Integrand size = 24, antiderivative size = 111 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {(3 b B-A c) x^2}{2 c^4}+\frac {B x^4}{4 c^3}-\frac {b^3 (b B-A c)}{4 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (4 b B-3 A c)}{2 c^5 \left (b+c x^2\right )}+\frac {3 b (2 b B-A c) \log \left (b+c x^2\right )}{2 c^5} \] Output:
-1/2*(-A*c+3*B*b)*x^2/c^4+1/4*B*x^4/c^3-1/4*b^3*(-A*c+B*b)/c^5/(c*x^2+b)^2 +1/2*b^2*(-3*A*c+4*B*b)/c^5/(c*x^2+b)+3/2*b*(-A*c+2*B*b)*ln(c*x^2+b)/c^5
Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 c (-3 b B+A c) x^2+B c^2 x^4+\frac {b^3 (-b B+A c)}{\left (b+c x^2\right )^2}+\frac {2 b^2 (4 b B-3 A c)}{b+c x^2}+6 b (2 b B-A c) \log \left (b+c x^2\right )}{4 c^5} \] Input:
Integrate[(x^13*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
Output:
(2*c*(-3*b*B + A*c)*x^2 + B*c^2*x^4 + (b^3*(-(b*B) + A*c))/(b + c*x^2)^2 + (2*b^2*(4*b*B - 3*A*c))/(b + c*x^2) + 6*b*(2*b*B - A*c)*Log[b + c*x^2])/( 4*c^5)
Time = 0.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.97, 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.
\(\displaystyle \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^7 \left (A+B x^2\right )}{\left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^6 \left (B x^2+A\right )}{\left (c x^2+b\right )^3}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {(b B-A c) b^3}{c^4 \left (c x^2+b\right )^3}-\frac {(4 b B-3 A c) b^2}{c^4 \left (c x^2+b\right )^2}+\frac {3 (2 b B-A c) b}{c^4 \left (c x^2+b\right )}+\frac {B x^2}{c^3}+\frac {A c-3 b B}{c^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {b^3 (b B-A c)}{2 c^5 \left (b+c x^2\right )^2}+\frac {b^2 (4 b B-3 A c)}{c^5 \left (b+c x^2\right )}+\frac {3 b (2 b B-A c) \log \left (b+c x^2\right )}{c^5}-\frac {x^2 (3 b B-A c)}{c^4}+\frac {B x^4}{2 c^3}\right )\) |
Input:
Int[(x^13*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
Output:
(-(((3*b*B - A*c)*x^2)/c^4) + (B*x^4)/(2*c^3) - (b^3*(b*B - A*c))/(2*c^5*( b + c*x^2)^2) + (b^2*(4*b*B - 3*A*c))/(c^5*(b + c*x^2)) + (3*b*(2*b*B - A* c)*Log[b + c*x^2])/c^5)/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 = 0.45 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\left (B c \,x^{2}+A c -3 B b \right )^{2}}{4 c^{5} B}-\frac {b \left (\frac {\left (3 A c -6 B b \right ) \ln \left (c \,x^{2}+b \right )}{c}+\frac {b \left (3 A c -4 B b \right )}{c \left (c \,x^{2}+b \right )}-\frac {b^{2} \left (A c -B b \right )}{2 c \left (c \,x^{2}+b \right )^{2}}\right )}{2 c^{4}}\) | \(102\) |
norman | \(\frac {\frac {B \,x^{13}}{4 c}+\frac {\left (A c -2 B b \right ) x^{11}}{2 c^{2}}-\frac {b \left (3 A b c -6 B \,b^{2}\right ) x^{7}}{c^{4}}-\frac {b^{2} \left (9 A b c -18 B \,b^{2}\right ) x^{5}}{4 c^{5}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}-\frac {3 b \left (A c -2 B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{5}}\) | \(107\) |
risch | \(\frac {B \,x^{4}}{4 c^{3}}+\frac {A \,x^{2}}{2 c^{3}}-\frac {3 B b \,x^{2}}{2 c^{4}}+\frac {A^{2}}{4 c^{3} B}-\frac {3 A b}{2 c^{4}}+\frac {9 B \,b^{2}}{4 c^{5}}+\frac {\left (-\frac {3}{2} A \,b^{2} c +2 B \,b^{3}\right ) x^{2}-\frac {b^{3} \left (5 A c -7 B b \right )}{4 c}}{c^{4} \left (c \,x^{2}+b \right )^{2}}-\frac {3 b \ln \left (c \,x^{2}+b \right ) A}{2 c^{4}}+\frac {3 b^{2} \ln \left (c \,x^{2}+b \right ) B}{c^{5}}\) | \(138\) |
parallelrisch | \(-\frac {-B \,x^{8} c^{4}-2 A \,x^{6} c^{4}+4 B \,x^{6} b \,c^{3}+6 A \ln \left (c \,x^{2}+b \right ) x^{4} b \,c^{3}-12 B \ln \left (c \,x^{2}+b \right ) x^{4} b^{2} c^{2}+12 A \ln \left (c \,x^{2}+b \right ) x^{2} b^{2} c^{2}-24 B \ln \left (c \,x^{2}+b \right ) x^{2} b^{3} c +12 A \,b^{2} c^{2} x^{2}-24 B \,b^{3} c \,x^{2}+6 A \ln \left (c \,x^{2}+b \right ) b^{3} c -12 B \ln \left (c \,x^{2}+b \right ) b^{4}+9 A \,b^{3} c -18 B \,b^{4}}{4 c^{5} \left (c \,x^{2}+b \right )^{2}}\) | \(184\) |
Input:
int(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
1/4*(B*c*x^2+A*c-3*B*b)^2/c^5/B-1/2*b/c^4*((3*A*c-6*B*b)/c*ln(c*x^2+b)+b*( 3*A*c-4*B*b)/c/(c*x^2+b)-1/2*b^2*(A*c-B*b)/c/(c*x^2+b)^2)
Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.61 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B c^{4} x^{8} - 2 \, {\left (2 \, B b c^{3} - A c^{4}\right )} x^{6} + 7 \, B b^{4} - 5 \, A b^{3} c - {\left (11 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{4} + 2 \, {\left (B b^{3} c - 2 \, A b^{2} c^{2}\right )} x^{2} + 6 \, {\left (2 \, B b^{4} - A b^{3} c + {\left (2 \, B b^{2} c^{2} - A b c^{3}\right )} x^{4} + 2 \, {\left (2 \, B b^{3} c - A b^{2} c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}} \] Input:
integrate(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
1/4*(B*c^4*x^8 - 2*(2*B*b*c^3 - A*c^4)*x^6 + 7*B*b^4 - 5*A*b^3*c - (11*B*b ^2*c^2 - 4*A*b*c^3)*x^4 + 2*(B*b^3*c - 2*A*b^2*c^2)*x^2 + 6*(2*B*b^4 - A*b ^3*c + (2*B*b^2*c^2 - A*b*c^3)*x^4 + 2*(2*B*b^3*c - A*b^2*c^2)*x^2)*log(c* x^2 + b))/(c^7*x^4 + 2*b*c^6*x^2 + b^2*c^5)
Time = 0.77 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.07 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^{4}}{4 c^{3}} + \frac {3 b \left (- A c + 2 B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{5}} + x^{2} \left (\frac {A}{2 c^{3}} - \frac {3 B b}{2 c^{4}}\right ) + \frac {- 5 A b^{3} c + 7 B b^{4} + x^{2} \left (- 6 A b^{2} c^{2} + 8 B b^{3} c\right )}{4 b^{2} c^{5} + 8 b c^{6} x^{2} + 4 c^{7} x^{4}} \] Input:
integrate(x**13*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
Output:
B*x**4/(4*c**3) + 3*b*(-A*c + 2*B*b)*log(b + c*x**2)/(2*c**5) + x**2*(A/(2 *c**3) - 3*B*b/(2*c**4)) + (-5*A*b**3*c + 7*B*b**4 + x**2*(-6*A*b**2*c**2 + 8*B*b**3*c))/(4*b**2*c**5 + 8*b*c**6*x**2 + 4*c**7*x**4)
Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {7 \, B b^{4} - 5 \, A b^{3} c + 2 \, {\left (4 \, B b^{3} c - 3 \, A b^{2} c^{2}\right )} x^{2}}{4 \, {\left (c^{7} x^{4} + 2 \, b c^{6} x^{2} + b^{2} c^{5}\right )}} + \frac {B c x^{4} - 2 \, {\left (3 \, B b - A c\right )} x^{2}}{4 \, c^{4}} + \frac {3 \, {\left (2 \, B b^{2} - A b c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{5}} \] Input:
integrate(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
1/4*(7*B*b^4 - 5*A*b^3*c + 2*(4*B*b^3*c - 3*A*b^2*c^2)*x^2)/(c^7*x^4 + 2*b *c^6*x^2 + b^2*c^5) + 1/4*(B*c*x^4 - 2*(3*B*b - A*c)*x^2)/c^4 + 3/2*(2*B*b ^2 - A*b*c)*log(c*x^2 + b)/c^5
Time = 0.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.19 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, {\left (2 \, B b^{2} - A b c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{5}} + \frac {B c^{3} x^{4} - 6 \, B b c^{2} x^{2} + 2 \, A c^{3} x^{2}}{4 \, c^{6}} - \frac {18 \, B b^{2} c^{2} x^{4} - 9 \, A b c^{3} x^{4} + 28 \, B b^{3} c x^{2} - 12 \, A b^{2} c^{2} x^{2} + 11 \, B b^{4} - 4 \, A b^{3} c}{4 \, {\left (c x^{2} + b\right )}^{2} c^{5}} \] Input:
integrate(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
3/2*(2*B*b^2 - A*b*c)*log(abs(c*x^2 + b))/c^5 + 1/4*(B*c^3*x^4 - 6*B*b*c^2 *x^2 + 2*A*c^3*x^2)/c^6 - 1/4*(18*B*b^2*c^2*x^4 - 9*A*b*c^3*x^4 + 28*B*b^3 *c*x^2 - 12*A*b^2*c^2*x^2 + 11*B*b^4 - 4*A*b^3*c)/((c*x^2 + b)^2*c^5)
Time = 9.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.06 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {7\,B\,b^4-5\,A\,b^3\,c}{4\,c}+x^2\,\left (2\,B\,b^3-\frac {3\,A\,b^2\,c}{2}\right )}{b^2\,c^4+2\,b\,c^5\,x^2+c^6\,x^4}+x^2\,\left (\frac {A}{2\,c^3}-\frac {3\,B\,b}{2\,c^4}\right )+\frac {\ln \left (c\,x^2+b\right )\,\left (6\,B\,b^2-3\,A\,b\,c\right )}{2\,c^5}+\frac {B\,x^4}{4\,c^3} \] Input:
int((x^13*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
Output:
((7*B*b^4 - 5*A*b^3*c)/(4*c) + x^2*(2*B*b^3 - (3*A*b^2*c)/2))/(b^2*c^4 + c ^6*x^4 + 2*b*c^5*x^2) + x^2*(A/(2*c^3) - (3*B*b)/(2*c^4)) + (log(b + c*x^2 )*(6*B*b^2 - 3*A*b*c))/(2*c^5) + (B*x^4)/(4*c^3)
Time = 0.21 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.70 \[ \int \frac {x^{13} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,b^{3} c -12 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,b^{2} c^{2} x^{2}-6 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b \,c^{3} x^{4}+12 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{5}+24 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{4} c \,x^{2}+12 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} c^{2} x^{4}-3 a \,b^{3} c +6 a b \,c^{3} x^{4}+2 a \,c^{4} x^{6}+6 b^{5}-12 b^{3} c^{2} x^{4}-4 b^{2} c^{3} x^{6}+b \,c^{4} x^{8}}{4 c^{5} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^13*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
Output:
( - 6*log(b + c*x**2)*a*b**3*c - 12*log(b + c*x**2)*a*b**2*c**2*x**2 - 6*l og(b + c*x**2)*a*b*c**3*x**4 + 12*log(b + c*x**2)*b**5 + 24*log(b + c*x**2 )*b**4*c*x**2 + 12*log(b + c*x**2)*b**3*c**2*x**4 - 3*a*b**3*c + 6*a*b*c** 3*x**4 + 2*a*c**4*x**6 + 6*b**5 - 12*b**3*c**2*x**4 - 4*b**2*c**3*x**6 + b *c**4*x**8)/(4*c**5*(b**2 + 2*b*c*x**2 + c**2*x**4))