Integrand size = 24, antiderivative size = 97 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{4 b^2 x^4}-\frac {b B-2 A c}{2 b^3 x^2}-\frac {c (b B-A c)}{2 b^3 \left (b+c x^2\right )}-\frac {c (2 b B-3 A c) \log (x)}{b^4}+\frac {c (2 b B-3 A c) \log \left (b+c x^2\right )}{2 b^4} \] Output:
-1/4*A/b^2/x^4-1/2*(-2*A*c+B*b)/b^3/x^2-1/2*c*(-A*c+B*b)/b^3/(c*x^2+b)-c*( -3*A*c+2*B*b)*ln(x)/b^4+1/2*c*(-3*A*c+2*B*b)*ln(c*x^2+b)/b^4
Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=-\frac {\frac {A b^2}{x^4}+\frac {2 b (b B-2 A c)}{x^2}+\frac {2 b c (b B-A c)}{b+c x^2}-4 c (-2 b B+3 A c) \log (x)+2 c (-2 b B+3 A c) \log \left (b+c x^2\right )}{4 b^4} \] Input:
Integrate[(A + B*x^2)/(x*(b*x^2 + c*x^4)^2),x]
Output:
-1/4*((A*b^2)/x^4 + (2*b*(b*B - 2*A*c))/x^2 + (2*b*c*(b*B - A*c))/(b + c*x ^2) - 4*c*(-2*b*B + 3*A*c)*Log[x] + 2*c*(-2*b*B + 3*A*c)*Log[b + c*x^2])/b ^4
Time = 0.44 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 354, 86, 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.
\(\displaystyle \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^5 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \left (c x^2+b\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {(2 b B-3 A c) c^2}{b^4 \left (c x^2+b\right )}+\frac {(b B-A c) c^2}{b^3 \left (c x^2+b\right )^2}-\frac {(2 b B-3 A c) c}{b^4 x^2}+\frac {b B-2 A c}{b^3 x^4}+\frac {A}{b^2 x^6}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {c \log \left (x^2\right ) (2 b B-3 A c)}{b^4}+\frac {c (2 b B-3 A c) \log \left (b+c x^2\right )}{b^4}-\frac {c (b B-A c)}{b^3 \left (b+c x^2\right )}-\frac {b B-2 A c}{b^3 x^2}-\frac {A}{2 b^2 x^4}\right )\) |
Input:
Int[(A + B*x^2)/(x*(b*x^2 + c*x^4)^2),x]
Output:
(-1/2*A/(b^2*x^4) - (b*B - 2*A*c)/(b^3*x^2) - (c*(b*B - A*c))/(b^3*(b + c* x^2)) - (c*(2*b*B - 3*A*c)*Log[x^2])/b^4 + (c*(2*b*B - 3*A*c)*Log[b + c*x^ 2])/b^4)/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_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 0.40 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99
method | result | size |
default | \(-\frac {A}{4 b^{2} x^{4}}-\frac {-2 A c +B b}{2 b^{3} x^{2}}+\frac {c \left (3 A c -2 B b \right ) \ln \left (x \right )}{b^{4}}-\frac {c^{2} \left (\frac {\left (3 A c -2 B b \right ) \ln \left (c \,x^{2}+b \right )}{c}-\frac {b \left (A c -B b \right )}{c \left (c \,x^{2}+b \right )}\right )}{2 b^{4}}\) | \(96\) |
norman | \(\frac {-\frac {A}{4 b}+\frac {\left (3 A c -2 B b \right ) x^{2}}{4 b^{2}}-\frac {c \left (3 A \,c^{2}-2 B b c \right ) x^{6}}{2 b^{4}}}{x^{4} \left (c \,x^{2}+b \right )}+\frac {c \left (3 A c -2 B b \right ) \ln \left (x \right )}{b^{4}}-\frac {c \left (3 A c -2 B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b^{4}}\) | \(99\) |
risch | \(\frac {\frac {c \left (3 A c -2 B b \right ) x^{4}}{2 b^{3}}+\frac {\left (3 A c -2 B b \right ) x^{2}}{4 b^{2}}-\frac {A}{4 b}}{x^{4} \left (c \,x^{2}+b \right )}+\frac {3 c^{2} \ln \left (x \right ) A}{b^{4}}-\frac {2 c \ln \left (x \right ) B}{b^{3}}-\frac {3 c^{2} \ln \left (c \,x^{2}+b \right ) A}{2 b^{4}}+\frac {c \ln \left (c \,x^{2}+b \right ) B}{b^{3}}\) | \(108\) |
parallelrisch | \(\frac {12 A \ln \left (x \right ) x^{6} c^{3}-6 A \ln \left (c \,x^{2}+b \right ) x^{6} c^{3}-8 B \ln \left (x \right ) x^{6} b \,c^{2}+4 B \ln \left (c \,x^{2}+b \right ) x^{6} b \,c^{2}-6 A \,c^{3} x^{6}+4 B b \,c^{2} x^{6}+12 A \ln \left (x \right ) x^{4} b \,c^{2}-6 A \ln \left (c \,x^{2}+b \right ) x^{4} b \,c^{2}-8 B \ln \left (x \right ) x^{4} b^{2} c +4 B \ln \left (c \,x^{2}+b \right ) x^{4} b^{2} c +3 A \,b^{2} c \,x^{2}-2 x^{2} B \,b^{3}-A \,b^{3}}{4 b^{4} x^{4} \left (c \,x^{2}+b \right )}\) | \(181\) |
Input:
int((B*x^2+A)/x/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/4*A/b^2/x^4-1/2*(-2*A*c+B*b)/b^3/x^2+c*(3*A*c-2*B*b)/b^4*ln(x)-1/2/b^4* c^2*((3*A*c-2*B*b)/c*ln(c*x^2+b)-b*(A*c-B*b)/c/(c*x^2+b))
Time = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=-\frac {2 \, {\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{4} + A b^{3} + {\left (2 \, B b^{3} - 3 \, A b^{2} c\right )} x^{2} - 2 \, {\left ({\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} x^{6} + {\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{4}\right )} \log \left (c x^{2} + b\right ) + 4 \, {\left ({\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} x^{6} + {\left (2 \, B b^{2} c - 3 \, A b c^{2}\right )} x^{4}\right )} \log \left (x\right )}{4 \, {\left (b^{4} c x^{6} + b^{5} x^{4}\right )}} \] Input:
integrate((B*x^2+A)/x/(c*x^4+b*x^2)^2,x, algorithm="fricas")
Output:
-1/4*(2*(2*B*b^2*c - 3*A*b*c^2)*x^4 + A*b^3 + (2*B*b^3 - 3*A*b^2*c)*x^2 - 2*((2*B*b*c^2 - 3*A*c^3)*x^6 + (2*B*b^2*c - 3*A*b*c^2)*x^4)*log(c*x^2 + b) + 4*((2*B*b*c^2 - 3*A*c^3)*x^6 + (2*B*b^2*c - 3*A*b*c^2)*x^4)*log(x))/(b^ 4*c*x^6 + b^5*x^4)
Time = 0.56 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=\frac {- A b^{2} + x^{4} \cdot \left (6 A c^{2} - 4 B b c\right ) + x^{2} \cdot \left (3 A b c - 2 B b^{2}\right )}{4 b^{4} x^{4} + 4 b^{3} c x^{6}} - \frac {c \left (- 3 A c + 2 B b\right ) \log {\left (x \right )}}{b^{4}} + \frac {c \left (- 3 A c + 2 B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \] Input:
integrate((B*x**2+A)/x/(c*x**4+b*x**2)**2,x)
Output:
(-A*b**2 + x**4*(6*A*c**2 - 4*B*b*c) + x**2*(3*A*b*c - 2*B*b**2))/(4*b**4* x**4 + 4*b**3*c*x**6) - c*(-3*A*c + 2*B*b)*log(x)/b**4 + c*(-3*A*c + 2*B*b )*log(b/c + x**2)/(2*b**4)
Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=-\frac {2 \, {\left (2 \, B b c - 3 \, A c^{2}\right )} x^{4} + A b^{2} + {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}}{4 \, {\left (b^{3} c x^{6} + b^{4} x^{4}\right )}} + \frac {{\left (2 \, B b c - 3 \, A c^{2}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{4}} - \frac {{\left (2 \, B b c - 3 \, A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} \] Input:
integrate((B*x^2+A)/x/(c*x^4+b*x^2)^2,x, algorithm="maxima")
Output:
-1/4*(2*(2*B*b*c - 3*A*c^2)*x^4 + A*b^2 + (2*B*b^2 - 3*A*b*c)*x^2)/(b^3*c* x^6 + b^4*x^4) + 1/2*(2*B*b*c - 3*A*c^2)*log(c*x^2 + b)/b^4 - 1/2*(2*B*b*c - 3*A*c^2)*log(x^2)/b^4
Time = 0.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (2 \, B b c - 3 \, A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{4}} + \frac {{\left (2 \, B b c^{2} - 3 \, A c^{3}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4} c} - \frac {2 \, B b c^{2} x^{2} - 3 \, A c^{3} x^{2} + 3 \, B b^{2} c - 4 \, A b c^{2}}{2 \, {\left (c x^{2} + b\right )} b^{4}} + \frac {6 \, B b c x^{4} - 9 \, A c^{2} x^{4} - 2 \, B b^{2} x^{2} + 4 \, A b c x^{2} - A b^{2}}{4 \, b^{4} x^{4}} \] Input:
integrate((B*x^2+A)/x/(c*x^4+b*x^2)^2,x, algorithm="giac")
Output:
-1/2*(2*B*b*c - 3*A*c^2)*log(x^2)/b^4 + 1/2*(2*B*b*c^2 - 3*A*c^3)*log(abs( c*x^2 + b))/(b^4*c) - 1/2*(2*B*b*c^2*x^2 - 3*A*c^3*x^2 + 3*B*b^2*c - 4*A*b *c^2)/((c*x^2 + b)*b^4) + 1/4*(6*B*b*c*x^4 - 9*A*c^2*x^4 - 2*B*b^2*x^2 + 4 *A*b*c*x^2 - A*b^2)/(b^4*x^4)
Time = 9.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {x^2\,\left (3\,A\,c-2\,B\,b\right )}{4\,b^2}-\frac {A}{4\,b}+\frac {c\,x^4\,\left (3\,A\,c-2\,B\,b\right )}{2\,b^3}}{c\,x^6+b\,x^4}-\frac {\ln \left (c\,x^2+b\right )\,\left (3\,A\,c^2-2\,B\,b\,c\right )}{2\,b^4}+\frac {\ln \left (x\right )\,\left (3\,A\,c^2-2\,B\,b\,c\right )}{b^4} \] Input:
int((A + B*x^2)/(x*(b*x^2 + c*x^4)^2),x)
Output:
((x^2*(3*A*c - 2*B*b))/(4*b^2) - A/(4*b) + (c*x^4*(3*A*c - 2*B*b))/(2*b^3) )/(b*x^4 + c*x^6) - (log(b + c*x^2)*(3*A*c^2 - 2*B*b*c))/(2*b^4) + (log(x) *(3*A*c^2 - 2*B*b*c))/b^4
Time = 0.20 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.86 \[ \int \frac {A+B x^2}{x \left (b x^2+c x^4\right )^2} \, dx=\frac {-6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b \,c^{2} x^{4}-6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,c^{3} x^{6}+4 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} c \,x^{4}+4 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{2} x^{6}+12 \,\mathrm {log}\left (x \right ) a b \,c^{2} x^{4}+12 \,\mathrm {log}\left (x \right ) a \,c^{3} x^{6}-8 \,\mathrm {log}\left (x \right ) b^{3} c \,x^{4}-8 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{6}-a \,b^{3}+3 a \,b^{2} c \,x^{2}-6 a \,c^{3} x^{6}-2 b^{4} x^{2}+4 b^{2} c^{2} x^{6}}{4 b^{4} x^{4} \left (c \,x^{2}+b \right )} \] Input:
int((B*x^2+A)/x/(c*x^4+b*x^2)^2,x)
Output:
( - 6*log(b + c*x**2)*a*b*c**2*x**4 - 6*log(b + c*x**2)*a*c**3*x**6 + 4*lo g(b + c*x**2)*b**3*c*x**4 + 4*log(b + c*x**2)*b**2*c**2*x**6 + 12*log(x)*a *b*c**2*x**4 + 12*log(x)*a*c**3*x**6 - 8*log(x)*b**3*c*x**4 - 8*log(x)*b** 2*c**2*x**6 - a*b**3 + 3*a*b**2*c*x**2 - 6*a*c**3*x**6 - 2*b**4*x**2 + 4*b **2*c**2*x**6)/(4*b**4*x**4*(b + c*x**2))