Integrand size = 24, antiderivative size = 89 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {(2 b B-A c) x}{c^3}+\frac {B x^3}{3 c^2}-\frac {b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}+\frac {\sqrt {b} (5 b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{7/2}} \]
-(-A*c+2*B*b)*x/c^3+1/3*B*x^3/c^2-1/2*b*(-A*c+B*b)*x/c^3/(c*x^2+b)+1/2*(-3 *A*c+5*B*b)*arctan(x*c^(1/2)/b^(1/2))*b^(1/2)/c^(7/2)
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {(-2 b B+A c) x}{c^3}+\frac {B x^3}{3 c^2}+\frac {\left (-b^2 B+A b c\right ) x}{2 c^3 \left (b+c x^2\right )}+\frac {\sqrt {b} (5 b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{7/2}} \]
((-2*b*B + A*c)*x)/c^3 + (B*x^3)/(3*c^2) + ((-(b^2*B) + A*b*c)*x)/(2*c^3*( b + c*x^2)) + (Sqrt[b]*(5*b*B - 3*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*c^( 7/2))
Time = 0.26 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {9, 360, 25, 1467, 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.
\(\displaystyle \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {x^4 \left (A+B x^2\right )}{\left (b+c x^2\right )^2}dx\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {\int -\frac {2 B c^2 x^4-2 c (b B-A c) x^2+b (b B-A c)}{c x^2+b}dx}{2 c^3}-\frac {b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {2 B c^2 x^4-2 c (b B-A c) x^2+b (b B-A c)}{c x^2+b}dx}{2 c^3}-\frac {b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 1467 |
\(\displaystyle \frac {\int \left (2 B c x^2-2 (2 b B-A c)+\frac {5 b^2 B-3 A b c}{c x^2+b}\right )dx}{2 c^3}-\frac {b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sqrt {b} (5 b B-3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {c}}-2 x (2 b B-A c)+\frac {2}{3} B c x^3}{2 c^3}-\frac {b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}\) |
-1/2*(b*(b*B - A*c)*x)/(c^3*(b + c*x^2)) + (-2*(2*b*B - A*c)*x + (2*B*c*x^ 3)/3 + (Sqrt[b]*(5*b*B - 3*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/Sqrt[c])/(2*c ^3)
3.1.62.3.1 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[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 1.78 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\frac {1}{3} B c \,x^{3}+A c x -2 b B x}{c^{3}}-\frac {b \left (\frac {\left (-\frac {A c}{2}+\frac {B b}{2}\right ) x}{c \,x^{2}+b}+\frac {\left (3 A c -5 B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}}\right )}{c^{3}}\) | \(75\) |
risch | \(\frac {B \,x^{3}}{3 c^{2}}+\frac {A x}{c^{2}}-\frac {2 b B x}{c^{3}}+\frac {\left (\frac {1}{2} A b c -\frac {1}{2} B \,b^{2}\right ) x}{c^{3} \left (c \,x^{2}+b \right )}+\frac {3 \sqrt {-b c}\, \ln \left (-\sqrt {-b c}\, x -b \right ) A}{4 c^{3}}-\frac {5 \sqrt {-b c}\, \ln \left (-\sqrt {-b c}\, x -b \right ) B b}{4 c^{4}}-\frac {3 \sqrt {-b c}\, \ln \left (\sqrt {-b c}\, x -b \right ) A}{4 c^{3}}+\frac {5 \sqrt {-b c}\, \ln \left (\sqrt {-b c}\, x -b \right ) B b}{4 c^{4}}\) | \(155\) |
1/c^3*(1/3*B*c*x^3+A*c*x-2*b*B*x)-b/c^3*((-1/2*A*c+1/2*B*b)*x/(c*x^2+b)+1/ 2*(3*A*c-5*B*b)/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.70 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\left [\frac {4 \, B c^{2} x^{5} - 4 \, {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} - 3 \, {\left (5 \, B b^{2} - 3 \, A b c + {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} - 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) - 6 \, {\left (5 \, B b^{2} - 3 \, A b c\right )} x}{12 \, {\left (c^{4} x^{2} + b c^{3}\right )}}, \frac {2 \, B c^{2} x^{5} - 2 \, {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} + 3 \, {\left (5 \, B b^{2} - 3 \, A b c + {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) - 3 \, {\left (5 \, B b^{2} - 3 \, A b c\right )} x}{6 \, {\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \]
[1/12*(4*B*c^2*x^5 - 4*(5*B*b*c - 3*A*c^2)*x^3 - 3*(5*B*b^2 - 3*A*b*c + (5 *B*b*c - 3*A*c^2)*x^2)*sqrt(-b/c)*log((c*x^2 - 2*c*x*sqrt(-b/c) - b)/(c*x^ 2 + b)) - 6*(5*B*b^2 - 3*A*b*c)*x)/(c^4*x^2 + b*c^3), 1/6*(2*B*c^2*x^5 - 2 *(5*B*b*c - 3*A*c^2)*x^3 + 3*(5*B*b^2 - 3*A*b*c + (5*B*b*c - 3*A*c^2)*x^2) *sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) - 3*(5*B*b^2 - 3*A*b*c)*x)/(c^4*x^2 + b *c^3)]
Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.45 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {B x^{3}}{3 c^{2}} + x \left (\frac {A}{c^{2}} - \frac {2 B b}{c^{3}}\right ) + \frac {x \left (A b c - B b^{2}\right )}{2 b c^{3} + 2 c^{4} x^{2}} - \frac {\sqrt {- \frac {b}{c^{7}}} \left (- 3 A c + 5 B b\right ) \log {\left (- c^{3} \sqrt {- \frac {b}{c^{7}}} + x \right )}}{4} + \frac {\sqrt {- \frac {b}{c^{7}}} \left (- 3 A c + 5 B b\right ) \log {\left (c^{3} \sqrt {- \frac {b}{c^{7}}} + x \right )}}{4} \]
B*x**3/(3*c**2) + x*(A/c**2 - 2*B*b/c**3) + x*(A*b*c - B*b**2)/(2*b*c**3 + 2*c**4*x**2) - sqrt(-b/c**7)*(-3*A*c + 5*B*b)*log(-c**3*sqrt(-b/c**7) + x )/4 + sqrt(-b/c**7)*(-3*A*c + 5*B*b)*log(c**3*sqrt(-b/c**7) + x)/4
Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (B b^{2} - A b c\right )} x}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{3}} + \frac {B c x^{3} - 3 \, {\left (2 \, B b - A c\right )} x}{3 \, c^{3}} \]
-1/2*(B*b^2 - A*b*c)*x/(c^4*x^2 + b*c^3) + 1/2*(5*B*b^2 - 3*A*b*c)*arctan( c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/3*(B*c*x^3 - 3*(2*B*b - A*c)*x)/c^3
Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{3}} - \frac {B b^{2} x - A b c x}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {B c^{4} x^{3} - 6 \, B b c^{3} x + 3 \, A c^{4} x}{3 \, c^{6}} \]
1/2*(5*B*b^2 - 3*A*b*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) - 1/2*(B*b^2 *x - A*b*c*x)/((c*x^2 + b)*c^3) + 1/3*(B*c^4*x^3 - 6*B*b*c^3*x + 3*A*c^4*x )/c^6
Time = 8.89 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17 \[ \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=x\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )-\frac {x\,\left (\frac {B\,b^2}{2}-\frac {A\,b\,c}{2}\right )}{c^4\,x^2+b\,c^3}+\frac {B\,x^3}{3\,c^2}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,x\,\left (3\,A\,c-5\,B\,b\right )}{5\,B\,b^2-3\,A\,b\,c}\right )\,\left (3\,A\,c-5\,B\,b\right )}{2\,c^{7/2}} \]