Integrand size = 24, antiderivative size = 89 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^2}{2 c^3}+\frac {b^2 (b B-A c)}{4 c^4 \left (b+c x^2\right )^2}-\frac {b (3 b B-2 A c)}{2 c^4 \left (b+c x^2\right )}-\frac {(3 b B-A c) \log \left (b+c x^2\right )}{2 c^4} \]
1/2*B*x^2/c^3+1/4*b^2*(-A*c+B*b)/c^4/(c*x^2+b)^2-1/2*b*(-2*A*c+3*B*b)/c^4/ (c*x^2+b)-1/2*(-A*c+3*B*b)*ln(c*x^2+b)/c^4
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^2}{2 c^3}+\frac {b^3 B-A b^2 c}{4 c^4 \left (b+c x^2\right )^2}+\frac {-3 b^2 B+2 A b c}{2 c^4 \left (b+c x^2\right )}+\frac {(-3 b B+A c) \log \left (b+c x^2\right )}{2 c^4} \]
(B*x^2)/(2*c^3) + (b^3*B - A*b^2*c)/(4*c^4*(b + c*x^2)^2) + (-3*b^2*B + 2* A*b*c)/(2*c^4*(b + c*x^2)) + ((-3*b*B + A*c)*Log[b + c*x^2])/(2*c^4)
Time = 0.24 (sec) , antiderivative size = 86, 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.
\(\displaystyle \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^5 \left (A+B x^2\right )}{\left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^4 \left (B x^2+A\right )}{\left (c x^2+b\right )^3}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {(b B-A c) b^2}{c^3 \left (c x^2+b\right )^3}+\frac {(3 b B-2 A c) b}{c^3 \left (c x^2+b\right )^2}+\frac {A c-3 b B}{c^3 \left (c x^2+b\right )}+\frac {B}{c^3}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {b^2 (b B-A c)}{2 c^4 \left (b+c x^2\right )^2}-\frac {b (3 b B-2 A c)}{c^4 \left (b+c x^2\right )}-\frac {(3 b B-A c) \log \left (b+c x^2\right )}{c^4}+\frac {B x^2}{c^3}\right )\) |
((B*x^2)/c^3 + (b^2*(b*B - A*c))/(2*c^4*(b + c*x^2)^2) - (b*(3*b*B - 2*A*c ))/(c^4*(b + c*x^2)) - ((3*b*B - A*c)*Log[b + c*x^2])/c^4)/2
3.1.76.3.1 Defintions of rubi rules used
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 = 1.76 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.92
method | result | size |
norman | \(\frac {\frac {b \left (A c -3 B b \right ) x^{7}}{c^{3}}+\frac {B \,x^{11}}{2 c}+\frac {b^{2} \left (3 A c -9 B b \right ) x^{5}}{4 c^{4}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {\left (A c -3 B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{4}}\) | \(82\) |
default | \(\frac {B \,x^{2}}{2 c^{3}}+\frac {\frac {\left (A c -3 B b \right ) \ln \left (c \,x^{2}+b \right )}{c}-\frac {b^{2} \left (A c -B b \right )}{2 c \left (c \,x^{2}+b \right )^{2}}+\frac {b \left (2 A c -3 B b \right )}{c \left (c \,x^{2}+b \right )}}{2 c^{3}}\) | \(85\) |
risch | \(\frac {B \,x^{2}}{2 c^{3}}+\frac {\left (A b c -\frac {3}{2} B \,b^{2}\right ) x^{2}+\frac {b^{2} \left (3 A c -5 B b \right )}{4 c}}{c^{3} \left (c \,x^{2}+b \right )^{2}}+\frac {\ln \left (c \,x^{2}+b \right ) A}{2 c^{3}}-\frac {3 \ln \left (c \,x^{2}+b \right ) B b}{2 c^{4}}\) | \(86\) |
parallelrisch | \(\frac {2 B \,c^{3} x^{6}+2 A \ln \left (c \,x^{2}+b \right ) x^{4} c^{3}-6 B \ln \left (c \,x^{2}+b \right ) x^{4} b \,c^{2}+4 A \ln \left (c \,x^{2}+b \right ) x^{2} b \,c^{2}-12 B \ln \left (c \,x^{2}+b \right ) x^{2} b^{2} c +4 A b \,c^{2} x^{2}-12 B \,b^{2} c \,x^{2}+2 A \ln \left (c \,x^{2}+b \right ) b^{2} c -6 B \ln \left (c \,x^{2}+b \right ) b^{3}+3 b^{2} A c -9 B \,b^{3}}{4 c^{4} \left (c \,x^{2}+b \right )^{2}}\) | \(158\) |
(b*(A*c-3*B*b)/c^3*x^7+1/2*B/c*x^11+1/4*b^2*(3*A*c-9*B*b)/c^4*x^5)/x^5/(c* x^2+b)^2+1/2*(A*c-3*B*b)/c^4*ln(c*x^2+b)
Time = 0.25 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.60 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 \, B c^{3} x^{6} + 4 \, B b c^{2} x^{4} - 5 \, B b^{3} + 3 \, A b^{2} c - 4 \, {\left (B b^{2} c - A b c^{2}\right )} x^{2} - 2 \, {\left ({\left (3 \, B b c^{2} - A c^{3}\right )} x^{4} + 3 \, B b^{3} - A b^{2} c + 2 \, {\left (3 \, B b^{2} c - A b c^{2}\right )} x^{2}\right )} \log \left (c x^{2} + b\right )}{4 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} \]
1/4*(2*B*c^3*x^6 + 4*B*b*c^2*x^4 - 5*B*b^3 + 3*A*b^2*c - 4*(B*b^2*c - A*b* c^2)*x^2 - 2*((3*B*b*c^2 - A*c^3)*x^4 + 3*B*b^3 - A*b^2*c + 2*(3*B*b^2*c - A*b*c^2)*x^2)*log(c*x^2 + b))/(c^6*x^4 + 2*b*c^5*x^2 + b^2*c^4)
Time = 0.69 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^{2}}{2 c^{3}} + \frac {3 A b^{2} c - 5 B b^{3} + x^{2} \cdot \left (4 A b c^{2} - 6 B b^{2} c\right )}{4 b^{2} c^{4} + 8 b c^{5} x^{2} + 4 c^{6} x^{4}} - \frac {\left (- A c + 3 B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{4}} \]
B*x**2/(2*c**3) + (3*A*b**2*c - 5*B*b**3 + x**2*(4*A*b*c**2 - 6*B*b**2*c)) /(4*b**2*c**4 + 8*b*c**5*x**2 + 4*c**6*x**4) - (-A*c + 3*B*b)*log(b + c*x* *2)/(2*c**4)
Time = 0.23 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {5 \, B b^{3} - 3 \, A b^{2} c + 2 \, {\left (3 \, B b^{2} c - 2 \, A b c^{2}\right )} x^{2}}{4 \, {\left (c^{6} x^{4} + 2 \, b c^{5} x^{2} + b^{2} c^{4}\right )}} + \frac {B x^{2}}{2 \, c^{3}} - \frac {{\left (3 \, B b - A c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{4}} \]
-1/4*(5*B*b^3 - 3*A*b^2*c + 2*(3*B*b^2*c - 2*A*b*c^2)*x^2)/(c^6*x^4 + 2*b* c^5*x^2 + b^2*c^4) + 1/2*B*x^2/c^3 - 1/2*(3*B*b - A*c)*log(c*x^2 + b)/c^4
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B x^{2}}{2 \, c^{3}} - \frac {{\left (3 \, B b - A c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{4}} + \frac {9 \, B b c^{2} x^{4} - 3 \, A c^{3} x^{4} + 12 \, B b^{2} c x^{2} - 2 \, A b c^{2} x^{2} + 4 \, B b^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} c^{4}} \]
1/2*B*x^2/c^3 - 1/2*(3*B*b - A*c)*log(abs(c*x^2 + b))/c^4 + 1/4*(9*B*b*c^2 *x^4 - 3*A*c^3*x^4 + 12*B*b^2*c*x^2 - 2*A*b*c^2*x^2 + 4*B*b^3)/((c*x^2 + b )^2*c^4)
Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07 \[ \int \frac {x^{11} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {B\,x^2}{2\,c^3}-\frac {x^2\,\left (\frac {3\,B\,b^2}{2}-A\,b\,c\right )+\frac {5\,B\,b^3-3\,A\,b^2\,c}{4\,c}}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {\ln \left (c\,x^2+b\right )\,\left (A\,c-3\,B\,b\right )}{2\,c^4} \]