Integrand size = 22, antiderivative size = 150 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}-\frac {\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (b d+5 a e)\right ) x}{16 d^3 e^2 \left (d+e x^2\right )}+\frac {\left (c d^2+e (b d+5 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \]
1/6*(a+d*(-b*e+c*d)/e^2)*x/d/(e*x^2+d)^3-1/24*(7*c*d^2-e*(5*a*e+b*d))*x/d^ 2/e^2/(e*x^2+d)^2+1/16*(c*d^2+e*(5*a*e+b*d))*x/d^3/e^2/(e*x^2+d)+1/16*(c*d ^2+e*(5*a*e+b*d))*arctan(x*e^(1/2)/d^(1/2))/d^(7/2)/e^(5/2)
Time = 0.09 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {x \left (c d^2 \left (-3 d^2-8 d e x^2+3 e^2 x^4\right )+e \left (b d \left (-3 d^2+8 d e x^2+3 e^2 x^4\right )+a e \left (33 d^2+40 d e x^2+15 e^2 x^4\right )\right )\right )}{48 d^3 e^2 \left (d+e x^2\right )^3}+\frac {\left (c d^2+e (b d+5 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \]
(x*(c*d^2*(-3*d^2 - 8*d*e*x^2 + 3*e^2*x^4) + e*(b*d*(-3*d^2 + 8*d*e*x^2 + 3*e^2*x^4) + a*e*(33*d^2 + 40*d*e*x^2 + 15*e^2*x^4))))/(48*d^3*e^2*(d + e* x^2)^3) + ((c*d^2 + e*(b*d + 5*a*e))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(16*d^(7 /2)*e^(5/2))
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1471, 25, 27, 298, 215, 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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx\) |
\(\Big \downarrow \) 1471 |
\(\displaystyle \frac {x \left (a e^2-b d e+c d^2\right )}{6 d e^2 \left (d+e x^2\right )^3}-\frac {\int -\frac {6 c d x^2+e \left (5 a-\frac {d (c d-b e)}{e^2}\right )}{e \left (e x^2+d\right )^3}dx}{6 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {-\frac {c d^2}{e}+6 c x^2 d+b d+5 a e}{e \left (e x^2+d\right )^3}dx}{6 d}+\frac {x \left (a e^2-b d e+c d^2\right )}{6 d e^2 \left (d+e x^2\right )^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-\frac {c d^2}{e}+6 c x^2 d+b d+5 a e}{\left (e x^2+d\right )^3}dx}{6 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{6 d e^2 \left (d+e x^2\right )^3}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {5 a e}{d}+b+\frac {c d}{e}\right ) \int \frac {1}{\left (e x^2+d\right )^2}dx+\frac {x \left (5 a e+b d-\frac {7 c d^2}{e}\right )}{4 d \left (d+e x^2\right )^2}}{6 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{6 d e^2 \left (d+e x^2\right )^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {5 a e}{d}+b+\frac {c d}{e}\right ) \left (\frac {\int \frac {1}{e x^2+d}dx}{2 d}+\frac {x}{2 d \left (d+e x^2\right )}\right )+\frac {x \left (5 a e+b d-\frac {7 c d^2}{e}\right )}{4 d \left (d+e x^2\right )^2}}{6 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{6 d e^2 \left (d+e x^2\right )^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} \sqrt {e}}+\frac {x}{2 d \left (d+e x^2\right )}\right ) \left (\frac {5 a e}{d}+b+\frac {c d}{e}\right )+\frac {x \left (5 a e+b d-\frac {7 c d^2}{e}\right )}{4 d \left (d+e x^2\right )^2}}{6 d e}+\frac {x \left (a e^2-b d e+c d^2\right )}{6 d e^2 \left (d+e x^2\right )^3}\) |
((c*d^2 - b*d*e + a*e^2)*x)/(6*d*e^2*(d + e*x^2)^3) + (((b*d - (7*c*d^2)/e + 5*a*e)*x)/(4*d*(d + e*x^2)^2) + (3*(b + (c*d)/e + (5*a*e)/d)*(x/(2*d*(d + e*x^2)) + ArcTan[(Sqrt[e]*x)/Sqrt[d]]/(2*d^(3/2)*Sqrt[e])))/4)/(6*d*e)
3.3.51.3.1 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)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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 = 0.57 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.87
method | result | size |
default | \(\frac {\frac {\left (5 a \,e^{2}+b d e +c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}+b d e -c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-b d e -c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}+\frac {\left (5 a \,e^{2}+b d e +c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{16 d^{3} e^{2} \sqrt {e d}}\) | \(130\) |
risch | \(\frac {\frac {\left (5 a \,e^{2}+b d e +c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}+b d e -c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-b d e -c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) a}{32 \sqrt {-e d}\, d^{3}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) b}{32 \sqrt {-e d}\, e \,d^{2}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) c}{32 \sqrt {-e d}\, e^{2} d}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) a}{32 \sqrt {-e d}\, d^{3}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) b}{32 \sqrt {-e d}\, e \,d^{2}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) c}{32 \sqrt {-e d}\, e^{2} d}\) | \(245\) |
(1/16*(5*a*e^2+b*d*e+c*d^2)/d^3*x^5+1/6*(5*a*e^2+b*d*e-c*d^2)/d^2/e*x^3+1/ 16*(11*a*e^2-b*d*e-c*d^2)/d/e^2*x)/(e*x^2+d)^3+1/16*(5*a*e^2+b*d*e+c*d^2)/ d^3/e^2/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2))
Time = 0.25 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.53 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=\left [\frac {6 \, {\left (c d^{3} e^{3} + b d^{2} e^{4} + 5 \, a d e^{5}\right )} x^{5} - 16 \, {\left (c d^{4} e^{2} - b d^{3} e^{3} - 5 \, a d^{2} e^{4}\right )} x^{3} - 3 \, {\left ({\left (c d^{2} e^{3} + b d e^{4} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + b d^{4} e + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + b d^{2} e^{3} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + b d^{3} e^{2} + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 6 \, {\left (c d^{5} e + b d^{4} e^{2} - 11 \, a d^{3} e^{3}\right )} x}{96 \, {\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}, \frac {3 \, {\left (c d^{3} e^{3} + b d^{2} e^{4} + 5 \, a d e^{5}\right )} x^{5} - 8 \, {\left (c d^{4} e^{2} - b d^{3} e^{3} - 5 \, a d^{2} e^{4}\right )} x^{3} + 3 \, {\left ({\left (c d^{2} e^{3} + b d e^{4} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + b d^{4} e + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + b d^{2} e^{3} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + b d^{3} e^{2} + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (c d^{5} e + b d^{4} e^{2} - 11 \, a d^{3} e^{3}\right )} x}{48 \, {\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}\right ] \]
[1/96*(6*(c*d^3*e^3 + b*d^2*e^4 + 5*a*d*e^5)*x^5 - 16*(c*d^4*e^2 - b*d^3*e ^3 - 5*a*d^2*e^4)*x^3 - 3*((c*d^2*e^3 + b*d*e^4 + 5*a*e^5)*x^6 + c*d^5 + b *d^4*e + 5*a*d^3*e^2 + 3*(c*d^3*e^2 + b*d^2*e^3 + 5*a*d*e^4)*x^4 + 3*(c*d^ 4*e + b*d^3*e^2 + 5*a*d^2*e^3)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 6*(c*d^5*e + b*d^4*e^2 - 11*a*d^3*e^3)*x)/(d^4*e^6*x^ 6 + 3*d^5*e^5*x^4 + 3*d^6*e^4*x^2 + d^7*e^3), 1/48*(3*(c*d^3*e^3 + b*d^2*e ^4 + 5*a*d*e^5)*x^5 - 8*(c*d^4*e^2 - b*d^3*e^3 - 5*a*d^2*e^4)*x^3 + 3*((c* d^2*e^3 + b*d*e^4 + 5*a*e^5)*x^6 + c*d^5 + b*d^4*e + 5*a*d^3*e^2 + 3*(c*d^ 3*e^2 + b*d^2*e^3 + 5*a*d*e^4)*x^4 + 3*(c*d^4*e + b*d^3*e^2 + 5*a*d^2*e^3) *x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) - 3*(c*d^5*e + b*d^4*e^2 - 11*a*d^3* e^3)*x)/(d^4*e^6*x^6 + 3*d^5*e^5*x^4 + 3*d^6*e^4*x^2 + d^7*e^3)]
Time = 1.18 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.61 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + b d e + c d^{2}\right ) \log {\left (- d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + b d e + c d^{2}\right ) \log {\left (d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a e^{4} + 3 b d e^{3} + 3 c d^{2} e^{2}\right ) + x^{3} \cdot \left (40 a d e^{3} + 8 b d^{2} e^{2} - 8 c d^{3} e\right ) + x \left (33 a d^{2} e^{2} - 3 b d^{3} e - 3 c d^{4}\right )}{48 d^{6} e^{2} + 144 d^{5} e^{3} x^{2} + 144 d^{4} e^{4} x^{4} + 48 d^{3} e^{5} x^{6}} \]
-sqrt(-1/(d**7*e**5))*(5*a*e**2 + b*d*e + c*d**2)*log(-d**4*e**2*sqrt(-1/( d**7*e**5)) + x)/32 + sqrt(-1/(d**7*e**5))*(5*a*e**2 + b*d*e + c*d**2)*log (d**4*e**2*sqrt(-1/(d**7*e**5)) + x)/32 + (x**5*(15*a*e**4 + 3*b*d*e**3 + 3*c*d**2*e**2) + x**3*(40*a*d*e**3 + 8*b*d**2*e**2 - 8*c*d**3*e) + x*(33*a *d**2*e**2 - 3*b*d**3*e - 3*c*d**4))/(48*d**6*e**2 + 144*d**5*e**3*x**2 + 144*d**4*e**4*x**4 + 48*d**3*e**5*x**6)
Exception generated. \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=\text {Exception raised: ValueError} \]
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 = 145, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {{\left (c d^{2} + b d e + 5 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{2}} + \frac {3 \, c d^{2} e^{2} x^{5} + 3 \, b d e^{3} x^{5} + 15 \, a e^{4} x^{5} - 8 \, c d^{3} e x^{3} + 8 \, b d^{2} e^{2} x^{3} + 40 \, a d e^{3} x^{3} - 3 \, c d^{4} x - 3 \, b d^{3} e x + 33 \, a d^{2} e^{2} x}{48 \, {\left (e x^{2} + d\right )}^{3} d^{3} e^{2}} \]
1/16*(c*d^2 + b*d*e + 5*a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^3*e^2) + 1/48*(3*c*d^2*e^2*x^5 + 3*b*d*e^3*x^5 + 15*a*e^4*x^5 - 8*c*d^3*e*x^3 + 8* b*d^2*e^2*x^3 + 40*a*d*e^3*x^3 - 3*c*d^4*x - 3*b*d^3*e*x + 33*a*d^2*e^2*x) /((e*x^2 + d)^3*d^3*e^2)
Time = 7.58 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.96 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\frac {x^5\,\left (c\,d^2+b\,d\,e+5\,a\,e^2\right )}{16\,d^3}-\frac {x\,\left (c\,d^2+b\,d\,e-11\,a\,e^2\right )}{16\,d\,e^2}+\frac {x^3\,\left (-c\,d^2+b\,d\,e+5\,a\,e^2\right )}{6\,d^2\,e}}{d^3+3\,d^2\,e\,x^2+3\,d\,e^2\,x^4+e^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+b\,d\,e+5\,a\,e^2\right )}{16\,d^{7/2}\,e^{5/2}} \]