Integrand size = 22, antiderivative size = 169 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}+\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 b^2 c^2+14 a b c d+15 a^2 d^2-10 b d (b c+a d) x\right )}{24 b^3 d^3}-\frac {(b c+a d) \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{7/2}} \]
-1/8*(a*d+b*c)*(5*a^2*d^2-2*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/ 2)/b^(1/2)/(d*x+c)^(1/2))/b^(7/2)/d^(7/2)+1/3*x^2*(b*x+a)^(1/2)*(d*x+c)^(1 /2)/b/d+1/24*(15*b^2*c^2+14*a*b*c*d+15*a^2*d^2-10*b*d*(a*d+b*c)*x)*(b*x+a) ^(1/2)*(d*x+c)^(1/2)/b^3/d^3
Time = 0.56 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2+2 a b d (7 c-5 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 b^3 d^3}-\frac {\left (5 b^3 c^3+3 a b^2 c^2 d+3 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{7/2} d^{7/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 + 2*a*b*d*(7*c - 5*d*x) + b^2*(15 *c^2 - 10*c*d*x + 8*d^2*x^2)))/(24*b^3*d^3) - ((5*b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt [a + b*x])])/(8*b^(7/2)*d^(7/2))
Time = 0.27 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {111, 27, 164, 66, 221}
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.
| \(\displaystyle \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {\int -\frac {x (4 a c+5 (b c+a d) x)}{2 \sqrt {a+b x} \sqrt {c+d x}}dx}{3 b d}+\frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\int \frac {x (4 a c+5 (b c+a d) x)}{\sqrt {a+b x} \sqrt {c+d x}}dx}{6 b d}\) |
| \(\Big \downarrow \) 164 |
| \(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\frac {3 (a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{8 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{4 b^2 d^2}}{6 b d}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\frac {3 (a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{4 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{4 b^2 d^2}}{6 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\frac {3 (a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{4 b^2 d^2}}{6 b d}\) |
(x^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(3*b*d) - (-1/4*(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*b^2*c^2 + 14*a*b*c*d + 15*a^2*d^2 - 10*b*d*(b*c + a*d)*x))/(b^2*d ^2) + (3*(b*c + a*d)*(5*b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]* Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(4*b^(5/2)*d^(5/2)))/(6*b*d)
3.8.33.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_ ))*((g_.) + (h_.)*(x_)), x_] :> Simp[(-(a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x))*(a + b*x)^(m + 1)*(( c + d*x)^(n + 1)/(b^2*d^2*(m + n + 2)*(m + n + 3))), x] + Simp[(a^2*d^2*f*h *(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)) Int[( a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Leaf count of result is larger than twice the leaf count of optimal. \(394\) vs. \(2(143)=286\).
Time = 1.67 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.34
| method | result | size |
| default | \(-\frac {\left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d +15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x +20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}-28 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right ) \sqrt {b x +a}\, \sqrt {d x +c}}{48 d^{3} b^{3} \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}\) | \(395\) |
-1/48*(-16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+15*ln(1/2*(2*b* d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^3*d^3+9* ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2) )*a^2*b*c*d^2+9*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(b*d)^(1/2)+a*d+ b*c)/(b*d)^(1/2))*a*b^2*c^2*d+15*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2) *(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^3*c^3+20*(b*d)^(1/2)*((b*x+a)*(d*x+c) )^(1/2)*a*b*d^2*x+20*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c*d*x-30*(b*d )^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2-28*(b*d)^(1/2)*((b*x+a)*(d*x+c))^( 1/2)*a*b*c*d-30*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2)*(b*x+a)^(1/2) *(d*x+c)^(1/2)/d^3/b^3/(b*d)^(1/2)/((b*x+a)*(d*x+c))^(1/2)
Time = 0.30 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.43 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} - 10 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} - 10 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{4}}\right ] \]
[1/96*(3*(5*b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(b*d) *log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 - 4*(2*b*d*x + b*c + a* d)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) + 4*(8 *b^3*d^3*x^2 + 15*b^3*c^2*d + 14*a*b^2*c*d^2 + 15*a^2*b*d^3 - 10*(b^3*c*d^ 2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^4), 1/48*(3*(5*b^3*c ^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2*b *d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a* b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(8*b^3*d^3*x^2 + 15*b^3*c^2*d + 14*a*b ^2*c*d^2 + 15*a^2*b*d^3 - 10*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt (d*x + c))/(b^4*d^4)]
\[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\int \frac {x^{3}}{\sqrt {a + b x} \sqrt {c + d x}}\, dx \]
Exception generated. \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.31 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.27 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{4} d} - \frac {5 \, b^{12} c d^{3} + 13 \, a b^{11} d^{4}}{b^{15} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{13} c^{2} d^{2} + 8 \, a b^{12} c d^{3} + 11 \, a^{2} b^{11} d^{4}\right )}}{b^{15} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{3} d^{3}}\right )} b}{24 \, {\left | b \right |}} \]
1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b *x + a)/(b^4*d) - (5*b^12*c*d^3 + 13*a*b^11*d^4)/(b^15*d^5)) + 3*(5*b^13*c ^2*d^2 + 8*a*b^12*c*d^3 + 11*a^2*b^11*d^4)/(b^15*d^5)) + 3*(5*b^3*c^3 + 3* a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^3*d^3))*b/abs(b)
Time = 41.13 (sec) , antiderivative size = 900, normalized size of antiderivative = 5.33 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=-\frac {\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {85\,a^3\,b\,d^3}{12}+\frac {17\,a^2\,b^2\,c\,d^2}{4}+\frac {17\,a\,b^3\,c^2\,d}{4}+\frac {85\,b^4\,c^3}{12}\right )}{d^8\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {5\,a^3\,b^2\,d^3}{4}+\frac {3\,a^2\,b^3\,c\,d^2}{4}+\frac {3\,a\,b^4\,c^2\,d}{4}+\frac {5\,b^5\,c^3}{4}\right )}{d^9\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {33\,a^3\,d^3}{2}+\frac {327\,a^2\,b\,c\,d^2}{2}+\frac {327\,a\,b^2\,c^2\,d}{2}+\frac {33\,b^3\,c^3}{2}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {64\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (\frac {5\,a^3\,d^3}{4}+\frac {3\,a^2\,b\,c\,d^2}{4}+\frac {3\,a\,b^2\,c^2\,d}{4}+\frac {5\,b^3\,c^3}{4}\right )}{b^3\,d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (\frac {85\,a^3\,d^3}{12}+\frac {17\,a^2\,b\,c\,d^2}{4}+\frac {17\,a\,b^2\,c^2\,d}{4}+\frac {85\,b^3\,c^3}{12}\right )}{b^2\,d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {33\,a^3\,d^3}{2}+\frac {327\,a^2\,b\,c\,d^2}{2}+\frac {327\,a\,b^2\,c^2\,d}{2}+\frac {33\,b^3\,c^3}{2}\right )}{b\,d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (128\,a^2\,d^2+\frac {896\,a\,b\,c\,d}{3}+128\,b^2\,c^2\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {64\,a^{3/2}\,b^2\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{12}}{{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{12}}+\frac {b^6}{d^6}-\frac {6\,b^5\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {15\,b^4\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}-\frac {20\,b^3\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}{d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {15\,b^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}-\frac {6\,b\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{d\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}\right )\,\left (a\,d+b\,c\right )\,\left (5\,a^2\,d^2-2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{4\,b^{7/2}\,d^{7/2}} \]
- ((((a + b*x)^(1/2) - a^(1/2))^3*((85*b^4*c^3)/12 + (85*a^3*b*d^3)/12 + ( 17*a^2*b^2*c*d^2)/4 + (17*a*b^3*c^2*d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2) )^3) - (((a + b*x)^(1/2) - a^(1/2))*((5*b^5*c^3)/4 + (5*a^3*b^2*d^3)/4 + ( 3*a^2*b^3*c*d^2)/4 + (3*a*b^4*c^2*d)/4))/(d^9*((c + d*x)^(1/2) - c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^5*((33*a^3*d^3)/2 + (33*b^3*c^3)/2 + (327* a*b^2*c^2*d)/2 + (327*a^2*b*c*d^2)/2))/(d^7*((c + d*x)^(1/2) - c^(1/2))^5) + (64*a^(3/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^8)/(d^4*((c + d*x)^(1/2 ) - c^(1/2))^8) - (((a + b*x)^(1/2) - a^(1/2))^11*((5*a^3*d^3)/4 + (5*b^3* c^3)/4 + (3*a*b^2*c^2*d)/4 + (3*a^2*b*c*d^2)/4))/(b^3*d^4*((c + d*x)^(1/2) - c^(1/2))^11) + (((a + b*x)^(1/2) - a^(1/2))^9*((85*a^3*d^3)/12 + (85*b^ 3*c^3)/12 + (17*a*b^2*c^2*d)/4 + (17*a^2*b*c*d^2)/4))/(b^2*d^5*((c + d*x)^ (1/2) - c^(1/2))^9) - (((a + b*x)^(1/2) - a^(1/2))^7*((33*a^3*d^3)/2 + (33 *b^3*c^3)/2 + (327*a*b^2*c^2*d)/2 + (327*a^2*b*c*d^2)/2))/(b*d^6*((c + d*x )^(1/2) - c^(1/2))^7) + (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(12 8*a^2*d^2 + 128*b^2*c^2 + (896*a*b*c*d)/3))/(d^6*((c + d*x)^(1/2) - c^(1/2 ))^6) + (64*a^(3/2)*b^2*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^4)/(d^6*((c + d*x)^(1/2) - c^(1/2))^4))/(((a + b*x)^(1/2) - a^(1/2))^12/((c + d*x)^(1/2) - c^(1/2))^12 + b^6/d^6 - (6*b^5*((a + b*x)^(1/2) - a^(1/2))^2)/(d^5*((c + d*x)^(1/2) - c^(1/2))^2) + (15*b^4*((a + b*x)^(1/2) - a^(1/2))^4)/(d^4*( (c + d*x)^(1/2) - c^(1/2))^4) - (20*b^3*((a + b*x)^(1/2) - a^(1/2))^6)/...