Integrand size = 20, antiderivative size = 87 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=a^3 A x+\frac {1}{3} a^2 (3 A b+a C) x^3+\frac {3}{5} a b (A b+a C) x^5+\frac {1}{7} b^2 (A b+3 a C) x^7+\frac {1}{9} b^3 C x^9+\frac {B \left (a+b x^2\right )^4}{8 b} \] Output:
a^3*A*x+1/3*a^2*(3*A*b+C*a)*x^3+3/5*a*b*(A*b+C*a)*x^5+1/7*b^2*(A*b+3*C*a)* x^7+1/9*b^3*C*x^9+1/8*B*(b*x^2+a)^4/b
Time = 0.02 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.15 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{6} a^3 x (6 A+x (3 B+2 C x))+\frac {1}{20} a^2 b x^3 (20 A+3 x (5 B+4 C x))+\frac {1}{70} a b^2 x^5 (42 A+5 x (7 B+6 C x))+\frac {1}{504} b^3 x^7 (72 A+7 x (9 B+8 C x)) \] Input:
Integrate[(a + b*x^2)^3*(A + B*x + C*x^2),x]
Output:
(a^3*x*(6*A + x*(3*B + 2*C*x)))/6 + (a^2*b*x^3*(20*A + 3*x*(5*B + 4*C*x))) /20 + (a*b^2*x^5*(42*A + 5*x*(7*B + 6*C*x)))/70 + (b^3*x^7*(72*A + 7*x*(9* B + 8*C*x)))/504
Time = 0.26 (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.
\(\displaystyle \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^2+a\right )^3 \left (C x^2+A\right )dx+\frac {B \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \int \left (b^3 C x^8+b^2 (A b+3 a C) x^6+3 a b (A b+a C) x^4+a^2 (3 A b+a C) x^2+a^3 A\right )dx+\frac {B \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 A x+\frac {1}{3} a^2 x^3 (a C+3 A b)+\frac {1}{7} b^2 x^7 (3 a C+A b)+\frac {3}{5} a b x^5 (a C+A b)+\frac {B \left (a+b x^2\right )^4}{8 b}+\frac {1}{9} b^3 C x^9\) |
Input:
Int[(a + b*x^2)^3*(A + B*x + C*x^2),x]
Output:
a^3*A*x + (a^2*(3*A*b + a*C)*x^3)/3 + (3*a*b*(A*b + a*C)*x^5)/5 + (b^2*(A* b + 3*a*C)*x^7)/7 + (b^3*C*x^9)/9 + (B*(a + b*x^2)^4)/(8*b)
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 = 0.43 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.25
method | result | size |
norman | \(\frac {b^{3} C \,x^{9}}{9}+\frac {b^{3} B \,x^{8}}{8}+\left (\frac {1}{7} b^{3} A +\frac {3}{7} a C \,b^{2}\right ) x^{7}+\frac {B a \,b^{2} x^{6}}{2}+\left (\frac {3}{5} a \,b^{2} A +\frac {3}{5} a^{2} b C \right ) x^{5}+\frac {3 B \,a^{2} b \,x^{4}}{4}+\left (a^{2} b A +\frac {1}{3} C \,a^{3}\right ) x^{3}+\frac {B \,a^{3} x^{2}}{2}+a^{3} A x\) | \(109\) |
default | \(\frac {b^{3} C \,x^{9}}{9}+\frac {b^{3} B \,x^{8}}{8}+\frac {\left (b^{3} A +3 a C \,b^{2}\right ) x^{7}}{7}+\frac {B a \,b^{2} x^{6}}{2}+\frac {\left (3 a \,b^{2} A +3 a^{2} b C \right ) x^{5}}{5}+\frac {3 B \,a^{2} b \,x^{4}}{4}+\frac {\left (3 a^{2} b A +C \,a^{3}\right ) x^{3}}{3}+\frac {B \,a^{3} x^{2}}{2}+a^{3} A x\) | \(111\) |
gosper | \(\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a C \,b^{2}+\frac {1}{2} B a \,b^{2} x^{6}+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b C +\frac {3}{4} B \,a^{2} b \,x^{4}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(112\) |
risch | \(\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a C \,b^{2}+\frac {1}{2} B a \,b^{2} x^{6}+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b C +\frac {3}{4} B \,a^{2} b \,x^{4}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(112\) |
parallelrisch | \(\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a C \,b^{2}+\frac {1}{2} B a \,b^{2} x^{6}+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b C +\frac {3}{4} B \,a^{2} b \,x^{4}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(112\) |
orering | \(\frac {x \left (280 b^{3} C \,x^{8}+315 b^{3} B \,x^{7}+360 A \,b^{3} x^{6}+1080 C a \,b^{2} x^{6}+1260 B a \,b^{2} x^{5}+1512 a A \,b^{2} x^{4}+1512 C \,a^{2} b \,x^{4}+1890 B \,a^{2} b \,x^{3}+2520 a^{2} A b \,x^{2}+840 C \,a^{3} x^{2}+1260 B \,a^{3} x +2520 a^{3} A \right )}{2520}\) | \(114\) |
Input:
int((b*x^2+a)^3*(C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
1/9*b^3*C*x^9+1/8*b^3*B*x^8+(1/7*b^3*A+3/7*a*C*b^2)*x^7+1/2*B*a*b^2*x^6+(3 /5*a*b^2*A+3/5*a^2*b*C)*x^5+3/4*B*a^2*b*x^4+(a^2*b*A+1/3*C*a^3)*x^3+1/2*B* a^3*x^2+a^3*A*x
Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, B b^{3} x^{8} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {1}{7} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate((b*x^2+a)^3*(C*x^2+B*x+A),x, algorithm="fricas")
Output:
1/9*C*b^3*x^9 + 1/8*B*b^3*x^8 + 1/2*B*a*b^2*x^6 + 3/4*B*a^2*b*x^4 + 1/7*(3 *C*a*b^2 + A*b^3)*x^7 + 1/2*B*a^3*x^2 + 3/5*(C*a^2*b + A*a*b^2)*x^5 + A*a^ 3*x + 1/3*(C*a^3 + 3*A*a^2*b)*x^3
Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.40 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=A a^{3} x + \frac {B a^{3} x^{2}}{2} + \frac {3 B a^{2} b x^{4}}{4} + \frac {B a b^{2} x^{6}}{2} + \frac {B b^{3} x^{8}}{8} + \frac {C b^{3} x^{9}}{9} + x^{7} \left (\frac {A b^{3}}{7} + \frac {3 C a b^{2}}{7}\right ) + x^{5} \cdot \left (\frac {3 A a b^{2}}{5} + \frac {3 C a^{2} b}{5}\right ) + x^{3} \left (A a^{2} b + \frac {C a^{3}}{3}\right ) \] Input:
integrate((b*x**2+a)**3*(C*x**2+B*x+A),x)
Output:
A*a**3*x + B*a**3*x**2/2 + 3*B*a**2*b*x**4/4 + B*a*b**2*x**6/2 + B*b**3*x* *8/8 + C*b**3*x**9/9 + x**7*(A*b**3/7 + 3*C*a*b**2/7) + x**5*(3*A*a*b**2/5 + 3*C*a**2*b/5) + x**3*(A*a**2*b + C*a**3/3)
Time = 0.03 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.24 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, B b^{3} x^{8} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {1}{7} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate((b*x^2+a)^3*(C*x^2+B*x+A),x, algorithm="maxima")
Output:
1/9*C*b^3*x^9 + 1/8*B*b^3*x^8 + 1/2*B*a*b^2*x^6 + 3/4*B*a^2*b*x^4 + 1/7*(3 *C*a*b^2 + A*b^3)*x^7 + 1/2*B*a^3*x^2 + 3/5*(C*a^2*b + A*a*b^2)*x^5 + A*a^ 3*x + 1/3*(C*a^3 + 3*A*a^2*b)*x^3
Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.28 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, B b^{3} x^{8} + \frac {3}{7} \, C a b^{2} x^{7} + \frac {1}{7} \, A b^{3} x^{7} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {3}{5} \, C a^{2} b x^{5} + \frac {3}{5} \, A a b^{2} x^{5} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {1}{3} \, C a^{3} x^{3} + A a^{2} b x^{3} + \frac {1}{2} \, B a^{3} x^{2} + A a^{3} x \] Input:
integrate((b*x^2+a)^3*(C*x^2+B*x+A),x, algorithm="giac")
Output:
1/9*C*b^3*x^9 + 1/8*B*b^3*x^8 + 3/7*C*a*b^2*x^7 + 1/7*A*b^3*x^7 + 1/2*B*a* b^2*x^6 + 3/5*C*a^2*b*x^5 + 3/5*A*a*b^2*x^5 + 3/4*B*a^2*b*x^4 + 1/3*C*a^3* x^3 + A*a^2*b*x^3 + 1/2*B*a^3*x^2 + A*a^3*x
Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.18 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=x^3\,\left (\frac {C\,a^3}{3}+A\,b\,a^2\right )+x^7\,\left (\frac {A\,b^3}{7}+\frac {3\,C\,a\,b^2}{7}\right )+\frac {B\,a^3\,x^2}{2}+\frac {B\,b^3\,x^8}{8}+\frac {C\,b^3\,x^9}{9}+A\,a^3\,x+\frac {3\,a\,b\,x^5\,\left (A\,b+C\,a\right )}{5}+\frac {3\,B\,a^2\,b\,x^4}{4}+\frac {B\,a\,b^2\,x^6}{2} \] Input:
int((a + b*x^2)^3*(A + B*x + C*x^2),x)
Output:
x^3*((C*a^3)/3 + A*a^2*b) + x^7*((A*b^3)/7 + (3*C*a*b^2)/7) + (B*a^3*x^2)/ 2 + (B*b^3*x^8)/8 + (C*b^3*x^9)/9 + A*a^3*x + (3*a*b*x^5*(A*b + C*a))/5 + (3*B*a^2*b*x^4)/4 + (B*a*b^2*x^6)/2
Time = 0.15 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.28 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2\right ) \, dx=\frac {x \left (280 b^{3} c \,x^{8}+315 b^{4} x^{7}+360 a \,b^{3} x^{6}+1080 a \,b^{2} c \,x^{6}+1260 a \,b^{3} x^{5}+1512 a^{2} b^{2} x^{4}+1512 a^{2} b c \,x^{4}+1890 a^{2} b^{2} x^{3}+2520 a^{3} b \,x^{2}+840 a^{3} c \,x^{2}+1260 a^{3} b x +2520 a^{4}\right )}{2520} \] Input:
int((b*x^2+a)^3*(C*x^2+B*x+A),x)
Output:
(x*(2520*a**4 + 2520*a**3*b*x**2 + 1260*a**3*b*x + 840*a**3*c*x**2 + 1512* a**2*b**2*x**4 + 1890*a**2*b**2*x**3 + 1512*a**2*b*c*x**4 + 360*a*b**3*x** 6 + 1260*a*b**3*x**5 + 1080*a*b**2*c*x**6 + 315*b**4*x**7 + 280*b**3*c*x** 8))/2520