Integrand size = 20, antiderivative size = 87 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=a^3 A x+\frac {1}{3} a^2 (3 A c+a C) x^3+\frac {3}{5} a c (A c+a C) x^5+\frac {1}{7} c^2 (A c+3 a C) x^7+\frac {1}{9} c^3 C x^9+\frac {B \left (a+c x^2\right )^4}{8 c} \]
a^3*A*x+1/3*a^2*(3*A*c+C*a)*x^3+3/5*a*c*(A*c+C*a)*x^5+1/7*c^2*(A*c+3*C*a)* x^7+1/9*c^3*C*x^9+1/8*B*(c*x^2+a)^4/c
Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.15 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{6} a^3 x (6 A+x (3 B+2 C x))+\frac {1}{20} a^2 c x^3 (20 A+3 x (5 B+4 C x))+\frac {1}{70} a c^2 x^5 (42 A+5 x (7 B+6 C x))+\frac {1}{504} c^3 x^7 (72 A+7 x (9 B+8 C x)) \]
(a^3*x*(6*A + x*(3*B + 2*C*x)))/6 + (a^2*c*x^3*(20*A + 3*x*(5*B + 4*C*x))) /20 + (a*c^2*x^5*(42*A + 5*x*(7*B + 6*C*x)))/70 + (c^3*x^7*(72*A + 7*x*(9* B + 8*C*x)))/504
Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2017, 290, 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.
\(\displaystyle \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (c x^2+a\right )^3 \left (C x^2+A\right )dx+\frac {B \left (a+c x^2\right )^4}{8 c}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \int \left (c^3 C x^8+c^2 (A c+3 a C) x^6+3 a c (A c+a C) x^4+a^2 (3 A c+a C) x^2+a^3 A\right )dx+\frac {B \left (a+c x^2\right )^4}{8 c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 A x+\frac {1}{3} a^2 x^3 (a C+3 A c)+\frac {1}{7} c^2 x^7 (3 a C+A c)+\frac {3}{5} a c x^5 (a C+A c)+\frac {B \left (a+c x^2\right )^4}{8 c}+\frac {1}{9} c^3 C x^9\) |
a^3*A*x + (a^2*(3*A*c + a*C)*x^3)/3 + (3*a*c*(A*c + a*C)*x^5)/5 + (c^2*(A* c + 3*a*C)*x^7)/7 + (c^3*C*x^9)/9 + (B*(a + c*x^2)^4)/(8*c)
3.1.35.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
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]]
Time = 0.54 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.25
method | result | size |
norman | \(\frac {c^{3} C \,x^{9}}{9}+\frac {B \,c^{3} x^{8}}{8}+\left (\frac {1}{7} A \,c^{3}+\frac {3}{7} a \,c^{2} C \right ) x^{7}+\frac {B a \,c^{2} x^{6}}{2}+\left (\frac {3}{5} A a \,c^{2}+\frac {3}{5} c \,a^{2} C \right ) x^{5}+\frac {3 B \,a^{2} c \,x^{4}}{4}+\left (A \,a^{2} c +\frac {1}{3} a^{3} C \right ) x^{3}+\frac {B \,a^{3} x^{2}}{2}+a^{3} A x\) | \(109\) |
default | \(\frac {c^{3} C \,x^{9}}{9}+\frac {B \,c^{3} x^{8}}{8}+\frac {\left (A \,c^{3}+3 a \,c^{2} C \right ) x^{7}}{7}+\frac {B a \,c^{2} x^{6}}{2}+\frac {\left (3 A a \,c^{2}+3 c \,a^{2} C \right ) x^{5}}{5}+\frac {3 B \,a^{2} c \,x^{4}}{4}+\frac {\left (3 A \,a^{2} c +a^{3} C \right ) x^{3}}{3}+\frac {B \,a^{3} x^{2}}{2}+a^{3} A x\) | \(111\) |
gosper | \(\frac {1}{9} c^{3} C \,x^{9}+\frac {1}{8} B \,c^{3} x^{8}+\frac {1}{7} A \,c^{3} x^{7}+\frac {3}{7} x^{7} a \,c^{2} C +\frac {1}{2} B a \,c^{2} x^{6}+\frac {3}{5} a A \,c^{2} x^{5}+\frac {3}{5} x^{5} c \,a^{2} C +\frac {3}{4} B \,a^{2} c \,x^{4}+a^{2} A c \,x^{3}+\frac {1}{3} x^{3} a^{3} C +\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(112\) |
risch | \(\frac {1}{9} c^{3} C \,x^{9}+\frac {1}{8} B \,c^{3} x^{8}+\frac {1}{7} A \,c^{3} x^{7}+\frac {3}{7} x^{7} a \,c^{2} C +\frac {1}{2} B a \,c^{2} x^{6}+\frac {3}{5} a A \,c^{2} x^{5}+\frac {3}{5} x^{5} c \,a^{2} C +\frac {3}{4} B \,a^{2} c \,x^{4}+a^{2} A c \,x^{3}+\frac {1}{3} x^{3} a^{3} C +\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(112\) |
parallelrisch | \(\frac {1}{9} c^{3} C \,x^{9}+\frac {1}{8} B \,c^{3} x^{8}+\frac {1}{7} A \,c^{3} x^{7}+\frac {3}{7} x^{7} a \,c^{2} C +\frac {1}{2} B a \,c^{2} x^{6}+\frac {3}{5} a A \,c^{2} x^{5}+\frac {3}{5} x^{5} c \,a^{2} C +\frac {3}{4} B \,a^{2} c \,x^{4}+a^{2} A c \,x^{3}+\frac {1}{3} x^{3} a^{3} C +\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(112\) |
1/9*c^3*C*x^9+1/8*B*c^3*x^8+(1/7*A*c^3+3/7*a*c^2*C)*x^7+1/2*B*a*c^2*x^6+(3 /5*A*a*c^2+3/5*c*a^2*C)*x^5+3/4*B*a^2*c*x^4+(A*a^2*c+1/3*a^3*C)*x^3+1/2*B* a^3*x^2+a^3*A*x
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{9} \, C c^{3} x^{9} + \frac {1}{8} \, B c^{3} x^{8} + \frac {1}{2} \, B a c^{2} x^{6} + \frac {3}{4} \, B a^{2} c x^{4} + \frac {1}{7} \, {\left (3 \, C a c^{2} + A c^{3}\right )} x^{7} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} c + A a c^{2}\right )} x^{5} + A a^{3} x + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} c\right )} x^{3} \]
1/9*C*c^3*x^9 + 1/8*B*c^3*x^8 + 1/2*B*a*c^2*x^6 + 3/4*B*a^2*c*x^4 + 1/7*(3 *C*a*c^2 + A*c^3)*x^7 + 1/2*B*a^3*x^2 + 3/5*(C*a^2*c + A*a*c^2)*x^5 + A*a^ 3*x + 1/3*(C*a^3 + 3*A*a^2*c)*x^3
Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.40 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=A a^{3} x + \frac {B a^{3} x^{2}}{2} + \frac {3 B a^{2} c x^{4}}{4} + \frac {B a c^{2} x^{6}}{2} + \frac {B c^{3} x^{8}}{8} + \frac {C c^{3} x^{9}}{9} + x^{7} \left (\frac {A c^{3}}{7} + \frac {3 C a c^{2}}{7}\right ) + x^{5} \cdot \left (\frac {3 A a c^{2}}{5} + \frac {3 C a^{2} c}{5}\right ) + x^{3} \left (A a^{2} c + \frac {C a^{3}}{3}\right ) \]
A*a**3*x + B*a**3*x**2/2 + 3*B*a**2*c*x**4/4 + B*a*c**2*x**6/2 + B*c**3*x* *8/8 + C*c**3*x**9/9 + x**7*(A*c**3/7 + 3*C*a*c**2/7) + x**5*(3*A*a*c**2/5 + 3*C*a**2*c/5) + x**3*(A*a**2*c + C*a**3/3)
Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{9} \, C c^{3} x^{9} + \frac {1}{8} \, B c^{3} x^{8} + \frac {1}{2} \, B a c^{2} x^{6} + \frac {3}{4} \, B a^{2} c x^{4} + \frac {1}{7} \, {\left (3 \, C a c^{2} + A c^{3}\right )} x^{7} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} c + A a c^{2}\right )} x^{5} + A a^{3} x + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} c\right )} x^{3} \]
1/9*C*c^3*x^9 + 1/8*B*c^3*x^8 + 1/2*B*a*c^2*x^6 + 3/4*B*a^2*c*x^4 + 1/7*(3 *C*a*c^2 + A*c^3)*x^7 + 1/2*B*a^3*x^2 + 3/5*(C*a^2*c + A*a*c^2)*x^5 + A*a^ 3*x + 1/3*(C*a^3 + 3*A*a^2*c)*x^3
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.28 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{9} \, C c^{3} x^{9} + \frac {1}{8} \, B c^{3} x^{8} + \frac {3}{7} \, C a c^{2} x^{7} + \frac {1}{7} \, A c^{3} x^{7} + \frac {1}{2} \, B a c^{2} x^{6} + \frac {3}{5} \, C a^{2} c x^{5} + \frac {3}{5} \, A a c^{2} x^{5} + \frac {3}{4} \, B a^{2} c x^{4} + \frac {1}{3} \, C a^{3} x^{3} + A a^{2} c x^{3} + \frac {1}{2} \, B a^{3} x^{2} + A a^{3} x \]
1/9*C*c^3*x^9 + 1/8*B*c^3*x^8 + 3/7*C*a*c^2*x^7 + 1/7*A*c^3*x^7 + 1/2*B*a* c^2*x^6 + 3/5*C*a^2*c*x^5 + 3/5*A*a*c^2*x^5 + 3/4*B*a^2*c*x^4 + 1/3*C*a^3* x^3 + A*a^2*c*x^3 + 1/2*B*a^3*x^2 + A*a^3*x
Time = 12.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.18 \[ \int \left (a+c x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=x^3\,\left (\frac {C\,a^3}{3}+A\,c\,a^2\right )+x^7\,\left (\frac {A\,c^3}{7}+\frac {3\,C\,a\,c^2}{7}\right )+\frac {B\,a^3\,x^2}{2}+\frac {B\,c^3\,x^8}{8}+\frac {C\,c^3\,x^9}{9}+A\,a^3\,x+\frac {3\,a\,c\,x^5\,\left (A\,c+C\,a\right )}{5}+\frac {3\,B\,a^2\,c\,x^4}{4}+\frac {B\,a\,c^2\,x^6}{2} \]