Integrand size = 22, antiderivative size = 73 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {A}{2 b^2 x^2}+\frac {b B-A c}{2 b^2 \left (b+c x^2\right )}+\frac {(b B-2 A c) \log (x)}{b^3}-\frac {(b B-2 A c) \log \left (b+c x^2\right )}{2 b^3} \] Output:
-1/2*A/b^2/x^2+1/2*(-A*c+B*b)/b^2/(c*x^2+b)+(-2*A*c+B*b)*ln(x)/b^3-1/2*(-2 *A*c+B*b)*ln(c*x^2+b)/b^3
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {-\frac {A b}{x^2}+\frac {b (b B-A c)}{b+c x^2}+2 (b B-2 A c) \log (x)+(-b B+2 A c) \log \left (b+c x^2\right )}{2 b^3} \] Input:
Integrate[(x*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
Output:
(-((A*b)/x^2) + (b*(b*B - A*c))/(b + c*x^2) + 2*(b*B - 2*A*c)*Log[x] + (-( b*B) + 2*A*c)*Log[b + c*x^2])/(2*b^3)
Time = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, 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.
\(\displaystyle \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^3 \left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^4 \left (c x^2+b\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {A}{b^2 x^4}-\frac {c (b B-2 A c)}{b^3 \left (c x^2+b\right )}+\frac {b B-2 A c}{b^3 x^2}-\frac {c (b B-A c)}{b^2 \left (c x^2+b\right )^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right ) (b B-2 A c)}{b^3}-\frac {(b B-2 A c) \log \left (b+c x^2\right )}{b^3}+\frac {b B-A c}{b^2 \left (b+c x^2\right )}-\frac {A}{b^2 x^2}\right )\) |
Input:
Int[(x*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
Output:
(-(A/(b^2*x^2)) + (b*B - A*c)/(b^2*(b + c*x^2)) + ((b*B - 2*A*c)*Log[x^2]) /b^3 - ((b*B - 2*A*c)*Log[b + c*x^2])/b^3)/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 = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04
method | result | size |
default | \(-\frac {A}{2 b^{2} x^{2}}+\frac {\left (-2 A c +B b \right ) \ln \left (x \right )}{b^{3}}+\frac {c \left (\frac {\left (2 A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{c}-\frac {b \left (A c -B b \right )}{c \left (c \,x^{2}+b \right )}\right )}{2 b^{3}}\) | \(76\) |
norman | \(\frac {-\frac {A x}{2 b}+\frac {c \left (2 A c -B b \right ) x^{5}}{2 b^{3}}}{\left (c \,x^{2}+b \right ) x^{3}}-\frac {\left (2 A c -B b \right ) \ln \left (x \right )}{b^{3}}+\frac {\left (2 A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b^{3}}\) | \(79\) |
risch | \(\frac {-\frac {\left (2 A c -B b \right ) x^{2}}{2 b^{2}}-\frac {A}{2 b}}{x^{2} \left (c \,x^{2}+b \right )}-\frac {2 \ln \left (x \right ) A c}{b^{3}}+\frac {\ln \left (x \right ) B}{b^{2}}+\frac {\ln \left (-c \,x^{2}-b \right ) A c}{b^{3}}-\frac {\ln \left (-c \,x^{2}-b \right ) B}{2 b^{2}}\) | \(89\) |
parallelrisch | \(-\frac {4 A \ln \left (x \right ) x^{4} c^{2}-2 A \ln \left (c \,x^{2}+b \right ) x^{4} c^{2}-2 B \ln \left (x \right ) x^{4} b c +B \ln \left (c \,x^{2}+b \right ) x^{4} b c -2 x^{4} A \,c^{2}+x^{4} B b c +4 A \ln \left (x \right ) x^{2} b c -2 A \ln \left (c \,x^{2}+b \right ) x^{2} b c -2 B \ln \left (x \right ) x^{2} b^{2}+B \ln \left (c \,x^{2}+b \right ) x^{2} b^{2}+b^{2} A}{2 b^{3} x^{2} \left (c \,x^{2}+b \right )}\) | \(146\) |
Input:
int(x*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/2*A/b^2/x^2+(-2*A*c+B*b)*ln(x)/b^3+1/2/b^3*c*((2*A*c-B*b)/c*ln(c*x^2+b) -b*(A*c-B*b)/c/(c*x^2+b))
Time = 0.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.60 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {A b^{2} - {\left (B b^{2} - 2 \, A b c\right )} x^{2} + {\left ({\left (B b c - 2 \, A c^{2}\right )} x^{4} + {\left (B b^{2} - 2 \, A b c\right )} x^{2}\right )} \log \left (c x^{2} + b\right ) - 2 \, {\left ({\left (B b c - 2 \, A c^{2}\right )} x^{4} + {\left (B b^{2} - 2 \, A b c\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left (b^{3} c x^{4} + b^{4} x^{2}\right )}} \] Input:
integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
Output:
-1/2*(A*b^2 - (B*b^2 - 2*A*b*c)*x^2 + ((B*b*c - 2*A*c^2)*x^4 + (B*b^2 - 2* A*b*c)*x^2)*log(c*x^2 + b) - 2*((B*b*c - 2*A*c^2)*x^4 + (B*b^2 - 2*A*b*c)* x^2)*log(x))/(b^3*c*x^4 + b^4*x^2)
Time = 0.50 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {- A b + x^{2} \left (- 2 A c + B b\right )}{2 b^{3} x^{2} + 2 b^{2} c x^{4}} + \frac {\left (- 2 A c + B b\right ) \log {\left (x \right )}}{b^{3}} - \frac {\left (- 2 A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{3}} \] Input:
integrate(x*(B*x**2+A)/(c*x**4+b*x**2)**2,x)
Output:
(-A*b + x**2*(-2*A*c + B*b))/(2*b**3*x**2 + 2*b**2*c*x**4) + (-2*A*c + B*b )*log(x)/b**3 - (-2*A*c + B*b)*log(b/c + x**2)/(2*b**3)
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (B b - 2 \, A c\right )} x^{2} - A b}{2 \, {\left (b^{2} c x^{4} + b^{3} x^{2}\right )}} - \frac {{\left (B b - 2 \, A c\right )} \log \left (c x^{2} + b\right )}{2 \, b^{3}} + \frac {{\left (B b - 2 \, A c\right )} \log \left (x^{2}\right )}{2 \, b^{3}} \] Input:
integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
Output:
1/2*((B*b - 2*A*c)*x^2 - A*b)/(b^2*c*x^4 + b^3*x^2) - 1/2*(B*b - 2*A*c)*lo g(c*x^2 + b)/b^3 + 1/2*(B*b - 2*A*c)*log(x^2)/b^3
Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.10 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (B b - 2 \, A c\right )} \log \left ({\left | x \right |}\right )}{b^{3}} + \frac {B b x^{2} - 2 \, A c x^{2} - A b}{2 \, {\left (c x^{4} + b x^{2}\right )} b^{2}} - \frac {{\left (B b c - 2 \, A c^{2}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{3} c} \] Input:
integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
Output:
(B*b - 2*A*c)*log(abs(x))/b^3 + 1/2*(B*b*x^2 - 2*A*c*x^2 - A*b)/((c*x^4 + b*x^2)*b^2) - 1/2*(B*b*c - 2*A*c^2)*log(abs(c*x^2 + b))/(b^3*c)
Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\ln \left (c\,x^2+b\right )\,\left (2\,A\,c-B\,b\right )}{2\,b^3}-\frac {\frac {A}{2\,b}+\frac {x^2\,\left (2\,A\,c-B\,b\right )}{2\,b^2}}{c\,x^4+b\,x^2}-\frac {\ln \left (x\right )\,\left (2\,A\,c-B\,b\right )}{b^3} \] Input:
int((x*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
Output:
(log(b + c*x^2)*(2*A*c - B*b))/(2*b^3) - (A/(2*b) + (x^2*(2*A*c - B*b))/(2 *b^2))/(b*x^2 + c*x^4) - (log(x)*(2*A*c - B*b))/b^3
Time = 0.19 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.05 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {2 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b c \,x^{2}+2 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,c^{2} x^{4}-\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} x^{2}-\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c \,x^{4}-4 \,\mathrm {log}\left (x \right ) a b c \,x^{2}-4 \,\mathrm {log}\left (x \right ) a \,c^{2} x^{4}+2 \,\mathrm {log}\left (x \right ) b^{3} x^{2}+2 \,\mathrm {log}\left (x \right ) b^{2} c \,x^{4}-a \,b^{2}+2 a \,c^{2} x^{4}-b^{2} c \,x^{4}}{2 b^{3} x^{2} \left (c \,x^{2}+b \right )} \] Input:
int(x*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
Output:
(2*log(b + c*x**2)*a*b*c*x**2 + 2*log(b + c*x**2)*a*c**2*x**4 - log(b + c* x**2)*b**3*x**2 - log(b + c*x**2)*b**2*c*x**4 - 4*log(x)*a*b*c*x**2 - 4*lo g(x)*a*c**2*x**4 + 2*log(x)*b**3*x**2 + 2*log(x)*b**2*c*x**4 - a*b**2 + 2* a*c**2*x**4 - b**2*c*x**4)/(2*b**3*x**2*(b + c*x**2))