3.1.74 \(\int \frac {(1+x^2) (1+2 x^2+x^4)^5}{x^3} \, dx\) [74]

Optimal. Leaf size=80 \[ -\frac {1}{2 x^2}+\frac {55 x^2}{2}+\frac {165 x^4}{4}+55 x^6+\frac {231 x^8}{4}+\frac {231 x^{10}}{5}+\frac {55 x^{12}}{2}+\frac {165 x^{14}}{14}+\frac {55 x^{16}}{16}+\frac {11 x^{18}}{18}+\frac {x^{20}}{20}+11 \log (x) \]

[Out]

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

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Rubi [A]
time = 0.02, 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, 272, 45} \begin {gather*} \frac {x^{20}}{20}+\frac {11 x^{18}}{18}+\frac {55 x^{16}}{16}+\frac {165 x^{14}}{14}+\frac {55 x^{12}}{2}+\frac {231 x^{10}}{5}+\frac {231 x^8}{4}+55 x^6+\frac {165 x^4}{4}+\frac {55 x^2}{2}-\frac {1}{2 x^2}+11 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)*(1 + 2*x^2 + x^4)^5)/x^3,x]

[Out]

-1/2*1/x^2 + (55*x^2)/2 + (165*x^4)/4 + 55*x^6 + (231*x^8)/4 + (231*x^10)/5 + (55*x^12)/2 + (165*x^14)/14 + (5
5*x^16)/16 + (11*x^18)/18 + x^20/20 + 11*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 45

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 272

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

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Mathematica [A]
time = 0.00, size = 80, normalized size = 1.00 \begin {gather*} -\frac {1}{2 x^2}+\frac {55 x^2}{2}+\frac {165 x^4}{4}+55 x^6+\frac {231 x^8}{4}+\frac {231 x^{10}}{5}+\frac {55 x^{12}}{2}+\frac {165 x^{14}}{14}+\frac {55 x^{16}}{16}+\frac {11 x^{18}}{18}+\frac {x^{20}}{20}+11 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^2)*(1 + 2*x^2 + x^4)^5)/x^3,x]

[Out]

-1/2*1/x^2 + (55*x^2)/2 + (165*x^4)/4 + 55*x^6 + (231*x^8)/4 + (231*x^10)/5 + (55*x^12)/2 + (165*x^14)/14 + (5
5*x^16)/16 + (11*x^18)/18 + x^20/20 + 11*Log[x]

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Maple [A]
time = 0.03, size = 61, normalized size = 0.76

method	result	size
default	\(-\frac {1}{2 x^{2}}+\frac {55 x^{2}}{2}+\frac {165 x^{4}}{4}+55 x^{6}+\frac {231 x^{8}}{4}+\frac {231 x^{10}}{5}+\frac {55 x^{12}}{2}+\frac {165 x^{14}}{14}+\frac {55 x^{16}}{16}+\frac {11 x^{18}}{18}+\frac {x^{20}}{20}+11 \ln \left (x \right )\)	\(61\)
risch	\(-\frac {1}{2 x^{2}}+\frac {55 x^{2}}{2}+\frac {165 x^{4}}{4}+55 x^{6}+\frac {231 x^{8}}{4}+\frac {231 x^{10}}{5}+\frac {55 x^{12}}{2}+\frac {165 x^{14}}{14}+\frac {55 x^{16}}{16}+\frac {11 x^{18}}{18}+\frac {x^{20}}{20}+11 \ln \left (x \right )\)	\(61\)
norman	\(\frac {-\frac {1}{2}+\frac {55}{2} x^{4}+\frac {165}{4} x^{6}+55 x^{8}+\frac {231}{4} x^{10}+\frac {231}{5} x^{12}+\frac {55}{2} x^{14}+\frac {165}{14} x^{16}+\frac {55}{16} x^{18}+\frac {11}{18} x^{20}+\frac {1}{20} x^{22}}{x^{2}}+11 \ln \left (x \right )\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*(x^4+2*x^2+1)^5/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 62, normalized size = 0.78 \begin {gather*} \frac {1}{20} \, x^{20} + \frac {11}{18} \, x^{18} + \frac {55}{16} \, x^{16} + \frac {165}{14} \, x^{14} + \frac {55}{2} \, x^{12} + \frac {231}{5} \, x^{10} + \frac {231}{4} \, x^{8} + 55 \, x^{6} + \frac {165}{4} \, x^{4} + \frac {55}{2} \, x^{2} - \frac {1}{2 \, x^{2}} + \frac {11}{2} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^4+2*x^2+1)^5/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.35, size = 64, normalized size = 0.80 \begin {gather*} \frac {252 \, x^{22} + 3080 \, x^{20} + 17325 \, x^{18} + 59400 \, x^{16} + 138600 \, x^{14} + 232848 \, x^{12} + 291060 \, x^{10} + 277200 \, x^{8} + 207900 \, x^{6} + 138600 \, x^{4} + 55440 \, x^{2} \log \left (x\right ) - 2520}{5040 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^4+2*x^2+1)^5/x^3,x, algorithm="fricas")

[Out]

1/5040*(252*x^22 + 3080*x^20 + 17325*x^18 + 59400*x^16 + 138600*x^14 + 232848*x^12 + 291060*x^10 + 277200*x^8
+ 207900*x^6 + 138600*x^4 + 55440*x^2*log(x) - 2520)/x^2

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Sympy [A]
time = 0.03, size = 75, normalized size = 0.94 \begin {gather*} \frac {x^{20}}{20} + \frac {11 x^{18}}{18} + \frac {55 x^{16}}{16} + \frac {165 x^{14}}{14} + \frac {55 x^{12}}{2} + \frac {231 x^{10}}{5} + \frac {231 x^{8}}{4} + 55 x^{6} + \frac {165 x^{4}}{4} + \frac {55 x^{2}}{2} + 11 \log {\left (x \right )} - \frac {1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+1)*(x**4+2*x**2+1)**5/x**3,x)

[Out]

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

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Giac [A]
time = 5.06, size = 69, normalized size = 0.86 \begin {gather*} \frac {1}{20} \, x^{20} + \frac {11}{18} \, x^{18} + \frac {55}{16} \, x^{16} + \frac {165}{14} \, x^{14} + \frac {55}{2} \, x^{12} + \frac {231}{5} \, x^{10} + \frac {231}{4} \, x^{8} + 55 \, x^{6} + \frac {165}{4} \, x^{4} + \frac {55}{2} \, x^{2} - \frac {11 \, x^{2} + 1}{2 \, x^{2}} + \frac {11}{2} \, \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^4+2*x^2+1)^5/x^3,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.06, size = 60, normalized size = 0.75 \begin {gather*} 11\,\ln \left (x\right )-\frac {1}{2\,x^2}+\frac {55\,x^2}{2}+\frac {165\,x^4}{4}+55\,x^6+\frac {231\,x^8}{4}+\frac {231\,x^{10}}{5}+\frac {55\,x^{12}}{2}+\frac {165\,x^{14}}{14}+\frac {55\,x^{16}}{16}+\frac {11\,x^{18}}{18}+\frac {x^{20}}{20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2 + 1)*(2*x^2 + x^4 + 1)^5)/x^3,x)

[Out]

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

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