Integrand size = 25, antiderivative size = 173 \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {\left (6 c d^2-e (3 b d-a e)\right ) x}{e^5}-\frac {(3 c d-b e) x^3}{3 e^4}+\frac {c x^5}{5 e^3}-\frac {d^2 \left (c d^2-b d e+a e^2\right ) x}{4 e^5 \left (d+e x^2\right )^2}+\frac {d \left (17 c d^2-e (13 b d-9 a e)\right ) x}{8 e^5 \left (d+e x^2\right )}-\frac {\sqrt {d} \left (63 c d^2-35 b d e+15 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 e^{11/2}} \]
(6*c*d^2-e*(-a*e+3*b*d))*x/e^5-1/3*(-b*e+3*c*d)*x^3/e^4+1/5*c*x^5/e^3-1/4* d^2*(a*e^2-b*d*e+c*d^2)*x/e^5/(e*x^2+d)^2+1/8*d*(17*c*d^2-e*(-9*a*e+13*b*d ))*x/e^5/(e*x^2+d)-1/8*(15*a*e^2-35*b*d*e+63*c*d^2)*arctan(x*e^(1/2)/d^(1/ 2))*d^(1/2)/e^(11/2)
Time = 0.08 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.98 \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {\left (6 c d^2+e (-3 b d+a e)\right ) x}{e^5}+\frac {(-3 c d+b e) x^3}{3 e^4}+\frac {c x^5}{5 e^3}-\frac {\left (c d^4+d^2 e (-b d+a e)\right ) x}{4 e^5 \left (d+e x^2\right )^2}+\frac {\left (17 c d^3+d e (-13 b d+9 a e)\right ) x}{8 e^5 \left (d+e x^2\right )}-\frac {\sqrt {d} \left (63 c d^2+5 e (-7 b d+3 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 e^{11/2}} \]
((6*c*d^2 + e*(-3*b*d + a*e))*x)/e^5 + ((-3*c*d + b*e)*x^3)/(3*e^4) + (c*x ^5)/(5*e^3) - ((c*d^4 + d^2*e*(-(b*d) + a*e))*x)/(4*e^5*(d + e*x^2)^2) + ( (17*c*d^3 + d*e*(-13*b*d + 9*a*e))*x)/(8*e^5*(d + e*x^2)) - (Sqrt[d]*(63*c *d^2 + 5*e*(-7*b*d + 3*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*e^(11/2))
Time = 0.68 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1580, 25, 2345, 2341, 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.
| \(\displaystyle \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1580 |
| \(\displaystyle -\frac {\int -\frac {4 c e^4 x^8-4 e^3 (c d-b e) x^6+4 e^2 \left (c d^2-b e d+a e^2\right ) x^4-4 d e \left (c d^2-b e d+a e^2\right ) x^2+d^2 \left (c d^2-b e d+a e^2\right )}{\left (e x^2+d\right )^2}dx}{4 e^5}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {4 c e^4 x^8-4 e^3 (c d-b e) x^6+4 e^2 \left (c d^2-b e d+a e^2\right ) x^4-4 d e \left (c d^2-b e d+a e^2\right ) x^2+d^2 \left (c d^2-b e d+a e^2\right )}{\left (e x^2+d\right )^2}dx}{4 e^5}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 2345 |
| \(\displaystyle \frac {\frac {d x \left (17 c d^2-e (13 b d-9 a e)\right )}{2 \left (d+e x^2\right )}-\frac {\int \frac {-8 c d e^3 x^6+8 d e^2 (2 c d-b e) x^4-8 d e \left (3 c d^2-e (2 b d-a e)\right ) x^2+d^2 \left (15 c d^2-e (11 b d-7 a e)\right )}{e x^2+d}dx}{2 d}}{4 e^5}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle \frac {\frac {d x \left (17 c d^2-e (13 b d-9 a e)\right )}{2 \left (d+e x^2\right )}-\frac {\int \left (-8 c d e^2 x^4+8 d e (3 c d-b e) x^2-8 d \left (6 c d^2-e (3 b d-a e)\right )+\frac {63 c d^4-35 b e d^3+15 a e^2 d^2}{e x^2+d}\right )dx}{2 d}}{4 e^5}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {d x \left (17 c d^2-e (13 b d-9 a e)\right )}{2 \left (d+e x^2\right )}-\frac {\frac {d^{3/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 a e^2-35 b d e+63 c d^2\right )}{\sqrt {e}}-8 d x \left (6 c d^2-e (3 b d-a e)\right )+\frac {8}{3} d e x^3 (3 c d-b e)-\frac {8}{5} c d e^2 x^5}{2 d}}{4 e^5}-\frac {d^2 x \left (a e^2-b d e+c d^2\right )}{4 e^5 \left (d+e x^2\right )^2}\) |
-1/4*(d^2*(c*d^2 - b*d*e + a*e^2)*x)/(e^5*(d + e*x^2)^2) + ((d*(17*c*d^2 - e*(13*b*d - 9*a*e))*x)/(2*(d + e*x^2)) - (-8*d*(6*c*d^2 - e*(3*b*d - a*e) )*x + (8*d*e*(3*c*d - b*e)*x^3)/3 - (8*c*d*e^2*x^5)/5 + (d^(3/2)*(63*c*d^2 - 35*b*d*e + 15*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqrt[e])/(2*d))/(4*e^ 5)
3.3.88.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) ^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* (q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b *d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e }, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.33 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.87
| method | result | size |
| default | \(\frac {\frac {1}{5} c \,x^{5} e^{2}+\frac {1}{3} b \,e^{2} x^{3}-d c \,x^{3} e +a \,e^{2} x -3 b d e x +6 c \,d^{2} x}{e^{5}}-\frac {d \left (\frac {\left (-\frac {9}{8} a \,e^{3}+\frac {13}{8} d \,e^{2} b -\frac {17}{8} c \,d^{2} e \right ) x^{3}-\frac {d \left (7 a \,e^{2}-11 b d e +15 c \,d^{2}\right ) x}{8}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (15 a \,e^{2}-35 b d e +63 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 \sqrt {e d}}\right )}{e^{5}}\) | \(151\) |
| risch | \(\frac {c \,x^{5}}{5 e^{3}}+\frac {b \,x^{3}}{3 e^{3}}-\frac {d c \,x^{3}}{e^{4}}+\frac {a x}{e^{3}}-\frac {3 b d x}{e^{4}}+\frac {6 c \,d^{2} x}{e^{5}}+\frac {\left (\frac {9}{8} d \,e^{3} a -\frac {13}{8} e^{2} d^{2} b +\frac {17}{8} d^{3} e c \right ) x^{3}+\frac {d^{2} \left (7 a \,e^{2}-11 b d e +15 c \,d^{2}\right ) x}{8}}{e^{5} \left (e \,x^{2}+d \right )^{2}}+\frac {15 \sqrt {-e d}\, \ln \left (-\sqrt {-e d}\, x -d \right ) a}{16 e^{4}}-\frac {35 \sqrt {-e d}\, \ln \left (-\sqrt {-e d}\, x -d \right ) b d}{16 e^{5}}+\frac {63 \sqrt {-e d}\, \ln \left (-\sqrt {-e d}\, x -d \right ) c \,d^{2}}{16 e^{6}}-\frac {15 \sqrt {-e d}\, \ln \left (\sqrt {-e d}\, x -d \right ) a}{16 e^{4}}+\frac {35 \sqrt {-e d}\, \ln \left (\sqrt {-e d}\, x -d \right ) b d}{16 e^{5}}-\frac {63 \sqrt {-e d}\, \ln \left (\sqrt {-e d}\, x -d \right ) c \,d^{2}}{16 e^{6}}\) | \(281\) |
1/e^5*(1/5*c*x^5*e^2+1/3*b*e^2*x^3-d*c*x^3*e+a*e^2*x-3*b*d*e*x+6*c*d^2*x)- d/e^5*(((-9/8*a*e^3+13/8*d*e^2*b-17/8*c*d^2*e)*x^3-1/8*d*(7*a*e^2-11*b*d*e +15*c*d^2)*x)/(e*x^2+d)^2+1/8*(15*a*e^2-35*b*d*e+63*c*d^2)/(e*d)^(1/2)*arc tan(e*x/(e*d)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.91 \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\left [\frac {48 \, c e^{4} x^{9} - 16 \, {\left (9 \, c d e^{3} - 5 \, b e^{4}\right )} x^{7} + 16 \, {\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{5} + 50 \, {\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{3} + 15 \, {\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2} + {\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{4} + 2 \, {\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-\frac {d}{e}} \log \left (\frac {e x^{2} - 2 \, e x \sqrt {-\frac {d}{e}} - d}{e x^{2} + d}\right ) + 30 \, {\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2}\right )} x}{240 \, {\left (e^{7} x^{4} + 2 \, d e^{6} x^{2} + d^{2} e^{5}\right )}}, \frac {24 \, c e^{4} x^{9} - 8 \, {\left (9 \, c d e^{3} - 5 \, b e^{4}\right )} x^{7} + 8 \, {\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{5} + 25 \, {\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{3} - 15 \, {\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2} + {\left (63 \, c d^{2} e^{2} - 35 \, b d e^{3} + 15 \, a e^{4}\right )} x^{4} + 2 \, {\left (63 \, c d^{3} e - 35 \, b d^{2} e^{2} + 15 \, a d e^{3}\right )} x^{2}\right )} \sqrt {\frac {d}{e}} \arctan \left (\frac {e x \sqrt {\frac {d}{e}}}{d}\right ) + 15 \, {\left (63 \, c d^{4} - 35 \, b d^{3} e + 15 \, a d^{2} e^{2}\right )} x}{120 \, {\left (e^{7} x^{4} + 2 \, d e^{6} x^{2} + d^{2} e^{5}\right )}}\right ] \]
[1/240*(48*c*e^4*x^9 - 16*(9*c*d*e^3 - 5*b*e^4)*x^7 + 16*(63*c*d^2*e^2 - 3 5*b*d*e^3 + 15*a*e^4)*x^5 + 50*(63*c*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^ 3 + 15*(63*c*d^4 - 35*b*d^3*e + 15*a*d^2*e^2 + (63*c*d^2*e^2 - 35*b*d*e^3 + 15*a*e^4)*x^4 + 2*(63*c*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^2)*sqrt(-d/ e)*log((e*x^2 - 2*e*x*sqrt(-d/e) - d)/(e*x^2 + d)) + 30*(63*c*d^4 - 35*b*d ^3*e + 15*a*d^2*e^2)*x)/(e^7*x^4 + 2*d*e^6*x^2 + d^2*e^5), 1/120*(24*c*e^4 *x^9 - 8*(9*c*d*e^3 - 5*b*e^4)*x^7 + 8*(63*c*d^2*e^2 - 35*b*d*e^3 + 15*a*e ^4)*x^5 + 25*(63*c*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^3 - 15*(63*c*d^4 - 35*b*d^3*e + 15*a*d^2*e^2 + (63*c*d^2*e^2 - 35*b*d*e^3 + 15*a*e^4)*x^4 + 2*(63*c*d^3*e - 35*b*d^2*e^2 + 15*a*d*e^3)*x^2)*sqrt(d/e)*arctan(e*x*sqrt( d/e)/d) + 15*(63*c*d^4 - 35*b*d^3*e + 15*a*d^2*e^2)*x)/(e^7*x^4 + 2*d*e^6* x^2 + d^2*e^5)]
Time = 1.92 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.36 \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c x^{5}}{5 e^{3}} + x^{3} \left (\frac {b}{3 e^{3}} - \frac {c d}{e^{4}}\right ) + x \left (\frac {a}{e^{3}} - \frac {3 b d}{e^{4}} + \frac {6 c d^{2}}{e^{5}}\right ) + \frac {\sqrt {- \frac {d}{e^{11}}} \cdot \left (15 a e^{2} - 35 b d e + 63 c d^{2}\right ) \log {\left (- e^{5} \sqrt {- \frac {d}{e^{11}}} + x \right )}}{16} - \frac {\sqrt {- \frac {d}{e^{11}}} \cdot \left (15 a e^{2} - 35 b d e + 63 c d^{2}\right ) \log {\left (e^{5} \sqrt {- \frac {d}{e^{11}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (9 a d e^{3} - 13 b d^{2} e^{2} + 17 c d^{3} e\right ) + x \left (7 a d^{2} e^{2} - 11 b d^{3} e + 15 c d^{4}\right )}{8 d^{2} e^{5} + 16 d e^{6} x^{2} + 8 e^{7} x^{4}} \]
c*x**5/(5*e**3) + x**3*(b/(3*e**3) - c*d/e**4) + x*(a/e**3 - 3*b*d/e**4 + 6*c*d**2/e**5) + sqrt(-d/e**11)*(15*a*e**2 - 35*b*d*e + 63*c*d**2)*log(-e* *5*sqrt(-d/e**11) + x)/16 - sqrt(-d/e**11)*(15*a*e**2 - 35*b*d*e + 63*c*d* *2)*log(e**5*sqrt(-d/e**11) + x)/16 + (x**3*(9*a*d*e**3 - 13*b*d**2*e**2 + 17*c*d**3*e) + x*(7*a*d**2*e**2 - 11*b*d**3*e + 15*c*d**4))/(8*d**2*e**5 + 16*d*e**6*x**2 + 8*e**7*x**4)
Exception generated. \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.99 \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {{\left (63 \, c d^{3} - 35 \, b d^{2} e + 15 \, a d e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} e^{5}} + \frac {17 \, c d^{3} e x^{3} - 13 \, b d^{2} e^{2} x^{3} + 9 \, a d e^{3} x^{3} + 15 \, c d^{4} x - 11 \, b d^{3} e x + 7 \, a d^{2} e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} e^{5}} + \frac {3 \, c e^{12} x^{5} - 15 \, c d e^{11} x^{3} + 5 \, b e^{12} x^{3} + 90 \, c d^{2} e^{10} x - 45 \, b d e^{11} x + 15 \, a e^{12} x}{15 \, e^{15}} \]
-1/8*(63*c*d^3 - 35*b*d^2*e + 15*a*d*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e) *e^5) + 1/8*(17*c*d^3*e*x^3 - 13*b*d^2*e^2*x^3 + 9*a*d*e^3*x^3 + 15*c*d^4* x - 11*b*d^3*e*x + 7*a*d^2*e^2*x)/((e*x^2 + d)^2*e^5) + 1/15*(3*c*e^12*x^5 - 15*c*d*e^11*x^3 + 5*b*e^12*x^3 + 90*c*d^2*e^10*x - 45*b*d*e^11*x + 15*a *e^12*x)/e^15
Time = 7.75 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.29 \[ \int \frac {x^6 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=x^3\,\left (\frac {b}{3\,e^3}-\frac {c\,d}{e^4}\right )-x\,\left (\frac {3\,c\,d^2}{e^5}-\frac {a}{e^3}+\frac {3\,d\,\left (\frac {b}{e^3}-\frac {3\,c\,d}{e^4}\right )}{e}\right )+\frac {\left (\frac {17\,c\,d^3\,e}{8}-\frac {13\,b\,d^2\,e^2}{8}+\frac {9\,a\,d\,e^3}{8}\right )\,x^3+\left (\frac {15\,c\,d^4}{8}-\frac {11\,b\,d^3\,e}{8}+\frac {7\,a\,d^2\,e^2}{8}\right )\,x}{d^2\,e^5+2\,d\,e^6\,x^2+e^7\,x^4}+\frac {c\,x^5}{5\,e^3}-\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,x\,\left (63\,c\,d^2-35\,b\,d\,e+15\,a\,e^2\right )}{63\,c\,d^3-35\,b\,d^2\,e+15\,a\,d\,e^2}\right )\,\left (63\,c\,d^2-35\,b\,d\,e+15\,a\,e^2\right )}{8\,e^{11/2}} \]
x^3*(b/(3*e^3) - (c*d)/e^4) - x*((3*c*d^2)/e^5 - a/e^3 + (3*d*(b/e^3 - (3* c*d)/e^4))/e) + (x^3*((9*a*d*e^3)/8 - (13*b*d^2*e^2)/8 + (17*c*d^3*e)/8) + x*((15*c*d^4)/8 + (7*a*d^2*e^2)/8 - (11*b*d^3*e)/8))/(d^2*e^5 + e^7*x^4 + 2*d*e^6*x^2) + (c*x^5)/(5*e^3) - (d^(1/2)*atan((d^(1/2)*e^(1/2)*x*(15*a*e ^2 + 63*c*d^2 - 35*b*d*e))/(63*c*d^3 + 15*a*d*e^2 - 35*b*d^2*e))*(15*a*e^2 + 63*c*d^2 - 35*b*d*e))/(8*e^(11/2))