3.6.32 \(\int \frac {(1+x) (1+2 x+x^2)^5}{x^6} \, dx\)

Optimal. Leaf size=72 \[ \frac {x^6}{6}+\frac {11 x^5}{5}-\frac {1}{5 x^5}+\frac {55 x^4}{4}-\frac {11}{4 x^4}+55 x^3-\frac {55}{3 x^3}+165 x^2-\frac {165}{2 x^2}+462 x-\frac {330}{x}+462 \log (x) \]

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Rubi [A]  time = 0.02, antiderivative size = 72, 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^6}{6}+\frac {11 x^5}{5}+\frac {55 x^4}{4}+55 x^3+165 x^2-\frac {165}{2 x^2}-\frac {55}{3 x^3}-\frac {11}{4 x^4}-\frac {1}{5 x^5}+462 x-\frac {330}{x}+462 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x)*(1 + 2*x + x^2)^5)/x^6,x]

[Out]

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

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Mathematica [A]  time = 0.00, size = 72, normalized size = 1.00 \begin {gather*} \frac {x^6}{6}+\frac {11 x^5}{5}-\frac {1}{5 x^5}+\frac {55 x^4}{4}-\frac {11}{4 x^4}+55 x^3-\frac {55}{3 x^3}+165 x^2-\frac {165}{2 x^2}+462 x-\frac {330}{x}+462 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x)*(1 + 2*x + x^2)^5)/x^6,x]

[Out]

-1/5*1/x^5 - 11/(4*x^4) - 55/(3*x^3) - 165/(2*x^2) - 330/x + 462*x + 165*x^2 + 55*x^3 + (55*x^4)/4 + (11*x^5)/
5 + x^6/6 + 462*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^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((1 + x)*(1 + 2*x + x^2)^5)/x^6,x]

[Out]

IntegrateAlgebraic[((1 + x)*(1 + 2*x + x^2)^5)/x^6, x]

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fricas [A]  time = 0.40, size = 62, normalized size = 0.86 \begin {gather*} \frac {10 \, x^{11} + 132 \, x^{10} + 825 \, x^{9} + 3300 \, x^{8} + 9900 \, x^{7} + 27720 \, x^{6} + 27720 \, x^{5} \log \relax (x) - 19800 \, x^{4} - 4950 \, x^{3} - 1100 \, x^{2} - 165 \, x - 12}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2+2*x+1)^5/x^6,x, algorithm="fricas")

[Out]

1/60*(10*x^11 + 132*x^10 + 825*x^9 + 3300*x^8 + 9900*x^7 + 27720*x^6 + 27720*x^5*log(x) - 19800*x^4 - 4950*x^3
- 1100*x^2 - 165*x - 12)/x^5

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giac [A]  time = 0.16, size = 59, normalized size = 0.82 \begin {gather*} \frac {1}{6} \, x^{6} + \frac {11}{5} \, x^{5} + \frac {55}{4} \, x^{4} + 55 \, x^{3} + 165 \, x^{2} + 462 \, x - \frac {19800 \, x^{4} + 4950 \, x^{3} + 1100 \, x^{2} + 165 \, x + 12}{60 \, x^{5}} + 462 \, \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^6,x, algorithm="giac")

[Out]

1/6*x^6 + 11/5*x^5 + 55/4*x^4 + 55*x^3 + 165*x^2 + 462*x - 1/60*(19800*x^4 + 4950*x^3 + 1100*x^2 + 165*x + 12)
/x^5 + 462*log(abs(x))

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maple [A]  time = 0.05, size = 59, normalized size = 0.82 \begin {gather*} \frac {x^{6}}{6}+\frac {11 x^{5}}{5}+\frac {55 x^{4}}{4}+55 x^{3}+165 x^{2}+462 x +462 \ln \relax (x )-\frac {330}{x}-\frac {165}{2 x^{2}}-\frac {55}{3 x^{3}}-\frac {11}{4 x^{4}}-\frac {1}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+1)*(x^2+2*x+1)^5/x^6,x)

[Out]

-1/5/x^5-11/4/x^4-55/3/x^3-165/2/x^2-330/x+462*x+165*x^2+55*x^3+55/4*x^4+11/5*x^5+1/6*x^6+462*ln(x)

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maxima [A]  time = 0.61, size = 58, normalized size = 0.81 \begin {gather*} \frac {1}{6} \, x^{6} + \frac {11}{5} \, x^{5} + \frac {55}{4} \, x^{4} + 55 \, x^{3} + 165 \, x^{2} + 462 \, x - \frac {19800 \, x^{4} + 4950 \, x^{3} + 1100 \, x^{2} + 165 \, x + 12}{60 \, x^{5}} + 462 \, \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^6,x, algorithm="maxima")

[Out]

1/6*x^6 + 11/5*x^5 + 55/4*x^4 + 55*x^3 + 165*x^2 + 462*x - 1/60*(19800*x^4 + 4950*x^3 + 1100*x^2 + 165*x + 12)
/x^5 + 462*log(x)

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mupad [B]  time = 0.03, size = 58, normalized size = 0.81 \begin {gather*} 462\,x+462\,\ln \relax (x)-\frac {330\,x^4+\frac {165\,x^3}{2}+\frac {55\,x^2}{3}+\frac {11\,x}{4}+\frac {1}{5}}{x^5}+165\,x^2+55\,x^3+\frac {55\,x^4}{4}+\frac {11\,x^5}{5}+\frac {x^6}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x + 1)*(2*x + x^2 + 1)^5)/x^6,x)

[Out]

462*x + 462*log(x) - ((11*x)/4 + (55*x^2)/3 + (165*x^3)/2 + 330*x^4 + 1/5)/x^5 + 165*x^2 + 55*x^3 + (55*x^4)/4
+ (11*x^5)/5 + x^6/6

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sympy [A]  time = 0.14, size = 63, normalized size = 0.88 \begin {gather*} \frac {x^{6}}{6} + \frac {11 x^{5}}{5} + \frac {55 x^{4}}{4} + 55 x^{3} + 165 x^{2} + 462 x + 462 \log {\relax (x )} + \frac {- 19800 x^{4} - 4950 x^{3} - 1100 x^{2} - 165 x - 12}{60 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x**2+2*x+1)**5/x**6,x)

[Out]

x**6/6 + 11*x**5/5 + 55*x**4/4 + 55*x**3 + 165*x**2 + 462*x + 462*log(x) + (-19800*x**4 - 4950*x**3 - 1100*x**
2 - 165*x - 12)/(60*x**5)

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