Integrand size = 24, antiderivative size = 92 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {(b B-A c) x}{4 b c \left (b+c x^2\right )^2}+\frac {(b B+3 A c) x}{8 b^2 c \left (b+c x^2\right )}+\frac {(b B+3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{5/2} c^{3/2}} \]
-1/4*(-A*c+B*b)*x/b/c/(c*x^2+b)^2+1/8*(3*A*c+B*b)*x/b^2/c/(c*x^2+b)+1/8*(3 *A*c+B*b)*arctan(x*c^(1/2)/b^(1/2))/b^(5/2)/c^(3/2)
Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.91 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {x \left (-b^2 B+3 A c^2 x^2+b c \left (5 A+B x^2\right )\right )}{8 b^2 c \left (b+c x^2\right )^2}+\frac {(b B+3 A c) \arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{8 b^{5/2} c^{3/2}} \]
(x*(-(b^2*B) + 3*A*c^2*x^2 + b*c*(5*A + B*x^2)))/(8*b^2*c*(b + c*x^2)^2) + ((b*B + 3*A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(8*b^(5/2)*c^(3/2))
Time = 0.20 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 298, 215, 218}
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.
\(\displaystyle \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {A+B x^2}{\left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {(3 A c+b B) \int \frac {1}{\left (c x^2+b\right )^2}dx}{4 b c}-\frac {x (b B-A c)}{4 b c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {(3 A c+b B) \left (\frac {\int \frac {1}{c x^2+b}dx}{2 b}+\frac {x}{2 b \left (b+c x^2\right )}\right )}{4 b c}-\frac {x (b B-A c)}{4 b c \left (b+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(3 A c+b B) \left (\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 b^{3/2} \sqrt {c}}+\frac {x}{2 b \left (b+c x^2\right )}\right )}{4 b c}-\frac {x (b B-A c)}{4 b c \left (b+c x^2\right )^2}\) |
-1/4*((b*B - A*c)*x)/(b*c*(b + c*x^2)^2) + ((b*B + 3*A*c)*(x/(2*b*(b + c*x ^2)) + ArcTan[(Sqrt[c]*x)/Sqrt[b]]/(2*b^(3/2)*Sqrt[c])))/(4*b*c)
3.1.81.3.1 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_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Time = 1.84 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\frac {\left (3 A c +B b \right ) x^{3}}{8 b^{2}}+\frac {\left (5 A c -B b \right ) x}{8 b c}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (3 A c +B b \right ) \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 b^{2} c \sqrt {b c}}\) | \(77\) |
risch | \(\frac {\frac {\left (3 A c +B b \right ) x^{3}}{8 b^{2}}+\frac {\left (5 A c -B b \right ) x}{8 b c}}{\left (c \,x^{2}+b \right )^{2}}-\frac {3 \ln \left (c x +\sqrt {-b c}\right ) A}{16 \sqrt {-b c}\, b^{2}}-\frac {\ln \left (c x +\sqrt {-b c}\right ) B}{16 \sqrt {-b c}\, c b}+\frac {3 \ln \left (-c x +\sqrt {-b c}\right ) A}{16 \sqrt {-b c}\, b^{2}}+\frac {\ln \left (-c x +\sqrt {-b c}\right ) B}{16 \sqrt {-b c}\, c b}\) | \(147\) |
(1/8*(3*A*c+B*b)/b^2*x^3+1/8*(5*A*c-B*b)/b/c*x)/(c*x^2+b)^2+1/8*(3*A*c+B*b )/b^2/c/(b*c)^(1/2)*arctan(c*x/(b*c)^(1/2))
Time = 0.40 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.26 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\left [\frac {2 \, {\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x^{3} - {\left ({\left (B b c^{2} + 3 \, A c^{3}\right )} x^{4} + B b^{3} + 3 \, A b^{2} c + 2 \, {\left (B b^{2} c + 3 \, A b c^{2}\right )} x^{2}\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right ) - 2 \, {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x}{16 \, {\left (b^{3} c^{4} x^{4} + 2 \, b^{4} c^{3} x^{2} + b^{5} c^{2}\right )}}, \frac {{\left (B b^{2} c^{2} + 3 \, A b c^{3}\right )} x^{3} + {\left ({\left (B b c^{2} + 3 \, A c^{3}\right )} x^{4} + B b^{3} + 3 \, A b^{2} c + 2 \, {\left (B b^{2} c + 3 \, A b c^{2}\right )} x^{2}\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right ) - {\left (B b^{3} c - 5 \, A b^{2} c^{2}\right )} x}{8 \, {\left (b^{3} c^{4} x^{4} + 2 \, b^{4} c^{3} x^{2} + b^{5} c^{2}\right )}}\right ] \]
[1/16*(2*(B*b^2*c^2 + 3*A*b*c^3)*x^3 - ((B*b*c^2 + 3*A*c^3)*x^4 + B*b^3 + 3*A*b^2*c + 2*(B*b^2*c + 3*A*b*c^2)*x^2)*sqrt(-b*c)*log((c*x^2 - 2*sqrt(-b *c)*x - b)/(c*x^2 + b)) - 2*(B*b^3*c - 5*A*b^2*c^2)*x)/(b^3*c^4*x^4 + 2*b^ 4*c^3*x^2 + b^5*c^2), 1/8*((B*b^2*c^2 + 3*A*b*c^3)*x^3 + ((B*b*c^2 + 3*A*c ^3)*x^4 + B*b^3 + 3*A*b^2*c + 2*(B*b^2*c + 3*A*b*c^2)*x^2)*sqrt(b*c)*arcta n(sqrt(b*c)*x/b) - (B*b^3*c - 5*A*b^2*c^2)*x)/(b^3*c^4*x^4 + 2*b^4*c^3*x^2 + b^5*c^2)]
Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.63 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{b^{5} c^{3}}} \cdot \left (3 A c + B b\right ) \log {\left (- b^{3} c \sqrt {- \frac {1}{b^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{b^{5} c^{3}}} \cdot \left (3 A c + B b\right ) \log {\left (b^{3} c \sqrt {- \frac {1}{b^{5} c^{3}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 A c^{2} + B b c\right ) + x \left (5 A b c - B b^{2}\right )}{8 b^{4} c + 16 b^{3} c^{2} x^{2} + 8 b^{2} c^{3} x^{4}} \]
-sqrt(-1/(b**5*c**3))*(3*A*c + B*b)*log(-b**3*c*sqrt(-1/(b**5*c**3)) + x)/ 16 + sqrt(-1/(b**5*c**3))*(3*A*c + B*b)*log(b**3*c*sqrt(-1/(b**5*c**3)) + x)/16 + (x**3*(3*A*c**2 + B*b*c) + x*(5*A*b*c - B*b**2))/(8*b**4*c + 16*b* *3*c**2*x**2 + 8*b**2*c**3*x**4)
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (B b c + 3 \, A c^{2}\right )} x^{3} - {\left (B b^{2} - 5 \, A b c\right )} x}{8 \, {\left (b^{2} c^{3} x^{4} + 2 \, b^{3} c^{2} x^{2} + b^{4} c\right )}} + \frac {{\left (B b + 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{2} c} \]
1/8*((B*b*c + 3*A*c^2)*x^3 - (B*b^2 - 5*A*b*c)*x)/(b^2*c^3*x^4 + 2*b^3*c^2 *x^2 + b^4*c) + 1/8*(B*b + 3*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2*c)
Time = 0.29 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.85 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (B b + 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{8 \, \sqrt {b c} b^{2} c} + \frac {B b c x^{3} + 3 \, A c^{2} x^{3} - B b^{2} x + 5 \, A b c x}{8 \, {\left (c x^{2} + b\right )}^{2} b^{2} c} \]
1/8*(B*b + 3*A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*b^2*c) + 1/8*(B*b*c*x^3 + 3*A*c^2*x^3 - B*b^2*x + 5*A*b*c*x)/((c*x^2 + b)^2*b^2*c)
Time = 8.97 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {x^3\,\left (3\,A\,c+B\,b\right )}{8\,b^2}+\frac {x\,\left (5\,A\,c-B\,b\right )}{8\,b\,c}}{b^2+2\,b\,c\,x^2+c^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (3\,A\,c+B\,b\right )}{8\,b^{5/2}\,c^{3/2}} \]