Integrand size = 30, antiderivative size = 316 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=-\frac {2 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}+\frac {10 b (b d-a e)^4 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^6 (a+b x)}-\frac {4 b^2 (b d-a e)^3 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}{e^6 (a+b x)}+\frac {20 b^3 (b d-a e)^2 (d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}{7 e^6 (a+b x)}-\frac {10 b^4 (b d-a e) (d+e x)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 e^6 (a+b x)}+\frac {2 b^5 (d+e x)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 e^6 (a+b x)} \]
10/3*b*(-a*e+b*d)^4*(e*x+d)^(3/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)-4*b^2*(-a* e+b*d)^3*(e*x+d)^(5/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)+20/7*b^3*(-a*e+b*d)^2 *(e*x+d)^(7/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)-10/9*b^4*(-a*e+b*d)*(e*x+d)^( 9/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)+2/11*b^5*(e*x+d)^(11/2)*((b*x+a)^2)^(1/ 2)/e^6/(b*x+a)-2*(-a*e+b*d)^5*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/e^6/(b*x+a)
Time = 0.11 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {(a+b x)^2} \sqrt {d+e x} \left (693 a^5 e^5+1155 a^4 b e^4 (-2 d+e x)+462 a^3 b^2 e^3 \left (8 d^2-4 d e x+3 e^2 x^2\right )+198 a^2 b^3 e^2 \left (-16 d^3+8 d^2 e x-6 d e^2 x^2+5 e^3 x^3\right )+11 a b^4 e \left (128 d^4-64 d^3 e x+48 d^2 e^2 x^2-40 d e^3 x^3+35 e^4 x^4\right )+b^5 \left (-256 d^5+128 d^4 e x-96 d^3 e^2 x^2+80 d^2 e^3 x^3-70 d e^4 x^4+63 e^5 x^5\right )\right )}{693 e^6 (a+b x)} \]
(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(693*a^5*e^5 + 1155*a^4*b*e^4*(-2*d + e *x) + 462*a^3*b^2*e^3*(8*d^2 - 4*d*e*x + 3*e^2*x^2) + 198*a^2*b^3*e^2*(-16 *d^3 + 8*d^2*e*x - 6*d*e^2*x^2 + 5*e^3*x^3) + 11*a*b^4*e*(128*d^4 - 64*d^3 *e*x + 48*d^2*e^2*x^2 - 40*d*e^3*x^3 + 35*e^4*x^4) + b^5*(-256*d^5 + 128*d ^4*e*x - 96*d^3*e^2*x^2 + 80*d^2*e^3*x^3 - 70*d*e^4*x^4 + 63*e^5*x^5)))/(6 93*e^6*(a + b*x))
Time = 0.31 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.58, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {b^5 (a+b x)^5}{\sqrt {d+e x}}dx}{b^5 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {(a+b x)^5}{\sqrt {d+e x}}dx}{a+b x}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(d+e x)^{9/2} b^5}{e^5}-\frac {5 (b d-a e) (d+e x)^{7/2} b^4}{e^5}+\frac {10 (b d-a e)^2 (d+e x)^{5/2} b^3}{e^5}-\frac {10 (b d-a e)^3 (d+e x)^{3/2} b^2}{e^5}+\frac {5 (b d-a e)^4 \sqrt {d+e x} b}{e^5}+\frac {(a e-b d)^5}{e^5 \sqrt {d+e x}}\right )dx}{a+b x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (-\frac {10 b^4 (d+e x)^{9/2} (b d-a e)}{9 e^6}+\frac {20 b^3 (d+e x)^{7/2} (b d-a e)^2}{7 e^6}-\frac {4 b^2 (d+e x)^{5/2} (b d-a e)^3}{e^6}+\frac {10 b (d+e x)^{3/2} (b d-a e)^4}{3 e^6}-\frac {2 \sqrt {d+e x} (b d-a e)^5}{e^6}+\frac {2 b^5 (d+e x)^{11/2}}{11 e^6}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((-2*(b*d - a*e)^5*Sqrt[d + e*x])/e^6 + (10 *b*(b*d - a*e)^4*(d + e*x)^(3/2))/(3*e^6) - (4*b^2*(b*d - a*e)^3*(d + e*x) ^(5/2))/e^6 + (20*b^3*(b*d - a*e)^2*(d + e*x)^(7/2))/(7*e^6) - (10*b^4*(b* d - a*e)*(d + e*x)^(9/2))/(9*e^6) + (2*b^5*(d + e*x)^(11/2))/(11*e^6)))/(a + b*x)
3.17.95.3.1 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 = 2.23 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.91
| method | result | size |
| gosper | \(\frac {2 \sqrt {e x +d}\, \left (63 x^{5} e^{5} b^{5}+385 x^{4} a \,b^{4} e^{5}-70 x^{4} b^{5} d \,e^{4}+990 x^{3} a^{2} b^{3} e^{5}-440 x^{3} a \,b^{4} d \,e^{4}+80 x^{3} b^{5} d^{2} e^{3}+1386 x^{2} a^{3} b^{2} e^{5}-1188 x^{2} a^{2} b^{3} d \,e^{4}+528 x^{2} a \,b^{4} d^{2} e^{3}-96 x^{2} b^{5} d^{3} e^{2}+1155 a^{4} b \,e^{5} x -1848 a^{3} b^{2} d \,e^{4} x +1584 x \,a^{2} b^{3} d^{2} e^{3}-704 x a \,b^{4} d^{3} e^{2}+128 b^{5} d^{4} e x +693 a^{5} e^{5}-2310 a^{4} b d \,e^{4}+3696 a^{3} b^{2} d^{2} e^{3}-3168 a^{2} b^{3} d^{3} e^{2}+1408 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{693 e^{6} \left (b x +a \right )^{5}}\) | \(289\) |
| default | \(\frac {2 \sqrt {e x +d}\, \left (63 x^{5} e^{5} b^{5}+385 x^{4} a \,b^{4} e^{5}-70 x^{4} b^{5} d \,e^{4}+990 x^{3} a^{2} b^{3} e^{5}-440 x^{3} a \,b^{4} d \,e^{4}+80 x^{3} b^{5} d^{2} e^{3}+1386 x^{2} a^{3} b^{2} e^{5}-1188 x^{2} a^{2} b^{3} d \,e^{4}+528 x^{2} a \,b^{4} d^{2} e^{3}-96 x^{2} b^{5} d^{3} e^{2}+1155 a^{4} b \,e^{5} x -1848 a^{3} b^{2} d \,e^{4} x +1584 x \,a^{2} b^{3} d^{2} e^{3}-704 x a \,b^{4} d^{3} e^{2}+128 b^{5} d^{4} e x +693 a^{5} e^{5}-2310 a^{4} b d \,e^{4}+3696 a^{3} b^{2} d^{2} e^{3}-3168 a^{2} b^{3} d^{3} e^{2}+1408 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{693 e^{6} \left (b x +a \right )^{5}}\) | \(289\) |
| risch | \(\frac {2 \sqrt {\left (b x +a \right )^{2}}\, \left (63 x^{5} e^{5} b^{5}+385 x^{4} a \,b^{4} e^{5}-70 x^{4} b^{5} d \,e^{4}+990 x^{3} a^{2} b^{3} e^{5}-440 x^{3} a \,b^{4} d \,e^{4}+80 x^{3} b^{5} d^{2} e^{3}+1386 x^{2} a^{3} b^{2} e^{5}-1188 x^{2} a^{2} b^{3} d \,e^{4}+528 x^{2} a \,b^{4} d^{2} e^{3}-96 x^{2} b^{5} d^{3} e^{2}+1155 a^{4} b \,e^{5} x -1848 a^{3} b^{2} d \,e^{4} x +1584 x \,a^{2} b^{3} d^{2} e^{3}-704 x a \,b^{4} d^{3} e^{2}+128 b^{5} d^{4} e x +693 a^{5} e^{5}-2310 a^{4} b d \,e^{4}+3696 a^{3} b^{2} d^{2} e^{3}-3168 a^{2} b^{3} d^{3} e^{2}+1408 a \,b^{4} d^{4} e -256 b^{5} d^{5}\right ) \sqrt {e x +d}}{693 \left (b x +a \right ) e^{6}}\) | \(289\) |
2/693*(e*x+d)^(1/2)*(63*b^5*e^5*x^5+385*a*b^4*e^5*x^4-70*b^5*d*e^4*x^4+990 *a^2*b^3*e^5*x^3-440*a*b^4*d*e^4*x^3+80*b^5*d^2*e^3*x^3+1386*a^3*b^2*e^5*x ^2-1188*a^2*b^3*d*e^4*x^2+528*a*b^4*d^2*e^3*x^2-96*b^5*d^3*e^2*x^2+1155*a^ 4*b*e^5*x-1848*a^3*b^2*d*e^4*x+1584*a^2*b^3*d^2*e^3*x-704*a*b^4*d^3*e^2*x+ 128*b^5*d^4*e*x+693*a^5*e^5-2310*a^4*b*d*e^4+3696*a^3*b^2*d^2*e^3-3168*a^2 *b^3*d^3*e^2+1408*a*b^4*d^4*e-256*b^5*d^5)*((b*x+a)^2)^(5/2)/e^6/(b*x+a)^5
Time = 0.33 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (63 \, b^{5} e^{5} x^{5} - 256 \, b^{5} d^{5} + 1408 \, a b^{4} d^{4} e - 3168 \, a^{2} b^{3} d^{3} e^{2} + 3696 \, a^{3} b^{2} d^{2} e^{3} - 2310 \, a^{4} b d e^{4} + 693 \, a^{5} e^{5} - 35 \, {\left (2 \, b^{5} d e^{4} - 11 \, a b^{4} e^{5}\right )} x^{4} + 10 \, {\left (8 \, b^{5} d^{2} e^{3} - 44 \, a b^{4} d e^{4} + 99 \, a^{2} b^{3} e^{5}\right )} x^{3} - 6 \, {\left (16 \, b^{5} d^{3} e^{2} - 88 \, a b^{4} d^{2} e^{3} + 198 \, a^{2} b^{3} d e^{4} - 231 \, a^{3} b^{2} e^{5}\right )} x^{2} + {\left (128 \, b^{5} d^{4} e - 704 \, a b^{4} d^{3} e^{2} + 1584 \, a^{2} b^{3} d^{2} e^{3} - 1848 \, a^{3} b^{2} d e^{4} + 1155 \, a^{4} b e^{5}\right )} x\right )} \sqrt {e x + d}}{693 \, e^{6}} \]
2/693*(63*b^5*e^5*x^5 - 256*b^5*d^5 + 1408*a*b^4*d^4*e - 3168*a^2*b^3*d^3* e^2 + 3696*a^3*b^2*d^2*e^3 - 2310*a^4*b*d*e^4 + 693*a^5*e^5 - 35*(2*b^5*d* e^4 - 11*a*b^4*e^5)*x^4 + 10*(8*b^5*d^2*e^3 - 44*a*b^4*d*e^4 + 99*a^2*b^3* e^5)*x^3 - 6*(16*b^5*d^3*e^2 - 88*a*b^4*d^2*e^3 + 198*a^2*b^3*d*e^4 - 231* a^3*b^2*e^5)*x^2 + (128*b^5*d^4*e - 704*a*b^4*d^3*e^2 + 1584*a^2*b^3*d^2*e ^3 - 1848*a^3*b^2*d*e^4 + 1155*a^4*b*e^5)*x)*sqrt(e*x + d)/e^6
\[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=\int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}{\sqrt {d + e x}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 338, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (63 \, b^{5} e^{6} x^{6} - 256 \, b^{5} d^{6} + 1408 \, a b^{4} d^{5} e - 3168 \, a^{2} b^{3} d^{4} e^{2} + 3696 \, a^{3} b^{2} d^{3} e^{3} - 2310 \, a^{4} b d^{2} e^{4} + 693 \, a^{5} d e^{5} - 7 \, {\left (b^{5} d e^{5} - 55 \, a b^{4} e^{6}\right )} x^{5} + 5 \, {\left (2 \, b^{5} d^{2} e^{4} - 11 \, a b^{4} d e^{5} + 198 \, a^{2} b^{3} e^{6}\right )} x^{4} - 2 \, {\left (8 \, b^{5} d^{3} e^{3} - 44 \, a b^{4} d^{2} e^{4} + 99 \, a^{2} b^{3} d e^{5} - 693 \, a^{3} b^{2} e^{6}\right )} x^{3} + {\left (32 \, b^{5} d^{4} e^{2} - 176 \, a b^{4} d^{3} e^{3} + 396 \, a^{2} b^{3} d^{2} e^{4} - 462 \, a^{3} b^{2} d e^{5} + 1155 \, a^{4} b e^{6}\right )} x^{2} - {\left (128 \, b^{5} d^{5} e - 704 \, a b^{4} d^{4} e^{2} + 1584 \, a^{2} b^{3} d^{3} e^{3} - 1848 \, a^{3} b^{2} d^{2} e^{4} + 1155 \, a^{4} b d e^{5} - 693 \, a^{5} e^{6}\right )} x\right )}}{693 \, \sqrt {e x + d} e^{6}} \]
2/693*(63*b^5*e^6*x^6 - 256*b^5*d^6 + 1408*a*b^4*d^5*e - 3168*a^2*b^3*d^4* e^2 + 3696*a^3*b^2*d^3*e^3 - 2310*a^4*b*d^2*e^4 + 693*a^5*d*e^5 - 7*(b^5*d *e^5 - 55*a*b^4*e^6)*x^5 + 5*(2*b^5*d^2*e^4 - 11*a*b^4*d*e^5 + 198*a^2*b^3 *e^6)*x^4 - 2*(8*b^5*d^3*e^3 - 44*a*b^4*d^2*e^4 + 99*a^2*b^3*d*e^5 - 693*a ^3*b^2*e^6)*x^3 + (32*b^5*d^4*e^2 - 176*a*b^4*d^3*e^3 + 396*a^2*b^3*d^2*e^ 4 - 462*a^3*b^2*d*e^5 + 1155*a^4*b*e^6)*x^2 - (128*b^5*d^5*e - 704*a*b^4*d ^4*e^2 + 1584*a^2*b^3*d^3*e^3 - 1848*a^3*b^2*d^2*e^4 + 1155*a^4*b*d*e^5 - 693*a^5*e^6)*x)/(sqrt(e*x + d)*e^6)
Time = 0.31 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (693 \, \sqrt {e x + d} a^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1155 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x + d} d\right )} a^{4} b \mathrm {sgn}\left (b x + a\right )}{e} + \frac {462 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{3} b^{2} \mathrm {sgn}\left (b x + a\right )}{e^{2}} + \frac {198 \, {\left (5 \, {\left (e x + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x + d} d^{3}\right )} a^{2} b^{3} \mathrm {sgn}\left (b x + a\right )}{e^{3}} + \frac {11 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} a b^{4} \mathrm {sgn}\left (b x + a\right )}{e^{4}} + \frac {{\left (63 \, {\left (e x + d\right )}^{\frac {11}{2}} - 385 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 990 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 1386 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} + 1155 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 693 \, \sqrt {e x + d} d^{5}\right )} b^{5} \mathrm {sgn}\left (b x + a\right )}{e^{5}}\right )}}{693 \, e} \]
2/693*(693*sqrt(e*x + d)*a^5*sgn(b*x + a) + 1155*((e*x + d)^(3/2) - 3*sqrt (e*x + d)*d)*a^4*b*sgn(b*x + a)/e + 462*(3*(e*x + d)^(5/2) - 10*(e*x + d)^ (3/2)*d + 15*sqrt(e*x + d)*d^2)*a^3*b^2*sgn(b*x + a)/e^2 + 198*(5*(e*x + d )^(7/2) - 21*(e*x + d)^(5/2)*d + 35*(e*x + d)^(3/2)*d^2 - 35*sqrt(e*x + d) *d^3)*a^2*b^3*sgn(b*x + a)/e^3 + 11*(35*(e*x + d)^(9/2) - 180*(e*x + d)^(7 /2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 315*sqrt(e*x + d)*d^4)*a*b^4*sgn(b*x + a)/e^4 + (63*(e*x + d)^(11/2) - 385*(e*x + d)^(9/ 2)*d + 990*(e*x + d)^(7/2)*d^2 - 1386*(e*x + d)^(5/2)*d^3 + 1155*(e*x + d) ^(3/2)*d^4 - 693*sqrt(e*x + d)*d^5)*b^5*sgn(b*x + a)/e^5)/e
Time = 10.18 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{\sqrt {d+e x}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (\frac {2\,b^4\,x^6}{11}-\frac {-1386\,a^5\,d\,e^5+4620\,a^4\,b\,d^2\,e^4-7392\,a^3\,b^2\,d^3\,e^3+6336\,a^2\,b^3\,d^4\,e^2-2816\,a\,b^4\,d^5\,e+512\,b^5\,d^6}{693\,b\,e^6}+\frac {2\,b^3\,x^5\,\left (55\,a\,e-b\,d\right )}{99\,e}+\frac {x\,\left (1386\,a^5\,e^6-2310\,a^4\,b\,d\,e^5+3696\,a^3\,b^2\,d^2\,e^4-3168\,a^2\,b^3\,d^3\,e^3+1408\,a\,b^4\,d^4\,e^2-256\,b^5\,d^5\,e\right )}{693\,b\,e^6}+\frac {x^2\,\left (2310\,a^4\,b\,e^6-924\,a^3\,b^2\,d\,e^5+792\,a^2\,b^3\,d^2\,e^4-352\,a\,b^4\,d^3\,e^3+64\,b^5\,d^4\,e^2\right )}{693\,b\,e^6}+\frac {10\,b^2\,x^4\,\left (198\,a^2\,e^2-11\,a\,b\,d\,e+2\,b^2\,d^2\right )}{693\,e^2}+\frac {4\,b\,x^3\,\left (693\,a^3\,e^3-99\,a^2\,b\,d\,e^2+44\,a\,b^2\,d^2\,e-8\,b^3\,d^3\right )}{693\,e^3}\right )}{x\,\sqrt {d+e\,x}+\frac {a\,\sqrt {d+e\,x}}{b}} \]
((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*((2*b^4*x^6)/11 - (512*b^5*d^6 - 1386*a^5 *d*e^5 + 4620*a^4*b*d^2*e^4 + 6336*a^2*b^3*d^4*e^2 - 7392*a^3*b^2*d^3*e^3 - 2816*a*b^4*d^5*e)/(693*b*e^6) + (2*b^3*x^5*(55*a*e - b*d))/(99*e) + (x*( 1386*a^5*e^6 - 256*b^5*d^5*e + 1408*a*b^4*d^4*e^2 - 3168*a^2*b^3*d^3*e^3 + 3696*a^3*b^2*d^2*e^4 - 2310*a^4*b*d*e^5))/(693*b*e^6) + (x^2*(2310*a^4*b* e^6 + 64*b^5*d^4*e^2 - 352*a*b^4*d^3*e^3 - 924*a^3*b^2*d*e^5 + 792*a^2*b^3 *d^2*e^4))/(693*b*e^6) + (10*b^2*x^4*(198*a^2*e^2 + 2*b^2*d^2 - 11*a*b*d*e ))/(693*e^2) + (4*b*x^3*(693*a^3*e^3 - 8*b^3*d^3 + 44*a*b^2*d^2*e - 99*a^2 *b*d*e^2))/(693*e^3)))/(x*(d + e*x)^(1/2) + (a*(d + e*x)^(1/2))/b)