Integrand size = 24, antiderivative size = 83 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {(2 b B-A c) x^2}{2 c^3}+\frac {B x^4}{4 c^2}+\frac {b^2 (b B-A c)}{2 c^4 \left (b+c x^2\right )}+\frac {b (3 b B-2 A c) \log \left (b+c x^2\right )}{2 c^4} \] Output:
-1/2*(-A*c+2*B*b)*x^2/c^3+1/4*B*x^4/c^2+1/2*b^2*(-A*c+B*b)/c^4/(c*x^2+b)+1 /2*b*(-2*A*c+3*B*b)*ln(c*x^2+b)/c^4
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {2 c (-2 b B+A c) x^2+B c^2 x^4+\frac {2 b^2 (b B-A c)}{b+c x^2}+2 b (3 b B-2 A c) \log \left (b+c x^2\right )}{4 c^4} \] Input:
Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
Output:
(2*c*(-2*b*B + A*c)*x^2 + B*c^2*x^4 + (2*b^2*(b*B - A*c))/(b + c*x^2) + 2* b*(3*b*B - 2*A*c)*Log[b + c*x^2])/(4*c^4)
Time = 0.40 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95, 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.
\(\displaystyle \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^5 \left (A+B x^2\right )}{\left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^4 \left (B x^2+A\right )}{\left (c x^2+b\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {(b B-A c) b^2}{c^3 \left (c x^2+b\right )^2}+\frac {(3 b B-2 A c) b}{c^3 \left (c x^2+b\right )}+\frac {B x^2}{c^2}+\frac {A c-2 b B}{c^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {b^2 (b B-A c)}{c^4 \left (b+c x^2\right )}+\frac {b (3 b B-2 A c) \log \left (b+c x^2\right )}{c^4}-\frac {x^2 (2 b B-A c)}{c^3}+\frac {B x^4}{2 c^2}\right )\) |
Input:
Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
Output:
(-(((2*b*B - A*c)*x^2)/c^3) + (B*x^4)/(2*c^2) + (b^2*(b*B - A*c))/(c^4*(b + c*x^2)) + (b*(3*b*B - 2*A*c)*Log[b + c*x^2])/c^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 = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\left (B c \,x^{2}+A c -2 B b \right )^{2}}{4 c^{4} B}-\frac {b \left (\frac {\left (2 A c -3 B b \right ) \ln \left (c \,x^{2}+b \right )}{c}+\frac {b \left (A c -B b \right )}{c \left (c \,x^{2}+b \right )}\right )}{2 c^{3}}\) | \(76\) |
norman | \(\frac {\frac {B \,x^{9}}{4 c}+\frac {\left (2 A c -3 B b \right ) x^{7}}{4 c^{2}}-\frac {b \left (2 A b c -3 B \,b^{2}\right ) x^{3}}{2 c^{4}}}{\left (c \,x^{2}+b \right ) x^{3}}-\frac {b \left (2 A c -3 B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{4}}\) | \(86\) |
parallelrisch | \(-\frac {-B \,c^{3} x^{6}-2 A \,c^{3} x^{4}+3 B b \,c^{2} x^{4}+4 A \ln \left (c \,x^{2}+b \right ) x^{2} b \,c^{2}-6 B \ln \left (c \,x^{2}+b \right ) x^{2} b^{2} c +4 A \ln \left (c \,x^{2}+b \right ) b^{2} c -6 B \ln \left (c \,x^{2}+b \right ) b^{3}+4 A \,b^{2} c -6 B \,b^{3}}{4 c^{4} \left (c \,x^{2}+b \right )}\) | \(122\) |
risch | \(\frac {B \,x^{4}}{4 c^{2}}+\frac {A \,x^{2}}{2 c^{2}}-\frac {B b \,x^{2}}{c^{3}}+\frac {A^{2}}{4 c^{2} B}-\frac {A b}{c^{3}}+\frac {B \,b^{2}}{c^{4}}-\frac {b^{2} A}{2 c^{3} \left (c \,x^{2}+b \right )}+\frac {b^{3} B}{2 c^{4} \left (c \,x^{2}+b \right )}-\frac {b \ln \left (c \,x^{2}+b \right ) A}{c^{3}}+\frac {3 b^{2} \ln \left (c \,x^{2}+b \right ) B}{2 c^{4}}\) | \(124\) |
Input:
int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
Output:
1/4*(B*c*x^2+A*c-2*B*b)^2/c^4/B-1/2*b/c^3*((2*A*c-3*B*b)/c*ln(c*x^2+b)+b*( A*c-B*b)/c/(c*x^2+b))
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.46 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {B c^{3} x^{6} - {\left (3 \, B b c^{2} - 2 \, A c^{3}\right )} x^{4} + 2 \, B b^{3} - 2 \, A b^{2} c - 2 \, {\left (2 \, B b^{2} c - A b c^{2}\right )} x^{2} + 2 \, {\left (3 \, B b^{3} - 2 \, A b^{2} c + {\left (3 \, B b^{2} c - 2 \, A b c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{5} x^{2} + b c^{4}\right )}} \] Input:
integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
Output:
1/4*(B*c^3*x^6 - (3*B*b*c^2 - 2*A*c^3)*x^4 + 2*B*b^3 - 2*A*b^2*c - 2*(2*B* b^2*c - A*b*c^2)*x^2 + 2*(3*B*b^3 - 2*A*b^2*c + (3*B*b^2*c - 2*A*b*c^2)*x^ 2)*log(c*x^2 + b))/(c^5*x^2 + b*c^4)
Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {B x^{4}}{4 c^{2}} + \frac {b \left (- 2 A c + 3 B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{4}} + x^{2} \left (\frac {A}{2 c^{2}} - \frac {B b}{c^{3}}\right ) + \frac {- A b^{2} c + B b^{3}}{2 b c^{4} + 2 c^{5} x^{2}} \] Input:
integrate(x**9*(B*x**2+A)/(c*x**4+b*x**2)**2,x)
Output:
B*x**4/(4*c**2) + b*(-2*A*c + 3*B*b)*log(b + c*x**2)/(2*c**4) + x**2*(A/(2 *c**2) - B*b/c**3) + (-A*b**2*c + B*b**3)/(2*b*c**4 + 2*c**5*x**2)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {B b^{3} - A b^{2} c}{2 \, {\left (c^{5} x^{2} + b c^{4}\right )}} + \frac {B c x^{4} - 2 \, {\left (2 \, B b - A c\right )} x^{2}}{4 \, c^{3}} + \frac {{\left (3 \, B b^{2} - 2 \, A b c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{4}} \] Input:
integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
Output:
1/2*(B*b^3 - A*b^2*c)/(c^5*x^2 + b*c^4) + 1/4*(B*c*x^4 - 2*(2*B*b - A*c)*x ^2)/c^3 + 1/2*(3*B*b^2 - 2*A*b*c)*log(c*x^2 + b)/c^4
Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.28 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (3 \, B b^{2} - 2 \, A b c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac {B c^{2} x^{4} - 4 \, B b c x^{2} + 2 \, A c^{2} x^{2}}{4 \, c^{4}} - \frac {3 \, B b^{2} c x^{2} - 2 \, A b c^{2} x^{2} + 2 \, B b^{3} - A b^{2} c}{2 \, {\left (c x^{2} + b\right )} c^{4}} \] Input:
integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
Output:
1/2*(3*B*b^2 - 2*A*b*c)*log(abs(c*x^2 + b))/c^4 + 1/4*(B*c^2*x^4 - 4*B*b*c *x^2 + 2*A*c^2*x^2)/c^4 - 1/2*(3*B*b^2*c*x^2 - 2*A*b*c^2*x^2 + 2*B*b^3 - A *b^2*c)/((c*x^2 + b)*c^4)
Time = 9.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=x^2\,\left (\frac {A}{2\,c^2}-\frac {B\,b}{c^3}\right )+\frac {\ln \left (c\,x^2+b\right )\,\left (3\,B\,b^2-2\,A\,b\,c\right )}{2\,c^4}+\frac {B\,x^4}{4\,c^2}+\frac {B\,b^3-A\,b^2\,c}{2\,c\,\left (c^4\,x^2+b\,c^3\right )} \] Input:
int((x^9*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
Output:
x^2*(A/(2*c^2) - (B*b)/c^3) + (log(b + c*x^2)*(3*B*b^2 - 2*A*b*c))/(2*c^4) + (B*x^4)/(4*c^2) + (B*b^3 - A*b^2*c)/(2*c*(b*c^3 + c^4*x^2))
Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.51 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {-4 \,\mathrm {log}\left (c \,x^{2}+b \right ) a \,b^{2} c -4 \,\mathrm {log}\left (c \,x^{2}+b \right ) a b \,c^{2} x^{2}+6 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{4}+6 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{3} c \,x^{2}+4 a b \,c^{2} x^{2}+2 a \,c^{3} x^{4}-6 b^{3} c \,x^{2}-3 b^{2} c^{2} x^{4}+b \,c^{3} x^{6}}{4 c^{4} \left (c \,x^{2}+b \right )} \] Input:
int(x^9*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
Output:
( - 4*log(b + c*x**2)*a*b**2*c - 4*log(b + c*x**2)*a*b*c**2*x**2 + 6*log(b + c*x**2)*b**4 + 6*log(b + c*x**2)*b**3*c*x**2 + 4*a*b*c**2*x**2 + 2*a*c* *3*x**4 - 6*b**3*c*x**2 - 3*b**2*c**2*x**4 + b*c**3*x**6)/(4*c**4*(b + c*x **2))