Integrand size = 28, antiderivative size = 124 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=-\frac {a^3 A}{x}+a^2 (3 A b+a C) x+\frac {3}{2} a^2 b B x^2+a b (A b+a C) x^3+\frac {3}{4} a b^2 B x^4+\frac {1}{5} b^2 (A b+3 a C) x^5+\frac {1}{6} b^3 B x^6+\frac {1}{7} b^3 C x^7+\frac {D \left (a+b x^2\right )^4}{8 b}+a^3 B \log (x) \]
-a^3*A/x+a^2*(3*A*b+C*a)*x+3/2*a^2*b*B*x^2+a*b*(A*b+C*a)*x^3+3/4*a*b^2*B*x ^4+1/5*b^2*(A*b+3*C*a)*x^5+1/6*b^3*B*x^6+1/7*b^3*C*x^7+1/8*D*(b*x^2+a)^4/b +a^3*B*ln(x)
Time = 0.05 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=a^3 \left (-\frac {A}{x}+C x+\frac {D x^2}{2}\right )+\frac {1}{4} a^2 b x (12 A+x (6 B+x (4 C+3 D x)))+\frac {1}{20} a b^2 x^3 (20 A+x (15 B+2 x (6 C+5 D x)))+\frac {1}{840} b^3 x^5 (168 A+5 x (28 B+3 x (8 C+7 D x)))+a^3 B \log (x) \]
a^3*(-(A/x) + C*x + (D*x^2)/2) + (a^2*b*x*(12*A + x*(6*B + x*(4*C + 3*D*x) )))/4 + (a*b^2*x^3*(20*A + x*(15*B + 2*x*(6*C + 5*D*x))))/20 + (b^3*x^5*(1 68*A + 5*x*(28*B + 3*x*(8*C + 7*D*x))))/840 + a^3*B*Log[x]
Time = 0.35 (sec) , antiderivative size = 124, 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, 2159, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2018 |
\(\displaystyle \int \frac {\left (b x^2+a\right )^3 \left (C x^2+B x+A\right )}{x^2}dx+\frac {D \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \int \left (b^3 C x^6+b^3 B x^5+b^2 (A b+3 a C) x^4+3 a b^2 B x^3+3 a b (A b+a C) x^2+3 a^2 b B x+a^2 (3 A b+a C)+\frac {a^3 B}{x}+\frac {a^3 A}{x^2}\right )dx+\frac {D \left (a+b x^2\right )^4}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 A}{x}+a^3 B \log (x)+a^2 x (a C+3 A b)+\frac {3}{2} a^2 b B x^2+\frac {1}{5} b^2 x^5 (3 a C+A b)+a b x^3 (a C+A b)+\frac {3}{4} a b^2 B x^4+\frac {D \left (a+b x^2\right )^4}{8 b}+\frac {1}{6} b^3 B x^6+\frac {1}{7} b^3 C x^7\) |
-((a^3*A)/x) + a^2*(3*A*b + a*C)*x + (3*a^2*b*B*x^2)/2 + a*b*(A*b + a*C)*x ^3 + (3*a*b^2*B*x^4)/4 + (b^2*(A*b + 3*a*C)*x^5)/5 + (b^3*B*x^6)/6 + (b^3* C*x^7)/7 + (D*(a + b*x^2)^4)/(8*b) + a^3*B*Log[x]
3.1.83.3.1 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_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 3.43 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17
method | result | size |
default | \(\frac {b^{3} D x^{8}}{8}+\frac {b^{3} C \,x^{7}}{7}+\frac {b^{3} B \,x^{6}}{6}+\frac {D a \,b^{2} x^{6}}{2}+\frac {A \,b^{3} x^{5}}{5}+\frac {3 C a \,b^{2} x^{5}}{5}+\frac {3 B a \,b^{2} x^{4}}{4}+\frac {3 D a^{2} b \,x^{4}}{4}+a A \,b^{2} x^{3}+C \,a^{2} b \,x^{3}+\frac {3 B \,a^{2} b \,x^{2}}{2}+\frac {D a^{3} x^{2}}{2}+3 a^{2} A b x +C \,a^{3} x +a^{3} B \ln \left (x \right )-\frac {a^{3} A}{x}\) | \(145\) |
norman | \(\frac {\left (\frac {1}{6} B \,b^{3}+\frac {1}{2} a \,b^{2} D\right ) x^{7}+\left (\frac {1}{5} b^{3} A +\frac {3}{5} C \,b^{2} a \right ) x^{6}+\left (\frac {3}{4} a \,b^{2} B +\frac {3}{4} D a^{2} b \right ) x^{5}+\left (\frac {3}{2} a^{2} b B +\frac {1}{2} D a^{3}\right ) x^{3}+\left (a \,b^{2} A +C \,a^{2} b \right ) x^{4}+\left (3 a^{2} b A +C \,a^{3}\right ) x^{2}-a^{3} A +\frac {b^{3} C \,x^{8}}{7}+\frac {b^{3} D x^{9}}{8}}{x}+a^{3} B \ln \left (x \right )\) | \(145\) |
parallelrisch | \(\frac {105 b^{3} D x^{9}+120 b^{3} C \,x^{8}+140 b^{3} B \,x^{7}+420 D a \,b^{2} x^{7}+168 x^{6} b^{3} A +504 C a \,b^{2} x^{6}+630 B a \,b^{2} x^{5}+630 D a^{2} b \,x^{5}+840 a A \,b^{2} x^{4}+840 C \,a^{2} b \,x^{4}+1260 B \,a^{2} b \,x^{3}+420 D a^{3} x^{3}+2520 a^{2} A b \,x^{2}+840 a^{3} B \ln \left (x \right ) x +840 C \,a^{3} x^{2}-840 a^{3} A}{840 x}\) | \(156\) |
1/8*b^3*D*x^8+1/7*b^3*C*x^7+1/6*b^3*B*x^6+1/2*D*a*b^2*x^6+1/5*A*b^3*x^5+3/ 5*C*a*b^2*x^5+3/4*B*a*b^2*x^4+3/4*D*a^2*b*x^4+a*A*b^2*x^3+C*a^2*b*x^3+3/2* B*a^2*b*x^2+1/2*D*a^3*x^2+3*a^2*A*b*x+C*a^3*x+a^3*B*ln(x)-a^3*A/x
Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {105 \, D b^{3} x^{9} + 120 \, C b^{3} x^{8} + 140 \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{7} + 168 \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{6} + 630 \, {\left (D a^{2} b + B a b^{2}\right )} x^{5} + 840 \, B a^{3} x \log \left (x\right ) + 840 \, {\left (C a^{2} b + A a b^{2}\right )} x^{4} - 840 \, A a^{3} + 420 \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{3} + 840 \, {\left (C a^{3} + 3 \, A a^{2} b\right )} x^{2}}{840 \, x} \]
1/840*(105*D*b^3*x^9 + 120*C*b^3*x^8 + 140*(3*D*a*b^2 + B*b^3)*x^7 + 168*( 3*C*a*b^2 + A*b^3)*x^6 + 630*(D*a^2*b + B*a*b^2)*x^5 + 840*B*a^3*x*log(x) + 840*(C*a^2*b + A*a*b^2)*x^4 - 840*A*a^3 + 420*(D*a^3 + 3*B*a^2*b)*x^3 + 840*(C*a^3 + 3*A*a^2*b)*x^2)/x
Time = 0.13 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.21 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=- \frac {A a^{3}}{x} + B a^{3} \log {\left (x \right )} + \frac {C b^{3} x^{7}}{7} + \frac {D b^{3} x^{8}}{8} + x^{6} \left (\frac {B b^{3}}{6} + \frac {D a b^{2}}{2}\right ) + x^{5} \left (\frac {A b^{3}}{5} + \frac {3 C a b^{2}}{5}\right ) + x^{4} \cdot \left (\frac {3 B a b^{2}}{4} + \frac {3 D a^{2} b}{4}\right ) + x^{3} \left (A a b^{2} + C a^{2} b\right ) + x^{2} \cdot \left (\frac {3 B a^{2} b}{2} + \frac {D a^{3}}{2}\right ) + x \left (3 A a^{2} b + C a^{3}\right ) \]
-A*a**3/x + B*a**3*log(x) + C*b**3*x**7/7 + D*b**3*x**8/8 + x**6*(B*b**3/6 + D*a*b**2/2) + x**5*(A*b**3/5 + 3*C*a*b**2/5) + x**4*(3*B*a*b**2/4 + 3*D *a**2*b/4) + x**3*(A*a*b**2 + C*a**2*b) + x**2*(3*B*a**2*b/2 + D*a**3/2) + x*(3*A*a**2*b + C*a**3)
Time = 0.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.12 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {1}{8} \, D b^{3} x^{8} + \frac {1}{7} \, C b^{3} x^{7} + \frac {1}{6} \, {\left (3 \, D a b^{2} + B b^{3}\right )} x^{6} + \frac {1}{5} \, {\left (3 \, C a b^{2} + A b^{3}\right )} x^{5} + \frac {3}{4} \, {\left (D a^{2} b + B a b^{2}\right )} x^{4} + B a^{3} \log \left (x\right ) + {\left (C a^{2} b + A a b^{2}\right )} x^{3} - \frac {A a^{3}}{x} + \frac {1}{2} \, {\left (D a^{3} + 3 \, B a^{2} b\right )} x^{2} + {\left (C a^{3} + 3 \, A a^{2} b\right )} x \]
1/8*D*b^3*x^8 + 1/7*C*b^3*x^7 + 1/6*(3*D*a*b^2 + B*b^3)*x^6 + 1/5*(3*C*a*b ^2 + A*b^3)*x^5 + 3/4*(D*a^2*b + B*a*b^2)*x^4 + B*a^3*log(x) + (C*a^2*b + A*a*b^2)*x^3 - A*a^3/x + 1/2*(D*a^3 + 3*B*a^2*b)*x^2 + (C*a^3 + 3*A*a^2*b) *x
Time = 0.30 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.17 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {1}{8} \, D b^{3} x^{8} + \frac {1}{7} \, C b^{3} x^{7} + \frac {1}{2} \, D a b^{2} x^{6} + \frac {1}{6} \, B b^{3} x^{6} + \frac {3}{5} \, C a b^{2} x^{5} + \frac {1}{5} \, A b^{3} x^{5} + \frac {3}{4} \, D a^{2} b x^{4} + \frac {3}{4} \, B a b^{2} x^{4} + C a^{2} b x^{3} + A a b^{2} x^{3} + \frac {1}{2} \, D a^{3} x^{2} + \frac {3}{2} \, B a^{2} b x^{2} + C a^{3} x + 3 \, A a^{2} b x + B a^{3} \log \left ({\left | x \right |}\right ) - \frac {A a^{3}}{x} \]
1/8*D*b^3*x^8 + 1/7*C*b^3*x^7 + 1/2*D*a*b^2*x^6 + 1/6*B*b^3*x^6 + 3/5*C*a* b^2*x^5 + 1/5*A*b^3*x^5 + 3/4*D*a^2*b*x^4 + 3/4*B*a*b^2*x^4 + C*a^2*b*x^3 + A*a*b^2*x^3 + 1/2*D*a^3*x^2 + 3/2*B*a^2*b*x^2 + C*a^3*x + 3*A*a^2*b*x + B*a^3*log(abs(x)) - A*a^3/x
Time = 6.02 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^3 \left (A+B x+C x^2+D x^3\right )}{x^2} \, dx=\frac {{\left (b\,x^2+a\right )}^4\,D}{8\,b}-\frac {A\,a^3}{x}+\frac {A\,b^3\,x^5}{5}+\frac {B\,b^3\,x^6}{6}+\frac {C\,b^3\,x^7}{7}+B\,a^3\,\ln \left (x\right )+C\,a^3\,x+3\,A\,a^2\,b\,x+A\,a\,b^2\,x^3+\frac {3\,B\,a^2\,b\,x^2}{2}+\frac {3\,B\,a\,b^2\,x^4}{4}+C\,a^2\,b\,x^3+\frac {3\,C\,a\,b^2\,x^5}{5} \]