3.615 \(\int \frac{(1+x) (1+2 x+x^2)^5}{x^{16}} \, dx\)

Optimal. Leaf size=49 \[ \frac{(x+1)^{12}}{5460 x^{12}}-\frac{(x+1)^{12}}{455 x^{13}}+\frac{(x+1)^{12}}{70 x^{14}}-\frac{(x+1)^{12}}{15 x^{15}} \]

[Out]

-(1 + x)^12/(15*x^15) + (1 + x)^12/(70*x^14) - (1 + x)^12/(455*x^13) + (1 + x)^12/(5460*x^12)

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Rubi [A]  time = 0.0094611, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {27, 45, 37} \[ \frac{(x+1)^{12}}{5460 x^{12}}-\frac{(x+1)^{12}}{455 x^{13}}+\frac{(x+1)^{12}}{70 x^{14}}-\frac{(x+1)^{12}}{15 x^{15}} \]

Antiderivative was successfully verified.

[In]

Int[((1 + x)*(1 + 2*x + x^2)^5)/x^16,x]

[Out]

-(1 + x)^12/(15*x^15) + (1 + x)^12/(70*x^14) - (1 + x)^12/(455*x^13) + (1 + x)^12/(5460*x^12)

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 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{(1+x) \left (1+2 x+x^2\right )^5}{x^{16}} \, dx &=\int \frac{(1+x)^{11}}{x^{16}} \, dx\\ &=-\frac{(1+x)^{12}}{15 x^{15}}-\frac{1}{5} \int \frac{(1+x)^{11}}{x^{15}} \, dx\\ &=-\frac{(1+x)^{12}}{15 x^{15}}+\frac{(1+x)^{12}}{70 x^{14}}+\frac{1}{35} \int \frac{(1+x)^{11}}{x^{14}} \, dx\\ &=-\frac{(1+x)^{12}}{15 x^{15}}+\frac{(1+x)^{12}}{70 x^{14}}-\frac{(1+x)^{12}}{455 x^{13}}-\frac{1}{455} \int \frac{(1+x)^{11}}{x^{13}} \, dx\\ &=-\frac{(1+x)^{12}}{15 x^{15}}+\frac{(1+x)^{12}}{70 x^{14}}-\frac{(1+x)^{12}}{455 x^{13}}+\frac{(1+x)^{12}}{5460 x^{12}}\\ \end{align*}

Mathematica [A]  time = 0.0024842, size = 83, normalized size = 1.69 \[ -\frac{1}{4 x^4}-\frac{11}{5 x^5}-\frac{55}{6 x^6}-\frac{165}{7 x^7}-\frac{165}{4 x^8}-\frac{154}{3 x^9}-\frac{231}{5 x^{10}}-\frac{30}{x^{11}}-\frac{55}{4 x^{12}}-\frac{55}{13 x^{13}}-\frac{11}{14 x^{14}}-\frac{1}{15 x^{15}} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 + x)*(1 + 2*x + x^2)^5)/x^16,x]

[Out]

-1/(15*x^15) - 11/(14*x^14) - 55/(13*x^13) - 55/(4*x^12) - 30/x^11 - 231/(5*x^10) - 154/(3*x^9) - 165/(4*x^8)
- 165/(7*x^7) - 55/(6*x^6) - 11/(5*x^5) - 1/(4*x^4)

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Maple [A]  time = 0.006, size = 62, normalized size = 1.3 \begin{align*} -{\frac{154}{3\,{x}^{9}}}-{\frac{1}{15\,{x}^{15}}}-{\frac{165}{4\,{x}^{8}}}-{\frac{165}{7\,{x}^{7}}}-{\frac{231}{5\,{x}^{10}}}-{\frac{55}{13\,{x}^{13}}}-{\frac{11}{5\,{x}^{5}}}-30\,{x}^{-11}-{\frac{1}{4\,{x}^{4}}}-{\frac{11}{14\,{x}^{14}}}-{\frac{55}{4\,{x}^{12}}}-{\frac{55}{6\,{x}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+x)*(x^2+2*x+1)^5/x^16,x)

[Out]

-154/3/x^9-1/15/x^15-165/4/x^8-165/7/x^7-231/5/x^10-55/13/x^13-11/5/x^5-30/x^11-1/4/x^4-11/14/x^14-55/4/x^12-5
5/6/x^6

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Maxima [A]  time = 1.01741, size = 81, normalized size = 1.65 \begin{align*} -\frac{1365 \, x^{11} + 12012 \, x^{10} + 50050 \, x^{9} + 128700 \, x^{8} + 225225 \, x^{7} + 280280 \, x^{6} + 252252 \, x^{5} + 163800 \, x^{4} + 75075 \, x^{3} + 23100 \, x^{2} + 4290 \, x + 364}{5460 \, x^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2+2*x+1)^5/x^16,x, algorithm="maxima")

[Out]

-1/5460*(1365*x^11 + 12012*x^10 + 50050*x^9 + 128700*x^8 + 225225*x^7 + 280280*x^6 + 252252*x^5 + 163800*x^4 +
75075*x^3 + 23100*x^2 + 4290*x + 364)/x^15

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Fricas [A]  time = 1.20059, size = 209, normalized size = 4.27 \begin{align*} -\frac{1365 \, x^{11} + 12012 \, x^{10} + 50050 \, x^{9} + 128700 \, x^{8} + 225225 \, x^{7} + 280280 \, x^{6} + 252252 \, x^{5} + 163800 \, x^{4} + 75075 \, x^{3} + 23100 \, x^{2} + 4290 \, x + 364}{5460 \, x^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2+2*x+1)^5/x^16,x, algorithm="fricas")

[Out]

-1/5460*(1365*x^11 + 12012*x^10 + 50050*x^9 + 128700*x^8 + 225225*x^7 + 280280*x^6 + 252252*x^5 + 163800*x^4 +
75075*x^3 + 23100*x^2 + 4290*x + 364)/x^15

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Sympy [A]  time = 0.205436, size = 61, normalized size = 1.24 \begin{align*} - \frac{1365 x^{11} + 12012 x^{10} + 50050 x^{9} + 128700 x^{8} + 225225 x^{7} + 280280 x^{6} + 252252 x^{5} + 163800 x^{4} + 75075 x^{3} + 23100 x^{2} + 4290 x + 364}{5460 x^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x**2+2*x+1)**5/x**16,x)

[Out]

-(1365*x**11 + 12012*x**10 + 50050*x**9 + 128700*x**8 + 225225*x**7 + 280280*x**6 + 252252*x**5 + 163800*x**4
+ 75075*x**3 + 23100*x**2 + 4290*x + 364)/(5460*x**15)

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Giac [A]  time = 1.12249, size = 81, normalized size = 1.65 \begin{align*} -\frac{1365 \, x^{11} + 12012 \, x^{10} + 50050 \, x^{9} + 128700 \, x^{8} + 225225 \, x^{7} + 280280 \, x^{6} + 252252 \, x^{5} + 163800 \, x^{4} + 75075 \, x^{3} + 23100 \, x^{2} + 4290 \, x + 364}{5460 \, x^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2+2*x+1)^5/x^16,x, algorithm="giac")

[Out]

-1/5460*(1365*x^11 + 12012*x^10 + 50050*x^9 + 128700*x^8 + 225225*x^7 + 280280*x^6 + 252252*x^5 + 163800*x^4 +
75075*x^3 + 23100*x^2 + 4290*x + 364)/x^15