Integrand size = 30, antiderivative size = 94 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=-\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{3 x^3}-\frac {A b^2+a (2 b B+a C)}{x}+\left (b^2 B+2 a b C+a^2 D\right ) x+\frac {1}{3} b (b C+2 a D) x^3+\frac {1}{5} b^2 D x^5 \] Output:
-1/5*a^2*A/x^5-1/3*a*(2*A*b+B*a)/x^3-(A*b^2+a*(2*B*b+C*a))/x+(B*b^2+2*C*a* b+D*a^2)*x+1/3*b*(C*b+2*D*a)*x^3+1/5*b^2*D*x^5
Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=\frac {10 a b x^2 \left (-A-3 B x^2+3 C x^4+D x^6\right )+b^2 x^4 \left (-15 A+15 B x^2+5 C x^4+3 D x^6\right )-a^2 \left (3 A+5 x^2 \left (B+3 C x^2-3 D x^4\right )\right )}{15 x^5} \] Input:
Integrate[((a + b*x^2)^2*(A + B*x^2 + C*x^4 + D*x^6))/x^6,x]
Output:
(10*a*b*x^2*(-A - 3*B*x^2 + 3*C*x^4 + D*x^6) + b^2*x^4*(-15*A + 15*B*x^2 + 5*C*x^4 + 3*D*x^6) - a^2*(3*A + 5*x^2*(B + 3*C*x^2 - 3*D*x^4)))/(15*x^5)
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2333, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \int \left (\frac {a^2 A}{x^6}+\frac {a (a C+2 b B)+A b^2}{x^2}+\frac {a (a B+2 A b)}{x^4}+b^2 B \left (\frac {a (a D+2 b C)}{b^2 B}+1\right )+b x^2 (2 a D+b C)+b^2 D x^4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 A}{5 x^5}+x \left (a^2 D+2 a b C+b^2 B\right )-\frac {a (a C+2 b B)+A b^2}{x}-\frac {a (a B+2 A b)}{3 x^3}+\frac {1}{3} b x^3 (2 a D+b C)+\frac {1}{5} b^2 D x^5\) |
Input:
Int[((a + b*x^2)^2*(A + B*x^2 + C*x^4 + D*x^6))/x^6,x]
Output:
-1/5*(a^2*A)/x^5 - (a*(2*A*b + a*B))/(3*x^3) - (A*b^2 + a*(2*b*B + a*C))/x + (b^2*B + 2*a*b*C + a^2*D)*x + (b*(b*C + 2*a*D)*x^3)/3 + (b^2*D*x^5)/5
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.96
method | result | size |
default | \(\frac {D x^{5} b^{2}}{5}+\frac {C \,b^{2} x^{3}}{3}+\frac {2 D a b \,x^{3}}{3}+b^{2} B x +2 C a b x +D a^{2} x -\frac {a \left (2 A b +B a \right )}{3 x^{3}}-\frac {a^{2} A}{5 x^{5}}-\frac {b^{2} A +2 a b B +a^{2} C}{x}\) | \(90\) |
norman | \(\frac {\frac {b^{2} x^{10} D}{5}+\left (\frac {1}{3} b^{2} C +\frac {2}{3} a b D\right ) x^{8}+\left (B \,b^{2}+2 C a b +D a^{2}\right ) x^{6}+\left (-b^{2} A -2 a b B -a^{2} C \right ) x^{4}+\left (-\frac {2}{3} a b A -\frac {1}{3} a^{2} B \right ) x^{2}-\frac {a^{2} A}{5}}{x^{5}}\) | \(95\) |
gosper | \(-\frac {-3 b^{2} x^{10} D-5 C \,b^{2} x^{8}-10 D a b \,x^{8}-15 b^{2} B \,x^{6}-30 b \,x^{6} C a -15 D a^{2} x^{6}+15 A \,b^{2} x^{4}+30 B a b \,x^{4}+15 C \,a^{2} x^{4}+10 a A b \,x^{2}+5 B \,a^{2} x^{2}+3 a^{2} A}{15 x^{5}}\) | \(108\) |
parallelrisch | \(-\frac {-3 b^{2} x^{10} D-5 C \,b^{2} x^{8}-10 D a b \,x^{8}-15 b^{2} B \,x^{6}-30 b \,x^{6} C a -15 D a^{2} x^{6}+15 A \,b^{2} x^{4}+30 B a b \,x^{4}+15 C \,a^{2} x^{4}+10 a A b \,x^{2}+5 B \,a^{2} x^{2}+3 a^{2} A}{15 x^{5}}\) | \(108\) |
orering | \(-\frac {-3 b^{2} x^{10} D-5 C \,b^{2} x^{8}-10 D a b \,x^{8}-15 b^{2} B \,x^{6}-30 b \,x^{6} C a -15 D a^{2} x^{6}+15 A \,b^{2} x^{4}+30 B a b \,x^{4}+15 C \,a^{2} x^{4}+10 a A b \,x^{2}+5 B \,a^{2} x^{2}+3 a^{2} A}{15 x^{5}}\) | \(108\) |
Input:
int((b*x^2+a)^2*(D*x^6+C*x^4+B*x^2+A)/x^6,x,method=_RETURNVERBOSE)
Output:
1/5*D*x^5*b^2+1/3*C*b^2*x^3+2/3*D*a*b*x^3+b^2*B*x+2*C*a*b*x+D*a^2*x-1/3*a* (2*A*b+B*a)/x^3-1/5*a^2*A/x^5-(A*b^2+2*B*a*b+C*a^2)/x
Time = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=\frac {3 \, D b^{2} x^{10} + 5 \, {\left (2 \, D a b + C b^{2}\right )} x^{8} + 15 \, {\left (D a^{2} + 2 \, C a b + B b^{2}\right )} x^{6} - 15 \, {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} x^{4} - 3 \, A a^{2} - 5 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x^{5}} \] Input:
integrate((b*x^2+a)^2*(D*x^6+C*x^4+B*x^2+A)/x^6,x, algorithm="fricas")
Output:
1/15*(3*D*b^2*x^10 + 5*(2*D*a*b + C*b^2)*x^8 + 15*(D*a^2 + 2*C*a*b + B*b^2 )*x^6 - 15*(C*a^2 + 2*B*a*b + A*b^2)*x^4 - 3*A*a^2 - 5*(B*a^2 + 2*A*a*b)*x ^2)/x^5
Time = 0.65 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=\frac {D b^{2} x^{5}}{5} + x^{3} \left (\frac {C b^{2}}{3} + \frac {2 D a b}{3}\right ) + x \left (B b^{2} + 2 C a b + D a^{2}\right ) + \frac {- 3 A a^{2} + x^{4} \left (- 15 A b^{2} - 30 B a b - 15 C a^{2}\right ) + x^{2} \left (- 10 A a b - 5 B a^{2}\right )}{15 x^{5}} \] Input:
integrate((b*x**2+a)**2*(D*x**6+C*x**4+B*x**2+A)/x**6,x)
Output:
D*b**2*x**5/5 + x**3*(C*b**2/3 + 2*D*a*b/3) + x*(B*b**2 + 2*C*a*b + D*a**2 ) + (-3*A*a**2 + x**4*(-15*A*b**2 - 30*B*a*b - 15*C*a**2) + x**2*(-10*A*a* b - 5*B*a**2))/(15*x**5)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=\frac {1}{5} \, D b^{2} x^{5} + \frac {1}{3} \, {\left (2 \, D a b + C b^{2}\right )} x^{3} + {\left (D a^{2} + 2 \, C a b + B b^{2}\right )} x - \frac {15 \, {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} x^{4} + 3 \, A a^{2} + 5 \, {\left (B a^{2} + 2 \, A a b\right )} x^{2}}{15 \, x^{5}} \] Input:
integrate((b*x^2+a)^2*(D*x^6+C*x^4+B*x^2+A)/x^6,x, algorithm="maxima")
Output:
1/5*D*b^2*x^5 + 1/3*(2*D*a*b + C*b^2)*x^3 + (D*a^2 + 2*C*a*b + B*b^2)*x - 1/15*(15*(C*a^2 + 2*B*a*b + A*b^2)*x^4 + 3*A*a^2 + 5*(B*a^2 + 2*A*a*b)*x^2 )/x^5
Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=\frac {1}{5} \, D b^{2} x^{5} + \frac {2}{3} \, D a b x^{3} + \frac {1}{3} \, C b^{2} x^{3} + D a^{2} x + 2 \, C a b x + B b^{2} x - \frac {15 \, C a^{2} x^{4} + 30 \, B a b x^{4} + 15 \, A b^{2} x^{4} + 5 \, B a^{2} x^{2} + 10 \, A a b x^{2} + 3 \, A a^{2}}{15 \, x^{5}} \] Input:
integrate((b*x^2+a)^2*(D*x^6+C*x^4+B*x^2+A)/x^6,x, algorithm="giac")
Output:
1/5*D*b^2*x^5 + 2/3*D*a*b*x^3 + 1/3*C*b^2*x^3 + D*a^2*x + 2*C*a*b*x + B*b^ 2*x - 1/15*(15*C*a^2*x^4 + 30*B*a*b*x^4 + 15*A*b^2*x^4 + 5*B*a^2*x^2 + 10* A*a*b*x^2 + 3*A*a^2)/x^5
Time = 1.48 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=a^2\,x\,D-\frac {\frac {A\,a^2}{5}+\frac {2\,A\,a\,b\,x^2}{3}+A\,b^2\,x^4}{x^5}-\frac {B\,\left (a^2+6\,a\,b\,x^2-3\,b^2\,x^4\right )}{3\,x^3}+\frac {b^2\,x^5\,D}{5}+\frac {C\,\left (-3\,a^2+6\,a\,b\,x^2+b^2\,x^4\right )}{3\,x}+\frac {2\,a\,b\,x^3\,D}{3} \] Input:
int(((a + b*x^2)^2*(A + B*x^2 + C*x^4 + x^6*D))/x^6,x)
Output:
a^2*x*D - ((A*a^2)/5 + A*b^2*x^4 + (2*A*a*b*x^2)/3)/x^5 - (B*(a^2 - 3*b^2* x^4 + 6*a*b*x^2))/(3*x^3) + (b^2*x^5*D)/5 + (C*(b^2*x^4 - 3*a^2 + 6*a*b*x^ 2))/(3*x) + (2*a*b*x^3*D)/3
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^2\right )^2 \left (A+B x^2+C x^4+D x^6\right )}{x^6} \, dx=\frac {3 b^{2} d \,x^{10}+10 a b d \,x^{8}+5 b^{2} c \,x^{8}+15 a^{2} d \,x^{6}+30 a b c \,x^{6}+15 b^{3} x^{6}-15 a^{2} c \,x^{4}-45 a \,b^{2} x^{4}-15 a^{2} b \,x^{2}-3 a^{3}}{15 x^{5}} \] Input:
int((b*x^2+a)^2*(D*x^6+C*x^4+B*x^2+A)/x^6,x)
Output:
( - 3*a**3 - 15*a**2*b*x**2 - 15*a**2*c*x**4 + 15*a**2*d*x**6 - 45*a*b**2* x**4 + 30*a*b*c*x**6 + 10*a*b*d*x**8 + 15*b**3*x**6 + 5*b**2*c*x**8 + 3*b* *2*d*x**10)/(15*x**5)