Integrand size = 23, antiderivative size = 121 \[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {\left (8 a^2+4 a b+3 b^2\right ) x}{8 b^3}-\frac {a^{5/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a-b} b^3 d}-\frac {(4 a+3 b) \cosh (c+d x) \sinh (c+d x)}{8 b^2 d}+\frac {\cosh (c+d x) \sinh ^3(c+d x)}{4 b d} \]
1/8*(8*a^2+4*a*b+3*b^2)*x/b^3-1/8*(4*a+3*b)*cosh(d*x+c)*sinh(d*x+c)/b^2/d+ 1/4*cosh(d*x+c)*sinh(d*x+c)^3/b/d-a^(5/2)*arctanh((a-b)^(1/2)*tanh(d*x+c)/ a^(1/2))/b^3/d/(a-b)^(1/2)
Time = 0.78 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {4 \left (8 a^2+4 a b+3 b^2\right ) (c+d x)-\frac {32 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a-b}}-8 b (a+b) \sinh (2 (c+d x))+b^2 \sinh (4 (c+d x))}{32 b^3 d} \]
(4*(8*a^2 + 4*a*b + 3*b^2)*(c + d*x) - (32*a^(5/2)*ArcTanh[(Sqrt[a - b]*Ta nh[c + d*x])/Sqrt[a]])/Sqrt[a - b] - 8*b*(a + b)*Sinh[2*(c + d*x)] + b^2*S inh[4*(c + d*x)])/(32*b^3*d)
Time = 0.43 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 25, 3666, 372, 440, 25, 397, 219, 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.
| \(\displaystyle \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i c+i d x)^6}{a-b \sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i c+i d x)^6}{a-b \sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\tanh ^6(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^3 \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tanh ^3(c+d x)}{4 b \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\int \frac {\tanh ^2(c+d x) \left ((a+3 b) \tanh ^2(c+d x)+3 a\right )}{\left (1-\tanh ^2(c+d x)\right )^2 \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{4 b}}{d}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\tanh ^3(c+d x)}{4 b \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {\left (4 a^2+b a+3 b^2\right ) \tanh ^2(c+d x)+a (4 a+3 b)}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{2 b}+\frac {(4 a+3 b) \tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}}{4 b}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\tanh ^3(c+d x)}{4 b \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(4 a+3 b) \tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\int \frac {\left (4 a^2+b a+3 b^2\right ) \tanh ^2(c+d x)+a (4 a+3 b)}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{2 b}}{4 b}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tanh ^3(c+d x)}{4 b \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(4 a+3 b) \tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (8 a^2+4 a b+3 b^2\right ) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{b}-\frac {8 a^3 \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{2 b}}{4 b}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\tanh ^3(c+d x)}{4 b \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(4 a+3 b) \tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (8 a^2+4 a b+3 b^2\right ) \text {arctanh}(\tanh (c+d x))}{b}-\frac {8 a^3 \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{2 b}}{4 b}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\tanh ^3(c+d x)}{4 b \left (1-\tanh ^2(c+d x)\right )^2}-\frac {\frac {(4 a+3 b) \tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right )}-\frac {\frac {\left (8 a^2+4 a b+3 b^2\right ) \text {arctanh}(\tanh (c+d x))}{b}-\frac {8 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{b \sqrt {a-b}}}{2 b}}{4 b}}{d}\) |
(Tanh[c + d*x]^3/(4*b*(1 - Tanh[c + d*x]^2)^2) - (-1/2*(((8*a^2 + 4*a*b + 3*b^2)*ArcTanh[Tanh[c + d*x]])/b - (8*a^(5/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a - b]*b))/b + ((4*a + 3*b)*Tanh[c + d*x])/(2*b*(1 - Tanh[c + d*x]^2)))/(4*b))/d
3.1.29.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 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]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(245\) vs. \(2(107)=214\).
Time = 0.71 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.03
| method | result | size |
| risch | \(\frac {x \,a^{2}}{b^{3}}+\frac {a x}{2 b^{2}}+\frac {3 x}{8 b}+\frac {{\mathrm e}^{4 d x +4 c}}{64 b d}-\frac {{\mathrm e}^{2 d x +2 c} a}{8 b^{2} d}-\frac {{\mathrm e}^{2 d x +2 c}}{8 b d}+\frac {{\mathrm e}^{-2 d x -2 c} a}{8 b^{2} d}+\frac {{\mathrm e}^{-2 d x -2 c}}{8 b d}-\frac {{\mathrm e}^{-4 d x -4 c}}{64 b d}+\frac {\sqrt {\left (a -b \right ) a}\, a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a +2 \sqrt {\left (a -b \right ) a}-b}{b}\right )}{2 \left (a -b \right ) d \,b^{3}}-\frac {\sqrt {\left (a -b \right ) a}\, a^{2} \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-2 a +2 \sqrt {\left (a -b \right ) a}+b}{b}\right )}{2 \left (a -b \right ) d \,b^{3}}\) | \(246\) |
| derivativedivides | \(\frac {\frac {1}{4 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4 a +b}{8 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {4 a +3 b}{8 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-8 a^{2}-4 a b -3 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 b^{3}}-\frac {1}{4 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {-4 a -b}{8 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 a +3 b}{8 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (8 a^{2}+4 a b +3 b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 b^{3}}+\frac {2 a^{4} \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{b^{3}}}{d}\) | \(417\) |
| default | \(\frac {\frac {1}{4 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {1}{2 b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4 a +b}{8 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {4 a +3 b}{8 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-8 a^{2}-4 a b -3 b^{2}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 b^{3}}-\frac {1}{4 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {1}{2 b \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {-4 a -b}{8 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 a +3 b}{8 b^{2} \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {\left (8 a^{2}+4 a b +3 b^{2}\right ) \ln \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 b^{3}}+\frac {2 a^{4} \left (\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{b^{3}}}{d}\) | \(417\) |
x/b^3*a^2+1/2*a*x/b^2+3/8*x/b+1/64/b/d*exp(4*d*x+4*c)-1/8/b^2/d*exp(2*d*x+ 2*c)*a-1/8/b/d*exp(2*d*x+2*c)+1/8/b^2/d*exp(-2*d*x-2*c)*a+1/8/b/d*exp(-2*d *x-2*c)-1/64/b/d*exp(-4*d*x-4*c)+1/2*((a-b)*a)^(1/2)/(a-b)*a^2/d/b^3*ln(ex p(2*d*x+2*c)+(2*a+2*((a-b)*a)^(1/2)-b)/b)-1/2*((a-b)*a)^(1/2)/(a-b)*a^2/d/ b^3*ln(exp(2*d*x+2*c)-(-2*a+2*((a-b)*a)^(1/2)+b)/b)
Leaf count of result is larger than twice the leaf count of optimal. 716 vs. \(2 (107) = 214\).
Time = 0.34 (sec) , antiderivative size = 1725, normalized size of antiderivative = 14.26 \[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Too large to display} \]
[1/64*(b^2*cosh(d*x + c)^8 + 8*b^2*cosh(d*x + c)*sinh(d*x + c)^7 + b^2*sin h(d*x + c)^8 + 8*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c)^4 - 8*(a*b + b^ 2)*cosh(d*x + c)^6 + 4*(7*b^2*cosh(d*x + c)^2 - 2*a*b - 2*b^2)*sinh(d*x + c)^6 + 8*(7*b^2*cosh(d*x + c)^3 - 6*(a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*b^2*cosh(d*x + c)^4 + 4*(8*a^2 + 4*a*b + 3*b^2)*d*x - 60*(a*b + b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*b^2*cosh(d*x + c)^5 + 4*(8 *a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c) - 20*(a*b + b^2)*cosh(d*x + c)^3)* sinh(d*x + c)^3 + 8*(a*b + b^2)*cosh(d*x + c)^2 + 4*(7*b^2*cosh(d*x + c)^6 + 12*(8*a^2 + 4*a*b + 3*b^2)*d*x*cosh(d*x + c)^2 - 30*(a*b + b^2)*cosh(d* x + c)^4 + 2*a*b + 2*b^2)*sinh(d*x + c)^2 + 32*(a^2*cosh(d*x + c)^4 + 4*a^ 2*cosh(d*x + c)^3*sinh(d*x + c) + 6*a^2*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4)*sqrt(a/(a - b)) *log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh (d*x + c)^4 + 2*(2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b - b^2)*sinh(d*x + c)^2 + 8*a^2 - 8*a*b + b^2 + 4*(b^2*cosh(d*x + c) ^3 + (2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c) + 4*((a*b - b^2)*cosh(d*x + c)^2 + 2*(a*b - b^2)*cosh(d*x + c)*sinh(d*x + c) + (a*b - b^2)*sinh(d*x + c)^2 + 2*a^2 - 3*a*b + b^2)*sqrt(a/(a - b)))/(b*cosh(d*x + c)^4 + 4*b*co sh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(2*a - b)*cosh(d*x + c )^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x...
Timed out. \[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
\[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{6}}{b \sinh \left (d x + c\right )^{2} + a} \,d x } \]
Time = 1.82 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.20 \[ \int \frac {\sinh ^6(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {x\,\left (8\,a^2+4\,a\,b+3\,b^2\right )}{8\,b^3}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,b\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,b\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+b\right )}{8\,b^2\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{8\,b^2\,d}+\frac {a^{5/2}\,\ln \left (\frac {4\,a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{b^4}-\frac {2\,a^{5/2}\,\left (b\,d+2\,a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^4\,d\,\sqrt {a-b}}\right )}{2\,b^3\,d\,\sqrt {a-b}}-\frac {a^{5/2}\,\ln \left (\frac {4\,a^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{b^4}+\frac {2\,a^{5/2}\,\left (b\,d+2\,a\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}-b\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}\right )}{b^4\,d\,\sqrt {a-b}}\right )}{2\,b^3\,d\,\sqrt {a-b}} \]
(x*(4*a*b + 8*a^2 + 3*b^2))/(8*b^3) - exp(- 4*c - 4*d*x)/(64*b*d) + exp(4* c + 4*d*x)/(64*b*d) + (exp(- 2*c - 2*d*x)*(a + b))/(8*b^2*d) - (exp(2*c + 2*d*x)*(a + b))/(8*b^2*d) + (a^(5/2)*log((4*a^3*exp(2*c + 2*d*x))/b^4 - (2 *a^(5/2)*(b*d + 2*a*d*exp(2*c + 2*d*x) - b*d*exp(2*c + 2*d*x)))/(b^4*d*(a - b)^(1/2))))/(2*b^3*d*(a - b)^(1/2)) - (a^(5/2)*log((4*a^3*exp(2*c + 2*d* x))/b^4 + (2*a^(5/2)*(b*d + 2*a*d*exp(2*c + 2*d*x) - b*d*exp(2*c + 2*d*x)) )/(b^4*d*(a - b)^(1/2))))/(2*b^3*d*(a - b)^(1/2))