Integrand size = 24, antiderivative size = 133 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {b^2 (4 b B-3 A c) x}{c^5}+\frac {b (3 b B-2 A c) x^3}{3 c^4}-\frac {(2 b B-A c) x^5}{5 c^3}+\frac {B x^7}{7 c^2}-\frac {b^3 (b B-A c) x}{2 c^5 \left (b+c x^2\right )}+\frac {b^{5/2} (9 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{11/2}} \]
-b^2*(-3*A*c+4*B*b)*x/c^5+1/3*b*(-2*A*c+3*B*b)*x^3/c^4-1/5*(-A*c+2*B*b)*x^ 5/c^3+1/7*B*x^7/c^2-1/2*b^3*(-A*c+B*b)*x/c^5/(c*x^2+b)+1/2*b^(5/2)*(-7*A*c +9*B*b)*arctan(x*c^(1/2)/b^(1/2))/c^(11/2)
Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.01 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {b^2 (4 b B-3 A c) x}{c^5}+\frac {b (3 b B-2 A c) x^3}{3 c^4}+\frac {(-2 b B+A c) x^5}{5 c^3}+\frac {B x^7}{7 c^2}+\frac {\left (-b^4 B+A b^3 c\right ) x}{2 c^5 \left (b+c x^2\right )}+\frac {b^{5/2} (9 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{11/2}} \]
-((b^2*(4*b*B - 3*A*c)*x)/c^5) + (b*(3*b*B - 2*A*c)*x^3)/(3*c^4) + ((-2*b* B + A*c)*x^5)/(5*c^3) + (B*x^7)/(7*c^2) + ((-(b^4*B) + A*b^3*c)*x)/(2*c^5* (b + c*x^2)) + (b^(5/2)*(9*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*c^ (11/2))
Time = 0.42 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {9, 360, 25, 2341, 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.
| \(\displaystyle \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 9 |
| \(\displaystyle \int \frac {x^8 \left (A+B x^2\right )}{\left (b+c x^2\right )^2}dx\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle -\frac {\int -\frac {2 B c^4 x^8-2 c^3 (b B-A c) x^6+2 b c^2 (b B-A c) x^4-2 b^2 c (b B-A c) x^2+b^3 (b B-A c)}{c x^2+b}dx}{2 c^5}-\frac {b^3 x (b B-A c)}{2 c^5 \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 B c^4 x^8-2 c^3 (b B-A c) x^6+2 b c^2 (b B-A c) x^4-2 b^2 c (b B-A c) x^2+b^3 (b B-A c)}{c x^2+b}dx}{2 c^5}-\frac {b^3 x (b B-A c)}{2 c^5 \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {\int \left (2 B c^3 x^6-2 c^2 (2 b B-A c) x^4+2 b c (3 b B-2 A c) x^2-2 b^2 (4 b B-3 A c)+\frac {9 b^4 B-7 A b^3 c}{c x^2+b}\right )dx}{2 c^5}-\frac {b^3 x (b B-A c)}{2 c^5 \left (b+c x^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^{5/2} (9 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {c}}-2 b^2 x (4 b B-3 A c)-\frac {2}{5} c^2 x^5 (2 b B-A c)+\frac {2}{3} b c x^3 (3 b B-2 A c)+\frac {2}{7} B c^3 x^7}{2 c^5}-\frac {b^3 x (b B-A c)}{2 c^5 \left (b+c x^2\right )}\) |
-1/2*(b^3*(b*B - A*c)*x)/(c^5*(b + c*x^2)) + (-2*b^2*(4*b*B - 3*A*c)*x + ( 2*b*c*(3*b*B - 2*A*c)*x^3)/3 - (2*c^2*(2*b*B - A*c)*x^5)/5 + (2*B*c^3*x^7) /7 + (b^(5/2)*(9*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/Sqrt[c])/(2*c^5 )
3.1.58.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 1.80 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(\frac {\frac {1}{7} B \,c^{3} x^{7}+\frac {1}{5} A \,c^{3} x^{5}-\frac {2}{5} B b \,c^{2} x^{5}-\frac {2}{3} A b \,c^{2} x^{3}+B \,b^{2} c \,x^{3}+3 A \,b^{2} c x -4 b^{3} B x}{c^{5}}-\frac {b^{3} \left (\frac {\left (-\frac {A c}{2}+\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (7 A c -9 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{c^{5}}\) | \(123\) |
| risch | \(\frac {B \,x^{7}}{7 c^{2}}+\frac {A \,x^{5}}{5 c^{2}}-\frac {2 B b \,x^{5}}{5 c^{3}}-\frac {2 A b \,x^{3}}{3 c^{3}}+\frac {B \,b^{2} x^{3}}{c^{4}}+\frac {3 A \,b^{2} x}{c^{4}}-\frac {4 b^{3} B x}{c^{5}}+\frac {\left (\frac {1}{2} A \,b^{3} c -\frac {1}{2} B \,b^{4}\right ) x}{c^{5} \left (c \,x^{2}+b \right )}+\frac {7 \sqrt {-b c}\, b^{2} \ln \left (-\sqrt {-b c}\, x -b \right ) A}{4 c^{5}}-\frac {9 \sqrt {-b c}\, b^{3} \ln \left (-\sqrt {-b c}\, x -b \right ) B}{4 c^{6}}-\frac {7 \sqrt {-b c}\, b^{2} \ln \left (\sqrt {-b c}\, x -b \right ) A}{4 c^{5}}+\frac {9 \sqrt {-b c}\, b^{3} \ln \left (\sqrt {-b c}\, x -b \right ) B}{4 c^{6}}\) | \(213\) |
1/c^5*(1/7*B*c^3*x^7+1/5*A*c^3*x^5-2/5*B*b*c^2*x^5-2/3*A*b*c^2*x^3+B*b^2*c *x^3+3*A*b^2*c*x-4*b^3*B*x)-b^3/c^5*((-1/2*A*c+1/2*B*b)*x/(c*x^2+b)+1/2*(7 *A*c-9*B*b)/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.55 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.63 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\left [\frac {60 \, B c^{4} x^{9} - 12 \, {\left (9 \, B b c^{3} - 7 \, A c^{4}\right )} x^{7} + 28 \, {\left (9 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{5} - 140 \, {\left (9 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3} - 105 \, {\left (9 \, B b^{4} - 7 \, A b^{3} c + {\left (9 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) - 210 \, {\left (9 \, B b^{4} - 7 \, A b^{3} c\right )} x}{420 \, {\left (c^{6} x^{2} + b c^{5}\right )}}, \frac {30 \, B c^{4} x^{9} - 6 \, {\left (9 \, B b c^{3} - 7 \, A c^{4}\right )} x^{7} + 14 \, {\left (9 \, B b^{2} c^{2} - 7 \, A b c^{3}\right )} x^{5} - 70 \, {\left (9 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{3} + 105 \, {\left (9 \, B b^{4} - 7 \, A b^{3} c + {\left (9 \, B b^{3} c - 7 \, A b^{2} c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) - 105 \, {\left (9 \, B b^{4} - 7 \, A b^{3} c\right )} x}{210 \, {\left (c^{6} x^{2} + b c^{5}\right )}}\right ] \]
[1/420*(60*B*c^4*x^9 - 12*(9*B*b*c^3 - 7*A*c^4)*x^7 + 28*(9*B*b^2*c^2 - 7* A*b*c^3)*x^5 - 140*(9*B*b^3*c - 7*A*b^2*c^2)*x^3 - 105*(9*B*b^4 - 7*A*b^3* c + (9*B*b^3*c - 7*A*b^2*c^2)*x^2)*sqrt(-b/c)*log((c*x^2 - 2*c*x*sqrt(-b/c ) - b)/(c*x^2 + b)) - 210*(9*B*b^4 - 7*A*b^3*c)*x)/(c^6*x^2 + b*c^5), 1/21 0*(30*B*c^4*x^9 - 6*(9*B*b*c^3 - 7*A*c^4)*x^7 + 14*(9*B*b^2*c^2 - 7*A*b*c^ 3)*x^5 - 70*(9*B*b^3*c - 7*A*b^2*c^2)*x^3 + 105*(9*B*b^4 - 7*A*b^3*c + (9* B*b^3*c - 7*A*b^2*c^2)*x^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) - 105*(9*B*b ^4 - 7*A*b^3*c)*x)/(c^6*x^2 + b*c^5)]
Time = 0.42 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.79 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {B x^{7}}{7 c^{2}} + x^{5} \left (\frac {A}{5 c^{2}} - \frac {2 B b}{5 c^{3}}\right ) + x^{3} \left (- \frac {2 A b}{3 c^{3}} + \frac {B b^{2}}{c^{4}}\right ) + x \left (\frac {3 A b^{2}}{c^{4}} - \frac {4 B b^{3}}{c^{5}}\right ) + \frac {x \left (A b^{3} c - B b^{4}\right )}{2 b c^{5} + 2 c^{6} x^{2}} - \frac {\sqrt {- \frac {b^{5}}{c^{11}}} \left (- 7 A c + 9 B b\right ) \log {\left (- \frac {c^{5} \sqrt {- \frac {b^{5}}{c^{11}}} \left (- 7 A c + 9 B b\right )}{- 7 A b^{2} c + 9 B b^{3}} + x \right )}}{4} + \frac {\sqrt {- \frac {b^{5}}{c^{11}}} \left (- 7 A c + 9 B b\right ) \log {\left (\frac {c^{5} \sqrt {- \frac {b^{5}}{c^{11}}} \left (- 7 A c + 9 B b\right )}{- 7 A b^{2} c + 9 B b^{3}} + x \right )}}{4} \]
B*x**7/(7*c**2) + x**5*(A/(5*c**2) - 2*B*b/(5*c**3)) + x**3*(-2*A*b/(3*c** 3) + B*b**2/c**4) + x*(3*A*b**2/c**4 - 4*B*b**3/c**5) + x*(A*b**3*c - B*b* *4)/(2*b*c**5 + 2*c**6*x**2) - sqrt(-b**5/c**11)*(-7*A*c + 9*B*b)*log(-c** 5*sqrt(-b**5/c**11)*(-7*A*c + 9*B*b)/(-7*A*b**2*c + 9*B*b**3) + x)/4 + sqr t(-b**5/c**11)*(-7*A*c + 9*B*b)*log(c**5*sqrt(-b**5/c**11)*(-7*A*c + 9*B*b )/(-7*A*b**2*c + 9*B*b**3) + x)/4
Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.02 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (B b^{4} - A b^{3} c\right )} x}{2 \, {\left (c^{6} x^{2} + b c^{5}\right )}} + \frac {{\left (9 \, B b^{4} - 7 \, A b^{3} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{5}} + \frac {15 \, B c^{3} x^{7} - 21 \, {\left (2 \, B b c^{2} - A c^{3}\right )} x^{5} + 35 \, {\left (3 \, B b^{2} c - 2 \, A b c^{2}\right )} x^{3} - 105 \, {\left (4 \, B b^{3} - 3 \, A b^{2} c\right )} x}{105 \, c^{5}} \]
-1/2*(B*b^4 - A*b^3*c)*x/(c^6*x^2 + b*c^5) + 1/2*(9*B*b^4 - 7*A*b^3*c)*arc tan(c*x/sqrt(b*c))/(sqrt(b*c)*c^5) + 1/105*(15*B*c^3*x^7 - 21*(2*B*b*c^2 - A*c^3)*x^5 + 35*(3*B*b^2*c - 2*A*b*c^2)*x^3 - 105*(4*B*b^3 - 3*A*b^2*c)*x )/c^5
Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.05 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (9 \, B b^{4} - 7 \, A b^{3} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{5}} - \frac {B b^{4} x - A b^{3} c x}{2 \, {\left (c x^{2} + b\right )} c^{5}} + \frac {15 \, B c^{12} x^{7} - 42 \, B b c^{11} x^{5} + 21 \, A c^{12} x^{5} + 105 \, B b^{2} c^{10} x^{3} - 70 \, A b c^{11} x^{3} - 420 \, B b^{3} c^{9} x + 315 \, A b^{2} c^{10} x}{105 \, c^{14}} \]
1/2*(9*B*b^4 - 7*A*b^3*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^5) - 1/2*(B*b ^4*x - A*b^3*c*x)/((c*x^2 + b)*c^5) + 1/105*(15*B*c^12*x^7 - 42*B*b*c^11*x ^5 + 21*A*c^12*x^5 + 105*B*b^2*c^10*x^3 - 70*A*b*c^11*x^3 - 420*B*b^3*c^9* x + 315*A*b^2*c^10*x)/c^14
Time = 8.88 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.53 \[ \int \frac {x^{12} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=x\,\left (\frac {2\,b\,\left (\frac {2\,b\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )}{c}+\frac {B\,b^2}{c^4}\right )}{c}-\frac {b^2\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )}{c^2}\right )+x^5\,\left (\frac {A}{5\,c^2}-\frac {2\,B\,b}{5\,c^3}\right )-x^3\,\left (\frac {2\,b\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )}{3\,c}+\frac {B\,b^2}{3\,c^4}\right )+\frac {B\,x^7}{7\,c^2}-\frac {x\,\left (\frac {B\,b^4}{2}-\frac {A\,b^3\,c}{2}\right )}{c^6\,x^2+b\,c^5}+\frac {b^{5/2}\,\mathrm {atan}\left (\frac {b^{5/2}\,\sqrt {c}\,x\,\left (7\,A\,c-9\,B\,b\right )}{9\,B\,b^4-7\,A\,b^3\,c}\right )\,\left (7\,A\,c-9\,B\,b\right )}{2\,c^{11/2}} \]
x*((2*b*((2*b*(A/c^2 - (2*B*b)/c^3))/c + (B*b^2)/c^4))/c - (b^2*(A/c^2 - ( 2*B*b)/c^3))/c^2) + x^5*(A/(5*c^2) - (2*B*b)/(5*c^3)) - x^3*((2*b*(A/c^2 - (2*B*b)/c^3))/(3*c) + (B*b^2)/(3*c^4)) + (B*x^7)/(7*c^2) - (x*((B*b^4)/2 - (A*b^3*c)/2))/(b*c^5 + c^6*x^2) + (b^(5/2)*atan((b^(5/2)*c^(1/2)*x*(7*A* c - 9*B*b))/(9*B*b^4 - 7*A*b^3*c))*(7*A*c - 9*B*b))/(2*c^(11/2))