Integrand size = 24, antiderivative size = 111 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{5 b^2 x^5}-\frac {b B-2 A c}{3 b^3 x^3}+\frac {c (2 b B-3 A c)}{b^4 x}+\frac {c^2 (b B-A c) x}{2 b^4 \left (b+c x^2\right )}+\frac {c^{3/2} (5 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \]
-1/5*A/b^2/x^5+1/3*(2*A*c-B*b)/b^3/x^3+c*(-3*A*c+2*B*b)/b^4/x+1/2*c^2*(-A* c+B*b)*x/b^4/(c*x^2+b)+1/2*c^(3/2)*(-7*A*c+5*B*b)*arctan(x*c^(1/2)/b^(1/2) )/b^(9/2)
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{5 b^2 x^5}+\frac {-b B+2 A c}{3 b^3 x^3}+\frac {c (2 b B-3 A c)}{b^4 x}+\frac {c^2 (b B-A c) x}{2 b^4 \left (b+c x^2\right )}+\frac {c^{3/2} (5 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{9/2}} \]
-1/5*A/(b^2*x^5) + (-(b*B) + 2*A*c)/(3*b^3*x^3) + (c*(2*b*B - 3*A*c))/(b^4 *x) + (c^2*(b*B - A*c)*x)/(2*b^4*(b + c*x^2)) + (c^(3/2)*(5*b*B - 7*A*c)*A rcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(9/2))
Time = 0.46 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {9, 361, 25, 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.
\(\displaystyle \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^6 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 361 |
\(\displaystyle \frac {c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}-\frac {1}{2} c^2 \int -\frac {\frac {(b B-A c) x^6}{b^4}-\frac {2 (b B-A c) x^4}{b^3 c}+\frac {2 (b B-A c) x^2}{b^2 c^2}+\frac {2 A}{b c^2}}{x^6 \left (c x^2+b\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} c^2 \int \frac {\frac {(b B-A c) x^6}{b^4}-\frac {2 (b B-A c) x^4}{b^3 c}+\frac {2 (b B-A c) x^2}{b^2 c^2}+\frac {2 A}{b c^2}}{x^6 \left (c x^2+b\right )}dx+\frac {c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 2333 |
\(\displaystyle \frac {1}{2} c^2 \int \left (\frac {2 A}{b^2 c^2 x^6}+\frac {5 b B-7 A c}{b^4 \left (c x^2+b\right )}-\frac {2 (2 b B-3 A c)}{b^4 c x^2}+\frac {2 (b B-2 A c)}{b^3 c^2 x^4}\right )dx+\frac {c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} c^2 \left (\frac {(5 b B-7 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{9/2} \sqrt {c}}+\frac {2 (2 b B-3 A c)}{b^4 c x}-\frac {2 (b B-2 A c)}{3 b^3 c^2 x^3}-\frac {2 A}{5 b^2 c^2 x^5}\right )+\frac {c^2 x (b B-A c)}{2 b^4 \left (b+c x^2\right )}\) |
(c^2*(b*B - A*c)*x)/(2*b^4*(b + c*x^2)) + (c^2*((-2*A)/(5*b^2*c^2*x^5) - ( 2*(b*B - 2*A*c))/(3*b^3*c^2*x^3) + (2*(2*b*B - 3*A*c))/(b^4*c*x) + ((5*b*B - 7*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(b^(9/2)*Sqrt[c])))/2
3.1.72.3.1 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[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[m/ 2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 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 = 1.83 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {A}{5 b^{2} x^{5}}-\frac {-2 A c +B b}{3 x^{3} b^{3}}-\frac {c \left (3 A c -2 B b \right )}{b^{4} x}-\frac {c^{2} \left (\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (7 A c -5 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{4}}\) | \(99\) |
risch | \(\frac {-\frac {c^{2} \left (7 A c -5 B b \right ) x^{6}}{2 b^{4}}-\frac {c \left (7 A c -5 B b \right ) x^{4}}{3 b^{3}}+\frac {\left (7 A c -5 B b \right ) x^{2}}{15 b^{2}}-\frac {A}{5 b}}{x^{5} \left (c \,x^{2}+b \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{9} \textit {\_Z}^{2}+49 A^{2} c^{5}-70 A B b \,c^{4}+25 B^{2} b^{2} c^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{9}+98 A^{2} c^{5}-140 A B b \,c^{4}+50 B^{2} b^{2} c^{3}\right ) x +\left (7 A \,b^{5} c^{2}-5 B \,b^{6} c \right ) \textit {\_R} \right )\right )}{4}\) | \(179\) |
-1/5*A/b^2/x^5-1/3*(-2*A*c+B*b)/x^3/b^3-c*(3*A*c-2*B*b)/b^4/x-1/b^4*c^2*(( 1/2*A*c-1/2*B*b)*x/(c*x^2+b)+1/2*(7*A*c-5*B*b)/(b*c)^(1/2)*arctan(c*x/(b*c )^(1/2)))
Time = 0.27 (sec) , antiderivative size = 308, normalized size of antiderivative = 2.77 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\left [\frac {30 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 20 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} - 12 \, A b^{3} - 4 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} - 15 \, {\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} - 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{60 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}, \frac {15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} - 6 \, A b^{3} - 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2} + 15 \, {\left ({\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{7} + {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{5}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}}\right ] \]
[1/60*(30*(5*B*b*c^2 - 7*A*c^3)*x^6 + 20*(5*B*b^2*c - 7*A*b*c^2)*x^4 - 12* A*b^3 - 4*(5*B*b^3 - 7*A*b^2*c)*x^2 - 15*((5*B*b*c^2 - 7*A*c^3)*x^7 + (5*B *b^2*c - 7*A*b*c^2)*x^5)*sqrt(-c/b)*log((c*x^2 - 2*b*x*sqrt(-c/b) - b)/(c* x^2 + b)))/(b^4*c*x^7 + b^5*x^5), 1/30*(15*(5*B*b*c^2 - 7*A*c^3)*x^6 + 10* (5*B*b^2*c - 7*A*b*c^2)*x^4 - 6*A*b^3 - 2*(5*B*b^3 - 7*A*b^2*c)*x^2 + 15*( (5*B*b*c^2 - 7*A*c^3)*x^7 + (5*B*b^2*c - 7*A*b*c^2)*x^5)*sqrt(c/b)*arctan( x*sqrt(c/b)))/(b^4*c*x^7 + b^5*x^5)]
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (104) = 208\).
Time = 0.36 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.96 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=- \frac {\sqrt {- \frac {c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right ) \log {\left (- \frac {b^{5} \sqrt {- \frac {c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right )}{- 7 A c^{3} + 5 B b c^{2}} + x \right )}}{4} + \frac {\sqrt {- \frac {c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right ) \log {\left (\frac {b^{5} \sqrt {- \frac {c^{3}}{b^{9}}} \left (- 7 A c + 5 B b\right )}{- 7 A c^{3} + 5 B b c^{2}} + x \right )}}{4} + \frac {- 6 A b^{3} + x^{6} \left (- 105 A c^{3} + 75 B b c^{2}\right ) + x^{4} \left (- 70 A b c^{2} + 50 B b^{2} c\right ) + x^{2} \cdot \left (14 A b^{2} c - 10 B b^{3}\right )}{30 b^{5} x^{5} + 30 b^{4} c x^{7}} \]
-sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)*log(-b**5*sqrt(-c**3/b**9)*(-7*A*c + 5* B*b)/(-7*A*c**3 + 5*B*b*c**2) + x)/4 + sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)*l og(b**5*sqrt(-c**3/b**9)*(-7*A*c + 5*B*b)/(-7*A*c**3 + 5*B*b*c**2) + x)/4 + (-6*A*b**3 + x**6*(-105*A*c**3 + 75*B*b*c**2) + x**4*(-70*A*b*c**2 + 50* B*b**2*c) + x**2*(14*A*b**2*c - 10*B*b**3))/(30*b**5*x**5 + 30*b**4*c*x**7 )
Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\frac {15 \, {\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} x^{6} + 10 \, {\left (5 \, B b^{2} c - 7 \, A b c^{2}\right )} x^{4} - 6 \, A b^{3} - 2 \, {\left (5 \, B b^{3} - 7 \, A b^{2} c\right )} x^{2}}{30 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}} + \frac {{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} \]
1/30*(15*(5*B*b*c^2 - 7*A*c^3)*x^6 + 10*(5*B*b^2*c - 7*A*b*c^2)*x^4 - 6*A* b^3 - 2*(5*B*b^3 - 7*A*b^2*c)*x^2)/(b^4*c*x^7 + b^5*x^5) + 1/2*(5*B*b*c^2 - 7*A*c^3)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4)
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.01 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (5 \, B b c^{2} - 7 \, A c^{3}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{4}} + \frac {B b c^{2} x - A c^{3} x}{2 \, {\left (c x^{2} + b\right )} b^{4}} + \frac {30 \, B b c x^{4} - 45 \, A c^{2} x^{4} - 5 \, B b^{2} x^{2} + 10 \, A b c x^{2} - 3 \, A b^{2}}{15 \, b^{4} x^{5}} \]
1/2*(5*B*b*c^2 - 7*A*c^3)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^4) + 1/2*(B*b *c^2*x - A*c^3*x)/((c*x^2 + b)*b^4) + 1/15*(30*B*b*c*x^4 - 45*A*c^2*x^4 - 5*B*b^2*x^2 + 10*A*b*c*x^2 - 3*A*b^2)/(b^4*x^5)
Time = 8.99 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{x^2 \left (b x^2+c x^4\right )^2} \, dx=-\frac {\frac {A}{5\,b}-\frac {x^2\,\left (7\,A\,c-5\,B\,b\right )}{15\,b^2}+\frac {c^2\,x^6\,\left (7\,A\,c-5\,B\,b\right )}{2\,b^4}+\frac {c\,x^4\,\left (7\,A\,c-5\,B\,b\right )}{3\,b^3}}{c\,x^7+b\,x^5}-\frac {c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (7\,A\,c-5\,B\,b\right )}{2\,b^{9/2}} \]