Integrand size = 28, antiderivative size = 129 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=a^3 B x+\frac {3}{2} a^2 A b x^2+\frac {1}{3} a^2 (3 b B+a D) x^3+\frac {3}{4} a A b^2 x^4+\frac {3}{5} a b (b B+a D) x^5+\frac {1}{6} A b^3 x^6+\frac {1}{7} b^2 (b B+3 a D) x^7+\frac {1}{9} b^3 D x^9+\frac {C \left (a+b x^2\right )^4}{8 b}+a^3 A \log (x) \]
a^3*B*x+3/2*a^2*A*b*x^2+1/3*a^2*(3*B*b+D*a)*x^3+3/4*a*A*b^2*x^4+3/5*a*b*(B *b+D*a)*x^5+1/6*A*b^3*x^6+1/7*b^2*(B*b+3*D*a)*x^7+1/9*b^3*D*x^9+1/8*C*(b*x ^2+a)^4/b+a^3*A*ln(x)
Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=\frac {x \left (420 a^3 (6 B+x (3 C+2 D x))+126 a^2 b x (30 A+x (20 B+3 x (5 C+4 D x)))+18 a b^2 x^3 (105 A+2 x (42 B+5 x (7 C+6 D x)))+5 b^3 x^5 (84 A+x (72 B+7 x (9 C+8 D x)))\right )}{2520}+a^3 A \log (x) \]
(x*(420*a^3*(6*B + x*(3*C + 2*D*x)) + 126*a^2*b*x*(30*A + x*(20*B + 3*x*(5 *C + 4*D*x))) + 18*a*b^2*x^3*(105*A + 2*x*(42*B + 5*x*(7*C + 6*D*x))) + 5* b^3*x^5*(84*A + x*(72*B + 7*x*(9*C + 8*D*x)))))/2520 + a^3*A*Log[x]
Time = 0.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2018, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx\) |
\(\Big \downarrow \) 2018 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^3 \left (D x^3+B x+A\right )}{x}dx+\frac {C \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \int \left (b^3 D x^8+b^2 (b B+3 a D) x^6+A b^3 x^5+3 a b (b B+a D) x^4+3 a A b^2 x^3+a^2 (3 b B+a D) x^2+3 a^2 A b x+a^3 B+\frac {a^3 A}{x}\right )dx+\frac {C \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^3 A \log (x)+a^3 B x+\frac {3}{2} a^2 A b x^2+\frac {1}{3} a^2 x^3 (a D+3 b B)+\frac {3}{4} a A b^2 x^4+\frac {1}{7} b^2 x^7 (3 a D+b B)+\frac {3}{5} a b x^5 (a D+b B)+\frac {C \left (a+b x^2\right )^4}{8 b}+\frac {1}{6} A b^3 x^6+\frac {1}{9} b^3 D x^9\) |
a^3*B*x + (3*a^2*A*b*x^2)/2 + (a^2*(3*b*B + a*D)*x^3)/3 + (3*a*A*b^2*x^4)/ 4 + (3*a*b*(b*B + a*D)*x^5)/5 + (A*b^3*x^6)/6 + (b^2*(b*B + 3*a*D)*x^7)/7 + (b^3*D*x^9)/9 + (C*(a + b*x^2)^4)/(8*b) + a^3*A*Log[x]
3.1.82.3.1 Defintions of rubi rules used
Int[(Px_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[Coef f[Px, x, n - m - 1]*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] + Int[(Px - Coe ff[Px, x, n - m - 1]*x^(n - m - 1))*x^m*(a + b*x^n)^p, x] /; FreeQ[{a, b, m , n}, x] && PolyQ[Px, x] && IGtQ[p, 1] && IGtQ[n - m, 0] && NeQ[Coeff[Px, x , n - m - 1], 0]
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 = 3.42 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.10
method | result | size |
norman | \(\left (\frac {1}{7} B \,b^{3}+\frac {3}{7} a \,b^{2} D\right ) x^{7}+\left (\frac {1}{6} b^{3} A +\frac {1}{2} C \,b^{2} a \right ) x^{6}+\left (\frac {3}{4} a \,b^{2} A +\frac {3}{4} C \,a^{2} b \right ) x^{4}+\left (\frac {3}{5} a \,b^{2} B +\frac {3}{5} D a^{2} b \right ) x^{5}+\left (\frac {3}{2} a^{2} b A +\frac {1}{2} C \,a^{3}\right ) x^{2}+\left (a^{2} b B +\frac {1}{3} D a^{3}\right ) x^{3}+a^{3} B x +\frac {b^{3} C \,x^{8}}{8}+\frac {b^{3} D x^{9}}{9}+a^{3} A \ln \left (x \right )\) | \(142\) |
default | \(\frac {b^{3} D x^{9}}{9}+\frac {b^{3} C \,x^{8}}{8}+\frac {b^{3} B \,x^{7}}{7}+\frac {3 D a \,b^{2} x^{7}}{7}+\frac {x^{6} b^{3} A}{6}+\frac {C a \,b^{2} x^{6}}{2}+\frac {3 B a \,b^{2} x^{5}}{5}+\frac {3 D a^{2} b \,x^{5}}{5}+\frac {3 a A \,b^{2} x^{4}}{4}+\frac {3 C \,a^{2} b \,x^{4}}{4}+B \,a^{2} b \,x^{3}+\frac {D a^{3} x^{3}}{3}+\frac {3 a^{2} A b \,x^{2}}{2}+\frac {C \,a^{3} x^{2}}{2}+a^{3} B x +a^{3} A \ln \left (x \right )\) | \(148\) |
parallelrisch | \(\frac {b^{3} D x^{9}}{9}+\frac {b^{3} C \,x^{8}}{8}+\frac {b^{3} B \,x^{7}}{7}+\frac {3 D a \,b^{2} x^{7}}{7}+\frac {x^{6} b^{3} A}{6}+\frac {C a \,b^{2} x^{6}}{2}+\frac {3 B a \,b^{2} x^{5}}{5}+\frac {3 D a^{2} b \,x^{5}}{5}+\frac {3 a A \,b^{2} x^{4}}{4}+\frac {3 C \,a^{2} b \,x^{4}}{4}+B \,a^{2} b \,x^{3}+\frac {D a^{3} x^{3}}{3}+\frac {3 a^{2} A b \,x^{2}}{2}+\frac {C \,a^{3} x^{2}}{2}+a^{3} B x +a^{3} A \ln \left (x \right )\) | \(148\) |
(1/7*B*b^3+3/7*a*b^2*D)*x^7+(1/6*b^3*A+1/2*C*b^2*a)*x^6+(3/4*a*b^2*A+3/4*C *a^2*b)*x^4+(3/5*a*b^2*B+3/5*D*a^2*b)*x^5+(3/2*a^2*b*A+1/2*C*a^3)*x^2+(a^2 *b*B+1/3*D*a^3)*x^3+a^3*B*x+1/8*b^3*C*x^8+1/9*b^3*D*x^9+a^3*A*ln(x)
Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=\frac {1}{9} \, D b^{3} x^{9} + \frac {1}{8} \, C b^{3} x^{8} + \frac {1}{7} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{7} + \frac {1}{6} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{6} + \frac {3}{5} \, {\left (D a^{2} b + B a b^{2}\right )} x^{5} + B a^{3} x + \frac {3}{4} \, {\left (C a^{2} b + A a b^{2}\right )} x^{4} + A a^{3} \log \left (x\right ) + \frac {1}{3} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{3} + \frac {1}{2} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2} \]
1/9*D*b^3*x^9 + 1/8*C*b^3*x^8 + 1/7*(3*D*a*b^2 + B*b^3)*x^7 + 1/6*(3*C*a*b ^2 + A*b^3)*x^6 + 3/5*(D*a^2*b + B*a*b^2)*x^5 + B*a^3*x + 3/4*(C*a^2*b + A *a*b^2)*x^4 + A*a^3*log(x) + 1/3*(D*a^3 + 3*B*a^2*b)*x^3 + 1/2*(C*a^3 + 3* A*a^2*b)*x^2
Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=A a^{3} \log {\left (x \right )} + B a^{3} x + \frac {C b^{3} x^{8}}{8} + \frac {D b^{3} x^{9}}{9} + x^{7} \left (\frac {B b^{3}}{7} + \frac {3 D a b^{2}}{7}\right ) + x^{6} \left (\frac {A b^{3}}{6} + \frac {C a b^{2}}{2}\right ) + x^{5} \cdot \left (\frac {3 B a b^{2}}{5} + \frac {3 D a^{2} b}{5}\right ) + x^{4} \cdot \left (\frac {3 A a b^{2}}{4} + \frac {3 C a^{2} b}{4}\right ) + x^{3} \left (B a^{2} b + \frac {D a^{3}}{3}\right ) + x^{2} \cdot \left (\frac {3 A a^{2} b}{2} + \frac {C a^{3}}{2}\right ) \]
A*a**3*log(x) + B*a**3*x + C*b**3*x**8/8 + D*b**3*x**9/9 + x**7*(B*b**3/7 + 3*D*a*b**2/7) + x**6*(A*b**3/6 + C*a*b**2/2) + x**5*(3*B*a*b**2/5 + 3*D* a**2*b/5) + x**4*(3*A*a*b**2/4 + 3*C*a**2*b/4) + x**3*(B*a**2*b + D*a**3/3 ) + x**2*(3*A*a**2*b/2 + C*a**3/2)
Time = 0.19 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=\frac {1}{9} \, D b^{3} x^{9} + \frac {1}{8} \, C b^{3} x^{8} + \frac {1}{7} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{7} + \frac {1}{6} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{6} + \frac {3}{5} \, {\left (D a^{2} b + B a b^{2}\right )} x^{5} + B a^{3} x + \frac {3}{4} \, {\left (C a^{2} b + A a b^{2}\right )} x^{4} + A a^{3} \log \left (x\right ) + \frac {1}{3} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{3} + \frac {1}{2} \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2} \]
1/9*D*b^3*x^9 + 1/8*C*b^3*x^8 + 1/7*(3*D*a*b^2 + B*b^3)*x^7 + 1/6*(3*C*a*b ^2 + A*b^3)*x^6 + 3/5*(D*a^2*b + B*a*b^2)*x^5 + B*a^3*x + 3/4*(C*a^2*b + A *a*b^2)*x^4 + A*a^3*log(x) + 1/3*(D*a^3 + 3*B*a^2*b)*x^3 + 1/2*(C*a^3 + 3* A*a^2*b)*x^2
Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=\frac {1}{9} \, D b^{3} x^{9} + \frac {1}{8} \, C b^{3} x^{8} + \frac {3}{7} \, D a b^{2} x^{7} + \frac {1}{7} \, B b^{3} x^{7} + \frac {1}{2} \, C a b^{2} x^{6} + \frac {1}{6} \, A b^{3} x^{6} + \frac {3}{5} \, D a^{2} b x^{5} + \frac {3}{5} \, B a b^{2} x^{5} + \frac {3}{4} \, C a^{2} b x^{4} + \frac {3}{4} \, A a b^{2} x^{4} + \frac {1}{3} \, D a^{3} x^{3} + B a^{2} b x^{3} + \frac {1}{2} \, C a^{3} x^{2} + \frac {3}{2} \, A a^{2} b x^{2} + B a^{3} x + A a^{3} \log \left ({\left | x \right |}\right ) \]
1/9*D*b^3*x^9 + 1/8*C*b^3*x^8 + 3/7*D*a*b^2*x^7 + 1/7*B*b^3*x^7 + 1/2*C*a* b^2*x^6 + 1/6*A*b^3*x^6 + 3/5*D*a^2*b*x^5 + 3/5*B*a*b^2*x^5 + 3/4*C*a^2*b* x^4 + 3/4*A*a*b^2*x^4 + 1/3*D*a^3*x^3 + B*a^2*b*x^3 + 1/2*C*a^3*x^2 + 3/2* A*a^2*b*x^2 + B*a^3*x + A*a^3*log(abs(x))
Time = 5.85 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x} \, dx=\frac {A\,b^3\,x^6}{6}+\frac {C\,a^3\,x^2}{2}+\frac {B\,b^3\,x^7}{7}+\frac {C\,b^3\,x^8}{8}+A\,a^3\,\ln \left (x\right )+\frac {a^3\,x^3\,D}{3}+\frac {b^3\,x^9\,D}{9}+B\,a^3\,x+\frac {3\,a^2\,b\,x^5\,D}{5}+\frac {3\,a\,b^2\,x^7\,D}{7}+\frac {3\,A\,a^2\,b\,x^2}{2}+\frac {3\,A\,a\,b^2\,x^4}{4}+B\,a^2\,b\,x^3+\frac {3\,B\,a\,b^2\,x^5}{5}+\frac {3\,C\,a^2\,b\,x^4}{4}+\frac {C\,a\,b^2\,x^6}{2} \]