Integrand size = 25, antiderivative size = 143 \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {(3 c d-b e) x}{e^4}+\frac {c x^3}{3 e^3}+\frac {d \left (c d^2-b d e+a e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac {\left (13 c d^2-e (9 b d-5 a e)\right ) x}{8 e^4 \left (d+e x^2\right )}+\frac {\left (35 c d^2-3 e (5 b d-a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {d} e^{9/2}} \]
-(-b*e+3*c*d)*x/e^4+1/3*c*x^3/e^3+1/4*d*(a*e^2-b*d*e+c*d^2)*x/e^4/(e*x^2+d )^2-1/8*(13*c*d^2-e*(-5*a*e+9*b*d))*x/e^4/(e*x^2+d)+1/8*(35*c*d^2-3*e*(-a* e+5*b*d))*arctan(x*e^(1/2)/d^(1/2))/e^(9/2)/d^(1/2)
Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {(-3 c d+b e) x}{e^4}+\frac {c x^3}{3 e^3}+\frac {\left (c d^3-b d^2 e+a d e^2\right ) x}{4 e^4 \left (d+e x^2\right )^2}-\frac {\left (13 c d^2-9 b d e+5 a e^2\right ) x}{8 e^4 \left (d+e x^2\right )}+\frac {\left (35 c d^2-15 b d e+3 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {d} e^{9/2}} \]
((-3*c*d + b*e)*x)/e^4 + (c*x^3)/(3*e^3) + ((c*d^3 - b*d^2*e + a*d*e^2)*x) /(4*e^4*(d + e*x^2)^2) - ((13*c*d^2 - 9*b*d*e + 5*a*e^2)*x)/(8*e^4*(d + e* x^2)) + ((35*c*d^2 - 15*b*d*e + 3*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*S qrt[d]*e^(9/2))
Time = 0.47 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1580, 2345, 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.
\(\displaystyle \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1580 |
\(\displaystyle \frac {d x \left (a e^2-b d e+c d^2\right )}{4 e^4 \left (d+e x^2\right )^2}-\frac {\int \frac {-4 c e^3 x^6+4 e^2 (c d-b e) x^4-4 e \left (c d^2-b e d+a e^2\right ) x^2+d \left (c d^2-b e d+a e^2\right )}{\left (e x^2+d\right )^2}dx}{4 e^4}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle \frac {d x \left (a e^2-b d e+c d^2\right )}{4 e^4 \left (d+e x^2\right )^2}-\frac {\frac {x \left (13 c d^2-e (9 b d-5 a e)\right )}{2 \left (d+e x^2\right )}-\frac {\int \frac {8 c d e^2 x^4-8 d e (2 c d-b e) x^2+d \left (11 c d^2-e (7 b d-3 a e)\right )}{e x^2+d}dx}{2 d}}{4 e^4}\) |
\(\Big \downarrow \) 1467 |
\(\displaystyle \frac {d x \left (a e^2-b d e+c d^2\right )}{4 e^4 \left (d+e x^2\right )^2}-\frac {\frac {x \left (13 c d^2-e (9 b d-5 a e)\right )}{2 \left (d+e x^2\right )}-\frac {\int \left (8 c d e x^2-8 d (3 c d-b e)+\frac {35 c d^3-15 b e d^2+3 a e^2 d}{e x^2+d}\right )dx}{2 d}}{4 e^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d x \left (a e^2-b d e+c d^2\right )}{4 e^4 \left (d+e x^2\right )^2}-\frac {\frac {x \left (13 c d^2-e (9 b d-5 a e)\right )}{2 \left (d+e x^2\right )}-\frac {\frac {\sqrt {d} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (35 c d^2-3 e (5 b d-a e)\right )}{\sqrt {e}}-8 d x (3 c d-b e)+\frac {8}{3} c d e x^3}{2 d}}{4 e^4}\) |
(d*(c*d^2 - b*d*e + a*e^2)*x)/(4*e^4*(d + e*x^2)^2) - (((13*c*d^2 - e*(9*b *d - 5*a*e))*x)/(2*(d + e*x^2)) - (-8*d*(3*c*d - b*e)*x + (8*c*d*e*x^3)/3 + (Sqrt[d]*(35*c*d^2 - 3*e*(5*b*d - a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/Sqr t[e])/(2*d))/(4*e^4)
3.3.89.3.1 Defintions of rubi rules used
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]
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] :> 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 = 120, normalized size of antiderivative = 0.84
method | result | size |
default | \(\frac {\frac {1}{3} c \,x^{3} e +b e x -3 c d x}{e^{4}}+\frac {\frac {\left (-\frac {5}{8} a \,e^{3}+\frac {9}{8} d \,e^{2} b -\frac {13}{8} c \,d^{2} e \right ) x^{3}-\frac {d \left (3 a \,e^{2}-7 b d e +11 c \,d^{2}\right ) x}{8}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (3 a \,e^{2}-15 b d e +35 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 \sqrt {e d}}}{e^{4}}\) | \(120\) |
risch | \(\frac {c \,x^{3}}{3 e^{3}}+\frac {b x}{e^{3}}-\frac {3 c d x}{e^{4}}+\frac {\left (-\frac {5}{8} a \,e^{3}+\frac {9}{8} d \,e^{2} b -\frac {13}{8} c \,d^{2} e \right ) x^{3}-\frac {d \left (3 a \,e^{2}-7 b d e +11 c \,d^{2}\right ) x}{8}}{e^{4} \left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a}{16 e^{2} \sqrt {-e d}}+\frac {15 \ln \left (e x +\sqrt {-e d}\right ) b d}{16 e^{3} \sqrt {-e d}}-\frac {35 \ln \left (e x +\sqrt {-e d}\right ) c \,d^{2}}{16 e^{4} \sqrt {-e d}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a}{16 e^{2} \sqrt {-e d}}-\frac {15 \ln \left (-e x +\sqrt {-e d}\right ) b d}{16 e^{3} \sqrt {-e d}}+\frac {35 \ln \left (-e x +\sqrt {-e d}\right ) c \,d^{2}}{16 e^{4} \sqrt {-e d}}\) | \(235\) |
1/e^4*(1/3*c*x^3*e+b*e*x-3*c*d*x)+1/e^4*(((-5/8*a*e^3+9/8*d*e^2*b-13/8*c*d ^2*e)*x^3-1/8*d*(3*a*e^2-7*b*d*e+11*c*d^2)*x)/(e*x^2+d)^2+1/8*(3*a*e^2-15* b*d*e+35*c*d^2)/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 462, normalized size of antiderivative = 3.23 \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\left [\frac {16 \, c d e^{4} x^{7} - 16 \, {\left (7 \, c d^{2} e^{3} - 3 \, b d e^{4}\right )} x^{5} - 10 \, {\left (35 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 3 \, a d e^{4}\right )} x^{3} - 3 \, {\left (35 \, c d^{4} - 15 \, b d^{3} e + 3 \, a d^{2} e^{2} + {\left (35 \, c d^{2} e^{2} - 15 \, b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (35 \, c d^{3} e - 15 \, b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 6 \, {\left (35 \, c d^{4} e - 15 \, b d^{3} e^{2} + 3 \, a d^{2} e^{3}\right )} x}{48 \, {\left (d e^{7} x^{4} + 2 \, d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}, \frac {8 \, c d e^{4} x^{7} - 8 \, {\left (7 \, c d^{2} e^{3} - 3 \, b d e^{4}\right )} x^{5} - 5 \, {\left (35 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 3 \, a d e^{4}\right )} x^{3} + 3 \, {\left (35 \, c d^{4} - 15 \, b d^{3} e + 3 \, a d^{2} e^{2} + {\left (35 \, c d^{2} e^{2} - 15 \, b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (35 \, c d^{3} e - 15 \, b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (35 \, c d^{4} e - 15 \, b d^{3} e^{2} + 3 \, a d^{2} e^{3}\right )} x}{24 \, {\left (d e^{7} x^{4} + 2 \, d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}\right ] \]
[1/48*(16*c*d*e^4*x^7 - 16*(7*c*d^2*e^3 - 3*b*d*e^4)*x^5 - 10*(35*c*d^3*e^ 2 - 15*b*d^2*e^3 + 3*a*d*e^4)*x^3 - 3*(35*c*d^4 - 15*b*d^3*e + 3*a*d^2*e^2 + (35*c*d^2*e^2 - 15*b*d*e^3 + 3*a*e^4)*x^4 + 2*(35*c*d^3*e - 15*b*d^2*e^ 2 + 3*a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d )) - 6*(35*c*d^4*e - 15*b*d^3*e^2 + 3*a*d^2*e^3)*x)/(d*e^7*x^4 + 2*d^2*e^6 *x^2 + d^3*e^5), 1/24*(8*c*d*e^4*x^7 - 8*(7*c*d^2*e^3 - 3*b*d*e^4)*x^5 - 5 *(35*c*d^3*e^2 - 15*b*d^2*e^3 + 3*a*d*e^4)*x^3 + 3*(35*c*d^4 - 15*b*d^3*e + 3*a*d^2*e^2 + (35*c*d^2*e^2 - 15*b*d*e^3 + 3*a*e^4)*x^4 + 2*(35*c*d^3*e - 15*b*d^2*e^2 + 3*a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(35*c *d^4*e - 15*b*d^3*e^2 + 3*a*d^2*e^3)*x)/(d*e^7*x^4 + 2*d^2*e^6*x^2 + d^3*e ^5)]
Time = 1.72 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.48 \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c x^{3}}{3 e^{3}} + x \left (\frac {b}{e^{3}} - \frac {3 c d}{e^{4}}\right ) - \frac {\sqrt {- \frac {1}{d e^{9}}} \cdot \left (3 a e^{2} - 15 b d e + 35 c d^{2}\right ) \log {\left (- d e^{4} \sqrt {- \frac {1}{d e^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d e^{9}}} \cdot \left (3 a e^{2} - 15 b d e + 35 c d^{2}\right ) \log {\left (d e^{4} \sqrt {- \frac {1}{d e^{9}}} + x \right )}}{16} + \frac {x^{3} \left (- 5 a e^{3} + 9 b d e^{2} - 13 c d^{2} e\right ) + x \left (- 3 a d e^{2} + 7 b d^{2} e - 11 c d^{3}\right )}{8 d^{2} e^{4} + 16 d e^{5} x^{2} + 8 e^{6} x^{4}} \]
c*x**3/(3*e**3) + x*(b/e**3 - 3*c*d/e**4) - sqrt(-1/(d*e**9))*(3*a*e**2 - 15*b*d*e + 35*c*d**2)*log(-d*e**4*sqrt(-1/(d*e**9)) + x)/16 + sqrt(-1/(d*e **9))*(3*a*e**2 - 15*b*d*e + 35*c*d**2)*log(d*e**4*sqrt(-1/(d*e**9)) + x)/ 16 + (x**3*(-5*a*e**3 + 9*b*d*e**2 - 13*c*d**2*e) + x*(-3*a*d*e**2 + 7*b*d **2*e - 11*c*d**3))/(8*d**2*e**4 + 16*d*e**5*x**2 + 8*e**6*x**4)
Exception generated. \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {{\left (35 \, c d^{2} - 15 \, b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} e^{4}} - \frac {13 \, c d^{2} e x^{3} - 9 \, b d e^{2} x^{3} + 5 \, a e^{3} x^{3} + 11 \, c d^{3} x - 7 \, b d^{2} e x + 3 \, a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} e^{4}} + \frac {c e^{6} x^{3} - 9 \, c d e^{5} x + 3 \, b e^{6} x}{3 \, e^{9}} \]
1/8*(35*c*d^2 - 15*b*d*e + 3*a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*e^4) - 1/8*(13*c*d^2*e*x^3 - 9*b*d*e^2*x^3 + 5*a*e^3*x^3 + 11*c*d^3*x - 7*b*d^2 *e*x + 3*a*d*e^2*x)/((e*x^2 + d)^2*e^4) + 1/3*(c*e^6*x^3 - 9*c*d*e^5*x + 3 *b*e^6*x)/e^9
Time = 7.71 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96 \[ \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=x\,\left (\frac {b}{e^3}-\frac {3\,c\,d}{e^4}\right )-\frac {\left (\frac {13\,c\,d^2\,e}{8}-\frac {9\,b\,d\,e^2}{8}+\frac {5\,a\,e^3}{8}\right )\,x^3+\left (\frac {11\,c\,d^3}{8}-\frac {7\,b\,d^2\,e}{8}+\frac {3\,a\,d\,e^2}{8}\right )\,x}{d^2\,e^4+2\,d\,e^5\,x^2+e^6\,x^4}+\frac {c\,x^3}{3\,e^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (35\,c\,d^2-15\,b\,d\,e+3\,a\,e^2\right )}{8\,\sqrt {d}\,e^{9/2}} \]