Integrand size = 25, antiderivative size = 133 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=a^3 A x+\frac {1}{3} a^2 (3 A b+a C) x^3+\frac {1}{4} a^3 D x^4+\frac {3}{5} a b (A b+a C) x^5+\frac {1}{2} a^2 b D x^6+\frac {1}{7} b^2 (A b+3 a C) x^7+\frac {3}{8} a b^2 D x^8+\frac {1}{9} b^3 C x^9+\frac {1}{10} b^3 D x^{10}+\frac {B \left (a+b x^2\right )^4}{8 b} \] Output:
a^3*A*x+1/3*a^2*(3*A*b+C*a)*x^3+1/4*a^3*D*x^4+3/5*a*b*(A*b+C*a)*x^5+1/2*a^ 2*b*D*x^6+1/7*b^2*(A*b+3*C*a)*x^7+3/8*a*b^2*D*x^8+1/9*b^3*C*x^9+1/10*b^3*D *x^10+1/8*B*(b*x^2+a)^4/b
Time = 0.06 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.91 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {210 a^3 x (12 A+x (6 B+x (4 C+3 D x)))+126 a^2 b x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+9 a b^2 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))+b^3 x^7 (360 A+7 x (45 B+4 x (10 C+9 D x)))}{2520} \] Input:
Integrate[(a + b*x^2)^3*(A + B*x + C*x^2 + D*x^3),x]
Output:
(210*a^3*x*(12*A + x*(6*B + x*(4*C + 3*D*x))) + 126*a^2*b*x^3*(20*A + x*(1 5*B + 2*x*(6*C + 5*D*x))) + 9*a*b^2*x^5*(168*A + 5*x*(28*B + 3*x*(8*C + 7* D*x))) + b^3*x^7*(360*A + 7*x*(45*B + 4*x*(10*C + 9*D*x))))/2520
Time = 0.37 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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.
\(\displaystyle \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^2+a\right )^3 \left (D x^3+C x^2+A\right )dx+\frac {B \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \int \left (b^3 D x^9+b^3 C x^8+3 a b^2 D x^7+b^2 (A b+3 a C) x^6+3 a^2 b D x^5+3 a b (A b+a C) x^4+a^3 D x^3+a^2 (3 A b+a C) x^2+a^3 A\right )dx+\frac {B \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 A x+\frac {1}{4} a^3 D x^4+\frac {1}{3} a^2 x^3 (a C+3 A b)+\frac {1}{2} a^2 b D x^6+\frac {1}{7} b^2 x^7 (3 a C+A b)+\frac {3}{5} a b x^5 (a C+A b)+\frac {3}{8} a b^2 D x^8+\frac {B \left (a+b x^2\right )^4}{8 b}+\frac {1}{9} b^3 C x^9+\frac {1}{10} b^3 D x^{10}\) |
Input:
Int[(a + b*x^2)^3*(A + B*x + C*x^2 + D*x^3),x]
Output:
a^3*A*x + (a^2*(3*A*b + a*C)*x^3)/3 + (a^3*D*x^4)/4 + (3*a*b*(A*b + a*C)*x ^5)/5 + (a^2*b*D*x^6)/2 + (b^2*(A*b + 3*a*C)*x^7)/7 + (3*a*b^2*D*x^8)/8 + (b^3*C*x^9)/9 + (b^3*D*x^10)/10 + (B*(a + b*x^2)^4)/(8*b)
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 = 0.37 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08
method | result | size |
norman | \(\frac {b^{3} D x^{10}}{10}+\frac {b^{3} C \,x^{9}}{9}+\left (\frac {1}{8} B \,b^{3}+\frac {3}{8} a \,b^{2} D\right ) x^{8}+\left (\frac {1}{7} b^{3} A +\frac {3}{7} a C \,b^{2}\right ) x^{7}+\left (\frac {1}{2} a \,b^{2} B +\frac {1}{2} a^{2} b D\right ) x^{6}+\left (\frac {3}{5} a \,b^{2} A +\frac {3}{5} a^{2} b C \right ) x^{5}+\left (\frac {3}{4} a^{2} b B +\frac {1}{4} a^{3} D\right ) x^{4}+\left (a^{2} b A +\frac {1}{3} C \,a^{3}\right ) x^{3}+\frac {B \,a^{3} x^{2}}{2}+a^{3} A x\) | \(144\) |
default | \(\frac {b^{3} D x^{10}}{10}+\frac {b^{3} C \,x^{9}}{9}+\frac {\left (B \,b^{3}+3 a \,b^{2} D\right ) x^{8}}{8}+\frac {\left (b^{3} A +3 a C \,b^{2}\right ) x^{7}}{7}+\frac {\left (3 a \,b^{2} B +3 a^{2} b D\right ) x^{6}}{6}+\frac {\left (3 a \,b^{2} A +3 a^{2} b C \right ) x^{5}}{5}+\frac {\left (3 a^{2} b B +a^{3} D\right ) x^{4}}{4}+\frac {\left (3 a^{2} b A +C \,a^{3}\right ) x^{3}}{3}+\frac {B \,a^{3} x^{2}}{2}+a^{3} A x\) | \(147\) |
gosper | \(\frac {1}{10} b^{3} D x^{10}+\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {3}{8} a \,b^{2} D x^{8}+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a C \,b^{2}+\frac {1}{2} B a \,b^{2} x^{6}+\frac {1}{2} a^{2} b D x^{6}+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b C +\frac {3}{4} B \,a^{2} b \,x^{4}+\frac {1}{4} a^{3} D x^{4}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(150\) |
parallelrisch | \(\frac {1}{10} b^{3} D x^{10}+\frac {1}{9} b^{3} C \,x^{9}+\frac {1}{8} b^{3} B \,x^{8}+\frac {3}{8} a \,b^{2} D x^{8}+\frac {1}{7} A \,b^{3} x^{7}+\frac {3}{7} x^{7} a C \,b^{2}+\frac {1}{2} B a \,b^{2} x^{6}+\frac {1}{2} a^{2} b D x^{6}+\frac {3}{5} a A \,b^{2} x^{5}+\frac {3}{5} x^{5} a^{2} b C +\frac {3}{4} B \,a^{2} b \,x^{4}+\frac {1}{4} a^{3} D x^{4}+a^{2} A b \,x^{3}+\frac {1}{3} x^{3} C \,a^{3}+\frac {1}{2} B \,a^{3} x^{2}+a^{3} A x\) | \(150\) |
orering | \(\frac {x \left (252 b^{3} D x^{9}+280 b^{3} C \,x^{8}+315 b^{3} B \,x^{7}+945 D a \,b^{2} x^{7}+360 A \,b^{3} x^{6}+1080 C a \,b^{2} x^{6}+1260 B a \,b^{2} x^{5}+1260 D a^{2} b \,x^{5}+1512 a A \,b^{2} x^{4}+1512 C \,a^{2} b \,x^{4}+1890 B \,a^{2} b \,x^{3}+630 D a^{3} x^{3}+2520 a^{2} A b \,x^{2}+840 C \,a^{3} x^{2}+1260 B \,a^{3} x +2520 a^{3} A \right )}{2520}\) | \(152\) |
Input:
int((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
1/10*b^3*D*x^10+1/9*b^3*C*x^9+(1/8*B*b^3+3/8*a*b^2*D)*x^8+(1/7*b^3*A+3/7*a *C*b^2)*x^7+(1/2*a*b^2*B+1/2*a^2*b*D)*x^6+(3/5*a*b^2*A+3/5*a^2*b*C)*x^5+(3 /4*a^2*b*B+1/4*a^3*D)*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 = 0.06 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{10} \, D b^{3} x^{10} + \frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{8} + \frac {1}{7} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, {\left (D a^{2} b + B a b^{2}\right )} x^{6} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{4} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{4} + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
1/10*D*b^3*x^10 + 1/9*C*b^3*x^9 + 1/8*(3*D*a*b^2 + B*b^3)*x^8 + 1/7*(3*C*a *b^2 + A*b^3)*x^7 + 1/2*(D*a^2*b + B*a*b^2)*x^6 + 1/2*B*a^3*x^2 + 3/5*(C*a ^2*b + A*a*b^2)*x^5 + A*a^3*x + 1/4*(D*a^3 + 3*B*a^2*b)*x^4 + 1/3*(C*a^3 + 3*A*a^2*b)*x^3
Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.19 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=A a^{3} x + \frac {B a^{3} x^{2}}{2} + \frac {C b^{3} x^{9}}{9} + \frac {D b^{3} x^{10}}{10} + x^{8} \left (\frac {B b^{3}}{8} + \frac {3 D a b^{2}}{8}\right ) + x^{7} \left (\frac {A b^{3}}{7} + \frac {3 C a b^{2}}{7}\right ) + x^{6} \left (\frac {B a b^{2}}{2} + \frac {D a^{2} b}{2}\right ) + x^{5} \cdot \left (\frac {3 A a b^{2}}{5} + \frac {3 C a^{2} b}{5}\right ) + x^{4} \cdot \left (\frac {3 B a^{2} b}{4} + \frac {D a^{3}}{4}\right ) + x^{3} \left (A a^{2} b + \frac {C a^{3}}{3}\right ) \] Input:
integrate((b*x**2+a)**3*(D*x**3+C*x**2+B*x+A),x)
Output:
A*a**3*x + B*a**3*x**2/2 + C*b**3*x**9/9 + D*b**3*x**10/10 + x**8*(B*b**3/ 8 + 3*D*a*b**2/8) + x**7*(A*b**3/7 + 3*C*a*b**2/7) + x**6*(B*a*b**2/2 + D* a**2*b/2) + x**5*(3*A*a*b**2/5 + 3*C*a**2*b/5) + x**4*(3*B*a**2*b/4 + D*a* *3/4) + x**3*(A*a**2*b + C*a**3/3)
Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{10} \, D b^{3} x^{10} + \frac {1}{9} \, C b^{3} x^{9} + \frac {1}{8} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{8} + \frac {1}{7} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{7} + \frac {1}{2} \, {\left (D a^{2} b + B a b^{2}\right )} x^{6} + \frac {1}{2} \, B a^{3} x^{2} + \frac {3}{5} \, {\left (C a^{2} b + A a b^{2}\right )} x^{5} + A a^{3} x + \frac {1}{4} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{4} + \frac {1}{3} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{3} \] Input:
integrate((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
1/10*D*b^3*x^10 + 1/9*C*b^3*x^9 + 1/8*(3*D*a*b^2 + B*b^3)*x^8 + 1/7*(3*C*a *b^2 + A*b^3)*x^7 + 1/2*(D*a^2*b + B*a*b^2)*x^6 + 1/2*B*a^3*x^2 + 3/5*(C*a ^2*b + A*a*b^2)*x^5 + A*a^3*x + 1/4*(D*a^3 + 3*B*a^2*b)*x^4 + 1/3*(C*a^3 + 3*A*a^2*b)*x^3
Time = 0.13 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{10} \, D b^{3} x^{10} + \frac {1}{9} \, C b^{3} x^{9} + \frac {3}{8} \, D a b^{2} x^{8} + \frac {1}{8} \, B b^{3} x^{8} + \frac {3}{7} \, C a b^{2} x^{7} + \frac {1}{7} \, A b^{3} x^{7} + \frac {1}{2} \, D a^{2} b x^{6} + \frac {1}{2} \, B a b^{2} x^{6} + \frac {3}{5} \, C a^{2} b x^{5} + \frac {3}{5} \, A a b^{2} x^{5} + \frac {1}{4} \, D a^{3} x^{4} + \frac {3}{4} \, B a^{2} b x^{4} + \frac {1}{3} \, C a^{3} x^{3} + A a^{2} b x^{3} + \frac {1}{2} \, B a^{3} x^{2} + A a^{3} x \] Input:
integrate((b*x^2+a)^3*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
1/10*D*b^3*x^10 + 1/9*C*b^3*x^9 + 3/8*D*a*b^2*x^8 + 1/8*B*b^3*x^8 + 3/7*C* a*b^2*x^7 + 1/7*A*b^3*x^7 + 1/2*D*a^2*b*x^6 + 1/2*B*a*b^2*x^6 + 3/5*C*a^2* b*x^5 + 3/5*A*a*b^2*x^5 + 1/4*D*a^3*x^4 + 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 = 0.69 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {B\,a^3\,x^2}{2}+\frac {A\,b^3\,x^7}{7}+\frac {C\,a^3\,x^3}{3}+\frac {B\,b^3\,x^8}{8}+\frac {C\,b^3\,x^9}{9}+\frac {a^3\,x^4\,D}{4}+\frac {b^3\,x^{10}\,D}{10}+A\,a^3\,x+\frac {a^2\,b\,x^6\,D}{2}+\frac {3\,a\,b^2\,x^8\,D}{8}+A\,a^2\,b\,x^3+\frac {3\,A\,a\,b^2\,x^5}{5}+\frac {3\,B\,a^2\,b\,x^4}{4}+\frac {B\,a\,b^2\,x^6}{2}+\frac {3\,C\,a^2\,b\,x^5}{5}+\frac {3\,C\,a\,b^2\,x^7}{7} \] Input:
int((a + b*x^2)^3*(A + B*x + C*x^2 + x^3*D),x)
Output:
(B*a^3*x^2)/2 + (A*b^3*x^7)/7 + (C*a^3*x^3)/3 + (B*b^3*x^8)/8 + (C*b^3*x^9 )/9 + (a^3*x^4*D)/4 + (b^3*x^10*D)/10 + A*a^3*x + (a^2*b*x^6*D)/2 + (3*a*b ^2*x^8*D)/8 + A*a^2*b*x^3 + (3*A*a*b^2*x^5)/5 + (3*B*a^2*b*x^4)/4 + (B*a*b ^2*x^6)/2 + (3*C*a^2*b*x^5)/5 + (3*C*a*b^2*x^7)/7
Time = 0.17 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.12 \[ \int \left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {x \left (252 b^{3} d \,x^{9}+280 b^{3} c \,x^{8}+945 a \,b^{2} d \,x^{7}+315 b^{4} x^{7}+360 a \,b^{3} x^{6}+1080 a \,b^{2} c \,x^{6}+1260 a^{2} b d \,x^{5}+1260 a \,b^{3} x^{5}+1512 a^{2} b^{2} x^{4}+1512 a^{2} b c \,x^{4}+630 a^{3} d \,x^{3}+1890 a^{2} b^{2} x^{3}+2520 a^{3} b \,x^{2}+840 a^{3} c \,x^{2}+1260 a^{3} b x +2520 a^{4}\right )}{2520} \] Input:
int((b*x^2+a)^3*(D*x^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 + 630*a **3*d*x**3 + 1512*a**2*b**2*x**4 + 1890*a**2*b**2*x**3 + 1512*a**2*b*c*x** 4 + 1260*a**2*b*d*x**5 + 360*a*b**3*x**6 + 1260*a*b**3*x**5 + 1080*a*b**2* c*x**6 + 945*a*b**2*d*x**7 + 315*b**4*x**7 + 280*b**3*c*x**8 + 252*b**3*d* x**9))/2520