Integrand size = 24, antiderivative size = 97 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {A}{2 b^3 x^2}+\frac {b B-A c}{4 b^2 \left (b+c x^2\right )^2}+\frac {b B-2 A c}{2 b^3 \left (b+c x^2\right )}+\frac {(b B-3 A c) \log (x)}{b^4}-\frac {(b B-3 A c) \log \left (b+c x^2\right )}{2 b^4} \] Output:
-1/2*A/b^3/x^2+1/4*(-A*c+B*b)/b^2/(c*x^2+b)^2+1/2*(-2*A*c+B*b)/b^3/(c*x^2+ b)+(-3*A*c+B*b)*ln(x)/b^4-1/2*(-3*A*c+B*b)*ln(c*x^2+b)/b^4
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.89 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {2 A b}{x^2}+\frac {b^2 (b B-A c)}{\left (b+c x^2\right )^2}+\frac {2 b (b B-2 A c)}{b+c x^2}+4 (b B-3 A c) \log (x)-2 (b B-3 A c) \log \left (b+c x^2\right )}{4 b^4} \] Input:
Integrate[(x^3*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
Output:
((-2*A*b)/x^2 + (b^2*(b*B - A*c))/(b + c*x^2)^2 + (2*b*(b*B - 2*A*c))/(b + c*x^2) + 4*(b*B - 3*A*c)*Log[x] - 2*(b*B - 3*A*c)*Log[b + c*x^2])/(4*b^4)
Time = 0.45 (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.
\(\displaystyle \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{x^3 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^4 \left (c x^2+b\right )^3}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {A}{b^3 x^4}-\frac {c (b B-3 A c)}{b^4 \left (c x^2+b\right )}+\frac {b B-3 A c}{b^4 x^2}-\frac {c (b B-2 A c)}{b^3 \left (c x^2+b\right )^2}-\frac {c (b B-A c)}{b^2 \left (c x^2+b\right )^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {\log \left (x^2\right ) (b B-3 A c)}{b^4}-\frac {(b B-3 A c) \log \left (b+c x^2\right )}{b^4}+\frac {b B-2 A c}{b^3 \left (b+c x^2\right )}-\frac {A}{b^3 x^2}+\frac {b B-A c}{2 b^2 \left (b+c x^2\right )^2}\right )\) |
Input:
Int[(x^3*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
Output:
(-(A/(b^3*x^2)) + (b*B - A*c)/(2*b^2*(b + c*x^2)^2) + (b*B - 2*A*c)/(b^3*( b + c*x^2)) + ((b*B - 3*A*c)*Log[x^2])/b^4 - ((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 = 100, normalized size of antiderivative = 1.03
method | result | size |
norman | \(\frac {\frac {c \left (3 A c -B b \right ) x^{7}}{b^{3}}-\frac {A \,x^{3}}{2 b}+\frac {c^{2} \left (9 A c -3 B b \right ) x^{9}}{4 b^{4}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}-\frac {\left (3 A c -B b \right ) \ln \left (x \right )}{b^{4}}+\frac {\left (3 A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 b^{4}}\) | \(100\) |
default | \(-\frac {A}{2 b^{3} x^{2}}+\frac {\left (-3 A c +B b \right ) \ln \left (x \right )}{b^{4}}+\frac {c \left (\frac {\left (3 A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{c}-\frac {b \left (2 A c -B b \right )}{c \left (c \,x^{2}+b \right )}-\frac {b^{2} \left (A c -B b \right )}{2 c \left (c \,x^{2}+b \right )^{2}}\right )}{2 b^{4}}\) | \(102\) |
risch | \(\frac {-\frac {c \left (3 A c -B b \right ) x^{4}}{2 b^{3}}-\frac {3 \left (3 A c -B b \right ) x^{2}}{4 b^{2}}-\frac {A}{2 b}}{\left (c \,x^{2}+b \right )^{2} x^{2}}-\frac {3 \ln \left (x \right ) A c}{b^{4}}+\frac {\ln \left (x \right ) B}{b^{3}}+\frac {3 \ln \left (-c \,x^{2}-b \right ) A c}{2 b^{4}}-\frac {\ln \left (-c \,x^{2}-b \right ) B}{2 b^{3}}\) | \(108\) |
parallelrisch | \(-\frac {12 A \ln \left (x \right ) x^{6} c^{3}-6 A \ln \left (c \,x^{2}+b \right ) x^{6} c^{3}-4 B \ln \left (x \right ) x^{6} b \,c^{2}+2 B \ln \left (c \,x^{2}+b \right ) x^{6} b \,c^{2}-9 A \,c^{3} x^{6}+3 B b \,c^{2} x^{6}+24 A \ln \left (x \right ) x^{4} b \,c^{2}-12 A \ln \left (c \,x^{2}+b \right ) x^{4} b \,c^{2}-8 B \ln \left (x \right ) x^{4} b^{2} c +4 B \ln \left (c \,x^{2}+b \right ) x^{4} b^{2} c -12 A b \,c^{2} x^{4}+4 x^{4} B \,b^{2} c +12 A \ln \left (x \right ) x^{2} b^{2} c -6 A \ln \left (c \,x^{2}+b \right ) x^{2} b^{2} c -4 B \ln \left (x \right ) x^{2} b^{3}+2 B \ln \left (c \,x^{2}+b \right ) x^{2} b^{3}+2 A \,b^{3}}{4 b^{4} x^{2} \left (c \,x^{2}+b \right )^{2}}\) | \(240\) |
Input:
int(x^3*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
(c*(3*A*c-B*b)/b^3*x^7-1/2*A/b*x^3+1/4*c^2*(9*A*c-3*B*b)/b^4*x^9)/x^5/(c*x ^2+b)^2-(3*A*c-B*b)/b^4*ln(x)+1/2*(3*A*c-B*b)/b^4*ln(c*x^2+b)
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (89) = 178\).
Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.03 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 \, {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} - 2 \, A b^{3} + 3 \, {\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2} - 2 \, {\left ({\left (B b c^{2} - 3 \, A c^{3}\right )} x^{6} + 2 \, {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} + {\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2}\right )} \log \left (c x^{2} + b\right ) + 4 \, {\left ({\left (B b c^{2} - 3 \, A c^{3}\right )} x^{6} + 2 \, {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} + {\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2}\right )} \log \left (x\right )}{4 \, {\left (b^{4} c^{2} x^{6} + 2 \, b^{5} c x^{4} + b^{6} x^{2}\right )}} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
1/4*(2*(B*b^2*c - 3*A*b*c^2)*x^4 - 2*A*b^3 + 3*(B*b^3 - 3*A*b^2*c)*x^2 - 2 *((B*b*c^2 - 3*A*c^3)*x^6 + 2*(B*b^2*c - 3*A*b*c^2)*x^4 + (B*b^3 - 3*A*b^2 *c)*x^2)*log(c*x^2 + b) + 4*((B*b*c^2 - 3*A*c^3)*x^6 + 2*(B*b^2*c - 3*A*b* c^2)*x^4 + (B*b^3 - 3*A*b^2*c)*x^2)*log(x))/(b^4*c^2*x^6 + 2*b^5*c*x^4 + b ^6*x^2)
Time = 0.58 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {- 2 A b^{2} + x^{4} \left (- 6 A c^{2} + 2 B b c\right ) + x^{2} \left (- 9 A b c + 3 B b^{2}\right )}{4 b^{5} x^{2} + 8 b^{4} c x^{4} + 4 b^{3} c^{2} x^{6}} + \frac {\left (- 3 A c + B b\right ) \log {\left (x \right )}}{b^{4}} - \frac {\left (- 3 A c + B b\right ) \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \] Input:
integrate(x**3*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
Output:
(-2*A*b**2 + x**4*(-6*A*c**2 + 2*B*b*c) + x**2*(-9*A*b*c + 3*B*b**2))/(4*b **5*x**2 + 8*b**4*c*x**4 + 4*b**3*c**2*x**6) + (-3*A*c + B*b)*log(x)/b**4 - (-3*A*c + B*b)*log(b/c + x**2)/(2*b**4)
Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.12 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 \, {\left (B b c - 3 \, A c^{2}\right )} x^{4} - 2 \, A b^{2} + 3 \, {\left (B b^{2} - 3 \, A b c\right )} x^{2}}{4 \, {\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )}} - \frac {{\left (B b - 3 \, A c\right )} \log \left (c x^{2} + b\right )}{2 \, b^{4}} + \frac {{\left (B b - 3 \, A c\right )} \log \left (x^{2}\right )}{2 \, b^{4}} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
1/4*(2*(B*b*c - 3*A*c^2)*x^4 - 2*A*b^2 + 3*(B*b^2 - 3*A*b*c)*x^2)/(b^3*c^2 *x^6 + 2*b^4*c*x^4 + b^5*x^2) - 1/2*(B*b - 3*A*c)*log(c*x^2 + b)/b^4 + 1/2 *(B*b - 3*A*c)*log(x^2)/b^4
Time = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.08 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (B b - 3 \, A c\right )} \log \left ({\left | x \right |}\right )}{b^{4}} - \frac {{\left (B b c - 3 \, A c^{2}\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4} c} + \frac {2 \, {\left (B b^{2} c - 3 \, A b c^{2}\right )} x^{4} - 2 \, A b^{3} + 3 \, {\left (B b^{3} - 3 \, A b^{2} c\right )} x^{2}}{4 \, {\left (c x^{2} + b\right )}^{2} b^{4} x^{2}} \] Input:
integrate(x^3*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
(B*b - 3*A*c)*log(abs(x))/b^4 - 1/2*(B*b*c - 3*A*c^2)*log(abs(c*x^2 + b))/ (b^4*c) + 1/4*(2*(B*b^2*c - 3*A*b*c^2)*x^4 - 2*A*b^3 + 3*(B*b^3 - 3*A*b^2* c)*x^2)/((c*x^2 + b)^2*b^4*x^2)
Time = 8.82 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\ln \left (c\,x^2+b\right )\,\left (3\,A\,c-B\,b\right )}{2\,b^4}-\frac {\frac {A}{2\,b}+\frac {3\,x^2\,\left (3\,A\,c-B\,b\right )}{4\,b^2}+\frac {c\,x^4\,\left (3\,A\,c-B\,b\right )}{2\,b^3}}{b^2\,x^2+2\,b\,c\,x^4+c^2\,x^6}-\frac {\ln \left (x\right )\,\left (3\,A\,c-B\,b\right )}{b^4} \] Input:
int((x^3*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
Output:
(log(b + c*x^2)*(3*A*c - B*b))/(2*b^4) - (A/(2*b) + (3*x^2*(3*A*c - B*b))/ (4*b^2) + (c*x^4*(3*A*c - B*b))/(2*b^3))/(b^2*x^2 + c^2*x^6 + 2*b*c*x^4) - (log(x)*(3*A*c - B*b))/b^4
Time = 0.20 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.55 \[ \int \frac {x^3 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,b^{2} c \,x^{2}+12 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b \,c^{2} x^{4}+6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,c^{3} x^{6}-2 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{4} x^{2}-4 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} c \,x^{4}-2 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{2} x^{6}-12 \,\mathrm {log}\left (x \right ) a \,b^{2} c \,x^{2}-24 \,\mathrm {log}\left (x \right ) a b \,c^{2} x^{4}-12 \,\mathrm {log}\left (x \right ) a \,c^{3} x^{6}+4 \,\mathrm {log}\left (x \right ) b^{4} x^{2}+8 \,\mathrm {log}\left (x \right ) b^{3} c \,x^{4}+4 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{6}-2 a \,b^{3}-6 a \,b^{2} c \,x^{2}+3 a \,c^{3} x^{6}+2 b^{4} x^{2}-b^{2} c^{2} x^{6}}{4 b^{4} x^{2} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^3*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
Output:
(6*log(b + c*x**2)*a*b**2*c*x**2 + 12*log(b + c*x**2)*a*b*c**2*x**4 + 6*lo g(b + c*x**2)*a*c**3*x**6 - 2*log(b + c*x**2)*b**4*x**2 - 4*log(b + c*x**2 )*b**3*c*x**4 - 2*log(b + c*x**2)*b**2*c**2*x**6 - 12*log(x)*a*b**2*c*x**2 - 24*log(x)*a*b*c**2*x**4 - 12*log(x)*a*c**3*x**6 + 4*log(x)*b**4*x**2 + 8*log(x)*b**3*c*x**4 + 4*log(x)*b**2*c**2*x**6 - 2*a*b**3 - 6*a*b**2*c*x** 2 + 3*a*c**3*x**6 + 2*b**4*x**2 - b**2*c**2*x**6)/(4*b**4*x**2*(b**2 + 2*b *c*x**2 + c**2*x**4))