Integrand size = 20, antiderivative size = 163 \[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\frac {1}{8} \left (\frac {a^2}{b^2}-\frac {c^2}{d^2}\right ) \sqrt {a+b x} \sqrt {c+d x}-\frac {(b c+a d) (a+b x)^{3/2} \sqrt {c+d x}}{4 b^2 d}+\frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}+\frac {(b c-a d)^2 (b c+a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{5/2} d^{5/2}} \]
1/3*(b*x+a)^(3/2)*(d*x+c)^(3/2)/b/d+1/8*(-a*d+b*c)^2*(a*d+b*c)*arctanh(d^( 1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(5/2)-1/4*(a*d+b*c)*(b *x+a)^(3/2)*(d*x+c)^(1/2)/b^2/d+1/8*(a^2/b^2-c^2/d^2)*(b*x+a)^(1/2)*(d*x+c )^(1/2)
Time = 0.42 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.81 \[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-3 a^2 d^2+2 a b d (c+d x)+b^2 \left (-3 c^2+2 c d x+8 d^2 x^2\right )\right )}{24 b^2 d^2}+\frac {(b c-a d)^2 (b c+a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{5/2} d^{5/2}} \]
(Sqrt[a + b*x]*Sqrt[c + d*x]*(-3*a^2*d^2 + 2*a*b*d*(c + d*x) + b^2*(-3*c^2 + 2*c*d*x + 8*d^2*x^2)))/(24*b^2*d^2) + ((b*c - a*d)^2*(b*c + a*d)*ArcTan h[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(8*b^(5/2)*d^(5/2))
Time = 0.23 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {90, 60, 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.
| \(\displaystyle \int x \sqrt {a+b x} \sqrt {c+d x} \, dx\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac {(a d+b c) \int \sqrt {a+b x} \sqrt {c+d x}dx}{2 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac {(a d+b c) \left (\frac {(b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 b}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac {(a d+b c) \left (\frac {(b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 b}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac {(a d+b c) \left (\frac {(b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 b}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 b}\right )}{2 b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x)^{3/2} (c+d x)^{3/2}}{3 b d}-\frac {(a d+b c) \left (\frac {(b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 b}+\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 b}\right )}{2 b d}\) |
((a + b*x)^(3/2)*(c + d*x)^(3/2))/(3*b*d) - ((b*c + a*d)*(((a + b*x)^(3/2) *Sqrt[c + d*x])/(2*b) + ((b*c - a*d)*((Sqrt[a + b*x]*Sqrt[c + d*x])/d - (( b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(Sqrt [b]*d^(3/2))))/(4*b)))/(2*b*d)
3.6.49.3.1 Defintions of rubi rules used
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)^(-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. \(2(131)=262\).
Time = 0.53 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.42
| method | result | size |
| default | \(\frac {\sqrt {b x +a}\, \sqrt {d x +c}\, \left (16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{3} d^{3}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a^{2} b c \,d^{2}-3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) a \,b^{2} c^{2} d +3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) b^{3} c^{3}+4 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b \,d^{2} x +4 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c d x -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a^{2} d^{2}+4 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a b c d -6 \sqrt {b d}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} c^{2}\right )}{48 \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b^{2} d^{2} \sqrt {b d}}\) | \(395\) |
1/48*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*( b*d)^(1/2)+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^3*d^3-3*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^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+4*(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-6*(b*d)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2+4*(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^2/d^2/(b*d)^(1/2)
Time = 0.25 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.49 \[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\left [\frac {3 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{3} d^{3}}, -\frac {3 \, {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (8 \, b^{3} d^{3} x^{2} - 3 \, b^{3} c^{2} d + 2 \, a b^{2} c d^{2} - 3 \, a^{2} b d^{3} + 2 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{3} d^{3}}\right ] \]
[1/96*(3*(b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + 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 - 3*b^3*c^2*d + 2*a*b^2*c*d^2 - 3*a^2*b*d^3 + 2*(b^3*c*d^2 + a*b^2*d^ 3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^3), -1/48*(3*(b^3*c^3 - a*b^2*c^ 2*d - a^2*b*c*d^2 + a^3*d^3)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*s qrt(-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 + 2*a*b^2*c*d^2 - 3*a^2*b*d^ 3 + 2*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^3*d^3)]
\[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\int x \sqrt {a + b x} \sqrt {c + d x}\, dx \]
Exception generated. \[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (131) = 262\).
Time = 0.33 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.12 \[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\frac {\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{2}} + \frac {b^{6} c d^{3} - 13 \, a b^{5} d^{4}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{7} c^{2} d^{2} + 2 \, a b^{6} c d^{3} - 11 \, a^{2} b^{5} d^{4}\right )}}{b^{7} d^{4}}\right )} - \frac {3 \, {\left (b^{3} c^{3} + a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b d^{2}}\right )} {\left | b \right |}}{b} + \frac {6 \, {\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} {\left (2 \, b x + 2 \, a + \frac {b c d - 5 \, a d^{2}}{d^{2}}\right )} \sqrt {b x + a} + \frac {{\left (b^{3} c^{2} + 2 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} d}\right )} a {\left | b \right |}}{b^{3}}}{24 \, b} \]
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 + 6*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) + (b^ 3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt (b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a*abs(b)/b^3)/b
Time = 77.10 (sec) , antiderivative size = 1077, normalized size of antiderivative = 6.61 \[ \int x \sqrt {a+b x} \sqrt {c+d x} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {d}\,\left (a\,d+b\,c\right )\,{\left (a\,d-b\,c\right )}^2\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {b}\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (a^3\,d^3-a^2\,b\,c\,d^2-a\,b^2\,c^2\,d+b^3\,c^3\right )}\right )\,\left (a\,d+b\,c\right )\,{\left (a\,d-b\,c\right )}^2}{4\,b^{5/2}\,d^{5/2}}-\frac {\frac {\left (\sqrt {a+b\,x}-\sqrt {a}\right )\,\left (\frac {a^3\,b^3\,d^3}{4}-\frac {a^2\,b^4\,c\,d^2}{4}-\frac {a\,b^5\,c^2\,d}{4}+\frac {b^6\,c^3}{4}\right )}{d^8\,\left (\sqrt {c+d\,x}-\sqrt {c}\right )}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (\frac {17\,a^3\,b^2\,d^3}{12}+\frac {101\,a^2\,b^3\,c\,d^2}{4}+\frac {101\,a\,b^4\,c^2\,d}{4}+\frac {17\,b^5\,c^3}{12}\right )}{d^7\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^3}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7\,\left (\frac {19\,a^3\,d^3}{2}+\frac {269\,a^2\,b\,c\,d^2}{2}+\frac {269\,a\,b^2\,c^2\,d}{2}+\frac {19\,b^3\,c^3}{2}\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^7}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (\frac {19\,a^3\,b\,d^3}{2}+\frac {269\,a^2\,b^2\,c\,d^2}{2}+\frac {269\,a\,b^3\,c^2\,d}{2}+\frac {19\,b^4\,c^3}{2}\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^5}+\frac {8\,a^{3/2}\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{10}}{d^2\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{10}}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^{11}\,\left (\frac {a^3\,d^3}{4}-\frac {a^2\,b\,c\,d^2}{4}-\frac {a\,b^2\,c^2\,d}{4}+\frac {b^3\,c^3}{4}\right )}{b^2\,d^3\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^{11}}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^9\,\left (\frac {17\,a^3\,d^3}{12}+\frac {101\,a^2\,b\,c\,d^2}{4}+\frac {101\,a\,b^2\,c^2\,d}{4}+\frac {17\,b^3\,c^3}{12}\right )}{b\,d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^9}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8\,\left (32\,a^2\,d^2+96\,a\,b\,c\,d+32\,b^2\,c^2\right )}{d^4\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^8}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (32\,a^2\,b^2\,d^2+96\,a\,b^3\,c\,d+32\,b^4\,c^2\right )}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^4}+\frac {8\,a^{3/2}\,b^4\,c^{3/2}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{d^6\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^2}+\frac {\sqrt {a}\,\sqrt {c}\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (64\,a^2\,b\,d^2+\frac {656\,a\,b^2\,c\,d}{3}+64\,b^3\,c^2\right )}{d^5\,{\left (\sqrt {c+d\,x}-\sqrt {c}\right )}^6}}{\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}}} \]
(atanh((d^(1/2)*(a*d + b*c)*(a*d - b*c)^2*((a + b*x)^(1/2) - a^(1/2)))/(b^ (1/2)*((c + d*x)^(1/2) - c^(1/2))*(a^3*d^3 + b^3*c^3 - a*b^2*c^2*d - a^2*b *c*d^2)))*(a*d + b*c)*(a*d - b*c)^2)/(4*b^(5/2)*d^(5/2)) - ((((a + b*x)^(1 /2) - a^(1/2))*((b^6*c^3)/4 + (a^3*b^3*d^3)/4 - (a^2*b^4*c*d^2)/4 - (a*b^5 *c^2*d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2))) - (((a + b*x)^(1/2) - a^(1/2 ))^3*((17*b^5*c^3)/12 + (17*a^3*b^2*d^3)/12 + (101*a^2*b^3*c*d^2)/4 + (101 *a*b^4*c^2*d)/4))/(d^7*((c + d*x)^(1/2) - c^(1/2))^3) - (((a + b*x)^(1/2) - a^(1/2))^7*((19*a^3*d^3)/2 + (19*b^3*c^3)/2 + (269*a*b^2*c^2*d)/2 + (269 *a^2*b*c*d^2)/2))/(d^5*((c + d*x)^(1/2) - c^(1/2))^7) - (((a + b*x)^(1/2) - a^(1/2))^5*((19*b^4*c^3)/2 + (19*a^3*b*d^3)/2 + (269*a^2*b^2*c*d^2)/2 + (269*a*b^3*c^2*d)/2))/(d^6*((c + d*x)^(1/2) - c^(1/2))^5) + (8*a^(3/2)*c^( 3/2)*((a + b*x)^(1/2) - a^(1/2))^10)/(d^2*((c + d*x)^(1/2) - c^(1/2))^10) + (((a + b*x)^(1/2) - a^(1/2))^11*((a^3*d^3)/4 + (b^3*c^3)/4 - (a*b^2*c^2* d)/4 - (a^2*b*c*d^2)/4))/(b^2*d^3*((c + d*x)^(1/2) - c^(1/2))^11) - (((a + b*x)^(1/2) - a^(1/2))^9*((17*a^3*d^3)/12 + (17*b^3*c^3)/12 + (101*a*b^2*c ^2*d)/4 + (101*a^2*b*c*d^2)/4))/(b*d^4*((c + d*x)^(1/2) - c^(1/2))^9) + (a ^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^8*(32*a^2*d^2 + 32*b^2*c^2 + 96 *a*b*c*d))/(d^4*((c + d*x)^(1/2) - c^(1/2))^8) + (a^(1/2)*c^(1/2)*((a + b* x)^(1/2) - a^(1/2))^4*(32*b^4*c^2 + 32*a^2*b^2*d^2 + 96*a*b^3*c*d))/(d^6*( (c + d*x)^(1/2) - c^(1/2))^4) + (8*a^(3/2)*b^4*c^(3/2)*((a + b*x)^(1/2)...