Integrand size = 22, antiderivative size = 191 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {\left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d^2}-\frac {(3 b c+5 a d) \sqrt {a+b x} (c+d x)^{3/2}}{12 b^2 d^2}+\frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}+\frac {(b c-a d) \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{5/2}} \]
1/8*(-a*d+b*c)*(5*a^2*d^2+2*a*b*c*d+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^(5/2)-1/12*(5*a*d+3*b*c)*(d*x+c)^(3/2)*( b*x+a)^(1/2)/b^2/d^2+1/3*x*(d*x+c)^(3/2)*(b*x+a)^(1/2)/b/d+1/8*(5*a^2*d^2+ 2*a*b*c*d+b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^2
Time = 10.24 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {\sqrt {c+d x} \left (\sqrt {d} \sqrt {a+b x} \left (15 a^2 d^2-2 a b d (2 c+5 d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )+\frac {3 \sqrt {b c-a d} \left (b^2 c^2+2 a b c d+5 a^2 d^2\right ) \text {arcsinh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\right )}{24 b^3 d^{5/2}} \]
(Sqrt[c + d*x]*(Sqrt[d]*Sqrt[a + b*x]*(15*a^2*d^2 - 2*a*b*d*(2*c + 5*d*x) + b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)) + (3*Sqrt[b*c - a*d]*(b^2*c^2 + 2*a* b*c*d + 5*a^2*d^2)*ArcSinh[(Sqrt[d]*Sqrt[a + b*x])/Sqrt[b*c - a*d]])/Sqrt[ (b*(c + d*x))/(b*c - a*d)]))/(24*b^3*d^(5/2))
Time = 0.28 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {101, 27, 90, 60, 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.
| \(\displaystyle \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx\) |
| \(\Big \downarrow \) 101 |
| \(\displaystyle \frac {\int -\frac {\sqrt {c+d x} (2 a c+(3 b c+5 a d) x)}{2 \sqrt {a+b x}}dx}{3 b d}+\frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\int \frac {\sqrt {c+d x} (2 a c+(3 b c+5 a d) x)}{\sqrt {a+b x}}dx}{6 b d}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \int \frac {\sqrt {c+d x}}{\sqrt {a+b x}}dx}{4 b d}}{6 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \left (\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}}{6 b d}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \left (\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}}{6 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {x \sqrt {a+b x} (c+d x)^{3/2}}{3 b d}-\frac {\frac {\sqrt {a+b x} (c+d x)^{3/2} (5 a d+3 b c)}{2 b d}-\frac {3 \left (5 a^2 d^2+2 a b c d+b^2 c^2\right ) \left (\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} \sqrt {d}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b}\right )}{4 b d}}{6 b d}\) |
(x*Sqrt[a + b*x]*(c + d*x)^(3/2))/(3*b*d) - (((3*b*c + 5*a*d)*Sqrt[a + b*x ]*(c + d*x)^(3/2))/(2*b*d) - (3*(b^2*c^2 + 2*a*b*c*d + 5*a^2*d^2)*((Sqrt[a + b*x]*Sqrt[c + d*x])/b + ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(S qrt[b]*Sqrt[c + d*x])])/(b^(3/2)*Sqrt[d])))/(4*b*d))/(6*b*d)
3.7.99.3.1 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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 3))), x] + Simp[1/(d*f*(n + p + 3)) Int[(c + d*x)^n*(e + f*x)^p*Simp [a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f *(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 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. \(2(159)=318\).
Time = 1.62 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.07
| method | result | size |
| default | \(-\frac {\sqrt {d x +c}\, \sqrt {b x +a}\, \left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}-9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d -3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+20 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x -4 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -30 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+8 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d +6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{3} d^{2} \sqrt {b d}}\) | \(395\) |
-1/48*(d*x+c)^(1/2)*(b*x+a)^(1/2)*(-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^2-3*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-3*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-4*(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+8*(b*d)^ (1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b*c*d+6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2 )*b^2*c^2)/((b*x+a)*(d*x+c))^(1/2)/b^3/d^2/(b*d)^(1/2)
Time = 0.24 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.14 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\left [-\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{3}}, -\frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d - 4 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{3}}\right ] \]
[-1/96*(3*(b^3*c^3 + a*b^2*c^2*d + 3*a^2*b*c*d^2 - 5*a^3*d^3)*sqrt(b*d)*lo g(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 - 3*b^3*c^2*d - 4*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(b^3*c*d^2 - 5* a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^3), -1/48*(3*(b^3*c^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 - 3*b^3*c^2*d - 4*a*b^2*c*d^2 + 15*a^2*b*d^3 + 2*(b^3*c*d^2 - 5*a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c ))/(b^4*d^3)]
\[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\int \frac {x^{2} \sqrt {c + d x}}{\sqrt {a + b x}}\, dx \]
Exception generated. \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\text {Exception raised: ValueError} \]
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 = 0.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.08 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} {\left | b \right |}}{24 \, b^{3}} \]
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^2 + (b^6*c*d^3 - 13*a*b^5*d^4)/(b^7*d^4)) - 3*(b^7*c^2*d^2 + 2*a *b^6*c*d^3 - 11*a^2*b^5*d^4)/(b^7*d^4)) - 3*(b^3*c^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*d^2))*abs(b)/b^3
Time = 50.67 (sec) , antiderivative size = 924, normalized size of antiderivative = 4.84 \[ \int \frac {x^2 \sqrt {c+d x}}{\sqrt {a+b x}} \, dx=-\frac {\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 {a\,b^4\,c^2\,d}{4}+\frac {b^5\,c^3}{4}\right )}{d^8\,\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 {313\,a^2\,b\,c\,d^2}{2}+\frac {275\,a\,b^2\,c^2\,d}{2}+\frac {19\,b^3\,c^3}{2}\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}-\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 {91\,a\,b^3\,c^2\,d}{4}+\frac {17\,b^4\,c^3}{12}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}+\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 {a\,b^2\,c^2\,d}{4}+\frac {b^3\,c^3}{4}\right )}{b^3\,d^3\,{\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 {91\,a\,b^2\,c^2\,d}{4}+\frac {17\,b^3\,c^3}{12}\right )}{b^2\,d^4\,{\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 {313\,a^2\,b\,c\,d^2}{2}+\frac {275\,a\,b^2\,c^2\,d}{2}+\frac {19\,b^3\,c^3}{2}\right )}{b\,d^5\,{\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 {704\,a\,b\,c\,d}{3}+64\,b^2\,c^2\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}+\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,b\,c^2+96\,a\,d\,c\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {\sqrt {a}\,\sqrt {c}\,\left (32\,b^3\,c^2+96\,a\,d\,b^2\,c\right )\,{\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+b^2\,c^2\right )}{4\,b^{7/2}\,d^{5/2}} \]
- ((((a + b*x)^(1/2) - a^(1/2))*((b^5*c^3)/4 - (5*a^3*b^2*d^3)/4 + (3*a^2* b^3*c*d^2)/4 + (a*b^4*c^2*d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2))^5*((33*a^3*d^3)/2 + (19*b^3*c^3)/2 + (275*a*b^2*c^ 2*d)/2 + (313*a^2*b*c*d^2)/2))/(d^6*((c + d*x)^(1/2) - c^(1/2))^5) - (((a + b*x)^(1/2) - a^(1/2))^3*((17*b^4*c^3)/12 - (85*a^3*b*d^3)/12 + (17*a^2*b ^2*c*d^2)/4 + (91*a*b^3*c^2*d)/4))/(d^7*((c + d*x)^(1/2) - c^(1/2))^3) + ( ((a + b*x)^(1/2) - a^(1/2))^11*((b^3*c^3)/4 - (5*a^3*d^3)/4 + (a*b^2*c^2*d )/4 + (3*a^2*b*c*d^2)/4))/(b^3*d^3*((c + d*x)^(1/2) - c^(1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^9*((17*b^3*c^3)/12 - (85*a^3*d^3)/12 + (91*a*b^2*c ^2*d)/4 + (17*a^2*b*c*d^2)/4))/(b^2*d^4*((c + d*x)^(1/2) - c^(1/2))^9) - ( ((a + b*x)^(1/2) - a^(1/2))^7*((33*a^3*d^3)/2 + (19*b^3*c^3)/2 + (275*a*b^ 2*c^2*d)/2 + (313*a^2*b*c*d^2)/2))/(b*d^5*((c + d*x)^(1/2) - c^(1/2))^7) + (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(128*a^2*d^2 + 64*b^2*c^2 + (704*a*b*c*d)/3))/(d^5*((c + d*x)^(1/2) - c^(1/2))^6) + (a^(1/2)*c^(1/2) *(32*b*c^2 + 96*a*c*d)*((a + b*x)^(1/2) - a^(1/2))^8)/(d^4*((c + d*x)^(1/2 ) - c^(1/2))^8) + (a^(1/2)*c^(1/2)*(32*b^3*c^2 + 96*a*b^2*c*d)*((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*...