Integrand size = 25, antiderivative size = 142 \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=-\frac {a}{3 d^3 x^3}-\frac {b d-3 a e}{d^4 x}+\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-e (7 b d-11 a e)\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\left (3 c d^2-15 b d e+35 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{9/2} \sqrt {e}} \]
-1/3*a/d^3/x^3+(3*a*e-b*d)/d^4/x+1/4*(a*e^2-b*d*e+c*d^2)*x/d^3/(e*x^2+d)^2 +1/8*(3*c*d^2-e*(-11*a*e+7*b*d))*x/d^4/(e*x^2+d)+1/8*(35*a*e^2-15*b*d*e+3* c*d^2)*arctan(x*e^(1/2)/d^(1/2))/d^(9/2)/e^(1/2)
Time = 0.06 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=-\frac {a}{3 d^3 x^3}+\frac {-b d+3 a e}{d^4 x}+\frac {\left (c d^2-b d e+a e^2\right ) x}{4 d^3 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2-7 b d e+11 a e^2\right ) x}{8 d^4 \left (d+e x^2\right )}+\frac {\left (3 c d^2-15 b d e+35 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{9/2} \sqrt {e}} \]
-1/3*a/(d^3*x^3) + (-(b*d) + 3*a*e)/(d^4*x) + ((c*d^2 - b*d*e + a*e^2)*x)/ (4*d^3*(d + e*x^2)^2) + ((3*c*d^2 - 7*b*d*e + 11*a*e^2)*x)/(8*d^4*(d + e*x ^2)) + ((3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^ (9/2)*Sqrt[e])
Time = 0.44 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.20, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1582, 1582, 1584, 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.
\(\displaystyle \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1582 |
\(\displaystyle \frac {\int \frac {3 e^2 \left (c d^2-b e d+a e^2\right ) x^4+4 d e^2 (b d-a e) x^2+4 a d^2 e^2}{x^4 \left (e x^2+d\right )^2}dx}{4 d^3 e^2}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 1582 |
\(\displaystyle \frac {\frac {\int \frac {8 a d^4 e^4+d^2 \left (3 c d^2-e (7 b d-11 a e)\right ) x^4 e^4+8 d^3 (b d-2 a e) x^2 e^4}{x^4 \left (e x^2+d\right )}dx}{2 d^3 e^2}+\frac {e^2 x \left (3 c d^2-e (7 b d-11 a e)\right )}{2 d \left (d+e x^2\right )}}{4 d^3 e^2}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 1584 |
\(\displaystyle \frac {\frac {\int \left (\frac {d^2 \left (3 c d^2-15 b e d+35 a e^2\right ) e^4}{e x^2+d}+\frac {8 d^2 (b d-3 a e) e^4}{x^2}+\frac {8 a d^3 e^4}{x^4}\right )dx}{2 d^3 e^2}+\frac {e^2 x \left (3 c d^2-e (7 b d-11 a e)\right )}{2 d \left (d+e x^2\right )}}{4 d^3 e^2}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {d^{3/2} e^{7/2} \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (35 a e^2-15 b d e+3 c d^2\right )-\frac {8 d^2 e^4 (b d-3 a e)}{x}-\frac {8 a d^3 e^4}{3 x^3}}{2 d^3 e^2}+\frac {e^2 x \left (3 c d^2-e (7 b d-11 a e)\right )}{2 d \left (d+e x^2\right )}}{4 d^3 e^2}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d^3 \left (d+e x^2\right )^2}\) |
((c*d^2 - b*d*e + a*e^2)*x)/(4*d^3*(d + e*x^2)^2) + ((e^2*(3*c*d^2 - e*(7* b*d - 11*a*e))*x)/(2*d*(d + e*x^2)) + ((-8*a*d^3*e^4)/(3*x^3) - (8*d^2*e^4 *(b*d - 3*a*e))/x + d^(3/2)*e^(7/2)*(3*c*d^2 - 15*b*d*e + 35*a*e^2)*ArcTan [(Sqrt[e]*x)/Sqrt[d]])/(2*d^3*e^2))/(4*d^3*e^2)
3.3.93.3.1 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[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.87
method | result | size |
default | \(-\frac {a}{3 d^{3} x^{3}}-\frac {-3 a e +b d}{d^{4} x}+\frac {\frac {\left (\frac {11}{8} a \,e^{3}-\frac {7}{8} d \,e^{2} b +\frac {3}{8} c \,d^{2} e \right ) x^{3}+\frac {d \left (13 a \,e^{2}-9 b d e +5 c \,d^{2}\right ) x}{8}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (35 a \,e^{2}-15 b d e +3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 \sqrt {e d}}}{d^{4}}\) | \(124\) |
risch | \(\frac {\frac {e \left (35 a \,e^{2}-15 b d e +3 c \,d^{2}\right ) x^{6}}{8 d^{4}}+\frac {5 \left (35 a \,e^{2}-15 b d e +3 c \,d^{2}\right ) x^{4}}{24 d^{3}}+\frac {\left (7 a e -3 b d \right ) x^{2}}{3 d^{2}}-\frac {a}{3 d}}{x^{3} \left (e \,x^{2}+d \right )^{2}}-\frac {35 \ln \left (-\sqrt {-e d}\, x +d \right ) a \,e^{2}}{16 \sqrt {-e d}\, d^{4}}+\frac {15 \ln \left (-\sqrt {-e d}\, x +d \right ) b e}{16 \sqrt {-e d}\, d^{3}}-\frac {3 \ln \left (-\sqrt {-e d}\, x +d \right ) c}{16 \sqrt {-e d}\, d^{2}}+\frac {35 \ln \left (-\sqrt {-e d}\, x -d \right ) a \,e^{2}}{16 \sqrt {-e d}\, d^{4}}-\frac {15 \ln \left (-\sqrt {-e d}\, x -d \right ) b e}{16 \sqrt {-e d}\, d^{3}}+\frac {3 \ln \left (-\sqrt {-e d}\, x -d \right ) c}{16 \sqrt {-e d}\, d^{2}}\) | \(250\) |
-1/3*a/d^3/x^3-(-3*a*e+b*d)/d^4/x+1/d^4*(((11/8*a*e^3-7/8*d*e^2*b+3/8*c*d^ 2*e)*x^3+1/8*d*(13*a*e^2-9*b*d*e+5*c*d^2)*x)/(e*x^2+d)^2+1/8*(35*a*e^2-15* b*d*e+3*c*d^2)/(e*d)^(1/2)*arctan(e*x/(e*d)^(1/2)))
Time = 0.27 (sec) , antiderivative size = 476, normalized size of antiderivative = 3.35 \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=\left [\frac {6 \, {\left (3 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 35 \, a d e^{4}\right )} x^{6} - 16 \, a d^{4} e + 10 \, {\left (3 \, c d^{4} e - 15 \, b d^{3} e^{2} + 35 \, a d^{2} e^{3}\right )} x^{4} - 16 \, {\left (3 \, b d^{4} e - 7 \, a d^{3} e^{2}\right )} x^{2} - 3 \, {\left ({\left (3 \, c d^{2} e^{2} - 15 \, b d e^{3} + 35 \, a e^{4}\right )} x^{7} + 2 \, {\left (3 \, c d^{3} e - 15 \, b d^{2} e^{2} + 35 \, a d e^{3}\right )} x^{5} + {\left (3 \, c d^{4} - 15 \, b d^{3} e + 35 \, a d^{2} e^{2}\right )} x^{3}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right )}{48 \, {\left (d^{5} e^{3} x^{7} + 2 \, d^{6} e^{2} x^{5} + d^{7} e x^{3}\right )}}, \frac {3 \, {\left (3 \, c d^{3} e^{2} - 15 \, b d^{2} e^{3} + 35 \, a d e^{4}\right )} x^{6} - 8 \, a d^{4} e + 5 \, {\left (3 \, c d^{4} e - 15 \, b d^{3} e^{2} + 35 \, a d^{2} e^{3}\right )} x^{4} - 8 \, {\left (3 \, b d^{4} e - 7 \, a d^{3} e^{2}\right )} x^{2} + 3 \, {\left ({\left (3 \, c d^{2} e^{2} - 15 \, b d e^{3} + 35 \, a e^{4}\right )} x^{7} + 2 \, {\left (3 \, c d^{3} e - 15 \, b d^{2} e^{2} + 35 \, a d e^{3}\right )} x^{5} + {\left (3 \, c d^{4} - 15 \, b d^{3} e + 35 \, a d^{2} e^{2}\right )} x^{3}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right )}{24 \, {\left (d^{5} e^{3} x^{7} + 2 \, d^{6} e^{2} x^{5} + d^{7} e x^{3}\right )}}\right ] \]
[1/48*(6*(3*c*d^3*e^2 - 15*b*d^2*e^3 + 35*a*d*e^4)*x^6 - 16*a*d^4*e + 10*( 3*c*d^4*e - 15*b*d^3*e^2 + 35*a*d^2*e^3)*x^4 - 16*(3*b*d^4*e - 7*a*d^3*e^2 )*x^2 - 3*((3*c*d^2*e^2 - 15*b*d*e^3 + 35*a*e^4)*x^7 + 2*(3*c*d^3*e - 15*b *d^2*e^2 + 35*a*d*e^3)*x^5 + (3*c*d^4 - 15*b*d^3*e + 35*a*d^2*e^2)*x^3)*sq rt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)))/(d^5*e^3*x^7 + 2*d ^6*e^2*x^5 + d^7*e*x^3), 1/24*(3*(3*c*d^3*e^2 - 15*b*d^2*e^3 + 35*a*d*e^4) *x^6 - 8*a*d^4*e + 5*(3*c*d^4*e - 15*b*d^3*e^2 + 35*a*d^2*e^3)*x^4 - 8*(3* b*d^4*e - 7*a*d^3*e^2)*x^2 + 3*((3*c*d^2*e^2 - 15*b*d*e^3 + 35*a*e^4)*x^7 + 2*(3*c*d^3*e - 15*b*d^2*e^2 + 35*a*d*e^3)*x^5 + (3*c*d^4 - 15*b*d^3*e + 35*a*d^2*e^2)*x^3)*sqrt(d*e)*arctan(sqrt(d*e)*x/d))/(d^5*e^3*x^7 + 2*d^6*e ^2*x^5 + d^7*e*x^3)]
Time = 1.35 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.51 \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{d^{9} e}} \left (35 a e^{2} - 15 b d e + 3 c d^{2}\right ) \log {\left (- d^{5} \sqrt {- \frac {1}{d^{9} e}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{9} e}} \left (35 a e^{2} - 15 b d e + 3 c d^{2}\right ) \log {\left (d^{5} \sqrt {- \frac {1}{d^{9} e}} + x \right )}}{16} + \frac {- 8 a d^{3} + x^{6} \cdot \left (105 a e^{3} - 45 b d e^{2} + 9 c d^{2} e\right ) + x^{4} \cdot \left (175 a d e^{2} - 75 b d^{2} e + 15 c d^{3}\right ) + x^{2} \cdot \left (56 a d^{2} e - 24 b d^{3}\right )}{24 d^{6} x^{3} + 48 d^{5} e x^{5} + 24 d^{4} e^{2} x^{7}} \]
-sqrt(-1/(d**9*e))*(35*a*e**2 - 15*b*d*e + 3*c*d**2)*log(-d**5*sqrt(-1/(d* *9*e)) + x)/16 + sqrt(-1/(d**9*e))*(35*a*e**2 - 15*b*d*e + 3*c*d**2)*log(d **5*sqrt(-1/(d**9*e)) + x)/16 + (-8*a*d**3 + x**6*(105*a*e**3 - 45*b*d*e** 2 + 9*c*d**2*e) + x**4*(175*a*d*e**2 - 75*b*d**2*e + 15*c*d**3) + x**2*(56 *a*d**2*e - 24*b*d**3))/(24*d**6*x**3 + 48*d**5*e*x**5 + 24*d**4*e**2*x**7 )
Exception generated. \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=\frac {{\left (3 \, c d^{2} - 15 \, b d e + 35 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{4}} + \frac {3 \, c d^{2} e x^{3} - 7 \, b d e^{2} x^{3} + 11 \, a e^{3} x^{3} + 5 \, c d^{3} x - 9 \, b d^{2} e x + 13 \, a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} d^{4}} - \frac {3 \, b d x^{2} - 9 \, a e x^{2} + a d}{3 \, d^{4} x^{3}} \]
1/8*(3*c*d^2 - 15*b*d*e + 35*a*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d^4) + 1/8*(3*c*d^2*e*x^3 - 7*b*d*e^2*x^3 + 11*a*e^3*x^3 + 5*c*d^3*x - 9*b*d^2* e*x + 13*a*d*e^2*x)/((e*x^2 + d)^2*d^4) - 1/3*(3*b*d*x^2 - 9*a*e*x^2 + a*d )/(d^4*x^3)
Time = 7.74 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x^2+c x^4}{x^4 \left (d+e x^2\right )^3} \, dx=\frac {\frac {x^2\,\left (7\,a\,e-3\,b\,d\right )}{3\,d^2}-\frac {a}{3\,d}+\frac {5\,x^4\,\left (3\,c\,d^2-15\,b\,d\,e+35\,a\,e^2\right )}{24\,d^3}+\frac {e\,x^6\,\left (3\,c\,d^2-15\,b\,d\,e+35\,a\,e^2\right )}{8\,d^4}}{d^2\,x^3+2\,d\,e\,x^5+e^2\,x^7}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2-15\,b\,d\,e+35\,a\,e^2\right )}{8\,d^{9/2}\,\sqrt {e}} \]