\(\int \frac {x (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [90]

Optimal result

Integrand size = 22, antiderivative size = 121 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{4 b^3 x^4}-\frac {b B-3 A c}{2 b^4 x^2}-\frac {c (b B-A c)}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (2 b B-3 A c)}{2 b^4 \left (b+c x^2\right )}-\frac {3 c (b B-2 A c) \log (x)}{b^5}+\frac {3 c (b B-2 A c) \log \left (b+c x^2\right )}{2 b^5} \] Output:

-1/4*A/b^3/x^4-1/2*(-3*A*c+B*b)/b^4/x^2-1/4*c*(-A*c+B*b)/b^3/(c*x^2+b)^2-1 
/2*c*(-3*A*c+2*B*b)/b^4/(c*x^2+b)-3*c*(-2*A*c+B*b)*ln(x)/b^5+3/2*c*(-2*A*c 
+B*b)*ln(c*x^2+b)/b^5
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {A b^2}{x^4}-\frac {2 b (b B-3 A c)}{x^2}+\frac {b^2 c (-b B+A c)}{\left (b+c x^2\right )^2}+\frac {2 b c (-2 b B+3 A c)}{b+c x^2}+12 c (-b B+2 A c) \log (x)+6 c (b B-2 A c) \log \left (b+c x^2\right )}{4 b^5} \] Input:

Integrate[(x*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

(-((A*b^2)/x^4) - (2*b*(b*B - 3*A*c))/x^2 + (b^2*c*(-(b*B) + A*c))/(b + c* 
x^2)^2 + (2*b*c*(-2*b*B + 3*A*c))/(b + c*x^2) + 12*c*(-(b*B) + 2*A*c)*Log[ 
x] + 6*c*(b*B - 2*A*c)*Log[b + c*x^2])/(4*b^5)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, 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.

\(\displaystyle \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {A+B x^2}{x^5 \left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 354

\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \left (c x^2+b\right )^3}dx^2\)

\(\Big \downarrow \) 86

\(\displaystyle \frac {1}{2} \int \left (\frac {3 (b B-2 A c) c^2}{b^5 \left (c x^2+b\right )}+\frac {(2 b B-3 A c) c^2}{b^4 \left (c x^2+b\right )^2}+\frac {(b B-A c) c^2}{b^3 \left (c x^2+b\right )^3}-\frac {3 (b B-2 A c) c}{b^5 x^2}+\frac {b B-3 A c}{b^4 x^4}+\frac {A}{b^3 x^6}\right )dx^2\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (-\frac {3 c \log \left (x^2\right ) (b B-2 A c)}{b^5}+\frac {3 c (b B-2 A c) \log \left (b+c x^2\right )}{b^5}-\frac {c (2 b B-3 A c)}{b^4 \left (b+c x^2\right )}-\frac {b B-3 A c}{b^4 x^2}-\frac {c (b B-A c)}{2 b^3 \left (b+c x^2\right )^2}-\frac {A}{2 b^3 x^4}\right )\)

Input:

Int[(x*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

(-1/2*A/(b^3*x^4) - (b*B - 3*A*c)/(b^4*x^2) - (c*(b*B - A*c))/(2*b^3*(b + 
c*x^2)^2) - (c*(2*b*B - 3*A*c))/(b^4*(b + c*x^2)) - (3*c*(b*B - 2*A*c)*Log 
[x^2])/b^5 + (3*c*(b*B - 2*A*c)*Log[b + c*x^2])/b^5)/2
 

Defintions of rubi rules used

rule 9

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]
 

rule 86

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]))))
 

rule 354

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]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {A}{4 b^{3} x^{4}}-\frac {-3 A c +B b}{2 b^{4} x^{2}}+\frac {3 c \left (2 A c -B b \right ) \ln \left (x \right )}{b^{5}}-\frac {c^{2} \left (\frac {\left (6 A c -3 B b \right ) \ln \left (c \,x^{2}+b \right )}{c}-\frac {b \left (3 A c -2 B b \right )}{c \left (c \,x^{2}+b \right )}-\frac {b^{2} \left (A c -B b \right )}{2 c \left (c \,x^{2}+b \right )^{2}}\right )}{2 b^{5}}\)	\(123\)
norman	\(\frac {-\frac {A x}{4 b}+\frac {\left (2 A c -B b \right ) x^{3}}{2 b^{2}}-\frac {c \left (6 A \,c^{2}-3 B b c \right ) x^{7}}{b^{4}}-\frac {c^{2} \left (18 A \,c^{2}-9 B b c \right ) x^{9}}{4 b^{5}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {3 c \left (2 A c -B b \right ) \ln \left (x \right )}{b^{5}}-\frac {3 c \left (2 A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b^{5}}\)	\(124\)
risch	\(\frac {\frac {3 c^{2} \left (2 A c -B b \right ) x^{6}}{2 b^{4}}+\frac {9 c \left (2 A c -B b \right ) x^{4}}{4 b^{3}}+\frac {\left (2 A c -B b \right ) x^{2}}{2 b^{2}}-\frac {A}{4 b}}{x^{4} \left (c \,x^{2}+b \right )^{2}}+\frac {6 c^{2} \ln \left (x \right ) A}{b^{5}}-\frac {3 c \ln \left (x \right ) B}{b^{4}}-\frac {3 c^{2} \ln \left (c \,x^{2}+b \right ) A}{b^{5}}+\frac {3 c \ln \left (c \,x^{2}+b \right ) B}{2 b^{4}}\)	\(129\)
parallelrisch	\(\frac {24 A \ln \left (x \right ) x^{8} c^{4}-12 A \ln \left (c \,x^{2}+b \right ) x^{8} c^{4}-12 B \ln \left (x \right ) x^{8} b \,c^{3}+6 B \ln \left (c \,x^{2}+b \right ) x^{8} b \,c^{3}-18 A \,x^{8} c^{4}+9 B \,x^{8} b \,c^{3}+48 A \ln \left (x \right ) x^{6} b \,c^{3}-24 A \ln \left (c \,x^{2}+b \right ) x^{6} b \,c^{3}-24 B \ln \left (x \right ) x^{6} b^{2} c^{2}+12 B \ln \left (c \,x^{2}+b \right ) x^{6} b^{2} c^{2}-24 A \,x^{6} b \,c^{3}+12 B \,x^{6} b^{2} c^{2}+24 A \ln \left (x \right ) x^{4} b^{2} c^{2}-12 A \ln \left (c \,x^{2}+b \right ) x^{4} b^{2} c^{2}-12 B \ln \left (x \right ) x^{4} b^{3} c +6 B \ln \left (c \,x^{2}+b \right ) x^{4} b^{3} c +4 A \,x^{2} b^{3} c -2 B \,b^{4} x^{2}-A \,b^{4}}{4 b^{5} x^{4} \left (c \,x^{2}+b \right )^{2}}\)	\(271\)

Input:

int(x*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

-1/4*A/b^3/x^4-1/2*(-3*A*c+B*b)/b^4/x^2+3*c*(2*A*c-B*b)/b^5*ln(x)-1/2/b^5* 
c^2*((6*A*c-3*B*b)/c*ln(c*x^2+b)-b*(3*A*c-2*B*b)/c/(c*x^2+b)-1/2*b^2*(A*c- 
B*b)/c/(c*x^2+b)^2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (111) = 222\).

Time = 0.09 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.89 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {6 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + A b^{4} + 9 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4} + 2 \, {\left (B b^{4} - 2 \, A b^{3} c\right )} x^{2} - 6 \, {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} x^{8} + 2 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4}\right )} \log \left (c x^{2} + b\right ) + 12 \, {\left ({\left (B b c^{3} - 2 \, A c^{4}\right )} x^{8} + 2 \, {\left (B b^{2} c^{2} - 2 \, A b c^{3}\right )} x^{6} + {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{4}\right )} \log \left (x\right )}{4 \, {\left (b^{5} c^{2} x^{8} + 2 \, b^{6} c x^{6} + b^{7} x^{4}\right )}} \] Input:

integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

-1/4*(6*(B*b^2*c^2 - 2*A*b*c^3)*x^6 + A*b^4 + 9*(B*b^3*c - 2*A*b^2*c^2)*x^ 
4 + 2*(B*b^4 - 2*A*b^3*c)*x^2 - 6*((B*b*c^3 - 2*A*c^4)*x^8 + 2*(B*b^2*c^2 
- 2*A*b*c^3)*x^6 + (B*b^3*c - 2*A*b^2*c^2)*x^4)*log(c*x^2 + b) + 12*((B*b* 
c^3 - 2*A*c^4)*x^8 + 2*(B*b^2*c^2 - 2*A*b*c^3)*x^6 + (B*b^3*c - 2*A*b^2*c^ 
2)*x^4)*log(x))/(b^5*c^2*x^8 + 2*b^6*c*x^6 + b^7*x^4)
 

Sympy [A] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {- A b^{3} + x^{6} \cdot \left (12 A c^{3} - 6 B b c^{2}\right ) + x^{4} \cdot \left (18 A b c^{2} - 9 B b^{2} c\right ) + x^{2} \cdot \left (4 A b^{2} c - 2 B b^{3}\right )}{4 b^{6} x^{4} + 8 b^{5} c x^{6} + 4 b^{4} c^{2} x^{8}} - \frac {3 c \left (- 2 A c + B b\right ) \log {\left (x \right )}}{b^{5}} + \frac {3 c \left (- 2 A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{5}} \] Input:

integrate(x*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

(-A*b**3 + x**6*(12*A*c**3 - 6*B*b*c**2) + x**4*(18*A*b*c**2 - 9*B*b**2*c) 
 + x**2*(4*A*b**2*c - 2*B*b**3))/(4*b**6*x**4 + 8*b**5*c*x**6 + 4*b**4*c** 
2*x**8) - 3*c*(-2*A*c + B*b)*log(x)/b**5 + 3*c*(-2*A*c + B*b)*log(b/c + x* 
*2)/(2*b**5)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.13 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {6 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} x^{6} + 9 \, {\left (B b^{2} c - 2 \, A b c^{2}\right )} x^{4} + A b^{3} + 2 \, {\left (B b^{3} - 2 \, A b^{2} c\right )} x^{2}}{4 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} + \frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left (c x^{2} + b\right )}{2 \, b^{5}} - \frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left (x^{2}\right )}{2 \, b^{5}} \] Input:

integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

-1/4*(6*(B*b*c^2 - 2*A*c^3)*x^6 + 9*(B*b^2*c - 2*A*b*c^2)*x^4 + A*b^3 + 2* 
(B*b^3 - 2*A*b^2*c)*x^2)/(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4) + 3/2*(B*b* 
c - 2*A*c^2)*log(c*x^2 + b)/b^5 - 3/2*(B*b*c - 2*A*c^2)*log(x^2)/b^5
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.09 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {3 \, {\left (B b c - 2 \, A c^{2}\right )} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {3 \, {\left (B b c^{2} - 2 \, A c^{3}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{5} c} - \frac {6 \, B b c^{2} x^{6} - 12 \, A c^{3} x^{6} + 9 \, B b^{2} c x^{4} - 18 \, A b c^{2} x^{4} + 2 \, B b^{3} x^{2} - 4 \, A b^{2} c x^{2} + A b^{3}}{4 \, {\left (c x^{4} + b x^{2}\right )}^{2} b^{4}} \] Input:

integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

-3*(B*b*c - 2*A*c^2)*log(abs(x))/b^5 + 3/2*(B*b*c^2 - 2*A*c^3)*log(abs(c*x 
^2 + b))/(b^5*c) - 1/4*(6*B*b*c^2*x^6 - 12*A*c^3*x^6 + 9*B*b^2*c*x^4 - 18* 
A*b*c^2*x^4 + 2*B*b^3*x^2 - 4*A*b^2*c*x^2 + A*b^3)/((c*x^4 + b*x^2)^2*b^4)
 

Mupad [B] (verification not implemented)

Time = 9.01 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.08 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {x^2\,\left (2\,A\,c-B\,b\right )}{2\,b^2}-\frac {A}{4\,b}+\frac {3\,c^2\,x^6\,\left (2\,A\,c-B\,b\right )}{2\,b^4}+\frac {9\,c\,x^4\,\left (2\,A\,c-B\,b\right )}{4\,b^3}}{b^2\,x^4+2\,b\,c\,x^6+c^2\,x^8}-\frac {\ln \left (c\,x^2+b\right )\,\left (6\,A\,c^2-3\,B\,b\,c\right )}{2\,b^5}+\frac {\ln \left (x\right )\,\left (6\,A\,c^2-3\,B\,b\,c\right )}{b^5} \] Input:

int((x*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

((x^2*(2*A*c - B*b))/(2*b^2) - A/(4*b) + (3*c^2*x^6*(2*A*c - B*b))/(2*b^4) 
 + (9*c*x^4*(2*A*c - B*b))/(4*b^3))/(b^2*x^4 + c^2*x^8 + 2*b*c*x^6) - (log 
(b + c*x^2)*(6*A*c^2 - 3*B*b*c))/(2*b^5) + (log(x)*(6*A*c^2 - 3*B*b*c))/b^ 
5
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.30 \[ \int \frac {x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-12 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,b^{2} c^{2} x^{4}-24 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b \,c^{3} x^{6}-12 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,c^{4} x^{8}+6 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{4} c \,x^{4}+12 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} c^{2} x^{6}+6 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{3} x^{8}+24 \,\mathrm {log}\left (x \right ) a \,b^{2} c^{2} x^{4}+48 \,\mathrm {log}\left (x \right ) a b \,c^{3} x^{6}+24 \,\mathrm {log}\left (x \right ) a \,c^{4} x^{8}-12 \,\mathrm {log}\left (x \right ) b^{4} c \,x^{4}-24 \,\mathrm {log}\left (x \right ) b^{3} c^{2} x^{6}-12 \,\mathrm {log}\left (x \right ) b^{2} c^{3} x^{8}-a \,b^{4}+4 a \,b^{3} c \,x^{2}+12 a \,b^{2} c^{2} x^{4}-6 a \,c^{4} x^{8}-2 b^{5} x^{2}-6 b^{4} c \,x^{4}+3 b^{2} c^{3} x^{8}}{4 b^{5} x^{4} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:

int(x*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 12*log(b + c*x**2)*a*b**2*c**2*x**4 - 24*log(b + c*x**2)*a*b*c**3*x**6 
 - 12*log(b + c*x**2)*a*c**4*x**8 + 6*log(b + c*x**2)*b**4*c*x**4 + 12*log 
(b + c*x**2)*b**3*c**2*x**6 + 6*log(b + c*x**2)*b**2*c**3*x**8 + 24*log(x) 
*a*b**2*c**2*x**4 + 48*log(x)*a*b*c**3*x**6 + 24*log(x)*a*c**4*x**8 - 12*l 
og(x)*b**4*c*x**4 - 24*log(x)*b**3*c**2*x**6 - 12*log(x)*b**2*c**3*x**8 - 
a*b**4 + 4*a*b**3*c*x**2 + 12*a*b**2*c**2*x**4 - 6*a*c**4*x**8 - 2*b**5*x* 
*2 - 6*b**4*c*x**4 + 3*b**2*c**3*x**8)/(4*b**5*x**4*(b**2 + 2*b*c*x**2 + c 
**2*x**4))