Integrand size = 24, antiderivative size = 70 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{b^2 x}+\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {c}} \] Output:
-A/b^2/x+1/2*(-A*c+B*b)*x/b^2/(c*x^2+b)+1/2*(-3*A*c+B*b)*arctan(c^(1/2)*x/ b^(1/2))/b^(5/2)/c^(1/2)
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{b^2 x}+\frac {(b B-A c) x}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{5/2} \sqrt {c}} \] Input:
Integrate[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
Output:
-(A/(b^2*x)) + ((b*B - A*c)*x)/(2*b^2*(b + c*x^2)) + ((b*B - 3*A*c)*ArcTan [(Sqrt[c]*x)/Sqrt[b]])/(2*b^(5/2)*Sqrt[c])
Time = 0.37 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {9, 361, 25, 27, 359, 218}
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.
\(\displaystyle \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^2 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 361 |
\(\displaystyle \frac {x (b B-A c)}{2 b^2 \left (b+c x^2\right )}-\frac {1}{2} \int -\frac {(b B-A c) x^2+2 A b}{b^2 x^2 \left (c x^2+b\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \frac {(b B-A c) x^2+2 A b}{b^2 x^2 \left (c x^2+b\right )}dx+\frac {x (b B-A c)}{2 b^2 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(b B-A c) x^2+2 A b}{x^2 \left (c x^2+b\right )}dx}{2 b^2}+\frac {x (b B-A c)}{2 b^2 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {(b B-3 A c) \int \frac {1}{c x^2+b}dx-\frac {2 A}{x}}{2 b^2}+\frac {x (b B-A c)}{2 b^2 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {(b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} \sqrt {c}}-\frac {2 A}{x}}{2 b^2}+\frac {x (b B-A c)}{2 b^2 \left (b+c x^2\right )}\) |
Input:
Int[(x^2*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
Output:
((b*B - A*c)*x)/(2*b^2*(b + c*x^2)) + ((-2*A)/x + ((b*B - 3*A*c)*ArcTan[(S qrt[c]*x)/Sqrt[b]])/(Sqrt[b]*Sqrt[c]))/(2*b^2)
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[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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])
Time = 0.42 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {A}{b^{2} x}-\frac {\frac {\left (\frac {A c}{2}-\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (3 A c -B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}}{b^{2}}\) | \(62\) |
risch | \(\frac {-\frac {\left (3 A c -B b \right ) x^{2}}{2 b^{2}}-\frac {A}{b}}{x \left (c \,x^{2}+b \right )}-\frac {3 \ln \left (-\sqrt {-b c}\, x -b \right ) A c}{4 \sqrt {-b c}\, b^{2}}+\frac {\ln \left (-\sqrt {-b c}\, x -b \right ) B}{4 \sqrt {-b c}\, b}+\frac {3 \ln \left (-\sqrt {-b c}\, x +b \right ) A c}{4 \sqrt {-b c}\, b^{2}}-\frac {\ln \left (-\sqrt {-b c}\, x +b \right ) B}{4 \sqrt {-b c}\, b}\) | \(141\) |
Input:
int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-A/b^2/x-1/b^2*((1/2*A*c-1/2*B*b)*x/(c*x^2+b)+1/2*(3*A*c-B*b)/(b*c)^(1/2)* arctan(c*x/(b*c)^(1/2)))
Time = 0.09 (sec) , antiderivative size = 210, normalized size of antiderivative = 3.00 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\left [-\frac {4 \, A b^{2} c - 2 \, {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{2} - {\left ({\left (B b c - 3 \, A c^{2}\right )} x^{3} + {\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {-b c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{4 \, {\left (b^{3} c^{2} x^{3} + b^{4} c x\right )}}, -\frac {2 \, A b^{2} c - {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{2} - {\left ({\left (B b c - 3 \, A c^{2}\right )} x^{3} + {\left (B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{2 \, {\left (b^{3} c^{2} x^{3} + b^{4} c x\right )}}\right ] \] Input:
integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
Output:
[-1/4*(4*A*b^2*c - 2*(B*b^2*c - 3*A*b*c^2)*x^2 - ((B*b*c - 3*A*c^2)*x^3 + (B*b^2 - 3*A*b*c)*x)*sqrt(-b*c)*log((c*x^2 + 2*sqrt(-b*c)*x - b)/(c*x^2 + b)))/(b^3*c^2*x^3 + b^4*c*x), -1/2*(2*A*b^2*c - (B*b^2*c - 3*A*b*c^2)*x^2 - ((B*b*c - 3*A*c^2)*x^3 + (B*b^2 - 3*A*b*c)*x)*sqrt(b*c)*arctan(sqrt(b*c) *x/b))/(b^3*c^2*x^3 + b^4*c*x)]
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.63 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=- \frac {\sqrt {- \frac {1}{b^{5} c}} \left (- 3 A c + B b\right ) \log {\left (- b^{3} \sqrt {- \frac {1}{b^{5} c}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{b^{5} c}} \left (- 3 A c + B b\right ) \log {\left (b^{3} \sqrt {- \frac {1}{b^{5} c}} + x \right )}}{4} + \frac {- 2 A b + x^{2} \left (- 3 A c + B b\right )}{2 b^{3} x + 2 b^{2} c x^{3}} \] Input:
integrate(x**2*(B*x**2+A)/(c*x**4+b*x**2)**2,x)
Output:
-sqrt(-1/(b**5*c))*(-3*A*c + B*b)*log(-b**3*sqrt(-1/(b**5*c)) + x)/4 + sqr t(-1/(b**5*c))*(-3*A*c + B*b)*log(b**3*sqrt(-1/(b**5*c)) + x)/4 + (-2*A*b + x**2*(-3*A*c + B*b))/(2*b**3*x + 2*b**2*c*x**3)
Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (B b - 3 \, A c\right )} x^{2} - 2 \, A b}{2 \, {\left (b^{2} c x^{3} + b^{3} x\right )}} + \frac {{\left (B b - 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{2}} \] Input:
integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
Output:
1/2*((B*b - 3*A*c)*x^2 - 2*A*b)/(b^2*c*x^3 + b^3*x) + 1/2*(B*b - 3*A*c)*ar ctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2)
Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (B b - 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} b^{2}} + \frac {B b x^{2} - 3 \, A c x^{2} - 2 \, A b}{2 \, {\left (c x^{3} + b x\right )} b^{2}} \] Input:
integrate(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
Output:
1/2*(B*b - 3*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2) + 1/2*(B*b*x^2 - 3 *A*c*x^2 - 2*A*b)/((c*x^3 + b*x)*b^2)
Time = 9.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {\frac {A}{b}+\frac {x^2\,\left (3\,A\,c-B\,b\right )}{2\,b^2}}{c\,x^3+b\,x}-\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (3\,A\,c-B\,b\right )}{2\,b^{5/2}\,\sqrt {c}} \] Input:
int((x^2*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
Output:
- (A/b + (x^2*(3*A*c - B*b))/(2*b^2))/(b*x + c*x^3) - (atan((c^(1/2)*x)/b^ (1/2))*(3*A*c - B*b))/(2*b^(5/2)*c^(1/2))
Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.97 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {-3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a b c x -3 \sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) a \,c^{2} x^{3}+\sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{3} x +\sqrt {c}\, \sqrt {b}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}\, \sqrt {b}}\right ) b^{2} c \,x^{3}-2 a \,b^{2} c -3 a b \,c^{2} x^{2}+b^{3} c \,x^{2}}{2 b^{3} c x \left (c \,x^{2}+b \right )} \] Input:
int(x^2*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
Output:
( - 3*sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*b*c*x - 3*sqrt(c)*sq rt(b)*atan((c*x)/(sqrt(c)*sqrt(b)))*a*c**2*x**3 + sqrt(c)*sqrt(b)*atan((c* x)/(sqrt(c)*sqrt(b)))*b**3*x + sqrt(c)*sqrt(b)*atan((c*x)/(sqrt(c)*sqrt(b) ))*b**2*c*x**3 - 2*a*b**2*c - 3*a*b*c**2*x**2 + b**3*c*x**2)/(2*b**3*c*x*( b + c*x**2))