Integrand size = 22, antiderivative size = 115 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}-\frac {\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]
1/4*(a+d*(-b*e+c*d)/e^2)*x/d/(e*x^2+d)^2-1/8*(5*c*d^2-e*(3*a*e+b*d))*x/d^2 /e^2/(e*x^2+d)+1/8*(3*c*d^2+e*(3*a*e+b*d))*arctan(x*e^(1/2)/d^(1/2))/d^(5/ 2)/e^(5/2)
Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {x \left (-c d^2 \left (3 d+5 e x^2\right )+e \left (b d \left (-d+e x^2\right )+a e \left (5 d+3 e x^2\right )\right )\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]
(x*(-(c*d^2*(3*d + 5*e*x^2)) + e*(b*d*(-d + e*x^2) + a*e*(5*d + 3*e*x^2))) )/(8*d^2*e^2*(d + e*x^2)^2) + ((3*c*d^2 + e*(b*d + 3*a*e))*ArcTan[(Sqrt[e] *x)/Sqrt[d]])/(8*d^(5/2)*e^(5/2))
Time = 0.26 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1471, 25, 27, 298, 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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2}-\frac {\int -\frac {4 c d x^2+e \left (3 a-\frac {d (c d-b e)}{e^2}\right )}{e \left (e x^2+d\right )^2}dx}{4 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {-\frac {c d^2}{e}+4 c x^2 d+b d+3 a e}{e \left (e x^2+d\right )^2}dx}{4 d}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-\frac {c d^2}{e}+4 c x^2 d+b d+3 a e}{\left (e x^2+d\right )^2}dx}{4 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {3 a e}{d}+b+\frac {3 c d}{e}\right ) \int \frac {1}{e x^2+d}dx+\frac {x \left (3 a e+b d-\frac {5 c d^2}{e}\right )}{2 d \left (d+e x^2\right )}}{4 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (\frac {3 a e}{d}+b+\frac {3 c d}{e}\right )}{2 \sqrt {d} \sqrt {e}}+\frac {x \left (3 a e+b d-\frac {5 c d^2}{e}\right )}{2 d \left (d+e x^2\right )}}{4 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2}\) |
((c*d^2 - b*d*e + a*e^2)*x)/(4*d*e^2*(d + e*x^2)^2) + (((b*d - (5*c*d^2)/e + 3*a*e)*x)/(2*d*(d + e*x^2)) + ((b + (3*c*d)/e + (3*a*e)/d)*ArcTan[(Sqrt [e]*x)/Sqrt[d]])/(2*Sqrt[d]*Sqrt[e]))/(4*d*e)
3.3.50.3.1 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[((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])
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]
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\frac {\left (3 a \,e^{2}+b d e -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-b d e -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (3 a \,e^{2}+b d e +3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 e^{2} d^{2} \sqrt {e d}}\) | \(107\) |
risch | \(\frac {\frac {\left (3 a \,e^{2}+b d e -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-b d e -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a}{16 \sqrt {-e d}\, d^{2}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) b}{16 \sqrt {-e d}\, e d}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) c}{16 \sqrt {-e d}\, e^{2}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a}{16 \sqrt {-e d}\, d^{2}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) b}{16 \sqrt {-e d}\, e d}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) c}{16 \sqrt {-e d}\, e^{2}}\) | \(215\) |
(1/8*(3*a*e^2+b*d*e-5*c*d^2)/d^2/e*x^3+1/8*(5*a*e^2-b*d*e-3*c*d^2)/d/e^2*x )/(e*x^2+d)^2+1/8*(3*a*e^2+b*d*e+3*c*d^2)/e^2/d^2/(e*d)^(1/2)*arctan(e*x/( e*d)^(1/2))
Time = 0.27 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.40 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} + {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{16 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}, -\frac {{\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} - {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{8 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}\right ] \]
[-1/16*(2*(5*c*d^3*e^2 - b*d^2*e^3 - 3*a*d*e^4)*x^3 + (3*c*d^4 + b*d^3*e + 3*a*d^2*e^2 + (3*c*d^2*e^2 + b*d*e^3 + 3*a*e^4)*x^4 + 2*(3*c*d^3*e + 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)) + 2*(3*c*d^4*e + b*d^3*e^2 - 5*a*d^2*e^3)*x)/(d^3*e^5*x^4 + 2*d^4*e ^4*x^2 + d^5*e^3), -1/8*((5*c*d^3*e^2 - b*d^2*e^3 - 3*a*d*e^4)*x^3 - (3*c* d^4 + b*d^3*e + 3*a*d^2*e^2 + (3*c*d^2*e^2 + b*d*e^3 + 3*a*e^4)*x^4 + 2*(3 *c*d^3*e + b*d^2*e^2 + 3*a*d*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + ( 3*c*d^4*e + b*d^3*e^2 - 5*a*d^2*e^3)*x)/(d^3*e^5*x^4 + 2*d^4*e^4*x^2 + d^5 *e^3)]
Time = 0.66 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.70 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \cdot \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (- d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \cdot \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a e^{3} + b d e^{2} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - b d^{2} e - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \]
-sqrt(-1/(d**5*e**5))*(3*a*e**2 + b*d*e + 3*c*d**2)*log(-d**3*e**2*sqrt(-1 /(d**5*e**5)) + x)/16 + sqrt(-1/(d**5*e**5))*(3*a*e**2 + b*d*e + 3*c*d**2) *log(d**3*e**2*sqrt(-1/(d**5*e**5)) + x)/16 + (x**3*(3*a*e**3 + b*d*e**2 - 5*c*d**2*e) + x*(5*a*d*e**2 - b*d**2*e - 3*c*d**3))/(8*d**4*e**2 + 16*d** 3*e**3*x**2 + 8*d**2*e**4*x**4)
Exception generated. \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {{\left (3 \, c d^{2} + b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} e^{2}} - \frac {5 \, c d^{2} e x^{3} - b d e^{2} x^{3} - 3 \, a e^{3} x^{3} + 3 \, c d^{3} x + b d^{2} e x - 5 \, a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} d^{2} e^{2}} \]
1/8*(3*c*d^2 + b*d*e + 3*a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^2*e^2) - 1/8*(5*c*d^2*e*x^3 - b*d*e^2*x^3 - 3*a*e^3*x^3 + 3*c*d^3*x + b*d^2*e*x - 5*a*d*e^2*x)/((e*x^2 + d)^2*d^2*e^2)
Time = 7.64 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^{5/2}\,e^{5/2}}-\frac {\frac {x\,\left (3\,c\,d^2+b\,d\,e-5\,a\,e^2\right )}{8\,d\,e^2}-\frac {x^3\,\left (-5\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^2\,e}}{d^2+2\,d\,e\,x^2+e^2\,x^4} \]