3.6.29 \(\int \frac {(1+x) (1+2 x+x^2)^5}{x^3} \, dx\)

Optimal. Leaf size=70 \[ \frac {x^9}{9}+\frac {11 x^8}{8}+\frac {55 x^7}{7}+\frac {55 x^6}{2}+66 x^5+\frac {231 x^4}{2}+154 x^3+165 x^2-\frac {1}{2 x^2}+165 x-\frac {11}{x}+55 \log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {27, 43} \begin {gather*} \frac {x^9}{9}+\frac {11 x^8}{8}+\frac {55 x^7}{7}+\frac {55 x^6}{2}+66 x^5+\frac {231 x^4}{2}+154 x^3+165 x^2-\frac {1}{2 x^2}+165 x-\frac {11}{x}+55 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*x^2) - 11/x + 165*x + 165*x^2 + 154*x^3 + (231*x^4)/2 + 66*x^5 + (55*x^6)/2 + (55*x^7)/7 + (11*x^8)/8 +
x^9/9 + 55*Log[x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && 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])

Rubi steps

\begin {align*} \int \frac {(1+x) \left (1+2 x+x^2\right )^5}{x^3} \, dx &=\int \frac {(1+x)^{11}}{x^3} \, dx\\ &=\int \left (165+\frac {1}{x^3}+\frac {11}{x^2}+\frac {55}{x}+330 x+462 x^2+462 x^3+330 x^4+165 x^5+55 x^6+11 x^7+x^8\right ) \, dx\\ &=-\frac {1}{2 x^2}-\frac {11}{x}+165 x+165 x^2+154 x^3+\frac {231 x^4}{2}+66 x^5+\frac {55 x^6}{2}+\frac {55 x^7}{7}+\frac {11 x^8}{8}+\frac {x^9}{9}+55 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 70, normalized size = 1.00 \begin {gather*} \frac {x^9}{9}+\frac {11 x^8}{8}+\frac {55 x^7}{7}+\frac {55 x^6}{2}+66 x^5+\frac {231 x^4}{2}+154 x^3+165 x^2-\frac {1}{2 x^2}+165 x-\frac {11}{x}+55 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*1/x^2 - 11/x + 165*x + 165*x^2 + 154*x^3 + (231*x^4)/2 + 66*x^5 + (55*x^6)/2 + (55*x^7)/7 + (11*x^8)/8 +
x^9/9 + 55*Log[x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(1+x) \left (1+2 x+x^2\right )^5}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 62, normalized size = 0.89 \begin {gather*} \frac {56 \, x^{11} + 693 \, x^{10} + 3960 \, x^{9} + 13860 \, x^{8} + 33264 \, x^{7} + 58212 \, x^{6} + 77616 \, x^{5} + 83160 \, x^{4} + 83160 \, x^{3} + 27720 \, x^{2} \log \relax (x) - 5544 \, x - 252}{504 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/504*(56*x^11 + 693*x^10 + 3960*x^9 + 13860*x^8 + 33264*x^7 + 58212*x^6 + 77616*x^5 + 83160*x^4 + 83160*x^3 +
27720*x^2*log(x) - 5544*x - 252)/x^2

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giac [A]  time = 0.17, size = 59, normalized size = 0.84 \begin {gather*} \frac {1}{9} \, x^{9} + \frac {11}{8} \, x^{8} + \frac {55}{7} \, x^{7} + \frac {55}{2} \, x^{6} + 66 \, x^{5} + \frac {231}{2} \, x^{4} + 154 \, x^{3} + 165 \, x^{2} + 165 \, x - \frac {22 \, x + 1}{2 \, x^{2}} + 55 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^9 + 11/8*x^8 + 55/7*x^7 + 55/2*x^6 + 66*x^5 + 231/2*x^4 + 154*x^3 + 165*x^2 + 165*x - 1/2*(22*x + 1)/x^2
+ 55*log(abs(x))

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maple [A]  time = 0.06, size = 59, normalized size = 0.84 \begin {gather*} \frac {x^{9}}{9}+\frac {11 x^{8}}{8}+\frac {55 x^{7}}{7}+\frac {55 x^{6}}{2}+66 x^{5}+\frac {231 x^{4}}{2}+154 x^{3}+165 x^{2}+165 x +55 \ln \relax (x )-\frac {11}{x}-\frac {1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2-11/x+165*x+165*x^2+154*x^3+231/2*x^4+66*x^5+55/2*x^6+55/7*x^7+11/8*x^8+1/9*x^9+55*ln(x)

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maxima [A]  time = 0.55, size = 58, normalized size = 0.83 \begin {gather*} \frac {1}{9} \, x^{9} + \frac {11}{8} \, x^{8} + \frac {55}{7} \, x^{7} + \frac {55}{2} \, x^{6} + 66 \, x^{5} + \frac {231}{2} \, x^{4} + 154 \, x^{3} + 165 \, x^{2} + 165 \, x - \frac {22 \, x + 1}{2 \, x^{2}} + 55 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*x^9 + 11/8*x^8 + 55/7*x^7 + 55/2*x^6 + 66*x^5 + 231/2*x^4 + 154*x^3 + 165*x^2 + 165*x - 1/2*(22*x + 1)/x^2
+ 55*log(x)

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mupad [B]  time = 0.05, size = 58, normalized size = 0.83 \begin {gather*} 165\,x+55\,\ln \relax (x)-\frac {11\,x+\frac {1}{2}}{x^2}+165\,x^2+154\,x^3+\frac {231\,x^4}{2}+66\,x^5+\frac {55\,x^6}{2}+\frac {55\,x^7}{7}+\frac {11\,x^8}{8}+\frac {x^9}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

165*x + 55*log(x) - (11*x + 1/2)/x^2 + 165*x^2 + 154*x^3 + (231*x^4)/2 + 66*x^5 + (55*x^6)/2 + (55*x^7)/7 + (1
1*x^8)/8 + x^9/9

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sympy [A]  time = 0.12, size = 66, normalized size = 0.94 \begin {gather*} \frac {x^{9}}{9} + \frac {11 x^{8}}{8} + \frac {55 x^{7}}{7} + \frac {55 x^{6}}{2} + 66 x^{5} + \frac {231 x^{4}}{2} + 154 x^{3} + 165 x^{2} + 165 x + 55 \log {\relax (x )} + \frac {- 22 x - 1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**9/9 + 11*x**8/8 + 55*x**7/7 + 55*x**6/2 + 66*x**5 + 231*x**4/2 + 154*x**3 + 165*x**2 + 165*x + 55*log(x) +
(-22*x - 1)/(2*x**2)

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