Integrand size = 22, antiderivative size = 127 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2 \sqrt {x}}{a^2}-\frac {4 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {2 b^2 \tanh \left (c+d \sqrt {x}\right )}{a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )} \]
-4*b*(2*a^2-b^2)*arctan((a-b)^(1/2)*tanh(1/2*c+1/2*d*x^(1/2))/(a+b)^(1/2)) /a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+2*x^(1/2)/a^2+2*b^2*tanh(c+d*x^(1/2))/a/(a^ 2-b^2)/d/(a+b*sech(c+d*x^(1/2)))
Time = 0.56 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.83 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2 \left (a \left (\left (a^2-b^2\right )^{3/2} \left (c+d \sqrt {x}\right )+\left (4 a^2 b-2 b^3\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \cosh \left (c+d \sqrt {x}\right )+b \left (\left (a^2-b^2\right )^{3/2} \left (c+d \sqrt {x}\right )+\left (4 a^2 b-2 b^3\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a^2-b^2}}\right )+a b \sqrt {a^2-b^2} \sinh \left (c+d \sqrt {x}\right )\right )\right )}{a^2 (a-b) (a+b) \sqrt {a^2-b^2} d \left (b+a \cosh \left (c+d \sqrt {x}\right )\right )} \]
(2*(a*((a^2 - b^2)^(3/2)*(c + d*Sqrt[x]) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tanh[(c + d*Sqrt[x])/2])/Sqrt[a^2 - b^2]])*Cosh[c + d*Sqrt[x]] + b*((a ^2 - b^2)^(3/2)*(c + d*Sqrt[x]) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tanh[ (c + d*Sqrt[x])/2])/Sqrt[a^2 - b^2]] + a*b*Sqrt[a^2 - b^2]*Sinh[c + d*Sqrt [x]])))/(a^2*(a - b)*(a + b)*Sqrt[a^2 - b^2]*d*(b + a*Cosh[c + d*Sqrt[x]]) )
Time = 0.72 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.20, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5959, 3042, 4272, 25, 3042, 4407, 3042, 4318, 3042, 3138, 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.
| \(\displaystyle \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 5959 |
| \(\displaystyle 2 \int \frac {1}{\left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2}d\sqrt {x}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int \frac {1}{\left (a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )\right )^2}d\sqrt {x}\) |
| \(\Big \downarrow \) 4272 |
| \(\displaystyle 2 \left (\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}-\frac {\int -\frac {a^2-b \text {sech}\left (c+d \sqrt {x}\right ) a-b^2}{a+b \text {sech}\left (c+d \sqrt {x}\right )}d\sqrt {x}}{a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \left (\frac {\int \frac {a^2-b \text {sech}\left (c+d \sqrt {x}\right ) a-b^2}{a+b \text {sech}\left (c+d \sqrt {x}\right )}d\sqrt {x}}{a \left (a^2-b^2\right )}+\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}+\frac {\int \frac {a^2-b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right ) a-b^2}{a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )}d\sqrt {x}}{a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle 2 \left (\frac {\frac {\sqrt {x} \left (a^2-b^2\right )}{a}-\frac {b \left (2 a^2-b^2\right ) \int \frac {\text {sech}\left (c+d \sqrt {x}\right )}{a+b \text {sech}\left (c+d \sqrt {x}\right )}d\sqrt {x}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}+\frac {\frac {\sqrt {x} \left (a^2-b^2\right )}{a}-\frac {b \left (2 a^2-b^2\right ) \int \frac {\csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )}{a+b \csc \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )}d\sqrt {x}}{a}}{a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle 2 \left (\frac {\frac {\sqrt {x} \left (a^2-b^2\right )}{a}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{\frac {a \cosh \left (c+d \sqrt {x}\right )}{b}+1}d\sqrt {x}}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \left (\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}+\frac {\frac {\sqrt {x} \left (a^2-b^2\right )}{a}-\frac {\left (2 a^2-b^2\right ) \int \frac {1}{\frac {a \sin \left (i c+i d \sqrt {x}+\frac {\pi }{2}\right )}{b}+1}d\sqrt {x}}{a}}{a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle 2 \left (\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}+\frac {\frac {\sqrt {x} \left (a^2-b^2\right )}{a}+\frac {2 i \left (2 a^2-b^2\right ) \int \frac {1}{\frac {a+b}{b}+\left (1-\frac {a}{b}\right ) x}d\left (i \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )\right )}{a d}}{a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 2 \left (\frac {\frac {\sqrt {x} \left (a^2-b^2\right )}{a}-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {b^2 \tanh \left (c+d \sqrt {x}\right )}{a d \left (a^2-b^2\right ) \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )}\right )\) |
2*((((a^2 - b^2)*Sqrt[x])/a - (2*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tanh[ (c + d*Sqrt[x])/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^2)) + (b^2*Tanh[c + d*Sqrt[x]])/(a*(a^2 - b^2)*d*(a + b*Sech[c + d*Sqrt [x]])))
3.1.69.3.1 Defintions of rubi rules used
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]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[ c + d*x]*((a + b*Csc[c + d*x])^(n + 1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Sim p[1/(a*(n + 1)*(a^2 - b^2)) Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x ], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ erQ[2*n]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] )^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Time = 0.27 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )-1\right )}{a^{2}}-\frac {4 b \left (-\frac {a b \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )^{2} a -\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}+\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+1\right )}{a^{2}}}{d}\) | \(177\) |
| default | \(\frac {-\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )-1\right )}{a^{2}}-\frac {4 b \left (-\frac {a b \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )^{2} a -\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}+\frac {2 \ln \left (\tanh \left (\frac {c}{2}+\frac {d \sqrt {x}}{2}\right )+1\right )}{a^{2}}}{d}\) | \(177\) |
2/d*(-1/a^2*ln(tanh(1/2*c+1/2*d*x^(1/2))-1)-2/a^2*b*(-a*b/(a^2-b^2)*tanh(1 /2*c+1/2*d*x^(1/2))/(tanh(1/2*c+1/2*d*x^(1/2))^2*a-tanh(1/2*c+1/2*d*x^(1/2 ))^2*b+a+b)+(2*a^2-b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh( 1/2*c+1/2*d*x^(1/2))/((a+b)*(a-b))^(1/2)))+1/a^2*ln(tanh(1/2*c+1/2*d*x^(1/ 2))+1))
Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (110) = 220\).
Time = 0.30 (sec) , antiderivative size = 1387, normalized size of antiderivative = 10.92 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Too large to display} \]
[-2*(2*a^3*b^2 - 2*a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*cosh(d*sqrt (x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*sinh(d*sqrt(x) + c)^2 - ( a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x) + 2*(a^2*b^3 - b^5 - (a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x))*cosh(d*sqrt(x) + c) + ((2*a^3*b - a*b^3)*sqrt(-a^2 + b^ 2)*cosh(d*sqrt(x) + c)^2 + (2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)*sinh(d*sqrt( x) + c)^2 + 2*(2*a^2*b^2 - b^4)*sqrt(-a^2 + b^2)*cosh(d*sqrt(x) + c) + 2*( (2*a^3*b - a*b^3)*sqrt(-a^2 + b^2)*cosh(d*sqrt(x) + c) + (2*a^2*b^2 - b^4) *sqrt(-a^2 + b^2))*sinh(d*sqrt(x) + c) + (2*a^3*b - a*b^3)*sqrt(-a^2 + b^2 ))*log((a*b + (b^2 + sqrt(-a^2 + b^2)*b)*cosh(d*sqrt(x) + c) + (a^2 - b^2 - sqrt(-a^2 + b^2)*b)*sinh(d*sqrt(x) + c) + sqrt(-a^2 + b^2)*a)/(a*cosh(d* sqrt(x) + c) + b)) + 2*(a^2*b^3 - b^5 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x )*cosh(d*sqrt(x) + c) - (a^4*b - 2*a^2*b^3 + b^5)*d*sqrt(x))*sinh(d*sqrt(x ) + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*sqrt(x) + c)^2 + (a^7 - 2*a^ 5*b^2 + a^3*b^4)*d*sinh(d*sqrt(x) + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5) *d*cosh(d*sqrt(x) + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 - 2*a^5*b ^2 + a^3*b^4)*d*cosh(d*sqrt(x) + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)*sin h(d*sqrt(x) + c)), -2*(2*a^3*b^2 - 2*a*b^4 - (a^5 - 2*a^3*b^2 + a*b^4)*d*s qrt(x)*cosh(d*sqrt(x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x)*sinh(d* sqrt(x) + c)^2 - (a^5 - 2*a^3*b^2 + a*b^4)*d*sqrt(x) - 2*((2*a^3*b - a*b^3 )*sqrt(a^2 - b^2)*cosh(d*sqrt(x) + c)^2 + (2*a^3*b - a*b^3)*sqrt(a^2 - ...
\[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\int \frac {1}{\sqrt {x} \left (a + b \operatorname {sech}{\left (c + d \sqrt {x} \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=-\frac {4 \, {\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{\left (d \sqrt {x} + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {a^{2} - b^{2}}} - \frac {4 \, {\left (b^{3} e^{\left (d \sqrt {x} + c\right )} + a b^{2}\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} {\left (a e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + 2 \, b e^{\left (d \sqrt {x} + c\right )} + a\right )}} + \frac {2 \, {\left (d \sqrt {x} + c\right )}}{a^{2} d} \]
-4*(2*a^2*b - b^3)*arctan((a*e^(d*sqrt(x) + c) + b)/sqrt(a^2 - b^2))/((a^4 *d - a^2*b^2*d)*sqrt(a^2 - b^2)) - 4*(b^3*e^(d*sqrt(x) + c) + a*b^2)/((a^4 *d - a^2*b^2*d)*(a*e^(2*d*sqrt(x) + 2*c) + 2*b*e^(d*sqrt(x) + c) + a)) + 2 *(d*sqrt(x) + c)/(a^2*d)
Time = 2.50 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.71 \[ \int \frac {1}{\sqrt {x} \left (a+b \text {sech}\left (c+d \sqrt {x}\right )\right )^2} \, dx=\frac {2\,\sqrt {x}}{a^2}-\frac {\frac {4\,b^2\,\sqrt {x}}{d\,\left (a^3\,\sqrt {x}-a\,b^2\,\sqrt {x}\right )}+\frac {4\,b^3\,\sqrt {x}\,{\mathrm {e}}^{c+d\,\sqrt {x}}}{a\,d\,\left (a^3\,\sqrt {x}-a\,b^2\,\sqrt {x}\right )}}{a+2\,b\,{\mathrm {e}}^{c+d\,\sqrt {x}}+a\,{\mathrm {e}}^{2\,c+2\,d\,\sqrt {x}}}+\frac {\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,\sqrt {x}}\,\left (2\,a^2\,b-b^3\right )}{a^3\,\sqrt {x}\,\left (a^2-b^2\right )}-\frac {\left (4\,a^2\,b-2\,b^3\right )\,\left (a+b\,{\mathrm {e}}^{c+d\,\sqrt {x}}\right )}{a^3\,\sqrt {x}\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (4\,a^2\,b-2\,b^3\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {2\,b\,\ln \left (\frac {2\,{\mathrm {e}}^{c+d\,\sqrt {x}}\,\left (2\,a^2\,b-b^3\right )}{a^3\,\sqrt {x}\,\left (a^2-b^2\right )}+\frac {2\,b\,\left (a+b\,{\mathrm {e}}^{c+d\,\sqrt {x}}\right )\,\left (2\,a^2-b^2\right )}{a^3\,\sqrt {x}\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]
(2*x^(1/2))/a^2 - ((4*b^2*x^(1/2))/(d*(a^3*x^(1/2) - a*b^2*x^(1/2))) + (4* b^3*x^(1/2)*exp(c + d*x^(1/2)))/(a*d*(a^3*x^(1/2) - a*b^2*x^(1/2))))/(a + 2*b*exp(c + d*x^(1/2)) + a*exp(2*c + 2*d*x^(1/2))) + (log((2*exp(c + d*x^( 1/2))*(2*a^2*b - b^3))/(a^3*x^(1/2)*(a^2 - b^2)) - ((4*a^2*b - 2*b^3)*(a + b*exp(c + d*x^(1/2))))/(a^3*x^(1/2)*(a + b)^(3/2)*(b - a)^(3/2)))*(4*a^2* b - 2*b^3))/(a^2*d*(a + b)^(3/2)*(b - a)^(3/2)) - (2*b*log((2*exp(c + d*x^ (1/2))*(2*a^2*b - b^3))/(a^3*x^(1/2)*(a^2 - b^2)) + (2*b*(a + b*exp(c + d* x^(1/2)))*(2*a^2 - b^2))/(a^3*x^(1/2)*(a + b)^(3/2)*(b - a)^(3/2)))*(2*a^2 - b^2))/(a^2*d*(a + b)^(3/2)*(b - a)^(3/2))