Integrand size = 26, antiderivative size = 104 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} a^2 B x^3+\frac {1}{4} a^2 C x^4+\frac {1}{5} a (2 b B+a D) x^5+\frac {1}{3} a b C x^6+\frac {1}{7} b (b B+2 a D) x^7+\frac {1}{8} b^2 C x^8+\frac {1}{9} b^2 D x^9+\frac {A \left (a+b x^2\right )^3}{6 b} \]
1/3*a^2*B*x^3+1/4*a^2*C*x^4+1/5*a*(2*B*b+D*a)*x^5+1/3*a*b*C*x^6+1/7*b*(B*b +2*D*a)*x^7+1/8*b^2*C*x^8+1/9*b^2*D*x^9+1/6*A*(b*x^2+a)^3/b
Time = 0.02 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {42 a^2 x^2 (30 A+x (20 B+3 x (5 C+4 D x)))+12 a b x^4 (105 A+2 x (42 B+5 x (7 C+6 D x)))+5 b^2 x^6 (84 A+x (72 B+7 x (9 C+8 D x)))}{2520} \]
(42*a^2*x^2*(30*A + x*(20*B + 3*x*(5*C + 4*D*x))) + 12*a*b*x^4*(105*A + 2* x*(42*B + 5*x*(7*C + 6*D*x))) + 5*b^2*x^6*(84*A + x*(72*B + 7*x*(9*C + 8*D *x))))/2520
Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2017, 2341, 2009}
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.
\(\displaystyle \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^2+a\right )^2 \left (x \left (D x^3+C x^2+B x+A\right )-A x\right )dx+\frac {A \left (a+b x^2\right )^3}{6 b}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \int \left (b^2 D x^8+b^2 C x^7+b (b B+2 a D) x^6+2 a b C x^5+a (2 b B+a D) x^4+a^2 C x^3+a^2 B x^2\right )dx+\frac {A \left (a+b x^2\right )^3}{6 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} a^2 B x^3+\frac {1}{4} a^2 C x^4+\frac {A \left (a+b x^2\right )^3}{6 b}+\frac {1}{7} b x^7 (2 a D+b B)+\frac {1}{5} a x^5 (a D+2 b B)+\frac {1}{3} a b C x^6+\frac {1}{8} b^2 C x^8+\frac {1}{9} b^2 D x^9\) |
(a^2*B*x^3)/3 + (a^2*C*x^4)/4 + (a*(2*b*B + a*D)*x^5)/5 + (a*b*C*x^6)/3 + (b*(b*B + 2*a*D)*x^7)/7 + (b^2*C*x^8)/8 + (b^2*D*x^9)/9 + (A*(a + b*x^2)^3 )/(6*b)
3.1.72.3.1 Defintions of rubi rules used
Int[(Px_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[Coeff[Px, x, n - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coeff[Px, x, n - 1] *x^(n - 1))*(a + b*x^n)^p, x] /; FreeQ[{a, b}, x] && PolyQ[Px, x] && IGtQ[p , 1] && IGtQ[n, 1] && NeQ[Coeff[Px, x, n - 1], 0] && NeQ[Px, Coeff[Px, x, n - 1]*x^(n - 1)] && !MatchQ[Px, (Qx_.)*((c_) + (d_.)*x^(m_))^(q_) /; FreeQ [{c, d}, x] && PolyQ[Qx, x] && IGtQ[q, 1] && IGtQ[m, 1] && NeQ[Coeff[Qx*(a + b*x^n)^p, x, m - 1], 0] && GtQ[m*q, n*p]]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 3.35 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {b^{2} D x^{9}}{9}+\frac {b^{2} C \,x^{8}}{8}+\frac {\left (B \,b^{2}+2 D a b \right ) x^{7}}{7}+\frac {\left (b^{2} A +2 C a b \right ) x^{6}}{6}+\frac {\left (2 a b B +D a^{2}\right ) x^{5}}{5}+\frac {\left (2 a b A +C \,a^{2}\right ) x^{4}}{4}+\frac {a^{2} B \,x^{3}}{3}+\frac {a^{2} A \,x^{2}}{2}\) | \(102\) |
norman | \(\frac {b^{2} D x^{9}}{9}+\frac {b^{2} C \,x^{8}}{8}+\left (\frac {1}{7} B \,b^{2}+\frac {2}{7} D a b \right ) x^{7}+\left (\frac {1}{6} b^{2} A +\frac {1}{3} C a b \right ) x^{6}+\left (\frac {2}{5} a b B +\frac {1}{5} D a^{2}\right ) x^{5}+\left (\frac {1}{2} a b A +\frac {1}{4} C \,a^{2}\right ) x^{4}+\frac {a^{2} B \,x^{3}}{3}+\frac {a^{2} A \,x^{2}}{2}\) | \(102\) |
gosper | \(\frac {1}{9} b^{2} D x^{9}+\frac {1}{8} b^{2} C \,x^{8}+\frac {1}{7} b^{2} B \,x^{7}+\frac {2}{7} x^{7} D a b +\frac {1}{6} x^{6} b^{2} A +\frac {1}{3} a b C \,x^{6}+\frac {2}{5} x^{5} a b B +\frac {1}{5} x^{5} D a^{2}+\frac {1}{2} x^{4} a b A +\frac {1}{4} a^{2} C \,x^{4}+\frac {1}{3} a^{2} B \,x^{3}+\frac {1}{2} a^{2} A \,x^{2}\) | \(106\) |
parallelrisch | \(\frac {1}{9} b^{2} D x^{9}+\frac {1}{8} b^{2} C \,x^{8}+\frac {1}{7} b^{2} B \,x^{7}+\frac {2}{7} x^{7} D a b +\frac {1}{6} x^{6} b^{2} A +\frac {1}{3} a b C \,x^{6}+\frac {2}{5} x^{5} a b B +\frac {1}{5} x^{5} D a^{2}+\frac {1}{2} x^{4} a b A +\frac {1}{4} a^{2} C \,x^{4}+\frac {1}{3} a^{2} B \,x^{3}+\frac {1}{2} a^{2} A \,x^{2}\) | \(106\) |
1/9*b^2*D*x^9+1/8*b^2*C*x^8+1/7*(B*b^2+2*D*a*b)*x^7+1/6*(A*b^2+2*C*a*b)*x^ 6+1/5*(2*B*a*b+D*a^2)*x^5+1/4*(2*A*a*b+C*a^2)*x^4+1/3*a^2*B*x^3+1/2*a^2*A* x^2
Time = 0.25 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{9} \, D b^{2} x^{9} + \frac {1}{8} \, C b^{2} x^{8} + \frac {1}{7} \, {\left (2 \, D a b + B b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (2 \, C a b + A b^{2}\right )} x^{6} + \frac {1}{3} \, B a^{2} x^{3} + \frac {1}{5} \, {\left (D a^{2} + 2 \, B a b\right )} x^{5} + \frac {1}{2} \, A a^{2} x^{2} + \frac {1}{4} \, {\left (C a^{2} + 2 \, A a b\right )} x^{4} \]
1/9*D*b^2*x^9 + 1/8*C*b^2*x^8 + 1/7*(2*D*a*b + B*b^2)*x^7 + 1/6*(2*C*a*b + A*b^2)*x^6 + 1/3*B*a^2*x^3 + 1/5*(D*a^2 + 2*B*a*b)*x^5 + 1/2*A*a^2*x^2 + 1/4*(C*a^2 + 2*A*a*b)*x^4
Time = 0.02 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.06 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A a^{2} x^{2}}{2} + \frac {B a^{2} x^{3}}{3} + \frac {C b^{2} x^{8}}{8} + \frac {D b^{2} x^{9}}{9} + x^{7} \left (\frac {B b^{2}}{7} + \frac {2 D a b}{7}\right ) + x^{6} \left (\frac {A b^{2}}{6} + \frac {C a b}{3}\right ) + x^{5} \cdot \left (\frac {2 B a b}{5} + \frac {D a^{2}}{5}\right ) + x^{4} \left (\frac {A a b}{2} + \frac {C a^{2}}{4}\right ) \]
A*a**2*x**2/2 + B*a**2*x**3/3 + C*b**2*x**8/8 + D*b**2*x**9/9 + x**7*(B*b* *2/7 + 2*D*a*b/7) + x**6*(A*b**2/6 + C*a*b/3) + x**5*(2*B*a*b/5 + D*a**2/5 ) + x**4*(A*a*b/2 + C*a**2/4)
Time = 0.19 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{9} \, D b^{2} x^{9} + \frac {1}{8} \, C b^{2} x^{8} + \frac {1}{7} \, {\left (2 \, D a b + B b^{2}\right )} x^{7} + \frac {1}{6} \, {\left (2 \, C a b + A b^{2}\right )} x^{6} + \frac {1}{3} \, B a^{2} x^{3} + \frac {1}{5} \, {\left (D a^{2} + 2 \, B a b\right )} x^{5} + \frac {1}{2} \, A a^{2} x^{2} + \frac {1}{4} \, {\left (C a^{2} + 2 \, A a b\right )} x^{4} \]
1/9*D*b^2*x^9 + 1/8*C*b^2*x^8 + 1/7*(2*D*a*b + B*b^2)*x^7 + 1/6*(2*C*a*b + A*b^2)*x^6 + 1/3*B*a^2*x^3 + 1/5*(D*a^2 + 2*B*a*b)*x^5 + 1/2*A*a^2*x^2 + 1/4*(C*a^2 + 2*A*a*b)*x^4
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{9} \, D b^{2} x^{9} + \frac {1}{8} \, C b^{2} x^{8} + \frac {2}{7} \, D a b x^{7} + \frac {1}{7} \, B b^{2} x^{7} + \frac {1}{3} \, C a b x^{6} + \frac {1}{6} \, A b^{2} x^{6} + \frac {1}{5} \, D a^{2} x^{5} + \frac {2}{5} \, B a b x^{5} + \frac {1}{4} \, C a^{2} x^{4} + \frac {1}{2} \, A a b x^{4} + \frac {1}{3} \, B a^{2} x^{3} + \frac {1}{2} \, A a^{2} x^{2} \]
1/9*D*b^2*x^9 + 1/8*C*b^2*x^8 + 2/7*D*a*b*x^7 + 1/7*B*b^2*x^7 + 1/3*C*a*b* x^6 + 1/6*A*b^2*x^6 + 1/5*D*a^2*x^5 + 2/5*B*a*b*x^5 + 1/4*C*a^2*x^4 + 1/2* A*a*b*x^4 + 1/3*B*a^2*x^3 + 1/2*A*a^2*x^2
Time = 5.73 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int x \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a^2\,x^5\,D}{5}+\frac {b^2\,x^9\,D}{9}+\frac {A\,x^2\,\left (3\,a^2+3\,a\,b\,x^2+b^2\,x^4\right )}{6}+\frac {B\,x^3\,\left (35\,a^2+42\,a\,b\,x^2+15\,b^2\,x^4\right )}{105}+\frac {C\,x^4\,\left (6\,a^2+8\,a\,b\,x^2+3\,b^2\,x^4\right )}{24}+\frac {2\,a\,b\,x^7\,D}{7} \]