Integrand size = 17, antiderivative size = 190 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=-\frac {3 d e^2 \left (2 c d^2-5 a e^2\right ) x}{2 a c^2}-\frac {e^3 \left (2 c d^2-a e^2\right ) x^2}{a c^2}-\frac {d e^4 x^3}{2 a c}-\frac {(a e-c d x) (d+e x)^4}{2 a c \left (a+c x^2\right )}+\frac {d \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}+\frac {e^3 \left (5 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{c^3} \]
-3/2*d*e^2*(-5*a*e^2+2*c*d^2)*x/a/c^2-e^3*(-a*e^2+2*c*d^2)*x^2/a/c^2-1/2*d *e^4*x^3/a/c-1/2*(-c*d*x+a*e)*(e*x+d)^4/a/c/(c*x^2+a)+1/2*d*(-15*a^2*e^4+1 0*a*c*d^2*e^2+c^2*d^4)*arctan(x*c^(1/2)/a^(1/2))/a^(3/2)/c^(5/2)+e^3*(-a*e ^2+5*c*d^2)*ln(c*x^2+a)/c^3
Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\frac {10 c d e^4 x+c e^5 x^2+\frac {-a^3 e^5+c^3 d^5 x+5 a^2 c d e^3 (2 d+e x)-5 a c^2 d^3 e (d+2 e x)}{a \left (a+c x^2\right )}+\frac {\sqrt {c} d \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2}}+2 \left (5 c d^2 e^3-a e^5\right ) \log \left (a+c x^2\right )}{2 c^3} \]
(10*c*d*e^4*x + c*e^5*x^2 + (-(a^3*e^5) + c^3*d^5*x + 5*a^2*c*d*e^3*(2*d + e*x) - 5*a*c^2*d^3*e*(d + 2*e*x))/(a*(a + c*x^2)) + (Sqrt[c]*d*(c^2*d^4 + 10*a*c*d^2*e^2 - 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 2*(5* c*d^2*e^3 - a*e^5)*Log[a + c*x^2])/(2*c^3)
Time = 0.36 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {495, 657, 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.
| \(\displaystyle \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 495 |
| \(\displaystyle \frac {\int \frac {(d+e x)^3 \left (c d^2-3 c e x d+4 a e^2\right )}{c x^2+a}dx}{2 a c}-\frac {(d+e x)^4 (a e-c d x)}{2 a c \left (a+c x^2\right )}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {\int \left (-3 d x^2 e^4-\frac {4 \left (2 c d^2-a e^2\right ) x e^3}{c}-3 d \left (2 d^2-\frac {5 a e^2}{c}\right ) e^2+\frac {c^2 d^5+10 a c e^2 d^3-15 a^2 e^4 d+4 a e^3 \left (5 c d^2-a e^2\right ) x}{c \left (c x^2+a\right )}\right )dx}{2 a c}-\frac {(d+e x)^4 (a e-c d x)}{2 a c \left (a+c x^2\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {d \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (-15 a^2 e^4+10 a c d^2 e^2+c^2 d^4\right )}{\sqrt {a} c^{3/2}}+\frac {2 a e^3 \left (5 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{c^2}-3 d e^2 x \left (2 d^2-\frac {5 a e^2}{c}\right )-\frac {2 e^3 x^2 \left (2 c d^2-a e^2\right )}{c}-d e^4 x^3}{2 a c}-\frac {(d+e x)^4 (a e-c d x)}{2 a c \left (a+c x^2\right )}\) |
-1/2*((a*e - c*d*x)*(d + e*x)^4)/(a*c*(a + c*x^2)) + (-3*d*e^2*(2*d^2 - (5 *a*e^2)/c)*x - (2*e^3*(2*c*d^2 - a*e^2)*x^2)/c - d*e^4*x^3 + (d*(c^2*d^4 + 10*a*c*d^2*e^2 - 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(3/2 )) + (2*a*e^3*(5*c*d^2 - a*e^2)*Log[a + c*x^2])/c^2)/(2*a*c)
3.6.5.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (a*d - b*c*x)*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1)*Simp[a* d^2*(n - 1) - b*c^2*(2*p + 3) - b*c*d*(n + 2*p + 2)*x, x], x], x] /; FreeQ[ {a, b, c, d}, x] && LtQ[p, -1] && GtQ[n, 1] && IntQuadraticQ[a, 0, b, c, d, n, p, x]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Time = 2.31 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {e^{4} \left (\frac {1}{2} e \,x^{2}+5 d x \right )}{c^{2}}-\frac {\frac {-\frac {d \left (5 a^{2} e^{4}-10 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x}{2 a}+\frac {e \left (a^{2} e^{4}-10 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right )}{2 c}}{c \,x^{2}+a}+\frac {\frac {\left (4 a^{2} e^{5}-20 a c \,d^{2} e^{3}\right ) \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {\left (15 a^{2} d \,e^{4}-10 a c \,d^{3} e^{2}-c^{2} d^{5}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a}}{c^{2}}\) | \(183\) |
| risch | \(\frac {e^{5} x^{2}}{2 c^{2}}+\frac {5 e^{4} d x}{c^{2}}+\frac {\frac {d \left (5 a^{2} e^{4}-10 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) x}{2 a}-\frac {e \left (a^{2} e^{4}-10 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right )}{2 c}}{c^{2} \left (c \,x^{2}+a \right )}-\frac {a \ln \left (-15 d \,e^{4} a^{3}+10 d^{3} e^{2} a^{2} c +d^{5} c^{2} a -\sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}\, x \right ) e^{5}}{c^{3}}+\frac {5 \ln \left (-15 d \,e^{4} a^{3}+10 d^{3} e^{2} a^{2} c +d^{5} c^{2} a -\sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}\, x \right ) d^{2} e^{3}}{c^{2}}+\frac {\ln \left (-15 d \,e^{4} a^{3}+10 d^{3} e^{2} a^{2} c +d^{5} c^{2} a -\sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}\, x \right ) \sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}}{4 c^{3} a^{2}}-\frac {a \ln \left (-15 d \,e^{4} a^{3}+10 d^{3} e^{2} a^{2} c +d^{5} c^{2} a +\sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}\, x \right ) e^{5}}{c^{3}}+\frac {5 \ln \left (-15 d \,e^{4} a^{3}+10 d^{3} e^{2} a^{2} c +d^{5} c^{2} a +\sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}\, x \right ) d^{2} e^{3}}{c^{2}}-\frac {\ln \left (-15 d \,e^{4} a^{3}+10 d^{3} e^{2} a^{2} c +d^{5} c^{2} a +\sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}\, x \right ) \sqrt {-a c \,d^{2} \left (15 a^{2} e^{4}-10 a c \,d^{2} e^{2}-c^{2} d^{4}\right )^{2}}}{4 c^{3} a^{2}}\) | \(663\) |
e^4/c^2*(1/2*e*x^2+5*d*x)-1/c^2*((-1/2*d*(5*a^2*e^4-10*a*c*d^2*e^2+c^2*d^4 )/a*x+1/2*e*(a^2*e^4-10*a*c*d^2*e^2+5*c^2*d^4)/c)/(c*x^2+a)+1/2/a*(1/2*(4* a^2*e^5-20*a*c*d^2*e^3)/c*ln(c*x^2+a)+(15*a^2*d*e^4-10*a*c*d^3*e^2-c^2*d^5 )/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))))
Time = 0.31 (sec) , antiderivative size = 561, normalized size of antiderivative = 2.95 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\left [\frac {2 \, a^{2} c^{2} e^{5} x^{4} + 20 \, a^{2} c^{2} d e^{4} x^{3} + 2 \, a^{3} c e^{5} x^{2} - 10 \, a^{2} c^{2} d^{4} e + 20 \, a^{3} c d^{2} e^{3} - 2 \, a^{4} e^{5} + {\left (a c^{2} d^{5} + 10 \, a^{2} c d^{3} e^{2} - 15 \, a^{3} d e^{4} + {\left (c^{3} d^{5} + 10 \, a c^{2} d^{3} e^{2} - 15 \, a^{2} c d e^{4}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (a c^{3} d^{5} - 10 \, a^{2} c^{2} d^{3} e^{2} + 15 \, a^{3} c d e^{4}\right )} x + 4 \, {\left (5 \, a^{3} c d^{2} e^{3} - a^{4} e^{5} + {\left (5 \, a^{2} c^{2} d^{2} e^{3} - a^{3} c e^{5}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac {a^{2} c^{2} e^{5} x^{4} + 10 \, a^{2} c^{2} d e^{4} x^{3} + a^{3} c e^{5} x^{2} - 5 \, a^{2} c^{2} d^{4} e + 10 \, a^{3} c d^{2} e^{3} - a^{4} e^{5} + {\left (a c^{2} d^{5} + 10 \, a^{2} c d^{3} e^{2} - 15 \, a^{3} d e^{4} + {\left (c^{3} d^{5} + 10 \, a c^{2} d^{3} e^{2} - 15 \, a^{2} c d e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (a c^{3} d^{5} - 10 \, a^{2} c^{2} d^{3} e^{2} + 15 \, a^{3} c d e^{4}\right )} x + 2 \, {\left (5 \, a^{3} c d^{2} e^{3} - a^{4} e^{5} + {\left (5 \, a^{2} c^{2} d^{2} e^{3} - a^{3} c e^{5}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}\right ] \]
[1/4*(2*a^2*c^2*e^5*x^4 + 20*a^2*c^2*d*e^4*x^3 + 2*a^3*c*e^5*x^2 - 10*a^2* c^2*d^4*e + 20*a^3*c*d^2*e^3 - 2*a^4*e^5 + (a*c^2*d^5 + 10*a^2*c*d^3*e^2 - 15*a^3*d*e^4 + (c^3*d^5 + 10*a*c^2*d^3*e^2 - 15*a^2*c*d*e^4)*x^2)*sqrt(-a *c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(a*c^3*d^5 - 10*a^2* c^2*d^3*e^2 + 15*a^3*c*d*e^4)*x + 4*(5*a^3*c*d^2*e^3 - a^4*e^5 + (5*a^2*c^ 2*d^2*e^3 - a^3*c*e^5)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3), 1/2*( a^2*c^2*e^5*x^4 + 10*a^2*c^2*d*e^4*x^3 + a^3*c*e^5*x^2 - 5*a^2*c^2*d^4*e + 10*a^3*c*d^2*e^3 - a^4*e^5 + (a*c^2*d^5 + 10*a^2*c*d^3*e^2 - 15*a^3*d*e^4 + (c^3*d^5 + 10*a*c^2*d^3*e^2 - 15*a^2*c*d*e^4)*x^2)*sqrt(a*c)*arctan(sqr t(a*c)*x/a) + (a*c^3*d^5 - 10*a^2*c^2*d^3*e^2 + 15*a^3*c*d*e^4)*x + 2*(5*a ^3*c*d^2*e^3 - a^4*e^5 + (5*a^2*c^2*d^2*e^3 - a^3*c*e^5)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3)]
Leaf count of result is larger than twice the leaf count of optimal. 515 vs. \(2 (175) = 350\).
Time = 1.16 (sec) , antiderivative size = 515, normalized size of antiderivative = 2.71 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} - \frac {d \sqrt {- a^{3} c^{7}} \cdot \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) \log {\left (x + \frac {- 4 a^{3} e^{5} - 4 a^{2} c^{3} \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} - \frac {d \sqrt {- a^{3} c^{7}} \cdot \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) + 20 a^{2} c d^{2} e^{3}}{15 a^{2} c d e^{4} - 10 a c^{2} d^{3} e^{2} - c^{3} d^{5}} \right )} + \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} + \frac {d \sqrt {- a^{3} c^{7}} \cdot \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) \log {\left (x + \frac {- 4 a^{3} e^{5} - 4 a^{2} c^{3} \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} + \frac {d \sqrt {- a^{3} c^{7}} \cdot \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) + 20 a^{2} c d^{2} e^{3}}{15 a^{2} c d e^{4} - 10 a c^{2} d^{3} e^{2} - c^{3} d^{5}} \right )} + \frac {- a^{3} e^{5} + 10 a^{2} c d^{2} e^{3} - 5 a c^{2} d^{4} e + x \left (5 a^{2} c d e^{4} - 10 a c^{2} d^{3} e^{2} + c^{3} d^{5}\right )}{2 a^{2} c^{3} + 2 a c^{4} x^{2}} + \frac {5 d e^{4} x}{c^{2}} + \frac {e^{5} x^{2}}{2 c^{2}} \]
(-e**3*(a*e**2 - 5*c*d**2)/c**3 - d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 10*a* c*d**2*e**2 - c**2*d**4)/(4*a**3*c**6))*log(x + (-4*a**3*e**5 - 4*a**2*c** 3*(-e**3*(a*e**2 - 5*c*d**2)/c**3 - d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 10* a*c*d**2*e**2 - c**2*d**4)/(4*a**3*c**6)) + 20*a**2*c*d**2*e**3)/(15*a**2* c*d*e**4 - 10*a*c**2*d**3*e**2 - c**3*d**5)) + (-e**3*(a*e**2 - 5*c*d**2)/ c**3 + d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 10*a*c*d**2*e**2 - c**2*d**4)/(4 *a**3*c**6))*log(x + (-4*a**3*e**5 - 4*a**2*c**3*(-e**3*(a*e**2 - 5*c*d**2 )/c**3 + d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 10*a*c*d**2*e**2 - c**2*d**4)/ (4*a**3*c**6)) + 20*a**2*c*d**2*e**3)/(15*a**2*c*d*e**4 - 10*a*c**2*d**3*e **2 - c**3*d**5)) + (-a**3*e**5 + 10*a**2*c*d**2*e**3 - 5*a*c**2*d**4*e + x*(5*a**2*c*d*e**4 - 10*a*c**2*d**3*e**2 + c**3*d**5))/(2*a**2*c**3 + 2*a* c**4*x**2) + 5*d*e**4*x/c**2 + e**5*x**2/(2*c**2)
Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=-\frac {5 \, a c^{2} d^{4} e - 10 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} - {\left (c^{3} d^{5} - 10 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4}\right )} x}{2 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}} + \frac {e^{5} x^{2} + 10 \, d e^{4} x}{2 \, c^{2}} + \frac {{\left (5 \, c d^{2} e^{3} - a e^{5}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (c^{2} d^{5} + 10 \, a c d^{3} e^{2} - 15 \, a^{2} d e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} \]
-1/2*(5*a*c^2*d^4*e - 10*a^2*c*d^2*e^3 + a^3*e^5 - (c^3*d^5 - 10*a*c^2*d^3 *e^2 + 5*a^2*c*d*e^4)*x)/(a*c^4*x^2 + a^2*c^3) + 1/2*(e^5*x^2 + 10*d*e^4*x )/c^2 + (5*c*d^2*e^3 - a*e^5)*log(c*x^2 + a)/c^3 + 1/2*(c^2*d^5 + 10*a*c*d ^3*e^2 - 15*a^2*d*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c^2)
Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\frac {{\left (5 \, c d^{2} e^{3} - a e^{5}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (c^{2} d^{5} + 10 \, a c d^{3} e^{2} - 15 \, a^{2} d e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} + \frac {c^{2} e^{5} x^{2} + 10 \, c^{2} d e^{4} x}{2 \, c^{4}} - \frac {5 \, a c^{2} d^{4} e - 10 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} - {\left (c^{3} d^{5} - 10 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4}\right )} x}{2 \, {\left (c x^{2} + a\right )} a c^{3}} \]
(5*c*d^2*e^3 - a*e^5)*log(c*x^2 + a)/c^3 + 1/2*(c^2*d^5 + 10*a*c*d^3*e^2 - 15*a^2*d*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c^2) + 1/2*(c^2*e^5*x^2 + 10*c^2*d*e^4*x)/c^4 - 1/2*(5*a*c^2*d^4*e - 10*a^2*c*d^2*e^3 + a^3*e^5 - (c^3*d^5 - 10*a*c^2*d^3*e^2 + 5*a^2*c*d*e^4)*x)/((c*x^2 + a)*a*c^3)
Time = 0.15 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx=\frac {e^5\,x^2}{2\,c^2}-\frac {\frac {a^2\,e^5-10\,a\,c\,d^2\,e^3+5\,c^2\,d^4\,e}{2\,c}-\frac {x\,\left (5\,a^2\,d\,e^4-10\,a\,c\,d^3\,e^2+c^2\,d^5\right )}{2\,a}}{c^3\,x^2+a\,c^2}-\frac {\ln \left (c\,x^2+a\right )\,\left (32\,a^4\,c^3\,e^5-160\,a^3\,c^4\,d^2\,e^3\right )}{32\,a^3\,c^6}+\frac {5\,d\,e^4\,x}{c^2}+\frac {d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-15\,a^2\,e^4+10\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{2\,a^{3/2}\,c^{5/2}} \]
(e^5*x^2)/(2*c^2) - ((a^2*e^5 + 5*c^2*d^4*e - 10*a*c*d^2*e^3)/(2*c) - (x*( c^2*d^5 + 5*a^2*d*e^4 - 10*a*c*d^3*e^2))/(2*a))/(a*c^2 + c^3*x^2) - (log(a + c*x^2)*(32*a^4*c^3*e^5 - 160*a^3*c^4*d^2*e^3))/(32*a^3*c^6) + (5*d*e^4* x)/c^2 + (d*atan((c^(1/2)*x)/a^(1/2))*(c^2*d^4 - 15*a^2*e^4 + 10*a*c*d^2*e ^2))/(2*a^(3/2)*c^(5/2))