Integrand size = 17, antiderivative size = 155 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3}-\frac {(a e-5 c d x) (d+e x)^3}{24 a^2 c \left (a+c x^2\right )^2}-\frac {\left (5 c d^2+a e^2\right ) (a e-c d x) (d+e x)}{16 a^3 c^2 \left (a+c x^2\right )}+\frac {\left (c d^2+a e^2\right ) \left (5 c d^2+a e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}} \]
1/6*x*(e*x+d)^4/a/(c*x^2+a)^3-1/24*(-5*c*d*x+a*e)*(e*x+d)^3/a^2/c/(c*x^2+a )^2-1/16*(a*e^2+5*c*d^2)*(-c*d*x+a*e)*(e*x+d)/a^3/c^2/(c*x^2+a)+1/16*(a*e^ 2+c*d^2)*(a*e^2+5*c*d^2)*arctan(x*c^(1/2)/a^(1/2))/a^(7/2)/c^(5/2)
Time = 0.15 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=\frac {15 c^4 d^4 x^5-a^4 e^3 (16 d+3 e x)+2 a c^3 d^2 x^3 \left (20 d^2+9 e^2 x^2\right )-2 a^3 c e \left (16 d^3+9 d^2 e x+24 d e^2 x^2+4 e^3 x^3\right )+3 a^2 c^2 x \left (11 d^4+16 d^2 e^2 x^2+e^4 x^4\right )}{48 a^3 c^2 \left (a+c x^2\right )^3}+\frac {\left (5 c^2 d^4+6 a c d^2 e^2+a^2 e^4\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}} \]
(15*c^4*d^4*x^5 - a^4*e^3*(16*d + 3*e*x) + 2*a*c^3*d^2*x^3*(20*d^2 + 9*e^2 *x^2) - 2*a^3*c*e*(16*d^3 + 9*d^2*e*x + 24*d*e^2*x^2 + 4*e^3*x^3) + 3*a^2* c^2*x*(11*d^4 + 16*d^2*e^2*x^2 + e^4*x^4))/(48*a^3*c^2*(a + c*x^2)^3) + (( 5*c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(16*a^(7 /2)*c^(5/2))
Time = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {494, 25, 678, 487, 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.
| \(\displaystyle \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx\) |
| \(\Big \downarrow \) 494 |
| \(\displaystyle \frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3}-\frac {\int -\frac {(d+e x)^3 (5 d+e x)}{\left (c x^2+a\right )^3}dx}{6 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {(d+e x)^3 (5 d+e x)}{\left (c x^2+a\right )^3}dx}{6 a}+\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 678 |
| \(\displaystyle \frac {\frac {3 \left (a e^2+5 c d^2\right ) \int \frac {(d+e x)^2}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {(d+e x)^3 (a e-5 c d x)}{4 a c \left (a+c x^2\right )^2}}{6 a}+\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 487 |
| \(\displaystyle \frac {\frac {3 \left (a e^2+5 c d^2\right ) \left (\frac {\left (a e^2+c d^2\right ) \int \frac {1}{c x^2+a}dx}{2 a c}-\frac {(d+e x) (a e-c d x)}{2 a c \left (a+c x^2\right )}\right )}{4 a c}-\frac {(d+e x)^3 (a e-5 c d x)}{4 a c \left (a+c x^2\right )^2}}{6 a}+\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {3 \left (a e^2+5 c d^2\right ) \left (\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a e^2+c d^2\right )}{2 a^{3/2} c^{3/2}}-\frac {(d+e x) (a e-c d x)}{2 a c \left (a+c x^2\right )}\right )}{4 a c}-\frac {(d+e x)^3 (a e-5 c d x)}{4 a c \left (a+c x^2\right )^2}}{6 a}+\frac {x (d+e x)^4}{6 a \left (a+c x^2\right )^3}\) |
(x*(d + e*x)^4)/(6*a*(a + c*x^2)^3) + (-1/4*((a*e - 5*c*d*x)*(d + e*x)^3)/ (a*c*(a + c*x^2)^2) + (3*(5*c*d^2 + a*e^2)*(-1/2*((a*e - c*d*x)*(d + e*x)) /(a*c*(a + c*x^2)) + ((c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(2*a^(3 /2)*c^(3/2))))/(4*a*c))/(6*a)
3.6.19.3.1 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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (c + d*x)^(n - 1)*(a*d - b*c*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[(2*p + 3)*((b*c^2 + a*d^2)/(2*a*b*(p + 1))) Int[(c + d*x)^(n - 2)*( a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[n + 2*p + 2, 0] && LtQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-x)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1)*(c*(2*p + 3) + d*(n + 2*p + 3)*x), x], x] /; FreeQ[{a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 0] && (Lt Q[n, 1] || (ILtQ[n + 2*p + 3, 0] && NeQ[n, 2])) && IntQuadraticQ[a, 0, b, c , d, n, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c *(p + 1))), x] - Simp[m*((c*d*f + a*e*g)/(2*a*c*(p + 1))) Int[(d + e*x)^( m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[S implify[m + 2*p + 3], 0] && LtQ[p, -1]
Time = 2.45 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.28
| method | result | size |
| default | \(\frac {\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right ) x^{5}}{16 a^{3}}-\frac {\left (a^{2} e^{4}-6 a c \,d^{2} e^{2}-5 c^{2} d^{4}\right ) x^{3}}{6 a^{2} c}-\frac {d \,e^{3} x^{2}}{c}-\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}-11 c^{2} d^{4}\right ) x}{16 c^{2} a}-\frac {d e \left (e^{2} a +2 c \,d^{2}\right )}{3 c^{2}}}{\left (c \,x^{2}+a \right )^{3}}+\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 a^{3} c^{2} \sqrt {a c}}\) | \(198\) |
| risch | \(\frac {\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right ) x^{5}}{16 a^{3}}-\frac {\left (a^{2} e^{4}-6 a c \,d^{2} e^{2}-5 c^{2} d^{4}\right ) x^{3}}{6 a^{2} c}-\frac {d \,e^{3} x^{2}}{c}-\frac {\left (a^{2} e^{4}+6 a c \,d^{2} e^{2}-11 c^{2} d^{4}\right ) x}{16 c^{2} a}-\frac {d e \left (e^{2} a +2 c \,d^{2}\right )}{3 c^{2}}}{\left (c \,x^{2}+a \right )^{3}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) e^{4}}{32 \sqrt {-a c}\, c^{2} a}-\frac {3 \ln \left (c x +\sqrt {-a c}\right ) d^{2} e^{2}}{16 \sqrt {-a c}\, c \,a^{2}}-\frac {5 \ln \left (c x +\sqrt {-a c}\right ) d^{4}}{32 \sqrt {-a c}\, a^{3}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) e^{4}}{32 \sqrt {-a c}\, c^{2} a}+\frac {3 \ln \left (-c x +\sqrt {-a c}\right ) d^{2} e^{2}}{16 \sqrt {-a c}\, c \,a^{2}}+\frac {5 \ln \left (-c x +\sqrt {-a c}\right ) d^{4}}{32 \sqrt {-a c}\, a^{3}}\) | \(321\) |
(1/16*(a^2*e^4+6*a*c*d^2*e^2+5*c^2*d^4)/a^3*x^5-1/6*(a^2*e^4-6*a*c*d^2*e^2 -5*c^2*d^4)/a^2/c*x^3-d*e^3*x^2/c-1/16*(a^2*e^4+6*a*c*d^2*e^2-11*c^2*d^4)/ c^2/a*x-1/3*d*e*(a*e^2+2*c*d^2)/c^2)/(c*x^2+a)^3+1/16*(a^2*e^4+6*a*c*d^2*e ^2+5*c^2*d^4)/a^3/c^2/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 363 vs. \(2 (140) = 280\).
Time = 0.64 (sec) , antiderivative size = 746, normalized size of antiderivative = 4.81 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=\left [-\frac {96 \, a^{4} c^{2} d e^{3} x^{2} + 64 \, a^{4} c^{2} d^{3} e + 32 \, a^{5} c d e^{3} - 6 \, {\left (5 \, a c^{5} d^{4} + 6 \, a^{2} c^{4} d^{2} e^{2} + a^{3} c^{3} e^{4}\right )} x^{5} - 16 \, {\left (5 \, a^{2} c^{4} d^{4} + 6 \, a^{3} c^{3} d^{2} e^{2} - a^{4} c^{2} e^{4}\right )} x^{3} + 3 \, {\left (5 \, a^{3} c^{2} d^{4} + 6 \, a^{4} c d^{2} e^{2} + a^{5} e^{4} + {\left (5 \, c^{5} d^{4} + 6 \, a c^{4} d^{2} e^{2} + a^{2} c^{3} e^{4}\right )} x^{6} + 3 \, {\left (5 \, a c^{4} d^{4} + 6 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4} + 3 \, {\left (5 \, a^{2} c^{3} d^{4} + 6 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 6 \, {\left (11 \, a^{3} c^{3} d^{4} - 6 \, a^{4} c^{2} d^{2} e^{2} - a^{5} c e^{4}\right )} x}{96 \, {\left (a^{4} c^{6} x^{6} + 3 \, a^{5} c^{5} x^{4} + 3 \, a^{6} c^{4} x^{2} + a^{7} c^{3}\right )}}, -\frac {48 \, a^{4} c^{2} d e^{3} x^{2} + 32 \, a^{4} c^{2} d^{3} e + 16 \, a^{5} c d e^{3} - 3 \, {\left (5 \, a c^{5} d^{4} + 6 \, a^{2} c^{4} d^{2} e^{2} + a^{3} c^{3} e^{4}\right )} x^{5} - 8 \, {\left (5 \, a^{2} c^{4} d^{4} + 6 \, a^{3} c^{3} d^{2} e^{2} - a^{4} c^{2} e^{4}\right )} x^{3} - 3 \, {\left (5 \, a^{3} c^{2} d^{4} + 6 \, a^{4} c d^{2} e^{2} + a^{5} e^{4} + {\left (5 \, c^{5} d^{4} + 6 \, a c^{4} d^{2} e^{2} + a^{2} c^{3} e^{4}\right )} x^{6} + 3 \, {\left (5 \, a c^{4} d^{4} + 6 \, a^{2} c^{3} d^{2} e^{2} + a^{3} c^{2} e^{4}\right )} x^{4} + 3 \, {\left (5 \, a^{2} c^{3} d^{4} + 6 \, a^{3} c^{2} d^{2} e^{2} + a^{4} c e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - 3 \, {\left (11 \, a^{3} c^{3} d^{4} - 6 \, a^{4} c^{2} d^{2} e^{2} - a^{5} c e^{4}\right )} x}{48 \, {\left (a^{4} c^{6} x^{6} + 3 \, a^{5} c^{5} x^{4} + 3 \, a^{6} c^{4} x^{2} + a^{7} c^{3}\right )}}\right ] \]
[-1/96*(96*a^4*c^2*d*e^3*x^2 + 64*a^4*c^2*d^3*e + 32*a^5*c*d*e^3 - 6*(5*a* c^5*d^4 + 6*a^2*c^4*d^2*e^2 + a^3*c^3*e^4)*x^5 - 16*(5*a^2*c^4*d^4 + 6*a^3 *c^3*d^2*e^2 - a^4*c^2*e^4)*x^3 + 3*(5*a^3*c^2*d^4 + 6*a^4*c*d^2*e^2 + a^5 *e^4 + (5*c^5*d^4 + 6*a*c^4*d^2*e^2 + a^2*c^3*e^4)*x^6 + 3*(5*a*c^4*d^4 + 6*a^2*c^3*d^2*e^2 + a^3*c^2*e^4)*x^4 + 3*(5*a^2*c^3*d^4 + 6*a^3*c^2*d^2*e^ 2 + a^4*c*e^4)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a )) - 6*(11*a^3*c^3*d^4 - 6*a^4*c^2*d^2*e^2 - a^5*c*e^4)*x)/(a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 + a^7*c^3), -1/48*(48*a^4*c^2*d*e^3*x^2 + 32 *a^4*c^2*d^3*e + 16*a^5*c*d*e^3 - 3*(5*a*c^5*d^4 + 6*a^2*c^4*d^2*e^2 + a^3 *c^3*e^4)*x^5 - 8*(5*a^2*c^4*d^4 + 6*a^3*c^3*d^2*e^2 - a^4*c^2*e^4)*x^3 - 3*(5*a^3*c^2*d^4 + 6*a^4*c*d^2*e^2 + a^5*e^4 + (5*c^5*d^4 + 6*a*c^4*d^2*e^ 2 + a^2*c^3*e^4)*x^6 + 3*(5*a*c^4*d^4 + 6*a^2*c^3*d^2*e^2 + a^3*c^2*e^4)*x ^4 + 3*(5*a^2*c^3*d^4 + 6*a^3*c^2*d^2*e^2 + a^4*c*e^4)*x^2)*sqrt(a*c)*arct an(sqrt(a*c)*x/a) - 3*(11*a^3*c^3*d^4 - 6*a^4*c^2*d^2*e^2 - a^5*c*e^4)*x)/ (a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 + a^7*c^3)]
Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (144) = 288\).
Time = 1.90 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.66 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right ) \log {\left (- \frac {a^{4} c^{2} \sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right )}{a^{2} e^{4} + 6 a c d^{2} e^{2} + 5 c^{2} d^{4}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right ) \log {\left (\frac {a^{4} c^{2} \sqrt {- \frac {1}{a^{7} c^{5}}} \left (a e^{2} + c d^{2}\right ) \left (a e^{2} + 5 c d^{2}\right )}{a^{2} e^{4} + 6 a c d^{2} e^{2} + 5 c^{2} d^{4}} + x \right )}}{32} + \frac {- 16 a^{4} d e^{3} - 32 a^{3} c d^{3} e - 48 a^{3} c d e^{3} x^{2} + x^{5} \cdot \left (3 a^{2} c^{2} e^{4} + 18 a c^{3} d^{2} e^{2} + 15 c^{4} d^{4}\right ) + x^{3} \left (- 8 a^{3} c e^{4} + 48 a^{2} c^{2} d^{2} e^{2} + 40 a c^{3} d^{4}\right ) + x \left (- 3 a^{4} e^{4} - 18 a^{3} c d^{2} e^{2} + 33 a^{2} c^{2} d^{4}\right )}{48 a^{6} c^{2} + 144 a^{5} c^{3} x^{2} + 144 a^{4} c^{4} x^{4} + 48 a^{3} c^{5} x^{6}} \]
-sqrt(-1/(a**7*c**5))*(a*e**2 + c*d**2)*(a*e**2 + 5*c*d**2)*log(-a**4*c**2 *sqrt(-1/(a**7*c**5))*(a*e**2 + c*d**2)*(a*e**2 + 5*c*d**2)/(a**2*e**4 + 6 *a*c*d**2*e**2 + 5*c**2*d**4) + x)/32 + sqrt(-1/(a**7*c**5))*(a*e**2 + c*d **2)*(a*e**2 + 5*c*d**2)*log(a**4*c**2*sqrt(-1/(a**7*c**5))*(a*e**2 + c*d* *2)*(a*e**2 + 5*c*d**2)/(a**2*e**4 + 6*a*c*d**2*e**2 + 5*c**2*d**4) + x)/3 2 + (-16*a**4*d*e**3 - 32*a**3*c*d**3*e - 48*a**3*c*d*e**3*x**2 + x**5*(3* a**2*c**2*e**4 + 18*a*c**3*d**2*e**2 + 15*c**4*d**4) + x**3*(-8*a**3*c*e** 4 + 48*a**2*c**2*d**2*e**2 + 40*a*c**3*d**4) + x*(-3*a**4*e**4 - 18*a**3*c *d**2*e**2 + 33*a**2*c**2*d**4))/(48*a**6*c**2 + 144*a**5*c**3*x**2 + 144* a**4*c**4*x**4 + 48*a**3*c**5*x**6)
Time = 0.27 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.52 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=-\frac {48 \, a^{3} c d e^{3} x^{2} + 32 \, a^{3} c d^{3} e + 16 \, a^{4} d e^{3} - 3 \, {\left (5 \, c^{4} d^{4} + 6 \, a c^{3} d^{2} e^{2} + a^{2} c^{2} e^{4}\right )} x^{5} - 8 \, {\left (5 \, a c^{3} d^{4} + 6 \, a^{2} c^{2} d^{2} e^{2} - a^{3} c e^{4}\right )} x^{3} - 3 \, {\left (11 \, a^{2} c^{2} d^{4} - 6 \, a^{3} c d^{2} e^{2} - a^{4} e^{4}\right )} x}{48 \, {\left (a^{3} c^{5} x^{6} + 3 \, a^{4} c^{4} x^{4} + 3 \, a^{5} c^{3} x^{2} + a^{6} c^{2}\right )}} + \frac {{\left (5 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} \]
-1/48*(48*a^3*c*d*e^3*x^2 + 32*a^3*c*d^3*e + 16*a^4*d*e^3 - 3*(5*c^4*d^4 + 6*a*c^3*d^2*e^2 + a^2*c^2*e^4)*x^5 - 8*(5*a*c^3*d^4 + 6*a^2*c^2*d^2*e^2 - a^3*c*e^4)*x^3 - 3*(11*a^2*c^2*d^4 - 6*a^3*c*d^2*e^2 - a^4*e^4)*x)/(a^3*c ^5*x^6 + 3*a^4*c^4*x^4 + 3*a^5*c^3*x^2 + a^6*c^2) + 1/16*(5*c^2*d^4 + 6*a* c*d^2*e^2 + a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c^2)
Time = 0.31 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=\frac {{\left (5 \, c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} + \frac {15 \, c^{4} d^{4} x^{5} + 18 \, a c^{3} d^{2} e^{2} x^{5} + 3 \, a^{2} c^{2} e^{4} x^{5} + 40 \, a c^{3} d^{4} x^{3} + 48 \, a^{2} c^{2} d^{2} e^{2} x^{3} - 8 \, a^{3} c e^{4} x^{3} - 48 \, a^{3} c d e^{3} x^{2} + 33 \, a^{2} c^{2} d^{4} x - 18 \, a^{3} c d^{2} e^{2} x - 3 \, a^{4} e^{4} x - 32 \, a^{3} c d^{3} e - 16 \, a^{4} d e^{3}}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \]
1/16*(5*c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c )*a^3*c^2) + 1/48*(15*c^4*d^4*x^5 + 18*a*c^3*d^2*e^2*x^5 + 3*a^2*c^2*e^4*x ^5 + 40*a*c^3*d^4*x^3 + 48*a^2*c^2*d^2*e^2*x^3 - 8*a^3*c*e^4*x^3 - 48*a^3* c*d*e^3*x^2 + 33*a^2*c^2*d^4*x - 18*a^3*c*d^2*e^2*x - 3*a^4*e^4*x - 32*a^3 *c*d^3*e - 16*a^4*d*e^3)/((c*x^2 + a)^3*a^3*c^2)
Time = 9.35 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.70 \[ \int \frac {(d+e x)^4}{\left (a+c x^2\right )^4} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x\,\left (c\,d^2+a\,e^2\right )\,\left (5\,c\,d^2+a\,e^2\right )}{\sqrt {a}\,\left (a^2\,e^4+6\,a\,c\,d^2\,e^2+5\,c^2\,d^4\right )}\right )\,\left (c\,d^2+a\,e^2\right )\,\left (5\,c\,d^2+a\,e^2\right )}{16\,a^{7/2}\,c^{5/2}}-\frac {\frac {d\,e^3\,x^2}{c}-\frac {x^5\,\left (a^2\,e^4+6\,a\,c\,d^2\,e^2+5\,c^2\,d^4\right )}{16\,a^3}+\frac {x\,\left (a^2\,e^4+6\,a\,c\,d^2\,e^2-11\,c^2\,d^4\right )}{16\,a\,c^2}+\frac {d\,e\,\left (2\,c\,d^2+a\,e^2\right )}{3\,c^2}-\frac {x^3\,\left (-a^2\,e^4+6\,a\,c\,d^2\,e^2+5\,c^2\,d^4\right )}{6\,a^2\,c}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6} \]
(atan((c^(1/2)*x*(a*e^2 + c*d^2)*(a*e^2 + 5*c*d^2))/(a^(1/2)*(a^2*e^4 + 5* c^2*d^4 + 6*a*c*d^2*e^2)))*(a*e^2 + c*d^2)*(a*e^2 + 5*c*d^2))/(16*a^(7/2)* c^(5/2)) - ((d*e^3*x^2)/c - (x^5*(a^2*e^4 + 5*c^2*d^4 + 6*a*c*d^2*e^2))/(1 6*a^3) + (x*(a^2*e^4 - 11*c^2*d^4 + 6*a*c*d^2*e^2))/(16*a*c^2) + (d*e*(a*e ^2 + 2*c*d^2))/(3*c^2) - (x^3*(5*c^2*d^4 - a^2*e^4 + 6*a*c*d^2*e^2))/(6*a^ 2*c))/(a^3 + c^3*x^6 + 3*a^2*c*x^2 + 3*a*c^2*x^4)