Integrand size = 35, antiderivative size = 42 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=-\frac {1}{15} \left (1-x^2-2 x^3-x^4\right )^{3/2} \left (2+3 x^2+6 x^3+3 x^4\right ) \]
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=\frac {1}{15} \left (-2-3 x^2-6 x^3-3 x^4\right ) \left (1-x^2-2 x^3-x^4\right )^{3/2} \]
Time = 0.44 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.76, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2459, 2029, 2069, 1576, 27, 1116, 1104}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^3 (x+1)^3 (2 x+1) \sqrt {-x^4-2 x^3-x^2+1} \, dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int \sqrt {-\left (x+\frac {1}{2}\right )^4+\frac {1}{2} \left (x+\frac {1}{2}\right )^2+\frac {15}{16}} \left (2 \left (x+\frac {1}{2}\right )^7-\frac {3}{2} \left (x+\frac {1}{2}\right )^5+\frac {3}{8} \left (x+\frac {1}{2}\right )^3+\frac {1}{32} \left (-x-\frac {1}{2}\right )\right )d\left (x+\frac {1}{2}\right )\) |
\(\Big \downarrow \) 2029 |
\(\displaystyle \int \left (x+\frac {1}{2}\right ) \sqrt {-\left (x+\frac {1}{2}\right )^4+\frac {1}{2} \left (x+\frac {1}{2}\right )^2+\frac {15}{16}} \left (2 \left (x+\frac {1}{2}\right )^6-\frac {3}{2} \left (x+\frac {1}{2}\right )^4+\frac {3}{8} \left (x+\frac {1}{2}\right )^2-\frac {1}{32}\right )d\left (x+\frac {1}{2}\right )\) |
\(\Big \downarrow \) 2069 |
\(\displaystyle \int \left (x+\frac {1}{2}\right ) \left (\sqrt [3]{2} \left (x+\frac {1}{2}\right )^2-\frac {1}{2\ 2^{2/3}}\right )^3 \sqrt {-\left (x+\frac {1}{2}\right )^4+\frac {1}{2} \left (x+\frac {1}{2}\right )^2+\frac {15}{16}}d\left (x+\frac {1}{2}\right )\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \frac {1}{2} \int -\frac {1}{128} \left (1-4 \left (x+\frac {1}{2}\right )^2\right )^3 \sqrt {-16 \left (x+\frac {1}{2}\right )^4+8 \left (x+\frac {1}{2}\right )^2+15}d\left (x+\frac {1}{2}\right )^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{256} \int \left (1-4 \left (x+\frac {1}{2}\right )^2\right )^3 \sqrt {-16 \left (x+\frac {1}{2}\right )^4+8 \left (x+\frac {1}{2}\right )^2+15}d\left (x+\frac {1}{2}\right )^2\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle \frac {1}{256} \left (-\frac {32}{5} \int \left (1-4 \left (x+\frac {1}{2}\right )^2\right ) \sqrt {-16 \left (x+\frac {1}{2}\right )^4+8 \left (x+\frac {1}{2}\right )^2+15}d\left (x+\frac {1}{2}\right )^2-\frac {1}{20} \left (-16 \left (x+\frac {1}{2}\right )^4+8 \left (x+\frac {1}{2}\right )^2+15\right )^{3/2} \left (1-4 \left (x+\frac {1}{2}\right )^2\right )^2\right )\) |
\(\Big \downarrow \) 1104 |
\(\displaystyle \frac {1}{256} \left (-\frac {1}{20} \left (-16 \left (x+\frac {1}{2}\right )^4+8 \left (x+\frac {1}{2}\right )^2+15\right )^{3/2} \left (1-4 \left (x+\frac {1}{2}\right )^2\right )^2-\frac {8}{15} \left (-16 \left (x+\frac {1}{2}\right )^4+8 \left (x+\frac {1}{2}\right )^2+15\right )^{3/2}\right )\) |
((-8*(15 + 8*(1/2 + x)^2 - 16*(1/2 + x)^4)^(3/2))/15 - ((1 - 4*(1/2 + x)^2 )^2*(15 + 8*(1/2 + x)^2 - 16*(1/2 + x)^4)^(3/2))/20)/256
3.8.62.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol ] :> Simp[d*((a + b*x + c*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] , x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
Int[(Fx_.)*((d_.)*(x_)^(q_.) + (a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)* (x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r) + d*x^(q - r))^p*Fx, x] /; FreeQ[{a, b, c, d, r, s, t, q}, x] && IntegerQ[p ] && PosQ[s - r] && PosQ[t - r] && PosQ[q - r] && !(EqQ[p, 1] && EqQ[u, 1] )
Int[(u_.)*(Px_), x_Symbol] :> With[{a = Rt[Coeff[Px, x^2, 0], Expon[Px, x^2 ]], b = Rt[Coeff[Px, x^2, Expon[Px, x^2]], Expon[Px, x^2]]}, Int[u*(a + b*x ^2)^Expon[Px, x^2], x] /; EqQ[Px, (a + b*x^2)^Expon[Px, x^2]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x^2], 1] && NeQ[Coeff[Px, x^2, 0], 0] && !MatchQ[Px , (a_.)*(v_)^Expon[Px, x^2] /; FreeQ[a, x] && BinomialQ[v, x, 2]]
Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 2.98 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(-\frac {\left (-x^{4}-2 x^{3}-x^{2}+1\right )^{\frac {3}{2}} \left (x^{4}+2 x^{3}+x^{2}+\frac {2}{3}\right )}{5}\) | \(35\) |
gosper | \(\frac {\left (x^{2}+x +1\right ) \left (x^{2}+x -1\right ) \left (3 x^{4}+6 x^{3}+3 x^{2}+2\right ) \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}\) | \(51\) |
trager | \(\left (\frac {1}{5} x^{8}+\frac {4}{5} x^{7}+\frac {6}{5} x^{6}+\frac {4}{5} x^{5}+\frac {2}{15} x^{4}-\frac {2}{15} x^{3}-\frac {1}{15} x^{2}-\frac {2}{15}\right ) \sqrt {-x^{4}-2 x^{3}-x^{2}+1}\) | \(58\) |
risch | \(-\frac {\left (3 x^{8}+12 x^{7}+18 x^{6}+12 x^{5}+2 x^{4}-2 x^{3}-x^{2}-2\right ) \left (x^{4}+2 x^{3}+x^{2}-1\right )}{15 \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}\) | \(72\) |
default | \(\frac {2 x^{4} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}-\frac {2 x^{3} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}-\frac {x^{2} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}-\frac {2 \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}+\frac {x^{8} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}+\frac {4 x^{7} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}+\frac {6 x^{6} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}+\frac {4 x^{5} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}\) | \(191\) |
elliptic | \(\frac {2 x^{4} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}-\frac {2 x^{3} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}-\frac {x^{2} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}-\frac {2 \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{15}+\frac {x^{8} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}+\frac {4 x^{7} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}+\frac {6 x^{6} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}+\frac {4 x^{5} \sqrt {-x^{4}-2 x^{3}-x^{2}+1}}{5}\) | \(191\) |
Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=\frac {1}{15} \, {\left (3 \, x^{8} + 12 \, x^{7} + 18 \, x^{6} + 12 \, x^{5} + 2 \, x^{4} - 2 \, x^{3} - x^{2} - 2\right )} \sqrt {-x^{4} - 2 \, x^{3} - x^{2} + 1} \]
1/15*(3*x^8 + 12*x^7 + 18*x^6 + 12*x^5 + 2*x^4 - 2*x^3 - x^2 - 2)*sqrt(-x^ 4 - 2*x^3 - x^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 182 vs. \(2 (36) = 72\).
Time = 0.16 (sec) , antiderivative size = 182, normalized size of antiderivative = 4.33 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=\frac {x^{8} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{5} + \frac {4 x^{7} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{5} + \frac {6 x^{6} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{5} + \frac {4 x^{5} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{5} + \frac {2 x^{4} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{15} - \frac {2 x^{3} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{15} - \frac {x^{2} \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{15} - \frac {2 \sqrt {- x^{4} - 2 x^{3} - x^{2} + 1}}{15} \]
x**8*sqrt(-x**4 - 2*x**3 - x**2 + 1)/5 + 4*x**7*sqrt(-x**4 - 2*x**3 - x**2 + 1)/5 + 6*x**6*sqrt(-x**4 - 2*x**3 - x**2 + 1)/5 + 4*x**5*sqrt(-x**4 - 2 *x**3 - x**2 + 1)/5 + 2*x**4*sqrt(-x**4 - 2*x**3 - x**2 + 1)/15 - 2*x**3*s qrt(-x**4 - 2*x**3 - x**2 + 1)/15 - x**2*sqrt(-x**4 - 2*x**3 - x**2 + 1)/1 5 - 2*sqrt(-x**4 - 2*x**3 - x**2 + 1)/15
Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.40 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=\frac {1}{15} \, {\left (3 \, x^{8} + 12 \, x^{7} + 18 \, x^{6} + 12 \, x^{5} + 2 \, x^{4} - 2 \, x^{3} - x^{2} - 2\right )} \sqrt {x^{2} + x + 1} \sqrt {-x^{2} - x + 1} \]
1/15*(3*x^8 + 12*x^7 + 18*x^6 + 12*x^5 + 2*x^4 - 2*x^3 - x^2 - 2)*sqrt(x^2 + x + 1)*sqrt(-x^2 - x + 1)
Time = 0.31 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=\frac {1}{5} \, {\left (x^{4} + 2 \, x^{3} + x^{2} - 1\right )}^{2} \sqrt {-x^{4} - 2 \, x^{3} - x^{2} + 1} - \frac {1}{3} \, {\left (-x^{4} - 2 \, x^{3} - x^{2} + 1\right )}^{\frac {3}{2}} \]
1/5*(x^4 + 2*x^3 + x^2 - 1)^2*sqrt(-x^4 - 2*x^3 - x^2 + 1) - 1/3*(-x^4 - 2 *x^3 - x^2 + 1)^(3/2)
Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int x^3 (1+x)^3 (1+2 x) \sqrt {1-x^2-2 x^3-x^4} \, dx=-\frac {\left (3\,x^4+6\,x^3+3\,x^2+2\right )\,{\left (-x^4-2\,x^3-x^2+1\right )}^{3/2}}{15} \]