Integrand size = 20, antiderivative size = 67 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=a^2 A x+\frac {1}{3} a (2 A b+a C) x^3+\frac {1}{5} b (A b+2 a C) x^5+\frac {1}{7} b^2 C x^7+\frac {B \left (a+b x^2\right )^3}{6 b} \] Output:
a^2*A*x+1/3*a*(2*A*b+C*a)*x^3+1/5*b*(A*b+2*C*a)*x^5+1/7*b^2*C*x^7+1/6*B*(b *x^2+a)^3/b
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=\frac {1}{210} x \left (35 a^2 (6 A+x (3 B+2 C x))+7 a b x^2 (20 A+3 x (5 B+4 C x))+b^2 x^4 (42 A+5 x (7 B+6 C x))\right ) \] Input:
Integrate[(a + b*x^2)^2*(A + B*x + C*x^2),x]
Output:
(x*(35*a^2*(6*A + x*(3*B + 2*C*x)) + 7*a*b*x^2*(20*A + 3*x*(5*B + 4*C*x)) + b^2*x^4*(42*A + 5*x*(7*B + 6*C*x))))/210
Time = 0.23 (sec) , antiderivative size = 67, 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.
\(\displaystyle \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx\) |
\(\Big \downarrow \) 2017 |
\(\displaystyle \int \left (b x^2+a\right )^2 \left (C x^2+A\right )dx+\frac {B \left (a+b x^2\right )^3}{6 b}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle \int \left (b^2 C x^6+b (A b+2 a C) x^4+a (2 A b+a C) x^2+a^2 A\right )dx+\frac {B \left (a+b x^2\right )^3}{6 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^2 A x+\frac {1}{5} b x^5 (2 a C+A b)+\frac {1}{3} a x^3 (a C+2 A b)+\frac {B \left (a+b x^2\right )^3}{6 b}+\frac {1}{7} b^2 C x^7\) |
Input:
Int[(a + b*x^2)^2*(A + B*x + C*x^2),x]
Output:
a^2*A*x + (a*(2*A*b + a*C)*x^3)/3 + (b*(A*b + 2*a*C)*x^5)/5 + (b^2*C*x^7)/ 7 + (B*(a + b*x^2)^3)/(6*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 = 0.56 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {b^{2} C \,x^{7}}{7}+\frac {b^{2} B \,x^{6}}{6}+\frac {\left (b^{2} A +2 C a b \right ) x^{5}}{5}+\frac {B a b \,x^{4}}{2}+\frac {\left (2 a b A +a^{2} C \right ) x^{3}}{3}+\frac {B \,a^{2} x^{2}}{2}+a^{2} A x\) | \(75\) |
norman | \(\frac {b^{2} C \,x^{7}}{7}+\frac {b^{2} B \,x^{6}}{6}+\left (\frac {1}{5} b^{2} A +\frac {2}{5} C a b \right ) x^{5}+\frac {B a b \,x^{4}}{2}+\left (\frac {2}{3} a b A +\frac {1}{3} a^{2} C \right ) x^{3}+\frac {B \,a^{2} x^{2}}{2}+a^{2} A x\) | \(75\) |
gosper | \(\frac {1}{7} b^{2} C \,x^{7}+\frac {1}{6} b^{2} B \,x^{6}+\frac {1}{5} A \,b^{2} x^{5}+\frac {2}{5} x^{5} C a b +\frac {1}{2} B a b \,x^{4}+\frac {2}{3} a A b \,x^{3}+\frac {1}{3} x^{3} a^{2} C +\frac {1}{2} B \,a^{2} x^{2}+a^{2} A x\) | \(77\) |
risch | \(\frac {1}{7} b^{2} C \,x^{7}+\frac {1}{6} b^{2} B \,x^{6}+\frac {1}{5} A \,b^{2} x^{5}+\frac {2}{5} x^{5} C a b +\frac {1}{2} B a b \,x^{4}+\frac {2}{3} a A b \,x^{3}+\frac {1}{3} x^{3} a^{2} C +\frac {1}{2} B \,a^{2} x^{2}+a^{2} A x\) | \(77\) |
parallelrisch | \(\frac {1}{7} b^{2} C \,x^{7}+\frac {1}{6} b^{2} B \,x^{6}+\frac {1}{5} A \,b^{2} x^{5}+\frac {2}{5} x^{5} C a b +\frac {1}{2} B a b \,x^{4}+\frac {2}{3} a A b \,x^{3}+\frac {1}{3} x^{3} a^{2} C +\frac {1}{2} B \,a^{2} x^{2}+a^{2} A x\) | \(77\) |
orering | \(\frac {x \left (30 b^{2} C \,x^{6}+35 b^{2} B \,x^{5}+42 A \,b^{2} x^{4}+84 C a b \,x^{4}+105 B a b \,x^{3}+140 a A b \,x^{2}+70 C \,a^{2} x^{2}+105 B \,a^{2} x +210 a^{2} A \right )}{210}\) | \(78\) |
Input:
int((b*x^2+a)^2*(C*x^2+B*x+A),x,method=_RETURNVERBOSE)
Output:
1/7*b^2*C*x^7+1/6*b^2*B*x^6+1/5*(A*b^2+2*C*a*b)*x^5+1/2*B*a*b*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 = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=\frac {1}{7} \, C b^{2} x^{7} + \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{2} \, B a b x^{4} + \frac {1}{5} \, {\left (2 \, C a b + A b^{2}\right )} x^{5} + \frac {1}{2} \, B a^{2} x^{2} + A a^{2} x + \frac {1}{3} \, {\left (C a^{2} + 2 \, A a b\right )} x^{3} \] Input:
integrate((b*x^2+a)^2*(C*x^2+B*x+A),x, algorithm="fricas")
Output:
1/7*C*b^2*x^7 + 1/6*B*b^2*x^6 + 1/2*B*a*b*x^4 + 1/5*(2*C*a*b + A*b^2)*x^5 + 1/2*B*a^2*x^2 + A*a^2*x + 1/3*(C*a^2 + 2*A*a*b)*x^3
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.24 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=A a^{2} x + \frac {B a^{2} x^{2}}{2} + \frac {B a b x^{4}}{2} + \frac {B b^{2} x^{6}}{6} + \frac {C b^{2} x^{7}}{7} + x^{5} \left (\frac {A b^{2}}{5} + \frac {2 C a b}{5}\right ) + x^{3} \cdot \left (\frac {2 A a b}{3} + \frac {C a^{2}}{3}\right ) \] Input:
integrate((b*x**2+a)**2*(C*x**2+B*x+A),x)
Output:
A*a**2*x + B*a**2*x**2/2 + B*a*b*x**4/2 + B*b**2*x**6/6 + C*b**2*x**7/7 + x**5*(A*b**2/5 + 2*C*a*b/5) + x**3*(2*A*a*b/3 + C*a**2/3)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=\frac {1}{7} \, C b^{2} x^{7} + \frac {1}{6} \, B b^{2} x^{6} + \frac {1}{2} \, B a b x^{4} + \frac {1}{5} \, {\left (2 \, C a b + A b^{2}\right )} x^{5} + \frac {1}{2} \, B a^{2} x^{2} + A a^{2} x + \frac {1}{3} \, {\left (C a^{2} + 2 \, A a b\right )} x^{3} \] Input:
integrate((b*x^2+a)^2*(C*x^2+B*x+A),x, algorithm="maxima")
Output:
1/7*C*b^2*x^7 + 1/6*B*b^2*x^6 + 1/2*B*a*b*x^4 + 1/5*(2*C*a*b + A*b^2)*x^5 + 1/2*B*a^2*x^2 + A*a^2*x + 1/3*(C*a^2 + 2*A*a*b)*x^3
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=\frac {1}{7} \, C b^{2} x^{7} + \frac {1}{6} \, B b^{2} x^{6} + \frac {2}{5} \, C a b x^{5} + \frac {1}{5} \, A b^{2} x^{5} + \frac {1}{2} \, B a b x^{4} + \frac {1}{3} \, C a^{2} x^{3} + \frac {2}{3} \, A a b x^{3} + \frac {1}{2} \, B a^{2} x^{2} + A a^{2} x \] Input:
integrate((b*x^2+a)^2*(C*x^2+B*x+A),x, algorithm="giac")
Output:
1/7*C*b^2*x^7 + 1/6*B*b^2*x^6 + 2/5*C*a*b*x^5 + 1/5*A*b^2*x^5 + 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 = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.10 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=x^3\,\left (\frac {C\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+x^5\,\left (\frac {A\,b^2}{5}+\frac {2\,C\,a\,b}{5}\right )+\frac {B\,a^2\,x^2}{2}+\frac {B\,b^2\,x^6}{6}+\frac {C\,b^2\,x^7}{7}+A\,a^2\,x+\frac {B\,a\,b\,x^4}{2} \] Input:
int((a + b*x^2)^2*(A + B*x + C*x^2),x)
Output:
x^3*((C*a^2)/3 + (2*A*a*b)/3) + x^5*((A*b^2)/5 + (2*C*a*b)/5) + (B*a^2*x^2 )/2 + (B*b^2*x^6)/6 + (C*b^2*x^7)/7 + A*a^2*x + (B*a*b*x^4)/2
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \left (a+b x^2\right )^2 \left (A+B x+C x^2\right ) \, dx=\frac {x \left (30 b^{2} c \,x^{6}+35 b^{3} x^{5}+42 a \,b^{2} x^{4}+84 a b c \,x^{4}+105 a \,b^{2} x^{3}+140 a^{2} b \,x^{2}+70 a^{2} c \,x^{2}+105 a^{2} b x +210 a^{3}\right )}{210} \] Input:
int((b*x^2+a)^2*(C*x^2+B*x+A),x)
Output:
(x*(210*a**3 + 140*a**2*b*x**2 + 105*a**2*b*x + 70*a**2*c*x**2 + 42*a*b**2 *x**4 + 105*a*b**2*x**3 + 84*a*b*c*x**4 + 35*b**3*x**5 + 30*b**2*c*x**6))/ 210