Integrand size = 18, antiderivative size = 400 \[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=-\frac {3 b \left (b^2-8 a c\right ) x}{8 c^2 \left (b^2-4 a c\right )^2}+\frac {\left (b^2-28 a c\right ) x^3}{8 c \left (b^2-4 a c\right )^2}+\frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {x^5 \left (12 a b-\left (b^2-28 a c\right ) x^2\right )}{8 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 \left (b^4-9 a b^2 c+28 a^2 c^2-\frac {b^5-11 a b^3 c+44 a^2 b c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{8 \sqrt {2} c^{5/2} \left (b^2-4 a c\right )^2 \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {3 \left (b^4-9 a b^2 c+28 a^2 c^2+\frac {b^5-11 a b^3 c+44 a^2 b c^2}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{8 \sqrt {2} c^{5/2} \left (b^2-4 a c\right )^2 \sqrt {b+\sqrt {b^2-4 a c}}} \]
-3/8*b*(-8*a*c+b^2)*x/c^2/(-4*a*c+b^2)^2+1/8*(-28*a*c+b^2)*x^3/c/(-4*a*c+b ^2)^2+1/4*x^7*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^2+1/8*x^5*(12*a*b-( -28*a*c+b^2)*x^2)/(-4*a*c+b^2)^2/(c*x^4+b*x^2+a)+3/16*arctan(x*2^(1/2)*c^( 1/2)/(b-(-4*a*c+b^2)^(1/2))^(1/2))*(b^4-9*a*b^2*c+28*a^2*c^2+(-44*a^2*b*c^ 2+11*a*b^3*c-b^5)/(-4*a*c+b^2)^(1/2))/c^(5/2)/(-4*a*c+b^2)^2*2^(1/2)/(b-(- 4*a*c+b^2)^(1/2))^(1/2)+3/16*arctan(x*2^(1/2)*c^(1/2)/(b+(-4*a*c+b^2)^(1/2 ))^(1/2))*(b^4-9*a*b^2*c+28*a^2*c^2+(44*a^2*b*c^2-11*a*b^3*c+b^5)/(-4*a*c+ b^2)^(1/2))/c^(5/2)/(-4*a*c+b^2)^2*2^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2)
Time = 0.69 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.14 \[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=\frac {\frac {2 x \left (2 b^5-17 a b^3 c+48 a^2 b c^2-5 b^4 c x^2+37 a b^2 c^2 x^2-44 a^2 c^3 x^2\right )}{\left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {4 \left (b^4 x^3+a b^2 x \left (b-4 c x^2\right )+a^2 c x \left (-3 b+2 c x^2\right )\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 \sqrt {2} \sqrt {c} \left (-b^5+11 a b^3 c-44 a^2 b c^2+b^4 \sqrt {b^2-4 a c}-9 a b^2 c \sqrt {b^2-4 a c}+28 a^2 c^2 \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{5/2} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {3 \sqrt {2} \sqrt {c} \left (b^5-11 a b^3 c+44 a^2 b c^2+b^4 \sqrt {b^2-4 a c}-9 a b^2 c \sqrt {b^2-4 a c}+28 a^2 c^2 \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\left (b^2-4 a c\right )^{5/2} \sqrt {b+\sqrt {b^2-4 a c}}}}{16 c^3} \]
((2*x*(2*b^5 - 17*a*b^3*c + 48*a^2*b*c^2 - 5*b^4*c*x^2 + 37*a*b^2*c^2*x^2 - 44*a^2*c^3*x^2))/((b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) - (4*(b^4*x^3 + a *b^2*x*(b - 4*c*x^2) + a^2*c*x*(-3*b + 2*c*x^2)))/((b^2 - 4*a*c)*(a + b*x^ 2 + c*x^4)^2) + (3*Sqrt[2]*Sqrt[c]*(-b^5 + 11*a*b^3*c - 44*a^2*b*c^2 + b^4 *Sqrt[b^2 - 4*a*c] - 9*a*b^2*c*Sqrt[b^2 - 4*a*c] + 28*a^2*c^2*Sqrt[b^2 - 4 *a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/((b^2 - 4* a*c)^(5/2)*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + (3*Sqrt[2]*Sqrt[c]*(b^5 - 11*a*b ^3*c + 44*a^2*b*c^2 + b^4*Sqrt[b^2 - 4*a*c] - 9*a*b^2*c*Sqrt[b^2 - 4*a*c] + 28*a^2*c^2*Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b ^2 - 4*a*c]]])/((b^2 - 4*a*c)^(5/2)*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/(16*c^3)
Time = 0.70 (sec) , antiderivative size = 392, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {1440, 1598, 27, 1602, 27, 1602, 1480, 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.
| \(\displaystyle \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx\) |
| \(\Big \downarrow \) 1440 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\int \frac {x^6 \left (b x^2+14 a\right )}{\left (c x^4+b x^2+a\right )^2}dx}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 1598 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {\int \frac {3 x^4 \left (20 a b-\left (b^2-28 a c\right ) x^2\right )}{c x^4+b x^2+a}dx}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {3 \int \frac {x^4 \left (20 a b-\left (b^2-28 a c\right ) x^2\right )}{c x^4+b x^2+a}dx}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {3 \left (-\frac {\int -\frac {3 x^2 \left (b \left (b^2-8 a c\right ) x^2+a \left (b^2-28 a c\right )\right )}{c x^4+b x^2+a}dx}{3 c}-\frac {x^3 \left (b^2-28 a c\right )}{3 c}\right )}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {3 \left (\frac {\int \frac {x^2 \left (b \left (b^2-8 a c\right ) x^2+a \left (b^2-28 a c\right )\right )}{c x^4+b x^2+a}dx}{c}-\frac {x^3 \left (b^2-28 a c\right )}{3 c}\right )}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 1602 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {3 \left (\frac {\frac {b x \left (b^2-8 a c\right )}{c}-\frac {\int \frac {\left (b^4-9 a c b^2+28 a^2 c^2\right ) x^2+a b \left (b^2-8 a c\right )}{c x^4+b x^2+a}dx}{c}}{c}-\frac {x^3 \left (b^2-28 a c\right )}{3 c}\right )}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {3 \left (\frac {\frac {b x \left (b^2-8 a c\right )}{c}-\frac {\frac {1}{2} \left (-\frac {b \left (44 a^2 c^2-11 a b^2 c+b^4\right )}{\sqrt {b^2-4 a c}}+28 a^2 c^2-9 a b^2 c+b^4\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} \left (\frac {44 a^2 b c^2-11 a b^3 c+b^5}{\sqrt {b^2-4 a c}}+28 a^2 c^2-9 a b^2 c+b^4\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{c}}{c}-\frac {x^3 \left (b^2-28 a c\right )}{3 c}\right )}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {x^7 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\frac {3 \left (\frac {\frac {b x \left (b^2-8 a c\right )}{c}-\frac {\frac {\left (-\frac {b \left (44 a^2 c^2-11 a b^2 c+b^4\right )}{\sqrt {b^2-4 a c}}+28 a^2 c^2-9 a b^2 c+b^4\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {44 a^2 b c^2-11 a b^3 c+b^5}{\sqrt {b^2-4 a c}}+28 a^2 c^2-9 a b^2 c+b^4\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}}{c}}{c}-\frac {x^3 \left (b^2-28 a c\right )}{3 c}\right )}{2 \left (b^2-4 a c\right )}-\frac {x^5 \left (12 a b-x^2 \left (b^2-28 a c\right )\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}}{4 \left (b^2-4 a c\right )}\) |
(x^7*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (-1/2*(x^5*( 12*a*b - (b^2 - 28*a*c)*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) + (3*(-1 /3*((b^2 - 28*a*c)*x^3)/c + ((b*(b^2 - 8*a*c)*x)/c - (((b^4 - 9*a*b^2*c + 28*a^2*c^2 - (b*(b^4 - 11*a*b^2*c + 44*a^2*c^2))/Sqrt[b^2 - 4*a*c])*ArcTan [(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((b^4 - 9*a*b^2*c + 28*a^2*c^2 + (b^5 - 11*a*b^3* c + 44*a^2*b*c^2)/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + S qrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]))/c)/c))/ (2*(b^2 - 4*a*c)))/(4*(b^2 - 4*a*c))
3.9.81.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-d^3)*(d*x)^(m - 3)*(2*a + b*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2* (p + 1)*(b^2 - 4*a*c))), x] + Simp[d^4/(2*(p + 1)*(b^2 - 4*a*c)) Int[(d*x )^(m - 4)*(2*a*(m - 3) + b*(m + 4*p + 3)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && Gt Q[m, 3] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_.), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1) *((b*d - 2*a*e - (b*e - 2*c*d)*x^2)/(2*(p + 1)*(b^2 - 4*a*c))), x] - Simp[f ^2/(2*(p + 1)*(b^2 - 4*a*c)) Int[(f*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p + 1 )*Simp[(m - 1)*(b*d - 2*a*e) - (4*p + 4 + m + 1)*(b*e - 2*c*d)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.14 (sec) , antiderivative size = 331, normalized size of antiderivative = 0.83
| method | result | size |
| risch | \(\frac {-\frac {\left (44 a^{2} c^{2}-37 a \,b^{2} c +5 b^{4}\right ) x^{7}}{8 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}+\frac {b \left (4 a^{2} c^{2}+20 a \,b^{2} c -3 b^{4}\right ) x^{5}}{8 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a \left (28 a^{2} c^{2}-49 a \,b^{2} c +6 b^{4}\right ) x^{3}}{8 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}+\frac {3 a^{2} b \left (8 a c -b^{2}\right ) x}{8 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}+\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} c +\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (\frac {\left (28 a^{2} c^{2}-9 a \,b^{2} c +b^{4}\right ) \textit {\_R}^{2}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {a b \left (8 a c -b^{2}\right )}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3} c +\textit {\_R} b}\right )}{16 c^{2}}\) | \(331\) |
| default | \(\frac {-\frac {\left (44 a^{2} c^{2}-37 a \,b^{2} c +5 b^{4}\right ) x^{7}}{8 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}+\frac {b \left (4 a^{2} c^{2}+20 a \,b^{2} c -3 b^{4}\right ) x^{5}}{8 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {a \left (28 a^{2} c^{2}-49 a \,b^{2} c +6 b^{4}\right ) x^{3}}{8 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}+\frac {3 a^{2} b \left (8 a c -b^{2}\right ) x}{8 c^{2} \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}+\frac {\frac {3 \left (28 a^{2} c^{2} \sqrt {-4 a c +b^{2}}-9 a \,b^{2} c \sqrt {-4 a c +b^{2}}+b^{4} \sqrt {-4 a c +b^{2}}+44 a^{2} b \,c^{2}-11 a \,b^{3} c +b^{5}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{16 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}-\frac {3 \left (28 a^{2} c^{2} \sqrt {-4 a c +b^{2}}-9 a \,b^{2} c \sqrt {-4 a c +b^{2}}+b^{4} \sqrt {-4 a c +b^{2}}-44 a^{2} b \,c^{2}+11 a \,b^{3} c -b^{5}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{16 c \sqrt {-4 a c +b^{2}}\, \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}}{c \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}\) | \(496\) |
(-1/8*(44*a^2*c^2-37*a*b^2*c+5*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)/c*x^7+1/8*b *(4*a^2*c^2+20*a*b^2*c-3*b^4)/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x^5-1/8*a/c^2 *(28*a^2*c^2-49*a*b^2*c+6*b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^3+3/8*a^2*b*(8 *a*c-b^2)/c^2/(16*a^2*c^2-8*a*b^2*c+b^4)*x)/(c*x^4+b*x^2+a)^2+3/16/c^2*sum (((28*a^2*c^2-9*a*b^2*c+b^4)/(16*a^2*c^2-8*a*b^2*c+b^4)*_R^2-a*b*(8*a*c-b^ 2)/(16*a^2*c^2-8*a*b^2*c+b^4))/(2*_R^3*c+_R*b)*ln(x-_R),_R=RootOf(_Z^4*c+_ Z^2*b+a))
Leaf count of result is larger than twice the leaf count of optimal. 4279 vs. \(2 (351) = 702\).
Time = 0.69 (sec) , antiderivative size = 4279, normalized size of antiderivative = 10.70 \[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \]
-1/16*(2*(5*b^4*c - 37*a*b^2*c^2 + 44*a^2*c^3)*x^7 + 2*(3*b^5 - 20*a*b^3*c - 4*a^2*b*c^2)*x^5 + 2*(6*a*b^4 - 49*a^2*b^2*c + 28*a^3*c^2)*x^3 + 3*sqrt (1/2)*((b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*x^8 + a^2*b^4*c^2 - 8*a^3*b^2* c^3 + 16*a^4*c^4 + 2*(b^5*c^3 - 8*a*b^3*c^4 + 16*a^2*b*c^5)*x^6 + (b^6*c^2 - 6*a*b^4*c^3 + 32*a^3*c^5)*x^4 + 2*(a*b^5*c^2 - 8*a^2*b^3*c^3 + 16*a^3*b *c^4)*x^2)*sqrt(-(b^9 - 21*a*b^7*c + 189*a^2*b^5*c^2 - 840*a^3*b^3*c^3 + 1 680*a^4*b*c^4 + (b^10*c^5 - 20*a*b^8*c^6 + 160*a^2*b^6*c^7 - 640*a^3*b^4*c ^8 + 1280*a^4*b^2*c^9 - 1024*a^5*c^10)*sqrt((b^8 - 22*a*b^6*c + 219*a^2*b^ 4*c^2 - 1078*a^3*b^2*c^3 + 2401*a^4*c^4)/(b^10*c^10 - 20*a*b^8*c^11 + 160* a^2*b^6*c^12 - 640*a^3*b^4*c^13 + 1280*a^4*b^2*c^14 - 1024*a^5*c^15)))/(b^ 10*c^5 - 20*a*b^8*c^6 + 160*a^2*b^6*c^7 - 640*a^3*b^4*c^8 + 1280*a^4*b^2*c ^9 - 1024*a^5*c^10))*log(27*(21*a^2*b^8 - 447*a^3*b^6*c + 4189*a^4*b^4*c^2 - 19208*a^5*b^2*c^3 + 38416*a^6*c^4)*x + 27/2*sqrt(1/2)*(b^13 - 31*a*b^11 *c + 413*a^2*b^9*c^2 - 3012*a^3*b^7*c^3 + 12496*a^4*b^5*c^4 - 27584*a^5*b^ 3*c^5 + 25088*a^6*b*c^6 - (b^14*c^5 - 30*a*b^12*c^6 + 416*a^2*b^10*c^7 - 3 360*a^3*b^8*c^8 + 16640*a^4*b^6*c^9 - 49664*a^5*b^4*c^10 + 81920*a^6*b^2*c ^11 - 57344*a^7*c^12)*sqrt((b^8 - 22*a*b^6*c + 219*a^2*b^4*c^2 - 1078*a^3* b^2*c^3 + 2401*a^4*c^4)/(b^10*c^10 - 20*a*b^8*c^11 + 160*a^2*b^6*c^12 - 64 0*a^3*b^4*c^13 + 1280*a^4*b^2*c^14 - 1024*a^5*c^15)))*sqrt(-(b^9 - 21*a*b^ 7*c + 189*a^2*b^5*c^2 - 840*a^3*b^3*c^3 + 1680*a^4*b*c^4 + (b^10*c^5 - ...
Timed out. \[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Timed out} \]
\[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=\int { \frac {x^{10}}{{\left (c x^{4} + b x^{2} + a\right )}^{3}} \,d x } \]
-1/8*((5*b^4*c - 37*a*b^2*c^2 + 44*a^2*c^3)*x^7 + (3*b^5 - 20*a*b^3*c - 4* a^2*b*c^2)*x^5 + (6*a*b^4 - 49*a^2*b^2*c + 28*a^3*c^2)*x^3 + 3*(a^2*b^3 - 8*a^3*b*c)*x)/((b^4*c^4 - 8*a*b^2*c^5 + 16*a^2*c^6)*x^8 + a^2*b^4*c^2 - 8* a^3*b^2*c^3 + 16*a^4*c^4 + 2*(b^5*c^3 - 8*a*b^3*c^4 + 16*a^2*b*c^5)*x^6 + (b^6*c^2 - 6*a*b^4*c^3 + 32*a^3*c^5)*x^4 + 2*(a*b^5*c^2 - 8*a^2*b^3*c^3 + 16*a^3*b*c^4)*x^2) + 3/8*integrate((a*b^3 - 8*a^2*b*c + (b^4 - 9*a*b^2*c + 28*a^2*c^2)*x^2)/(c*x^4 + b*x^2 + a), x)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2* c^4)
Leaf count of result is larger than twice the leaf count of optimal. 2430 vs. \(2 (351) = 702\).
Time = 2.33 (sec) , antiderivative size = 2430, normalized size of antiderivative = 6.08 \[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \]
3/32*(sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^7 - 16*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^5*c - 2*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b ^6*c - 2*b^7*c + 80*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^3*c^2 + 24*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4*c^2 + sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5*c^2 + 32*a*b^5*c^2 - 2*b^6*c^2 - 128*sqrt(2)*sqrt (b*c + sqrt(b^2 - 4*a*c)*c)*a^3*b*c^3 - 64*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4 *a*c)*c)*a^2*b^2*c^3 - 12*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^ 3 - 160*a^2*b^3*c^3 + 28*a*b^4*c^3 + 32*sqrt(2)*sqrt(b*c + sqrt(b^2 - 4*a* c)*c)*a^2*b*c^4 + 256*a^3*b*c^4 - 192*a^2*b^2*c^4 + 448*a^3*c^5 + sqrt(2)* sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^6 - 14*sqrt(2)*sqrt(b^ 2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^4*c - 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^5*c + 96*sqrt(2)*sqrt(b^2 - 4*a*c )*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b^2*c^2 + 20*sqrt(2)*sqrt(b^2 - 4*a* c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^3*c^2 + sqrt(2)*sqrt(b^2 - 4*a*c)*s qrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c^2 - 224*sqrt(2)*sqrt(b^2 - 4*a*c)*sqr t(b*c + sqrt(b^2 - 4*a*c)*c)*a^3*c^3 - 112*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt( b*c + sqrt(b^2 - 4*a*c)*c)*a^2*b*c^3 - 10*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b *c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^3 + 56*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b* c + sqrt(b^2 - 4*a*c)*c)*a^2*c^4 + 2*(b^2 - 4*a*c)*b^5*c - 24*(b^2 - 4*a*c )*a*b^3*c^2 + 2*(b^2 - 4*a*c)*b^4*c^2 + 64*(b^2 - 4*a*c)*a^2*b*c^3 - 20...
Time = 18.28 (sec) , antiderivative size = 10912, normalized size of antiderivative = 27.28 \[ \int \frac {x^{10}}{\left (a+b x^2+c x^4\right )^3} \, dx=\text {Too large to display} \]
- ((x^3*(6*a*b^4 + 28*a^3*c^2 - 49*a^2*b^2*c))/(8*c^2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^7*(5*b^4 + 44*a^2*c^2 - 37*a*b^2*c))/(8*c*(b^4 + 16*a^2*c ^2 - 8*a*b^2*c)) - (b*x^5*(4*a^2*c^2 - 3*b^4 + 20*a*b^2*c))/(8*c^2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) - (3*a^2*b*x*(8*a*c - b^2))/(8*c^2*(b^4 + 16*a^2* c^2 - 8*a*b^2*c)))/(x^4*(2*a*c + b^2) + a^2 + c^2*x^8 + 2*a*b*x^2 + 2*b*c* x^6) - atan(((((3*(256*a*b^13*c^3 + 2097152*a^7*b*c^9 - 7168*a^2*b^11*c^4 + 81920*a^3*b^9*c^5 - 491520*a^4*b^7*c^6 + 1638400*a^5*b^5*c^7 - 2883584*a ^6*b^3*c^8))/(512*(4096*a^6*c^9 + b^12*c^3 - 24*a*b^10*c^4 + 240*a^2*b^8*c ^5 - 1280*a^3*b^6*c^6 + 3840*a^4*b^4*c^7 - 6144*a^5*b^2*c^8)) - (x*((9*(b^ 4*(-(4*a*c - b^2)^15)^(1/2) - b^19 + 1720320*a^9*b*c^9 - 769*a^2*b^15*c^2 + 8620*a^3*b^13*c^3 - 63440*a^4*b^11*c^4 + 316864*a^5*b^9*c^5 - 1069824*a^ 6*b^7*c^6 + 2343936*a^7*b^5*c^7 - 3010560*a^8*b^3*c^8 + 49*a^2*c^2*(-(4*a* c - b^2)^15)^(1/2) + 41*a*b^17*c - 11*a*b^2*c*(-(4*a*c - b^2)^15)^(1/2)))/ (512*(1048576*a^10*c^15 + b^20*c^5 - 40*a*b^18*c^6 + 720*a^2*b^16*c^7 - 76 80*a^3*b^14*c^8 + 53760*a^4*b^12*c^9 - 258048*a^5*b^10*c^10 + 860160*a^6*b ^8*c^11 - 1966080*a^7*b^6*c^12 + 2949120*a^8*b^4*c^13 - 2621440*a^9*b^2*c^ 14)))^(1/2)*(256*b^11*c^5 - 5120*a*b^9*c^6 - 262144*a^5*b*c^10 + 40960*a^2 *b^7*c^7 - 163840*a^3*b^5*c^8 + 327680*a^4*b^3*c^9))/(32*(256*a^4*c^7 + b^ 8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6)))*((9*(b^4*(-(4*a *c - b^2)^15)^(1/2) - b^19 + 1720320*a^9*b*c^9 - 769*a^2*b^15*c^2 + 862...