\(\int \frac {\cosh ^6(c+d x)}{(a+b \sinh ^2(c+d x))^2} \, dx\) [293]

Optimal result

Integrand size = 23, antiderivative size = 158 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=-\frac {(4 a-5 b) x}{2 b^3}+\frac {(a-b)^{3/2} (4 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} b^3 d}+\frac {\cosh (c+d x) \sinh (c+d x)}{2 b d \left (a-(a-b) \tanh ^2(c+d x)\right )}+\frac {(a-b) (2 a-b) \tanh (c+d x)}{2 a b^2 d \left (a-(a-b) \tanh ^2(c+d x)\right )} \] Output:

-1/2*(4*a-5*b)*x/b^3+1/2*(a-b)^(3/2)*(4*a+b)*arctanh((a-b)^(1/2)*tanh(d*x+ 
c)/a^(1/2))/a^(3/2)/b^3/d+1/2*cosh(d*x+c)*sinh(d*x+c)/b/d/(a-(a-b)*tanh(d* 
x+c)^2)+1/2*(a-b)*(2*a-b)*tanh(d*x+c)/a/b^2/d/(a-(a-b)*tanh(d*x+c)^2)
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.75 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\frac {2 (-4 a+5 b) (c+d x)+\frac {2 (a-b)^{3/2} (4 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}+b \sinh (2 (c+d x))+\frac {2 (a-b)^2 b \sinh (2 (c+d x))}{a (2 a-b+b \cosh (2 (c+d x)))}}{4 b^3 d} \] Input:

Integrate[Cosh[c + d*x]^6/(a + b*Sinh[c + d*x]^2)^2,x]
 

Output:

(2*(-4*a + 5*b)*(c + d*x) + (2*(a - b)^(3/2)*(4*a + b)*ArcTanh[(Sqrt[a - b 
]*Tanh[c + d*x])/Sqrt[a]])/a^(3/2) + b*Sinh[2*(c + d*x)] + (2*(a - b)^2*b* 
Sinh[2*(c + d*x)])/(a*(2*a - b + b*Cosh[2*(c + d*x)])))/(4*b^3*d)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3670, 316, 25, 402, 27, 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.

\(\displaystyle \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (i c+i d x)^6}{\left (a-b \sin (i c+i d x)^2\right )^2}dx\)

\(\Big \downarrow \) 3670

\(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^2 \left (a-(a-b) \tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {\frac {\int -\frac {3 (a-b) \tanh ^2(c+d x)+a-2 b}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{2 b}+\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {\int \frac {3 (a-b) \tanh ^2(c+d x)+a-2 b}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{2 b}}{d}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {-\frac {\int -\frac {2 \left (2 a^2-2 b a-b^2+(a-b) (2 a-b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{2 a b}-\frac {(a-b) (2 a-b) \tanh (c+d x)}{a b \left (a-(a-b) \tanh ^2(c+d x)\right )}}{2 b}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {\frac {\int \frac {2 a^2-2 b a-b^2+(a-b) (2 a-b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}d\tanh (c+d x)}{a b}-\frac {(a-b) (2 a-b) \tanh (c+d x)}{a b \left (a-(a-b) \tanh ^2(c+d x)\right )}}{2 b}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {\frac {\frac {a (4 a-5 b) \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{b}-\frac {(a-b)^2 (4 a+b) \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{a b}-\frac {(a-b) (2 a-b) \tanh (c+d x)}{a b \left (a-(a-b) \tanh ^2(c+d x)\right )}}{2 b}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {\frac {\frac {a (4 a-5 b) \text {arctanh}(\tanh (c+d x))}{b}-\frac {(a-b)^2 (4 a+b) \int \frac {1}{a-(a-b) \tanh ^2(c+d x)}d\tanh (c+d x)}{b}}{a b}-\frac {(a-b) (2 a-b) \tanh (c+d x)}{a b \left (a-(a-b) \tanh ^2(c+d x)\right )}}{2 b}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\tanh (c+d x)}{2 b \left (1-\tanh ^2(c+d x)\right ) \left (a-(a-b) \tanh ^2(c+d x)\right )}-\frac {\frac {\frac {a (4 a-5 b) \text {arctanh}(\tanh (c+d x))}{b}-\frac {(a-b)^{3/2} (4 a+b) \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} b}}{a b}-\frac {(a-b) (2 a-b) \tanh (c+d x)}{a b \left (a-(a-b) \tanh ^2(c+d x)\right )}}{2 b}}{d}\)

Input:

Int[Cosh[c + d*x]^6/(a + b*Sinh[c + d*x]^2)^2,x]
 

Output:

(Tanh[c + d*x]/(2*b*(1 - Tanh[c + d*x]^2)*(a - (a - b)*Tanh[c + d*x]^2)) - 
 (((a*(4*a - 5*b)*ArcTanh[Tanh[c + d*x]])/b - ((a - b)^(3/2)*(4*a + b)*Arc 
Tanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/(Sqrt[a]*b))/(a*b) - ((a - b)*( 
2*a - b)*Tanh[c + d*x])/(a*b*(a - (a - b)*Tanh[c + d*x]^2)))/(2*b))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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])
 

rule 221

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]
 

rule 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

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]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3670

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Su 
bst[Int[(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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs. \(2(142)=284\).

Time = 0.24 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.69

\[\frac {\frac {1}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (4 a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{3}}-\frac {2 \left (\frac {-\frac {b \left (a^{2}-2 a b +b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {b \left (a^{2}-2 a b +b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +4 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a}+\frac {\left (4 a^{3}-7 a^{2} b +2 b^{2} a +b^{3}\right ) \left (-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2}\right )}{b^{3}}-\frac {1}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-4 a +5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{3}}}{d}\]

Input:

int(cosh(d*x+c)^6/(a+b*sinh(d*x+c)^2)^2,x)
 

Output:

1/d*(1/2/b^2/(tanh(1/2*d*x+1/2*c)-1)^2+1/2/b^2/(tanh(1/2*d*x+1/2*c)-1)+1/2 
*(4*a-5*b)/b^3*ln(tanh(1/2*d*x+1/2*c)-1)-2/b^3*((-1/2*b*(a^2-2*a*b+b^2)/a* 
tanh(1/2*d*x+1/2*c)^3-1/2*b*(a^2-2*a*b+b^2)/a*tanh(1/2*d*x+1/2*c))/(tanh(1 
/2*d*x+1/2*c)^4*a-2*tanh(1/2*d*x+1/2*c)^2*a+4*b*tanh(1/2*d*x+1/2*c)^2+a)+1 
/2*(4*a^3-7*a^2*b+2*a*b^2+b^3)*(-1/2*((-b*(a-b))^(1/2)-b)/a/(-b*(a-b))^(1/ 
2)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*arctanh(tanh(1/2*d*x+1/2*c)*a/((2* 
(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/2*((-b*(a-b))^(1/2)+b)/a/(-b*(a-b))^(1 
/2)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(tanh(1/2*d*x+1/2*c)*a/((2* 
(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))))-1/2/b^2/(tanh(1/2*d*x+1/2*c)+1)^2+1/2/ 
b^2/(tanh(1/2*d*x+1/2*c)+1)+1/2/b^3*(-4*a+5*b)*ln(tanh(1/2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1682 vs. \(2 (144) = 288\).

Time = 0.14 (sec) , antiderivative size = 3629, normalized size of antiderivative = 22.97 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

integrate(cosh(d*x+c)^6/(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(cosh(d*x+c)**6/(a+b*sinh(d*x+c)**2)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(cosh(d*x+c)^6/(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")
 

Output:

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
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (144) = 288\).

Time = 1.23 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.93 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=-\frac {\frac {12 \, {\left (d x + c\right )} {\left (4 \, a - 5 \, b\right )}}{b^{3}} - \frac {3 \, e^{\left (2 \, d x + 2 \, c\right )}}{b^{2}} - \frac {12 \, {\left (4 \, a^{3} - 7 \, a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} a b^{3}} - \frac {8 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 10 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 16 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 64 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 79 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 28 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 44 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 24 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, a b^{2}}{{\left (b e^{\left (6 \, d x + 6 \, c\right )} + 4 \, a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, b e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )}\right )} a b^{3}}}{24 \, d} \] Input:

integrate(cosh(d*x+c)^6/(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

-1/24*(12*(d*x + c)*(4*a - 5*b)/b^3 - 3*e^(2*d*x + 2*c)/b^2 - 12*(4*a^3 - 
7*a^2*b + 2*a*b^2 + b^3)*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^ 
2 + a*b))/(sqrt(-a^2 + a*b)*a*b^3) - (8*a^2*b*e^(6*d*x + 6*c) - 10*a*b^2*e 
^(6*d*x + 6*c) - 16*a^3*e^(4*d*x + 4*c) + 64*a^2*b*e^(4*d*x + 4*c) - 79*a* 
b^2*e^(4*d*x + 4*c) + 24*b^3*e^(4*d*x + 4*c) - 28*a^2*b*e^(2*d*x + 2*c) + 
44*a*b^2*e^(2*d*x + 2*c) - 24*b^3*e^(2*d*x + 2*c) - 3*a*b^2)/((b*e^(6*d*x 
+ 6*c) + 4*a*e^(4*d*x + 4*c) - 2*b*e^(4*d*x + 4*c) + b*e^(2*d*x + 2*c))*a* 
b^3))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6}{{\left (b\,{\mathrm {sinh}\left (c+d\,x\right )}^2+a\right )}^2} \,d x \] Input:

int(cosh(c + d*x)^6/(a + b*sinh(c + d*x)^2)^2,x)
 

Output:

int(cosh(c + d*x)^6/(a + b*sinh(c + d*x)^2)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1910, normalized size of antiderivative = 12.09 \[ \int \frac {\cosh ^6(c+d x)}{\left (a+b \sinh ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(cosh(d*x+c)^6/(a+b*sinh(d*x+c)^2)^2,x)
 

Output:

(8*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a - b)*log( - sqrt(2*sqrt(a)*sqrt(a - b) 
- 2*a + b) + e**(c + d*x)*sqrt(b))*a**2*b - 6*e**(6*c + 6*d*x)*sqrt(a)*sqr 
t(a - b)*log( - sqrt(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt( 
b))*a*b**2 - 2*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a - b)*log( - sqrt(2*sqrt(a)* 
sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt(b))*b**3 + 8*e**(6*c + 6*d*x)*s 
qrt(a)*sqrt(a - b)*log(sqrt(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c + d*x 
)*sqrt(b))*a**2*b - 6*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a - b)*log(sqrt(2*sqrt 
(a)*sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt(b))*a*b**2 - 2*e**(6*c + 6* 
d*x)*sqrt(a)*sqrt(a - b)*log(sqrt(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c 
 + d*x)*sqrt(b))*b**3 - 8*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a - b)*log(2*sqrt( 
a)*sqrt(a - b) + e**(2*c + 2*d*x)*b + 2*a - b)*a**2*b + 6*e**(6*c + 6*d*x) 
*sqrt(a)*sqrt(a - b)*log(2*sqrt(a)*sqrt(a - b) + e**(2*c + 2*d*x)*b + 2*a 
- b)*a*b**2 + 2*e**(6*c + 6*d*x)*sqrt(a)*sqrt(a - b)*log(2*sqrt(a)*sqrt(a 
- b) + e**(2*c + 2*d*x)*b + 2*a - b)*b**3 + 32*e**(4*c + 4*d*x)*sqrt(a)*sq 
rt(a - b)*log( - sqrt(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt 
(b))*a**3 - 40*e**(4*c + 4*d*x)*sqrt(a)*sqrt(a - b)*log( - sqrt(2*sqrt(a)* 
sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt(b))*a**2*b + 4*e**(4*c + 4*d*x) 
*sqrt(a)*sqrt(a - b)*log( - sqrt(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c 
+ d*x)*sqrt(b))*a*b**2 + 4*e**(4*c + 4*d*x)*sqrt(a)*sqrt(a - b)*log( - sqr 
t(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt(b))*b**3 + 32*e*...