Integrand size = 28, antiderivative size = 139 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a^3 A}{3 x^3}-\frac {a^3 B}{2 x^2}-\frac {a^2 (3 A b+a C)}{x}+3 a b (A b+a C) x+\frac {3}{2} a b (b B+a D) x^2+\frac {1}{3} b^2 (A b+3 a C) x^3+\frac {1}{4} b^2 (b B+3 a D) x^4+\frac {1}{5} b^3 C x^5+\frac {1}{6} b^3 D x^6+a^2 (3 b B+a D) \log (x) \]
-1/3*a^3*A/x^3-1/2*a^3*B/x^2-a^2*(3*A*b+C*a)/x+3*a*b*(A*b+C*a)*x+3/2*a*b*( B*b+D*a)*x^2+1/3*b^2*(A*b+3*C*a)*x^3+1/4*b^2*(B*b+3*D*a)*x^4+1/5*b^3*C*x^5 +1/6*b^3*D*x^6+a^2*(3*B*b+D*a)*ln(x)
Time = 0.03 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=-\frac {a^3 (2 A+3 x (B+2 C x))}{6 x^3}+\frac {3 a^2 b \left (-2 A+x^2 (2 C+D x)\right )}{2 x}+\frac {1}{4} a b^2 x (12 A+x (6 B+x (4 C+3 D x)))+\frac {1}{60} b^3 x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+a^2 (3 b B+a D) \log (x) \]
-1/6*(a^3*(2*A + 3*x*(B + 2*C*x)))/x^3 + (3*a^2*b*(-2*A + x^2*(2*C + D*x)) )/(2*x) + (a*b^2*x*(12*A + x*(6*B + x*(4*C + 3*D*x))))/4 + (b^3*x^3*(20*A + x*(15*B + 2*x*(6*C + 5*D*x))))/60 + a^2*(3*b*B + a*D)*Log[x]
Time = 0.35 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \int \left (\frac {a^3 A}{x^4}+\frac {a^3 B}{x^3}+\frac {a^2 (a C+3 A b)}{x^2}+\frac {a^2 (a D+3 b B)}{x}+b^2 x^2 (3 a C+A b)+3 a b (a C+A b)+b^2 x^3 (3 a D+b B)+3 a b x (a D+b B)+b^3 C x^4+b^3 D x^5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 A}{3 x^3}-\frac {a^3 B}{2 x^2}-\frac {a^2 (a C+3 A b)}{x}+a^2 \log (x) (a D+3 b B)+\frac {1}{3} b^2 x^3 (3 a C+A b)+3 a b x (a C+A b)+\frac {1}{4} b^2 x^4 (3 a D+b B)+\frac {3}{2} a b x^2 (a D+b B)+\frac {1}{5} b^3 C x^5+\frac {1}{6} b^3 D x^6\) |
-1/3*(a^3*A)/x^3 - (a^3*B)/(2*x^2) - (a^2*(3*A*b + a*C))/x + 3*a*b*(A*b + a*C)*x + (3*a*b*(b*B + a*D)*x^2)/2 + (b^2*(A*b + 3*a*C)*x^3)/3 + (b^2*(b*B + 3*a*D)*x^4)/4 + (b^3*C*x^5)/5 + (b^3*D*x^6)/6 + a^2*(3*b*B + a*D)*Log[x ]
3.1.85.3.1 Defintions of rubi rules used
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 = 3.43 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {b^{3} D x^{6}}{6}+\frac {b^{3} C \,x^{5}}{5}+\frac {b^{3} B \,x^{4}}{4}+\frac {3 D a \,b^{2} x^{4}}{4}+\frac {A \,b^{3} x^{3}}{3}+C a \,b^{2} x^{3}+\frac {3 B a \,b^{2} x^{2}}{2}+\frac {3 D a^{2} b \,x^{2}}{2}+3 a \,b^{2} A x +3 C \,a^{2} b x +a^{2} \left (3 B b +D a \right ) \ln \left (x \right )-\frac {a^{3} A}{3 x^{3}}-\frac {a^{2} \left (3 A b +C a \right )}{x}-\frac {a^{3} B}{2 x^{2}}\) | \(141\) |
norman | \(\frac {\left (\frac {1}{4} B \,b^{3}+\frac {3}{4} a \,b^{2} D\right ) x^{7}+\left (\frac {1}{3} b^{3} A +C \,b^{2} a \right ) x^{6}+\left (\frac {3}{2} a \,b^{2} B +\frac {3}{2} D a^{2} b \right ) x^{5}+\left (3 a \,b^{2} A +3 C \,a^{2} b \right ) x^{4}+\left (-3 a^{2} b A -C \,a^{3}\right ) x^{2}-\frac {a^{3} A}{3}-\frac {a^{3} B x}{2}+\frac {b^{3} C \,x^{8}}{5}+\frac {b^{3} D x^{9}}{6}}{x^{3}}+\left (3 a^{2} b B +D a^{3}\right ) \ln \left (x \right )\) | \(145\) |
parallelrisch | \(\frac {10 b^{3} D x^{9}+12 b^{3} C \,x^{8}+15 b^{3} B \,x^{7}+45 D a \,b^{2} x^{7}+20 x^{6} b^{3} A +60 C a \,b^{2} x^{6}+90 B a \,b^{2} x^{5}+90 D a^{2} b \,x^{5}+180 a A \,b^{2} x^{4}+180 B \ln \left (x \right ) x^{3} a^{2} b +180 C \,a^{2} b \,x^{4}+60 D \ln \left (x \right ) x^{3} a^{3}-180 a^{2} A b \,x^{2}-60 C \,a^{3} x^{2}-30 a^{3} B x -20 a^{3} A}{60 x^{3}}\) | \(158\) |
1/6*b^3*D*x^6+1/5*b^3*C*x^5+1/4*b^3*B*x^4+3/4*D*a*b^2*x^4+1/3*A*b^3*x^3+C* a*b^2*x^3+3/2*B*a*b^2*x^2+3/2*D*a^2*b*x^2+3*a*b^2*A*x+3*C*a^2*b*x+a^2*(3*B *b+D*a)*ln(x)-1/3*a^3*A/x^3-a^2*(3*A*b+C*a)/x-1/2*a^3*B/x^2
Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {10 \, D b^{3} x^{9} + 12 \, C b^{3} x^{8} + 15 \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{7} + 20 \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{6} + 90 \, {\left (D a^{2} b + B a b^{2}\right )} x^{5} - 30 \, B a^{3} x + 180 \, {\left (C a^{2} b + A a b^{2}\right )} x^{4} + 60 \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{3} \log \left (x\right ) - 20 \, A a^{3} - 60 \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2}}{60 \, x^{3}} \]
1/60*(10*D*b^3*x^9 + 12*C*b^3*x^8 + 15*(3*D*a*b^2 + B*b^3)*x^7 + 20*(3*C*a *b^2 + A*b^3)*x^6 + 90*(D*a^2*b + B*a*b^2)*x^5 - 30*B*a^3*x + 180*(C*a^2*b + A*a*b^2)*x^4 + 60*(D*a^3 + 3*B*a^2*b)*x^3*log(x) - 20*A*a^3 - 60*(C*a^3 + 3*A*a^2*b)*x^2)/x^3
Time = 0.48 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {C b^{3} x^{5}}{5} + \frac {D b^{3} x^{6}}{6} + a^{2} \cdot \left (3 B b + D a\right ) \log {\left (x \right )} + x^{4} \left (\frac {B b^{3}}{4} + \frac {3 D a b^{2}}{4}\right ) + x^{3} \left (\frac {A b^{3}}{3} + C a b^{2}\right ) + x^{2} \cdot \left (\frac {3 B a b^{2}}{2} + \frac {3 D a^{2} b}{2}\right ) + x \left (3 A a b^{2} + 3 C a^{2} b\right ) + \frac {- 2 A a^{3} - 3 B a^{3} x + x^{2} \left (- 18 A a^{2} b - 6 C a^{3}\right )}{6 x^{3}} \]
C*b**3*x**5/5 + D*b**3*x**6/6 + a**2*(3*B*b + D*a)*log(x) + x**4*(B*b**3/4 + 3*D*a*b**2/4) + x**3*(A*b**3/3 + C*a*b**2) + x**2*(3*B*a*b**2/2 + 3*D*a **2*b/2) + x*(3*A*a*b**2 + 3*C*a**2*b) + (-2*A*a**3 - 3*B*a**3*x + x**2*(- 18*A*a**2*b - 6*C*a**3))/(6*x**3)
Time = 0.19 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {1}{6} \, D b^{3} x^{6} + \frac {1}{5} \, C b^{3} x^{5} + \frac {1}{4} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{4} + \frac {1}{3} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{3} + \frac {3}{2} \, {\left (D a^{2} b + B a b^{2}\right )} x^{2} + 3 \, {\left (C a^{2} b + A a b^{2}\right )} x + {\left (D a^{3} + 3 \, B a^{2} b\right )} \log \left (x\right ) - \frac {3 \, B a^{3} x + 2 \, A a^{3} + 6 \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2}}{6 \, x^{3}} \]
1/6*D*b^3*x^6 + 1/5*C*b^3*x^5 + 1/4*(3*D*a*b^2 + B*b^3)*x^4 + 1/3*(3*C*a*b ^2 + A*b^3)*x^3 + 3/2*(D*a^2*b + B*a*b^2)*x^2 + 3*(C*a^2*b + A*a*b^2)*x + (D*a^3 + 3*B*a^2*b)*log(x) - 1/6*(3*B*a^3*x + 2*A*a^3 + 6*(C*a^3 + 3*A*a^2 *b)*x^2)/x^3
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {1}{6} \, D b^{3} x^{6} + \frac {1}{5} \, C b^{3} x^{5} + \frac {3}{4} \, D a b^{2} x^{4} + \frac {1}{4} \, B b^{3} x^{4} + C a b^{2} x^{3} + \frac {1}{3} \, A b^{3} x^{3} + \frac {3}{2} \, D a^{2} b x^{2} + \frac {3}{2} \, B a b^{2} x^{2} + 3 \, C a^{2} b x + 3 \, A a b^{2} x + {\left (D a^{3} + 3 \, B a^{2} b\right )} \log \left ({\left | x \right |}\right ) - \frac {3 \, B a^{3} x + 2 \, A a^{3} + 6 \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2}}{6 \, x^{3}} \]
1/6*D*b^3*x^6 + 1/5*C*b^3*x^5 + 3/4*D*a*b^2*x^4 + 1/4*B*b^3*x^4 + C*a*b^2* x^3 + 1/3*A*b^3*x^3 + 3/2*D*a^2*b*x^2 + 3/2*B*a*b^2*x^2 + 3*C*a^2*b*x + 3* A*a*b^2*x + (D*a^3 + 3*B*a^2*b)*log(abs(x)) - 1/6*(3*B*a^3*x + 2*A*a^3 + 6 *(C*a^3 + 3*A*a^2*b)*x^2)/x^3
Time = 6.22 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^4} \, dx=\frac {B\,b^3\,x^4}{4}-\frac {C\,a^3}{x}-\frac {B\,a^3}{2\,x^2}+\frac {C\,b^3\,x^5}{5}+\frac {b^3\,x^6\,D}{6}-\frac {A\,\left (a^3+9\,a^2\,b\,x^2-9\,a\,b^2\,x^4-b^3\,x^6\right )}{3\,x^3}+\frac {a^3\,\ln \left (x^2\right )\,D}{2}+\frac {3\,a^2\,b\,x^2\,D}{2}+3\,C\,a^2\,b\,x+\frac {3\,a\,b^2\,x^4\,D}{4}+\frac {3\,B\,a\,b^2\,x^2}{2}+C\,a\,b^2\,x^3+3\,B\,a^2\,b\,\ln \left (x\right ) \]