Integrand size = 30, antiderivative size = 314 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {2 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^6 (a+b x) (d+e x)^{9/2}}-\frac {10 b (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^6 (a+b x) (d+e x)^{7/2}}+\frac {4 b^2 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) (d+e x)^{5/2}}-\frac {20 b^3 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x) (d+e x)^{3/2}}+\frac {10 b^4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x) \sqrt {d+e x}}+\frac {2 b^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)} \]
2/9*(-a*e+b*d)^5*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)^(9/2)-10/7*b*(-a*e+ b*d)^4*((b*x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)^(7/2)+4*b^2*(-a*e+b*d)^3*((b* x+a)^2)^(1/2)/e^6/(b*x+a)/(e*x+d)^(5/2)-20/3*b^3*(-a*e+b*d)^2*((b*x+a)^2)^ (1/2)/e^6/(b*x+a)/(e*x+d)^(3/2)+10*b^4*(-a*e+b*d)*((b*x+a)^2)^(1/2)/e^6/(b *x+a)/(e*x+d)^(1/2)+2*b^5*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)
Time = 0.13 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=-\frac {2 \sqrt {(a+b x)^2} \left (7 a^5 e^5+5 a^4 b e^4 (2 d+9 e x)+2 a^3 b^2 e^3 \left (8 d^2+36 d e x+63 e^2 x^2\right )+2 a^2 b^3 e^2 \left (16 d^3+72 d^2 e x+126 d e^2 x^2+105 e^3 x^3\right )+a b^4 e \left (128 d^4+576 d^3 e x+1008 d^2 e^2 x^2+840 d e^3 x^3+315 e^4 x^4\right )-b^5 \left (256 d^5+1152 d^4 e x+2016 d^3 e^2 x^2+1680 d^2 e^3 x^3+630 d e^4 x^4+63 e^5 x^5\right )\right )}{63 e^6 (a+b x) (d+e x)^{9/2}} \]
(-2*Sqrt[(a + b*x)^2]*(7*a^5*e^5 + 5*a^4*b*e^4*(2*d + 9*e*x) + 2*a^3*b^2*e ^3*(8*d^2 + 36*d*e*x + 63*e^2*x^2) + 2*a^2*b^3*e^2*(16*d^3 + 72*d^2*e*x + 126*d*e^2*x^2 + 105*e^3*x^3) + a*b^4*e*(128*d^4 + 576*d^3*e*x + 1008*d^2*e ^2*x^2 + 840*d*e^3*x^3 + 315*e^4*x^4) - b^5*(256*d^5 + 1152*d^4*e*x + 2016 *d^3*e^2*x^2 + 1680*d^2*e^3*x^3 + 630*d*e^4*x^4 + 63*e^5*x^5)))/(63*e^6*(a + b*x)*(d + e*x)^(9/2))
Time = 0.31 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.57, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1102, 27, 53, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{(d+e x)^{11/2}}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{(d+e x)^{11/2}}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^5}{e^5 \sqrt {d+e x}}-\frac {5 (b d-a e) b^4}{e^5 (d+e x)^{3/2}}+\frac {10 (b d-a e)^2 b^3}{e^5 (d+e x)^{5/2}}-\frac {10 (b d-a e)^3 b^2}{e^5 (d+e x)^{7/2}}+\frac {5 (b d-a e)^4 b}{e^5 (d+e x)^{9/2}}+\frac {(a e-b d)^5}{e^5 (d+e x)^{11/2}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {10 b^4 (b d-a e)}{e^6 \sqrt {d+e x}}-\frac {20 b^3 (b d-a e)^2}{3 e^6 (d+e x)^{3/2}}+\frac {4 b^2 (b d-a e)^3}{e^6 (d+e x)^{5/2}}-\frac {10 b (b d-a e)^4}{7 e^6 (d+e x)^{7/2}}+\frac {2 (b d-a e)^5}{9 e^6 (d+e x)^{9/2}}+\frac {2 b^5 \sqrt {d+e x}}{e^6}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((2*(b*d - a*e)^5)/(9*e^6*(d + e*x)^(9/2)) - (10*b*(b*d - a*e)^4)/(7*e^6*(d + e*x)^(7/2)) + (4*b^2*(b*d - a*e)^3)/(e^ 6*(d + e*x)^(5/2)) - (20*b^3*(b*d - a*e)^2)/(3*e^6*(d + e*x)^(3/2)) + (10* b^4*(b*d - a*e))/(e^6*Sqrt[d + e*x]) + (2*b^5*Sqrt[d + e*x])/e^6))/(a + b* x)
3.17.100.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.27 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.90
| method | result | size |
| risch | \(\frac {2 b^{5} \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{e^{6} \left (b x +a \right )}-\frac {2 \left (315 b^{4} x^{4} e^{4}+210 x^{3} a \,b^{3} e^{4}+1050 x^{3} b^{4} d \,e^{3}+126 x^{2} a^{2} b^{2} e^{4}+378 x^{2} a \,b^{3} d \,e^{3}+1386 x^{2} b^{4} d^{2} e^{2}+45 x \,a^{3} b \,e^{4}+117 x \,a^{2} b^{2} d \,e^{3}+261 x a \,b^{3} d^{2} e^{2}+837 x \,b^{4} d^{3} e +7 e^{4} a^{4}+17 b \,e^{3} d \,a^{3}+33 b^{2} e^{2} d^{2} a^{2}+65 a \,b^{3} d^{3} e +193 b^{4} d^{4}\right ) \left (a e -b d \right ) \sqrt {\left (b x +a \right )^{2}}}{63 e^{6} \sqrt {e x +d}\, \left (e^{4} x^{4}+4 d \,e^{3} x^{3}+6 d^{2} e^{2} x^{2}+4 d^{3} e x +d^{4}\right ) \left (b x +a \right )}\) | \(282\) |
| gosper | \(-\frac {2 \left (-63 x^{5} e^{5} b^{5}+315 x^{4} a \,b^{4} e^{5}-630 x^{4} b^{5} d \,e^{4}+210 x^{3} a^{2} b^{3} e^{5}+840 x^{3} a \,b^{4} d \,e^{4}-1680 x^{3} b^{5} d^{2} e^{3}+126 x^{2} a^{3} b^{2} e^{5}+252 x^{2} a^{2} b^{3} d \,e^{4}+1008 x^{2} a \,b^{4} d^{2} e^{3}-2016 x^{2} b^{5} d^{3} e^{2}+45 a^{4} b \,e^{5} x +72 a^{3} b^{2} d \,e^{4} x +144 x \,a^{2} b^{3} d^{2} e^{3}+576 x a \,b^{4} d^{3} e^{2}-1152 b^{5} d^{4} e x +7 a^{5} e^{5}+10 a^{4} b d \,e^{4}+16 a^{3} b^{2} d^{2} e^{3}+32 a^{2} b^{3} d^{3} e^{2}+128 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{63 \left (e x +d \right )^{\frac {9}{2}} e^{6} \left (b x +a \right )^{5}}\) | \(289\) |
| default | \(-\frac {2 \left (-63 x^{5} e^{5} b^{5}+315 x^{4} a \,b^{4} e^{5}-630 x^{4} b^{5} d \,e^{4}+210 x^{3} a^{2} b^{3} e^{5}+840 x^{3} a \,b^{4} d \,e^{4}-1680 x^{3} b^{5} d^{2} e^{3}+126 x^{2} a^{3} b^{2} e^{5}+252 x^{2} a^{2} b^{3} d \,e^{4}+1008 x^{2} a \,b^{4} d^{2} e^{3}-2016 x^{2} b^{5} d^{3} e^{2}+45 a^{4} b \,e^{5} x +72 a^{3} b^{2} d \,e^{4} x +144 x \,a^{2} b^{3} d^{2} e^{3}+576 x a \,b^{4} d^{3} e^{2}-1152 b^{5} d^{4} e x +7 a^{5} e^{5}+10 a^{4} b d \,e^{4}+16 a^{3} b^{2} d^{2} e^{3}+32 a^{2} b^{3} d^{3} e^{2}+128 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{63 \left (e x +d \right )^{\frac {9}{2}} e^{6} \left (b x +a \right )^{5}}\) | \(289\) |
2*b^5*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)-2/63*(315*b^4*e^4*x^4+21 0*a*b^3*e^4*x^3+1050*b^4*d*e^3*x^3+126*a^2*b^2*e^4*x^2+378*a*b^3*d*e^3*x^2 +1386*b^4*d^2*e^2*x^2+45*a^3*b*e^4*x+117*a^2*b^2*d*e^3*x+261*a*b^3*d^2*e^2 *x+837*b^4*d^3*e*x+7*a^4*e^4+17*a^3*b*d*e^3+33*a^2*b^2*d^2*e^2+65*a*b^3*d^ 3*e+193*b^4*d^4)*(a*e-b*d)/e^6/(e*x+d)^(1/2)/(e^4*x^4+4*d*e^3*x^3+6*d^2*e^ 2*x^2+4*d^3*e*x+d^4)*((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.29 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {2 \, {\left (63 \, b^{5} e^{5} x^{5} + 256 \, b^{5} d^{5} - 128 \, a b^{4} d^{4} e - 32 \, a^{2} b^{3} d^{3} e^{2} - 16 \, a^{3} b^{2} d^{2} e^{3} - 10 \, a^{4} b d e^{4} - 7 \, a^{5} e^{5} + 315 \, {\left (2 \, b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 210 \, {\left (8 \, b^{5} d^{2} e^{3} - 4 \, a b^{4} d e^{4} - a^{2} b^{3} e^{5}\right )} x^{3} + 126 \, {\left (16 \, b^{5} d^{3} e^{2} - 8 \, a b^{4} d^{2} e^{3} - 2 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 9 \, {\left (128 \, b^{5} d^{4} e - 64 \, a b^{4} d^{3} e^{2} - 16 \, a^{2} b^{3} d^{2} e^{3} - 8 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x\right )} \sqrt {e x + d}}{63 \, {\left (e^{11} x^{5} + 5 \, d e^{10} x^{4} + 10 \, d^{2} e^{9} x^{3} + 10 \, d^{3} e^{8} x^{2} + 5 \, d^{4} e^{7} x + d^{5} e^{6}\right )}} \]
2/63*(63*b^5*e^5*x^5 + 256*b^5*d^5 - 128*a*b^4*d^4*e - 32*a^2*b^3*d^3*e^2 - 16*a^3*b^2*d^2*e^3 - 10*a^4*b*d*e^4 - 7*a^5*e^5 + 315*(2*b^5*d*e^4 - a*b ^4*e^5)*x^4 + 210*(8*b^5*d^2*e^3 - 4*a*b^4*d*e^4 - a^2*b^3*e^5)*x^3 + 126* (16*b^5*d^3*e^2 - 8*a*b^4*d^2*e^3 - 2*a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 9 *(128*b^5*d^4*e - 64*a*b^4*d^3*e^2 - 16*a^2*b^3*d^2*e^3 - 8*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x)*sqrt(e*x + d)/(e^11*x^5 + 5*d*e^10*x^4 + 10*d^2*e^9*x^3 + 10*d^3*e^8*x^2 + 5*d^4*e^7*x + d^5*e^6)
Timed out. \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {2 \, {\left (63 \, b^{5} e^{5} x^{5} + 256 \, b^{5} d^{5} - 128 \, a b^{4} d^{4} e - 32 \, a^{2} b^{3} d^{3} e^{2} - 16 \, a^{3} b^{2} d^{2} e^{3} - 10 \, a^{4} b d e^{4} - 7 \, a^{5} e^{5} + 315 \, {\left (2 \, b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 210 \, {\left (8 \, b^{5} d^{2} e^{3} - 4 \, a b^{4} d e^{4} - a^{2} b^{3} e^{5}\right )} x^{3} + 126 \, {\left (16 \, b^{5} d^{3} e^{2} - 8 \, a b^{4} d^{2} e^{3} - 2 \, a^{2} b^{3} d e^{4} - a^{3} b^{2} e^{5}\right )} x^{2} + 9 \, {\left (128 \, b^{5} d^{4} e - 64 \, a b^{4} d^{3} e^{2} - 16 \, a^{2} b^{3} d^{2} e^{3} - 8 \, a^{3} b^{2} d e^{4} - 5 \, a^{4} b e^{5}\right )} x\right )}}{63 \, {\left (e^{10} x^{4} + 4 \, d e^{9} x^{3} + 6 \, d^{2} e^{8} x^{2} + 4 \, d^{3} e^{7} x + d^{4} e^{6}\right )} \sqrt {e x + d}} \]
2/63*(63*b^5*e^5*x^5 + 256*b^5*d^5 - 128*a*b^4*d^4*e - 32*a^2*b^3*d^3*e^2 - 16*a^3*b^2*d^2*e^3 - 10*a^4*b*d*e^4 - 7*a^5*e^5 + 315*(2*b^5*d*e^4 - a*b ^4*e^5)*x^4 + 210*(8*b^5*d^2*e^3 - 4*a*b^4*d*e^4 - a^2*b^3*e^5)*x^3 + 126* (16*b^5*d^3*e^2 - 8*a*b^4*d^2*e^3 - 2*a^2*b^3*d*e^4 - a^3*b^2*e^5)*x^2 + 9 *(128*b^5*d^4*e - 64*a*b^4*d^3*e^2 - 16*a^2*b^3*d^2*e^3 - 8*a^3*b^2*d*e^4 - 5*a^4*b*e^5)*x)/((e^10*x^4 + 4*d*e^9*x^3 + 6*d^2*e^8*x^2 + 4*d^3*e^7*x + d^4*e^6)*sqrt(e*x + d))
Time = 0.30 (sec) , antiderivative size = 441, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=\frac {2 \, \sqrt {e x + d} b^{5} \mathrm {sgn}\left (b x + a\right )}{e^{6}} + \frac {2 \, {\left (315 \, {\left (e x + d\right )}^{4} b^{5} d \mathrm {sgn}\left (b x + a\right ) - 210 \, {\left (e x + d\right )}^{3} b^{5} d^{2} \mathrm {sgn}\left (b x + a\right ) + 126 \, {\left (e x + d\right )}^{2} b^{5} d^{3} \mathrm {sgn}\left (b x + a\right ) - 45 \, {\left (e x + d\right )} b^{5} d^{4} \mathrm {sgn}\left (b x + a\right ) + 7 \, b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 315 \, {\left (e x + d\right )}^{4} a b^{4} e \mathrm {sgn}\left (b x + a\right ) + 420 \, {\left (e x + d\right )}^{3} a b^{4} d e \mathrm {sgn}\left (b x + a\right ) - 378 \, {\left (e x + d\right )}^{2} a b^{4} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 180 \, {\left (e x + d\right )} a b^{4} d^{3} e \mathrm {sgn}\left (b x + a\right ) - 35 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) - 210 \, {\left (e x + d\right )}^{3} a^{2} b^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 378 \, {\left (e x + d\right )}^{2} a^{2} b^{3} d e^{2} \mathrm {sgn}\left (b x + a\right ) - 270 \, {\left (e x + d\right )} a^{2} b^{3} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 70 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 126 \, {\left (e x + d\right )}^{2} a^{3} b^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 180 \, {\left (e x + d\right )} a^{3} b^{2} d e^{3} \mathrm {sgn}\left (b x + a\right ) - 70 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 45 \, {\left (e x + d\right )} a^{4} b e^{4} \mathrm {sgn}\left (b x + a\right ) + 35 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - 7 \, a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )}}{63 \, {\left (e x + d\right )}^{\frac {9}{2}} e^{6}} \]
2*sqrt(e*x + d)*b^5*sgn(b*x + a)/e^6 + 2/63*(315*(e*x + d)^4*b^5*d*sgn(b*x + a) - 210*(e*x + d)^3*b^5*d^2*sgn(b*x + a) + 126*(e*x + d)^2*b^5*d^3*sgn (b*x + a) - 45*(e*x + d)*b^5*d^4*sgn(b*x + a) + 7*b^5*d^5*sgn(b*x + a) - 3 15*(e*x + d)^4*a*b^4*e*sgn(b*x + a) + 420*(e*x + d)^3*a*b^4*d*e*sgn(b*x + a) - 378*(e*x + d)^2*a*b^4*d^2*e*sgn(b*x + a) + 180*(e*x + d)*a*b^4*d^3*e* sgn(b*x + a) - 35*a*b^4*d^4*e*sgn(b*x + a) - 210*(e*x + d)^3*a^2*b^3*e^2*s gn(b*x + a) + 378*(e*x + d)^2*a^2*b^3*d*e^2*sgn(b*x + a) - 270*(e*x + d)*a ^2*b^3*d^2*e^2*sgn(b*x + a) + 70*a^2*b^3*d^3*e^2*sgn(b*x + a) - 126*(e*x + d)^2*a^3*b^2*e^3*sgn(b*x + a) + 180*(e*x + d)*a^3*b^2*d*e^3*sgn(b*x + a) - 70*a^3*b^2*d^2*e^3*sgn(b*x + a) - 45*(e*x + d)*a^4*b*e^4*sgn(b*x + a) + 35*a^4*b*d*e^4*sgn(b*x + a) - 7*a^5*e^5*sgn(b*x + a))/((e*x + d)^(9/2)*e^6 )
Time = 10.63 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.32 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{(d+e x)^{11/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {14\,a^5\,e^5+20\,a^4\,b\,d\,e^4+32\,a^3\,b^2\,d^2\,e^3+64\,a^2\,b^3\,d^3\,e^2+256\,a\,b^4\,d^4\,e-512\,b^5\,d^5}{63\,b\,e^{10}}-\frac {2\,b^4\,x^5}{e^5}+\frac {10\,b^3\,x^4\,\left (a\,e-2\,b\,d\right )}{e^6}+\frac {x\,\left (90\,a^4\,b\,e^5+144\,a^3\,b^2\,d\,e^4+288\,a^2\,b^3\,d^2\,e^3+1152\,a\,b^4\,d^3\,e^2-2304\,b^5\,d^4\,e\right )}{63\,b\,e^{10}}+\frac {20\,b^2\,x^3\,\left (a^2\,e^2+4\,a\,b\,d\,e-8\,b^2\,d^2\right )}{3\,e^7}+\frac {4\,b\,x^2\,\left (a^3\,e^3+2\,a^2\,b\,d\,e^2+8\,a\,b^2\,d^2\,e-16\,b^3\,d^3\right )}{e^8}\right )}{x^5\,\sqrt {d+e\,x}+\frac {a\,d^4\,\sqrt {d+e\,x}}{b\,e^4}+\frac {x^4\,\left (63\,a\,e^{10}+252\,b\,d\,e^9\right )\,\sqrt {d+e\,x}}{63\,b\,e^{10}}+\frac {2\,d\,x^3\,\left (2\,a\,e+3\,b\,d\right )\,\sqrt {d+e\,x}}{b\,e^2}+\frac {d^3\,x\,\left (4\,a\,e+b\,d\right )\,\sqrt {d+e\,x}}{b\,e^4}+\frac {2\,d^2\,x^2\,\left (3\,a\,e+2\,b\,d\right )\,\sqrt {d+e\,x}}{b\,e^3}} \]
-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((14*a^5*e^5 - 512*b^5*d^5 + 64*a^2*b^3* d^3*e^2 + 32*a^3*b^2*d^2*e^3 + 256*a*b^4*d^4*e + 20*a^4*b*d*e^4)/(63*b*e^1 0) - (2*b^4*x^5)/e^5 + (10*b^3*x^4*(a*e - 2*b*d))/e^6 + (x*(90*a^4*b*e^5 - 2304*b^5*d^4*e + 1152*a*b^4*d^3*e^2 + 144*a^3*b^2*d*e^4 + 288*a^2*b^3*d^2 *e^3))/(63*b*e^10) + (20*b^2*x^3*(a^2*e^2 - 8*b^2*d^2 + 4*a*b*d*e))/(3*e^7 ) + (4*b*x^2*(a^3*e^3 - 16*b^3*d^3 + 8*a*b^2*d^2*e + 2*a^2*b*d*e^2))/e^8)) /(x^5*(d + e*x)^(1/2) + (a*d^4*(d + e*x)^(1/2))/(b*e^4) + (x^4*(63*a*e^10 + 252*b*d*e^9)*(d + e*x)^(1/2))/(63*b*e^10) + (2*d*x^3*(2*a*e + 3*b*d)*(d + e*x)^(1/2))/(b*e^2) + (d^3*x*(4*a*e + b*d)*(d + e*x)^(1/2))/(b*e^4) + (2 *d^2*x^2*(3*a*e + 2*b*d)*(d + e*x)^(1/2))/(b*e^3))