Integrand size = 21, antiderivative size = 90 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{3 b^2 x^3}-\frac {b B-2 A c}{b^3 x}-\frac {c (b B-A c) x}{2 b^3 \left (b+c x^2\right )}-\frac {\sqrt {c} (3 b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}} \] Output:
-1/3*A/b^2/x^3-(-2*A*c+B*b)/b^3/x-1/2*c*(-A*c+B*b)*x/b^3/(c*x^2+b)-1/2*c^( 1/2)*(-5*A*c+3*B*b)*arctan(c^(1/2)*x/b^(1/2))/b^(7/2)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{3 b^2 x^3}+\frac {-b B+2 A c}{b^3 x}-\frac {c (b B-A c) x}{2 b^3 \left (b+c x^2\right )}-\frac {\sqrt {c} (3 b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{7/2}} \] Input:
Integrate[(A + B*x^2)/(b*x^2 + c*x^4)^2,x]
Output:
-1/3*A/(b^2*x^3) + (-(b*B) + 2*A*c)/(b^3*x) - (c*(b*B - A*c)*x)/(2*b^3*(b + c*x^2)) - (Sqrt[c]*(3*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*b^(7/ 2))
Time = 0.55 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2026, 361, 25, 1584, 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.
\(\displaystyle \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {A+B x^2}{x^4 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 361 |
\(\displaystyle -\frac {1}{2} c \int -\frac {-\frac {(b B-A c) x^4}{b^3}+\frac {2 (b B-A c) x^2}{b^2 c}+\frac {2 A}{b c}}{x^4 \left (c x^2+b\right )}dx-\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} c \int \frac {-\frac {(b B-A c) x^4}{b^3}+\frac {2 (b B-A c) x^2}{b^2 c}+\frac {2 A}{b c}}{x^4 \left (c x^2+b\right )}dx-\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {1}{2} c \int \left (\frac {2 A}{b^2 c x^4}+\frac {5 A c-3 b B}{b^3 \left (c x^2+b\right )}+\frac {2 (b B-2 A c)}{b^3 c x^2}\right )dx-\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} c \left (-\frac {(3 b B-5 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{b^{7/2} \sqrt {c}}-\frac {2 (b B-2 A c)}{b^3 c x}-\frac {2 A}{3 b^2 c x^3}\right )-\frac {c x (b B-A c)}{2 b^3 \left (b+c x^2\right )}\) |
Input:
Int[(A + B*x^2)/(b*x^2 + c*x^4)^2,x]
Output:
-1/2*(c*(b*B - A*c)*x)/(b^3*(b + c*x^2)) + (c*((-2*A)/(3*b^2*c*x^3) - (2*( b*B - 2*A*c))/(b^3*c*x) - ((3*b*B - 5*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(b ^(7/2)*Sqrt[c])))/2
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[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {A}{3 b^{2} x^{3}}-\frac {-2 A c +B b}{b^{3} x}+\frac {c \left (\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (5 A c -3 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{b^{3}}\) | \(78\) |
risch | \(\frac {\frac {c \left (5 A c -3 B b \right ) x^{4}}{2 b^{3}}+\frac {\left (5 A c -3 B b \right ) x^{2}}{3 b^{2}}-\frac {A}{3 b}}{\left (c \,x^{2}+b \right ) x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{7} \textit {\_Z}^{2}+25 A^{2} c^{3}-30 A B b \,c^{2}+9 B^{2} b^{2} c \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{7}+50 A^{2} c^{3}-60 A B b \,c^{2}+18 B^{2} b^{2} c \right ) x +\left (-5 A \,b^{4} c +3 b^{5} B \right ) \textit {\_R} \right )\right )}{4}\) | \(152\) |
Input:
int((B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/3*A/b^2/x^3-(-2*A*c+B*b)/b^3/x+1/b^3*c*((1/2*A*c-1/2*B*b)*x/(c*x^2+b)+1 /2*(5*A*c-3*B*b)/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.78 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=\left [-\frac {6 \, {\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 4 \, A b^{2} + 4 \, {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2} + 3 \, {\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{5} + {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{3}\right )} \sqrt {-\frac {c}{b}} \log \left (\frac {c x^{2} + 2 \, b x \sqrt {-\frac {c}{b}} - b}{c x^{2} + b}\right )}{12 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}, -\frac {3 \, {\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 2 \, A b^{2} + 2 \, {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2} + 3 \, {\left ({\left (3 \, B b c - 5 \, A c^{2}\right )} x^{5} + {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{3}\right )} \sqrt {\frac {c}{b}} \arctan \left (x \sqrt {\frac {c}{b}}\right )}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}}\right ] \] Input:
integrate((B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
Output:
[-1/12*(6*(3*B*b*c - 5*A*c^2)*x^4 + 4*A*b^2 + 4*(3*B*b^2 - 5*A*b*c)*x^2 + 3*((3*B*b*c - 5*A*c^2)*x^5 + (3*B*b^2 - 5*A*b*c)*x^3)*sqrt(-c/b)*log((c*x^ 2 + 2*b*x*sqrt(-c/b) - b)/(c*x^2 + b)))/(b^3*c*x^5 + b^4*x^3), -1/6*(3*(3* B*b*c - 5*A*c^2)*x^4 + 2*A*b^2 + 2*(3*B*b^2 - 5*A*b*c)*x^2 + 3*((3*B*b*c - 5*A*c^2)*x^5 + (3*B*b^2 - 5*A*b*c)*x^3)*sqrt(c/b)*arctan(x*sqrt(c/b)))/(b ^3*c*x^5 + b^4*x^3)]
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (82) = 164\).
Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.04 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right ) \log {\left (- \frac {b^{4} \sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right )}{- 5 A c^{2} + 3 B b c} + x \right )}}{4} - \frac {\sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right ) \log {\left (\frac {b^{4} \sqrt {- \frac {c}{b^{7}}} \left (- 5 A c + 3 B b\right )}{- 5 A c^{2} + 3 B b c} + x \right )}}{4} + \frac {- 2 A b^{2} + x^{4} \cdot \left (15 A c^{2} - 9 B b c\right ) + x^{2} \cdot \left (10 A b c - 6 B b^{2}\right )}{6 b^{4} x^{3} + 6 b^{3} c x^{5}} \] Input:
integrate((B*x**2+A)/(c*x**4+b*x**2)**2,x)
Output:
sqrt(-c/b**7)*(-5*A*c + 3*B*b)*log(-b**4*sqrt(-c/b**7)*(-5*A*c + 3*B*b)/(- 5*A*c**2 + 3*B*b*c) + x)/4 - sqrt(-c/b**7)*(-5*A*c + 3*B*b)*log(b**4*sqrt( -c/b**7)*(-5*A*c + 3*B*b)/(-5*A*c**2 + 3*B*b*c) + x)/4 + (-2*A*b**2 + x**4 *(15*A*c**2 - 9*B*b*c) + x**2*(10*A*b*c - 6*B*b**2))/(6*b**4*x**3 + 6*b**3 *c*x**5)
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {3 \, {\left (3 \, B b c - 5 \, A c^{2}\right )} x^{4} + 2 \, A b^{2} + 2 \, {\left (3 \, B b^{2} - 5 \, A b c\right )} x^{2}}{6 \, {\left (b^{3} c x^{5} + b^{4} x^{3}\right )}} - \frac {{\left (3 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} \] Input:
integrate((B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
Output:
-1/6*(3*(3*B*b*c - 5*A*c^2)*x^4 + 2*A*b^2 + 2*(3*B*b^2 - 5*A*b*c)*x^2)/(b^ 3*c*x^5 + b^4*x^3) - 1/2*(3*B*b*c - 5*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b *c)*b^3)
Time = 0.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (3 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{3}} - \frac {B b c x - A c^{2} x}{2 \, {\left (c x^{2} + b\right )} b^{3}} - \frac {3 \, B b x^{2} - 6 \, A c x^{2} + A b}{3 \, b^{3} x^{3}} \] Input:
integrate((B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
Output:
-1/2*(3*B*b*c - 5*A*c^2)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^3) - 1/2*(B*b* c*x - A*c^2*x)/((c*x^2 + b)*b^3) - 1/3*(3*B*b*x^2 - 6*A*c*x^2 + A*b)/(b^3* x^3)
Time = 9.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {x^2\,\left (5\,A\,c-3\,B\,b\right )}{3\,b^2}-\frac {A}{3\,b}+\frac {c\,x^4\,\left (5\,A\,c-3\,B\,b\right )}{2\,b^3}}{c\,x^5+b\,x^3}+\frac {\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (5\,A\,c-3\,B\,b\right )}{2\,b^{7/2}} \] Input:
int((A + B*x^2)/(b*x^2 + c*x^4)^2,x)
Output:
((x^2*(5*A*c - 3*B*b))/(3*b^2) - A/(3*b) + (c*x^4*(5*A*c - 3*B*b))/(2*b^3) )/(b*x^3 + c*x^5) + (c^(1/2)*atan((c^(1/2)*x)/b^(1/2))*(5*A*c - 3*B*b))/(2 *b^(7/2))
Time = 0.24 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.77 \[ \int \frac {A+B x^2}{\left (b x^2+c x^4\right )^2} \, dx=\frac {15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b c \,x^{3}+15 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{2} x^{5}-9 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} x^{3}-9 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c \,x^{5}-2 a \,b^{3}+10 a \,b^{2} c \,x^{2}+15 a b \,c^{2} x^{4}-6 b^{4} x^{2}-9 b^{3} c \,x^{4}}{6 b^{4} x^{3} \left (c \,x^{2}+b \right )} \] Input:
int((B*x^2+A)/(c*x^4+b*x^2)^2,x)
Output:
(15*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c*x**3 + 15*sqrt(c)* sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**2*x**5 - 9*sqrt(c)*sqrt(b)*atan ((c*x)/(sqrt(c)*sqrt(b)))*b**3*x**3 - 9*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c )*sqrt(b)))*b**2*c*x**5 - 2*a*b**3 + 10*a*b**2*c*x**2 + 15*a*b*c**2*x**4 - 6*b**4*x**2 - 9*b**3*c*x**4)/(6*b**4*x**3*(b + c*x**2))