 231/5*x^10 + 165/4*x^8 + 55/2*x^6
+ 55/4*x^4 + 11/2*x^2 + 1/2*log(x^2)

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mupad [B]  time = 0.06, size = 58, normalized size = 0.72 \[ \ln \relax (x)+\frac {11\,x^2}{2}+\frac {55\,x^4}{4}+\frac {55\,x^6}{2}+\frac {165\,x^8}{4}+\frac {231\,x^{10}}{5}+\frac {77\,x^{12}}{2}+\frac {165\,x^{14}}{7}+\frac {165\,x^{16}}{16}+\frac {55\,x^{18}}{18}+\frac {11\,x^{20}}{20}+\frac {x^{22}}{22} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) + (11*x^2)/2 + (55*x^4)/4 + (55*x^6)/2 + (165*x^8)/4 + (231*x^10)/5 + (77*x^12)/2 + (165*x^14)/7 + (165
*x^16)/16 + (55*x^18)/18 + (11*x^20)/20 + x^22/22

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sympy [A]  time = 0.11, size = 75, normalized size = 0.94 \[ \frac {x^{22}}{22} + \frac {11 x^{20}}{20} + \frac {55 x^{18}}{18} + \frac {165 x^{16}}{16} + \frac {165 x^{14}}{7} + \frac {77 x^{12}}{2} + \frac {231 x^{10}}{5} + \frac {165 x^{8}}{4} + \frac {55 x^{6}}{2} + \frac {55 x^{4}}{4} + \frac {11 x^{2}}{2} + \log {\relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**22/22 + 11*x**20/20 + 55*x**18/18 + 165*x**16/16 + 165*x**14/7 + 77*x**12/2 + 231*x**10/5 + 165*x**8/4 + 55
*x**6/2 + 55*x**4/4 + 11*x**2/2 + log(x)

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