Integrand size = 25, antiderivative size = 99 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=a^2 A x+\frac {1}{3} a (2 A b+a C) x^3+\frac {1}{4} a^2 D x^4+\frac {1}{5} b (A b+2 a C) x^5+\frac {1}{3} a b D x^6+\frac {1}{7} b^2 C x^7+\frac {1}{8} b^2 D x^8+\frac {B \left (a+b x^2\right )^3}{6 b} \] Output:
a^2*A*x+1/3*a*(2*A*b+C*a)*x^3+1/4*a^2*D*x^4+1/5*b*(A*b+2*C*a)*x^5+1/3*a*b* D*x^6+1/7*b^2*C*x^7+1/8*b^2*D*x^8+1/6*B*(b*x^2+a)^3/b
Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.89 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{840} \left (70 a^2 x (12 A+x (6 B+x (4 C+3 D x)))+28 a b x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+b^2 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))\right ) \] Input:
Integrate[(a + b*x^2)^2*(A + B*x + C*x^2 + D*x^3),x]
Output:
(70*a^2*x*(12*A + x*(6*B + x*(4*C + 3*D*x))) + 28*a*b*x^3*(20*A + x*(15*B + 2*x*(6*C + 5*D*x))) + b^2*x^5*(168*A + 5*x*(28*B + 3*x*(8*C + 7*D*x))))/ 840
Time = 0.30 (sec) , antiderivative size = 99, 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.
\(\displaystyle \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^2+a\right )^2 \left (D x^3+C x^2+A\right )dx+\frac {B \left (a+b x^2\right )^3}{6 b}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle \int \left (b^2 D x^7+b^2 C x^6+2 a b D x^5+b (A b+2 a C) x^4+a^2 D x^3+a (2 A b+a C) x^2+a^2 A\right )dx+\frac {B \left (a+b x^2\right )^3}{6 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^2 A x+\frac {1}{4} a^2 D x^4+\frac {1}{5} b x^5 (2 a C+A b)+\frac {1}{3} a x^3 (a C+2 A b)+\frac {B \left (a+b x^2\right )^3}{6 b}+\frac {1}{3} a b D x^6+\frac {1}{7} b^2 C x^7+\frac {1}{8} b^2 D x^8\) |
Input:
Int[(a + b*x^2)^2*(A + B*x + C*x^2 + D*x^3),x]
Output:
a^2*A*x + (a*(2*A*b + a*C)*x^3)/3 + (a^2*D*x^4)/4 + (b*(A*b + 2*a*C)*x^5)/ 5 + (a*b*D*x^6)/3 + (b^2*C*x^7)/7 + (b^2*D*x^8)/8 + (B*(a + b*x^2)^3)/(6*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 = 99, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {b^{2} D x^{8}}{8}+\frac {b^{2} C \,x^{7}}{7}+\frac {\left (B \,b^{2}+2 a b D\right ) x^{6}}{6}+\frac {\left (b^{2} A +2 C a b \right ) x^{5}}{5}+\frac {\left (2 a b B +D a^{2}\right ) x^{4}}{4}+\frac {\left (2 a b A +a^{2} C \right ) x^{3}}{3}+\frac {B \,a^{2} x^{2}}{2}+a^{2} A x\) | \(99\) |
norman | \(\frac {b^{2} D x^{8}}{8}+\frac {b^{2} C \,x^{7}}{7}+\left (\frac {1}{6} B \,b^{2}+\frac {1}{3} a b D\right ) x^{6}+\left (\frac {1}{5} b^{2} A +\frac {2}{5} C a b \right ) x^{5}+\left (\frac {1}{2} a b B +\frac {1}{4} D a^{2}\right ) x^{4}+\left (\frac {2}{3} a b A +\frac {1}{3} a^{2} C \right ) x^{3}+\frac {B \,a^{2} x^{2}}{2}+a^{2} A x\) | \(99\) |
gosper | \(\frac {1}{8} b^{2} D x^{8}+\frac {1}{7} b^{2} C \,x^{7}+\frac {1}{6} b^{2} B \,x^{6}+\frac {1}{3} a b D x^{6}+\frac {1}{5} A \,b^{2} x^{5}+\frac {2}{5} x^{5} C a b +\frac {1}{2} B a b \,x^{4}+\frac {1}{4} a^{2} D x^{4}+\frac {2}{3} a A b \,x^{3}+\frac {1}{3} x^{3} a^{2} C +\frac {1}{2} B \,a^{2} x^{2}+a^{2} A x\) | \(103\) |
parallelrisch | \(\frac {1}{8} b^{2} D x^{8}+\frac {1}{7} b^{2} C \,x^{7}+\frac {1}{6} b^{2} B \,x^{6}+\frac {1}{3} a b D x^{6}+\frac {1}{5} A \,b^{2} x^{5}+\frac {2}{5} x^{5} C a b +\frac {1}{2} B a b \,x^{4}+\frac {1}{4} a^{2} D x^{4}+\frac {2}{3} a A b \,x^{3}+\frac {1}{3} x^{3} a^{2} C +\frac {1}{2} B \,a^{2} x^{2}+a^{2} A x\) | \(103\) |
orering | \(\frac {x \left (105 b^{2} D x^{7}+120 b^{2} C \,x^{6}+140 b^{2} B \,x^{5}+280 D a b \,x^{5}+168 A \,b^{2} x^{4}+336 C a b \,x^{4}+420 B a b \,x^{3}+210 D a^{2} x^{3}+560 a A b \,x^{2}+280 C \,a^{2} x^{2}+420 B \,a^{2} x +840 a^{2} A \right )}{840}\) | \(104\) |
Input:
int((b*x^2+a)^2*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
1/8*b^2*D*x^8+1/7*b^2*C*x^7+1/6*(B*b^2+2*D*a*b)*x^6+1/5*(A*b^2+2*C*a*b)*x^ 5+1/4*(2*B*a*b+D*a^2)*x^4+1/3*(2*A*a*b+C*a^2)*x^3+1/2*B*a^2*x^2+a^2*A*x
Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \, D b^{2} x^{8} + \frac {1}{7} \, C b^{2} x^{7} + \frac {1}{6} \, {\left (2 \, D a b + B b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (2 \, C a b + A b^{2}\right )} x^{5} + \frac {1}{2} \, B a^{2} x^{2} + \frac {1}{4} \, {\left (D a^{2} + 2 \, B a b\right )} x^{4} + A a^{2} x + \frac {1}{3} \, {\left (C a^{2} + 2 \, A a b\right )} x^{3} \] Input:
integrate((b*x^2+a)^2*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")
Output:
1/8*D*b^2*x^8 + 1/7*C*b^2*x^7 + 1/6*(2*D*a*b + B*b^2)*x^6 + 1/5*(2*C*a*b + A*b^2)*x^5 + 1/2*B*a^2*x^2 + 1/4*(D*a^2 + 2*B*a*b)*x^4 + A*a^2*x + 1/3*(C *a^2 + 2*A*a*b)*x^3
Time = 0.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=A a^{2} x + \frac {B a^{2} x^{2}}{2} + \frac {C b^{2} x^{7}}{7} + \frac {D b^{2} x^{8}}{8} + x^{6} \left (\frac {B b^{2}}{6} + \frac {D a b}{3}\right ) + x^{5} \left (\frac {A b^{2}}{5} + \frac {2 C a b}{5}\right ) + x^{4} \left (\frac {B a b}{2} + \frac {D a^{2}}{4}\right ) + x^{3} \cdot \left (\frac {2 A a b}{3} + \frac {C a^{2}}{3}\right ) \] Input:
integrate((b*x**2+a)**2*(D*x**3+C*x**2+B*x+A),x)
Output:
A*a**2*x + B*a**2*x**2/2 + C*b**2*x**7/7 + D*b**2*x**8/8 + x**6*(B*b**2/6 + D*a*b/3) + x**5*(A*b**2/5 + 2*C*a*b/5) + x**4*(B*a*b/2 + D*a**2/4) + x** 3*(2*A*a*b/3 + C*a**2/3)
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \, D b^{2} x^{8} + \frac {1}{7} \, C b^{2} x^{7} + \frac {1}{6} \, {\left (2 \, D a b + B b^{2}\right )} x^{6} + \frac {1}{5} \, {\left (2 \, C a b + A b^{2}\right )} x^{5} + \frac {1}{2} \, B a^{2} x^{2} + \frac {1}{4} \, {\left (D a^{2} + 2 \, B a b\right )} x^{4} + A a^{2} x + \frac {1}{3} \, {\left (C a^{2} + 2 \, A a b\right )} x^{3} \] Input:
integrate((b*x^2+a)^2*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")
Output:
1/8*D*b^2*x^8 + 1/7*C*b^2*x^7 + 1/6*(2*D*a*b + B*b^2)*x^6 + 1/5*(2*C*a*b + A*b^2)*x^5 + 1/2*B*a^2*x^2 + 1/4*(D*a^2 + 2*B*a*b)*x^4 + A*a^2*x + 1/3*(C *a^2 + 2*A*a*b)*x^3
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.03 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \, D b^{2} x^{8} + \frac {1}{7} \, C b^{2} x^{7} + \frac {1}{3} \, D a b x^{6} + \frac {1}{6} \, B b^{2} x^{6} + \frac {2}{5} \, C a b x^{5} + \frac {1}{5} \, A b^{2} x^{5} + \frac {1}{4} \, D a^{2} x^{4} + \frac {1}{2} \, B a b x^{4} + \frac {1}{3} \, C a^{2} x^{3} + \frac {2}{3} \, A a b x^{3} + \frac {1}{2} \, B a^{2} x^{2} + A a^{2} x \] Input:
integrate((b*x^2+a)^2*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")
Output:
1/8*D*b^2*x^8 + 1/7*C*b^2*x^7 + 1/3*D*a*b*x^6 + 1/6*B*b^2*x^6 + 2/5*C*a*b* x^5 + 1/5*A*b^2*x^5 + 1/4*D*a^2*x^4 + 1/2*B*a*b*x^4 + 1/3*C*a^2*x^3 + 2/3* A*a*b*x^3 + 1/2*B*a^2*x^2 + A*a^2*x
Time = 0.50 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.06 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A\,x\,\left (15\,a^2+10\,a\,b\,x^2+3\,b^2\,x^4\right )}{15}+\frac {a^2\,x^4\,D}{4}+\frac {b^2\,x^8\,D}{8}+\frac {B\,x^2\,\left (3\,a^2+3\,a\,b\,x^2+b^2\,x^4\right )}{6}+\frac {C\,x^3\,\left (35\,a^2+42\,a\,b\,x^2+15\,b^2\,x^4\right )}{105}+\frac {a\,b\,x^6\,D}{3} \] Input:
int((a + b*x^2)^2*(A + B*x + C*x^2 + x^3*D),x)
Output:
(A*x*(15*a^2 + 3*b^2*x^4 + 10*a*b*x^2))/15 + (a^2*x^4*D)/4 + (b^2*x^8*D)/8 + (B*x^2*(3*a^2 + b^2*x^4 + 3*a*b*x^2))/6 + (C*x^3*(35*a^2 + 15*b^2*x^4 + 42*a*b*x^2))/105 + (a*b*x^6*D)/3
Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.04 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {x \left (105 b^{2} d \,x^{7}+120 b^{2} c \,x^{6}+280 a b d \,x^{5}+140 b^{3} x^{5}+168 a \,b^{2} x^{4}+336 a b c \,x^{4}+210 a^{2} d \,x^{3}+420 a \,b^{2} x^{3}+560 a^{2} b \,x^{2}+280 a^{2} c \,x^{2}+420 a^{2} b x +840 a^{3}\right )}{840} \] Input:
int((b*x^2+a)^2*(D*x^3+C*x^2+B*x+A),x)
Output:
(x*(840*a**3 + 560*a**2*b*x**2 + 420*a**2*b*x + 280*a**2*c*x**2 + 210*a**2 *d*x**3 + 168*a*b**2*x**4 + 420*a*b**2*x**3 + 336*a*b*c*x**4 + 280*a*b*d*x **5 + 140*b**3*x**5 + 120*b**2*c*x**6 + 105*b**2*d*x**7))/840