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

Optimal. Leaf size=70 \[ \frac {x^8}{8}+\frac {11 x^7}{7}+\frac {55 x^6}{6}+33 x^5+\frac {165 x^4}{2}+154 x^3-\frac {1}{3 x^3}+231 x^2-\frac {11}{2 x^2}+330 x-\frac {55}{x}+165 \log (x) \]

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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^8}{8}+\frac {11 x^7}{7}+\frac {55 x^6}{6}+33 x^5+\frac {165 x^4}{2}+154 x^3+231 x^2-\frac {11}{2 x^2}-\frac {1}{3 x^3}+330 x-\frac {55}{x}+165 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*x^3) - 11/(2*x^2) - 55/x + 330*x + 231*x^2 + 154*x^3 + (165*x^4)/2 + 33*x^5 + (55*x^6)/6 + (11*x^7)/7 +
x^8/8 + 165*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^4} \, dx &=\int \frac {(1+x)^{11}}{x^4} \, dx\\ &=\int \left (330+\frac {1}{x^4}+\frac {11}{x^3}+\frac {55}{x^2}+\frac {165}{x}+462 x+462 x^2+330 x^3+165 x^4+55 x^5+11 x^6+x^7\right ) \, dx\\ &=-\frac {1}{3 x^3}-\frac {11}{2 x^2}-\frac {55}{x}+330 x+231 x^2+154 x^3+\frac {165 x^4}{2}+33 x^5+\frac {55 x^6}{6}+\frac {11 x^7}{7}+\frac {x^8}{8}+165 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 70, normalized size = 1.00 \begin {gather*} \frac {x^8}{8}+\frac {11 x^7}{7}+\frac {55 x^6}{6}+33 x^5+\frac {165 x^4}{2}+154 x^3-\frac {1}{3 x^3}+231 x^2-\frac {11}{2 x^2}+330 x-\frac {55}{x}+165 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*1/x^3 - 11/(2*x^2) - 55/x + 330*x + 231*x^2 + 154*x^3 + (165*x^4)/2 + 33*x^5 + (55*x^6)/6 + (11*x^7)/7 +
x^8/8 + 165*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^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 62, normalized size = 0.89 \begin {gather*} \frac {21 \, x^{11} + 264 \, x^{10} + 1540 \, x^{9} + 5544 \, x^{8} + 13860 \, x^{7} + 25872 \, x^{6} + 38808 \, x^{5} + 55440 \, x^{4} + 27720 \, x^{3} \log \relax (x) - 9240 \, x^{2} - 924 \, x - 56}{168 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(21*x^11 + 264*x^10 + 1540*x^9 + 5544*x^8 + 13860*x^7 + 25872*x^6 + 38808*x^5 + 55440*x^4 + 27720*x^3*lo
g(x) - 9240*x^2 - 924*x - 56)/x^3

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giac [A]  time = 0.15, size = 59, normalized size = 0.84 \begin {gather*} \frac {1}{8} \, x^{8} + \frac {11}{7} \, x^{7} + \frac {55}{6} \, x^{6} + 33 \, x^{5} + \frac {165}{2} \, x^{4} + 154 \, x^{3} + 231 \, x^{2} + 330 \, x - \frac {330 \, x^{2} + 33 \, x + 2}{6 \, x^{3}} + 165 \, \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^4,x, algorithm="giac")

[Out]

1/8*x^8 + 11/7*x^7 + 55/6*x^6 + 33*x^5 + 165/2*x^4 + 154*x^3 + 231*x^2 + 330*x - 1/6*(330*x^2 + 33*x + 2)/x^3
+ 165*log(abs(x))

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maple [A]  time = 0.06, size = 59, normalized size = 0.84 \begin {gather*} \frac {x^{8}}{8}+\frac {11 x^{7}}{7}+\frac {55 x^{6}}{6}+33 x^{5}+\frac {165 x^{4}}{2}+154 x^{3}+231 x^{2}+330 x +165 \ln \relax (x )-\frac {55}{x}-\frac {11}{2 x^{2}}-\frac {1}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/x^3-11/2/x^2-55/x+330*x+231*x^2+154*x^3+165/2*x^4+33*x^5+55/6*x^6+11/7*x^7+1/8*x^8+165*ln(x)

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maxima [A]  time = 0.52, size = 58, normalized size = 0.83 \begin {gather*} \frac {1}{8} \, x^{8} + \frac {11}{7} \, x^{7} + \frac {55}{6} \, x^{6} + 33 \, x^{5} + \frac {165}{2} \, x^{4} + 154 \, x^{3} + 231 \, x^{2} + 330 \, x - \frac {330 \, x^{2} + 33 \, x + 2}{6 \, x^{3}} + 165 \, \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^4,x, algorithm="maxima")

[Out]

1/8*x^8 + 11/7*x^7 + 55/6*x^6 + 33*x^5 + 165/2*x^4 + 154*x^3 + 231*x^2 + 330*x - 1/6*(330*x^2 + 33*x + 2)/x^3
+ 165*log(x)

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mupad [B]  time = 0.04, size = 58, normalized size = 0.83 \begin {gather*} 330\,x+165\,\ln \relax (x)-\frac {55\,x^2+\frac {11\,x}{2}+\frac {1}{3}}{x^3}+231\,x^2+154\,x^3+\frac {165\,x^4}{2}+33\,x^5+\frac {55\,x^6}{6}+\frac {11\,x^7}{7}+\frac {x^8}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

330*x + 165*log(x) - ((11*x)/2 + 55*x^2 + 1/3)/x^3 + 231*x^2 + 154*x^3 + (165*x^4)/2 + 33*x^5 + (55*x^6)/6 + (
11*x^7)/7 + x^8/8

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sympy [A]  time = 0.13, size = 65, normalized size = 0.93 \begin {gather*} \frac {x^{8}}{8} + \frac {11 x^{7}}{7} + \frac {55 x^{6}}{6} + 33 x^{5} + \frac {165 x^{4}}{2} + 154 x^{3} + 231 x^{2} + 330 x + 165 \log {\relax (x )} + \frac {- 330 x^{2} - 33 x - 2}{6 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**8/8 + 11*x**7/7 + 55*x**6/6 + 33*x**5 + 165*x**4/2 + 154*x**3 + 231*x**2 + 330*x + 165*log(x) + (-330*x**2
- 33*x - 2)/(6*x**3)

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