Integrand size = 25, antiderivative size = 124 \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c x}{e^3}-\frac {\left (c d^2-b d e+a e^2\right ) x}{4 e^3 \left (d+e x^2\right )^2}+\frac {\left (9 c d^2-e (5 b d-a e)\right ) x}{8 d e^3 \left (d+e x^2\right )}-\frac {\left (15 c d^2-e (3 b d+a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{3/2} e^{7/2}} \]
c*x/e^3-1/4*(a*e^2-b*d*e+c*d^2)*x/e^3/(e*x^2+d)^2+1/8*(9*c*d^2-e*(-a*e+5*b *d))*x/d/e^3/(e*x^2+d)-1/8*(15*c*d^2-e*(a*e+3*b*d))*arctan(x*e^(1/2)/d^(1/ 2))/d^(3/2)/e^(7/2)
Time = 0.07 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98 \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c x}{e^3}-\frac {\left (c d^2-b d e+a e^2\right ) x}{4 e^3 \left (d+e x^2\right )^2}+\frac {\left (9 c d^2-5 b d e+a e^2\right ) x}{8 d e^3 \left (d+e x^2\right )}-\frac {\left (15 c d^2-3 b d e-a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{3/2} e^{7/2}} \]
(c*x)/e^3 - ((c*d^2 - b*d*e + a*e^2)*x)/(4*e^3*(d + e*x^2)^2) + ((9*c*d^2 - 5*b*d*e + a*e^2)*x)/(8*d*e^3*(d + e*x^2)) - ((15*c*d^2 - 3*b*d*e - a*e^2 )*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(3/2)*e^(7/2))
Time = 0.32 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1580, 25, 1471, 299, 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.
\(\displaystyle \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1580 |
\(\displaystyle -\frac {\int -\frac {4 c e^2 x^4-4 e (c d-b e) x^2+c d^2+a e^2-b d e}{\left (e x^2+d\right )^2}dx}{4 e^3}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {4 c e^2 x^4-4 e (c d-b e) x^2+c d^2+a e^2-b d e}{\left (e x^2+d\right )^2}dx}{4 e^3}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {\frac {x \left (9 c d^2-e (5 b d-a e)\right )}{2 d \left (d+e x^2\right )}-\frac {\int \frac {7 c d^2-8 c e x^2 d-e (3 b d+a e)}{e x^2+d}dx}{2 d}}{4 e^3}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {x \left (9 c d^2-e (5 b d-a e)\right )}{2 d \left (d+e x^2\right )}-\frac {\left (15 c d^2-e (a e+3 b d)\right ) \int \frac {1}{e x^2+d}dx-8 c d x}{2 d}}{4 e^3}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {x \left (9 c d^2-e (5 b d-a e)\right )}{2 d \left (d+e x^2\right )}-\frac {\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (15 c d^2-e (a e+3 b d)\right )}{\sqrt {d} \sqrt {e}}-8 c d x}{2 d}}{4 e^3}-\frac {x \left (a e^2-b d e+c d^2\right )}{4 e^3 \left (d+e x^2\right )^2}\) |
-1/4*((c*d^2 - b*d*e + a*e^2)*x)/(e^3*(d + e*x^2)^2) + (((9*c*d^2 - e*(5*b *d - a*e))*x)/(2*d*(d + e*x^2)) - (-8*c*d*x + ((15*c*d^2 - e*(3*b*d + a*e) )*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*Sqrt[e]))/(2*d))/(4*e^3)
3.3.90.3.1 Defintions of rubi rules used
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[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, d + e*x^2 , x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x , 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], 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] && LtQ[q, -1]
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]
Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.85
method | result | size |
default | \(\frac {c x}{e^{3}}+\frac {\frac {\frac {e \left (a \,e^{2}-5 b d e +9 c \,d^{2}\right ) x^{3}}{8 d}+\left (-\frac {1}{8} a \,e^{2}-\frac {3}{8} b d e +\frac {7}{8} c \,d^{2}\right ) x}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (a \,e^{2}+3 b d e -15 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 d \sqrt {e d}}}{e^{3}}\) | \(106\) |
risch | \(\frac {c x}{e^{3}}+\frac {\frac {e \left (a \,e^{2}-5 b d e +9 c \,d^{2}\right ) x^{3}}{8 d}+\left (-\frac {1}{8} a \,e^{2}-\frac {3}{8} b d e +\frac {7}{8} c \,d^{2}\right ) x}{e^{3} \left (e \,x^{2}+d \right )^{2}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) a}{16 e \sqrt {-e d}\, d}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) b}{16 e^{2} \sqrt {-e d}}+\frac {15 d \ln \left (e x +\sqrt {-e d}\right ) c}{16 e^{3} \sqrt {-e d}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) a}{16 e \sqrt {-e d}\, d}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) b}{16 e^{2} \sqrt {-e d}}-\frac {15 d \ln \left (-e x +\sqrt {-e d}\right ) c}{16 e^{3} \sqrt {-e d}}\) | \(217\) |
c*x/e^3+1/e^3*((1/8*e*(a*e^2-5*b*d*e+9*c*d^2)/d*x^3+(-1/8*a*e^2-3/8*b*d*e+ 7/8*c*d^2)*x)/(e*x^2+d)^2+1/8*(a*e^2+3*b*d*e-15*c*d^2)/d/(e*d)^(1/2)*arcta n(e*x/(e*d)^(1/2)))
Time = 0.28 (sec) , antiderivative size = 421, normalized size of antiderivative = 3.40 \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\left [\frac {16 \, c d^{2} e^{3} x^{5} + 2 \, {\left (25 \, c d^{3} e^{2} - 5 \, b d^{2} e^{3} + a d e^{4}\right )} x^{3} + {\left (15 \, c d^{4} - 3 \, b d^{3} e - a d^{2} e^{2} + {\left (15 \, c d^{2} e^{2} - 3 \, b d e^{3} - a e^{4}\right )} x^{4} + 2 \, {\left (15 \, c d^{3} e - 3 \, b d^{2} e^{2} - a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (15 \, c d^{4} e - 3 \, b d^{3} e^{2} - a d^{2} e^{3}\right )} x}{16 \, {\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}}, \frac {8 \, c d^{2} e^{3} x^{5} + {\left (25 \, c d^{3} e^{2} - 5 \, b d^{2} e^{3} + a d e^{4}\right )} x^{3} - {\left (15 \, c d^{4} - 3 \, b d^{3} e - a d^{2} e^{2} + {\left (15 \, c d^{2} e^{2} - 3 \, b d e^{3} - a e^{4}\right )} x^{4} + 2 \, {\left (15 \, c d^{3} e - 3 \, b d^{2} e^{2} - a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (15 \, c d^{4} e - 3 \, b d^{3} e^{2} - a d^{2} e^{3}\right )} x}{8 \, {\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}}\right ] \]
[1/16*(16*c*d^2*e^3*x^5 + 2*(25*c*d^3*e^2 - 5*b*d^2*e^3 + a*d*e^4)*x^3 + ( 15*c*d^4 - 3*b*d^3*e - a*d^2*e^2 + (15*c*d^2*e^2 - 3*b*d*e^3 - a*e^4)*x^4 + 2*(15*c*d^3*e - 3*b*d^2*e^2 - a*d*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sq rt(-d*e)*x - d)/(e*x^2 + d)) + 2*(15*c*d^4*e - 3*b*d^3*e^2 - a*d^2*e^3)*x) /(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^4*e^4), 1/8*(8*c*d^2*e^3*x^5 + (25*c*d^3 *e^2 - 5*b*d^2*e^3 + a*d*e^4)*x^3 - (15*c*d^4 - 3*b*d^3*e - a*d^2*e^2 + (1 5*c*d^2*e^2 - 3*b*d*e^3 - a*e^4)*x^4 + 2*(15*c*d^3*e - 3*b*d^2*e^2 - a*d*e ^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + (15*c*d^4*e - 3*b*d^3*e^2 - a*d ^2*e^3)*x)/(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^4*e^4)]
Time = 1.29 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.62 \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c x}{e^{3}} - \frac {\sqrt {- \frac {1}{d^{3} e^{7}}} \left (a e^{2} + 3 b d e - 15 c d^{2}\right ) \log {\left (- d^{2} e^{3} \sqrt {- \frac {1}{d^{3} e^{7}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{3} e^{7}}} \left (a e^{2} + 3 b d e - 15 c d^{2}\right ) \log {\left (d^{2} e^{3} \sqrt {- \frac {1}{d^{3} e^{7}}} + x \right )}}{16} + \frac {x^{3} \left (a e^{3} - 5 b d e^{2} + 9 c d^{2} e\right ) + x \left (- a d e^{2} - 3 b d^{2} e + 7 c d^{3}\right )}{8 d^{3} e^{3} + 16 d^{2} e^{4} x^{2} + 8 d e^{5} x^{4}} \]
c*x/e**3 - sqrt(-1/(d**3*e**7))*(a*e**2 + 3*b*d*e - 15*c*d**2)*log(-d**2*e **3*sqrt(-1/(d**3*e**7)) + x)/16 + sqrt(-1/(d**3*e**7))*(a*e**2 + 3*b*d*e - 15*c*d**2)*log(d**2*e**3*sqrt(-1/(d**3*e**7)) + x)/16 + (x**3*(a*e**3 - 5*b*d*e**2 + 9*c*d**2*e) + x*(-a*d*e**2 - 3*b*d**2*e + 7*c*d**3))/(8*d**3* e**3 + 16*d**2*e**4*x**2 + 8*d*e**5*x**4)
Exception generated. \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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 = 116, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c x}{e^{3}} - \frac {{\left (15 \, c d^{2} - 3 \, b d e - a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d e^{3}} + \frac {9 \, c d^{2} e x^{3} - 5 \, b d e^{2} x^{3} + a e^{3} x^{3} + 7 \, c d^{3} x - 3 \, b d^{2} e x - a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} d e^{3}} \]
c*x/e^3 - 1/8*(15*c*d^2 - 3*b*d*e - a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e )*d*e^3) + 1/8*(9*c*d^2*e*x^3 - 5*b*d*e^2*x^3 + a*e^3*x^3 + 7*c*d^3*x - 3* b*d^2*e*x - a*d*e^2*x)/((e*x^2 + d)^2*d*e^3)
Time = 7.76 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.95 \[ \int \frac {x^2 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^3} \, dx=\frac {c\,x}{e^3}-\frac {x\,\left (-\frac {7\,c\,d^2}{8}+\frac {3\,b\,d\,e}{8}+\frac {a\,e^2}{8}\right )-\frac {x^3\,\left (9\,c\,d^2\,e-5\,b\,d\,e^2+a\,e^3\right )}{8\,d}}{d^2\,e^3+2\,d\,e^4\,x^2+e^5\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (-15\,c\,d^2+3\,b\,d\,e+a\,e^2\right )}{8\,d^{3/2}\,e^{7/2}} \]