\(\int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\) [290]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{5/2} d}+\frac {(a-2 b) \tanh (c+d x)}{(a-b)^2 d}-\frac {\tanh ^3(c+d x)}{3 (a-b) d} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)^{5/2}}+\frac {\left (2 a-5 b+(a-b) \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{(a-b)^2}}{3 d} \] Input:

Integrate[Sech[c + d*x]^4/(a + b*Sinh[c + d*x]^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 300, 2009}

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.

\(\displaystyle \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (i c+i d x)^4 \left (a-b \sin (i c+i d x)^2\right )}dx\)

\(\Big \downarrow \) 3670

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 2009

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

Input:

Int[Sech[c + d*x]^4/(a + b*Sinh[c + d*x]^2),x]
 

Output:

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

Defintions of rubi rules used

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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. \(241\) vs. \(2(78)=156\).

Time = 283.41 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.75

method	result	size
risch	\(-\frac {2 \left (-3 b \,{\mathrm e}^{4 d x +4 c}+6 \,{\mathrm e}^{2 d x +2 c} a -12 \,{\mathrm e}^{2 d x +2 c} b +2 a -5 b \right )}{3 d \left (a -b \right )^{2} \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}+\frac {b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \sqrt {a^{2}-a b}-b \sqrt {a^{2}-a b}-2 a^{2}+2 a b}{b \sqrt {a^{2}-a b}}\right )}{2 \sqrt {a^{2}-a b}\, \left (a -b \right )^{2} d}-\frac {b^{2} \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 a \sqrt {a^{2}-a b}-b \sqrt {a^{2}-a b}+2 a^{2}-2 a b}{b \sqrt {a^{2}-a b}}\right )}{2 \sqrt {a^{2}-a b}\, \left (a -b \right )^{2} d}\)	\(242\)
derivativedivides	\(\frac {-\frac {2 b^{2} a \left (-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{\left (a -b \right )^{2}}-\frac {2 \left (\left (-a +2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2 a}{3}+\frac {8 b}{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a +2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{3}}}{d}\)	\(268\)
default	\(\frac {-\frac {2 b^{2} a \left (-\frac {\left (\sqrt {-b \left (a -b \right )}-b \right ) \operatorname {arctanh}\left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}+a -2 b \right ) a}}+\frac {\left (\sqrt {-b \left (a -b \right )}+b \right ) \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) a}{\sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{2 a \sqrt {-b \left (a -b \right )}\, \sqrt {\left (2 \sqrt {-b \left (a -b \right )}-a +2 b \right ) a}}\right )}{\left (a -b \right )^{2}}-\frac {2 \left (\left (-a +2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {2 a}{3}+\frac {8 b}{3}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-a +2 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )^{3}}}{d}\)	\(268\)

Input:

int(sech(d*x+c)^4/(a+b*sinh(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(-3*b*exp(4*d*x+4*c)+6*exp(2*d*x+2*c)*a-12*exp(2*d*x+2*c)*b+2*a-5*b)/ 
d/(a-b)^2/(exp(2*d*x+2*c)+1)^3+1/2/(a^2-a*b)^(1/2)*b^2/(a-b)^2/d*ln(exp(2* 
d*x+2*c)+(2*a*(a^2-a*b)^(1/2)-b*(a^2-a*b)^(1/2)-2*a^2+2*a*b)/b/(a^2-a*b)^( 
1/2))-1/2/(a^2-a*b)^(1/2)*b^2/(a-b)^2/d*ln(exp(2*d*x+2*c)+(2*a*(a^2-a*b)^( 
1/2)-b*(a^2-a*b)^(1/2)+2*a^2-2*a*b)/b/(a^2-a*b)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1094 vs. \(2 (78) = 156\).

Time = 0.12 (sec) , antiderivative size = 2444, normalized size of antiderivative = 27.77 \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^4/(a+b*sinh(d*x+c)^2),x, algorithm="fricas")
 

Output:

[1/6*(12*(a^2*b - a*b^2)*cosh(d*x + c)^4 + 48*(a^2*b - a*b^2)*cosh(d*x + c 
)*sinh(d*x + c)^3 + 12*(a^2*b - a*b^2)*sinh(d*x + c)^4 - 8*a^3 + 28*a^2*b 
- 20*a*b^2 - 24*(a^3 - 3*a^2*b + 2*a*b^2)*cosh(d*x + c)^2 - 24*(a^3 - 3*a^ 
2*b + 2*a*b^2 - 3*(a^2*b - a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 3*(b^ 
2*cosh(d*x + c)^6 + 6*b^2*cosh(d*x + c)*sinh(d*x + c)^5 + b^2*sinh(d*x + c 
)^6 + 3*b^2*cosh(d*x + c)^4 + 3*(5*b^2*cosh(d*x + c)^2 + b^2)*sinh(d*x + c 
)^4 + 3*b^2*cosh(d*x + c)^2 + 4*(5*b^2*cosh(d*x + c)^3 + 3*b^2*cosh(d*x + 
c))*sinh(d*x + c)^3 + 3*(5*b^2*cosh(d*x + c)^4 + 6*b^2*cosh(d*x + c)^2 + b 
^2)*sinh(d*x + c)^2 + b^2 + 6*(b^2*cosh(d*x + c)^5 + 2*b^2*cosh(d*x + c)^3 
 + b^2*cosh(d*x + c))*sinh(d*x + c))*sqrt(a^2 - 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)*cos 
h(d*x + c))*sinh(d*x + c) - 4*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh( 
d*x + c) + b*sinh(d*x + c)^2 + 2*a - b)*sqrt(a^2 - a*b))/(b*cosh(d*x + c)^ 
4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(2*a - b)*co 
sh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*c 
osh(d*x + c)^3 + (2*a - b)*cosh(d*x + c))*sinh(d*x + c) + b)) + 48*((a^2*b 
 - a*b^2)*cosh(d*x + c)^3 - (a^3 - 3*a^2*b + 2*a*b^2)*cosh(d*x + c))*sinh( 
d*x + c))/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d*cosh(d*x + c)^6 + 6*(a...
 

Sympy [F]

\[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\int \frac {\operatorname {sech}^{4}{\left (c + d x \right )}}{a + b \sinh ^{2}{\left (c + d x \right )}}\, dx \] Input:

integrate(sech(d*x+c)**4/(a+b*sinh(d*x+c)**2),x)
 

Output:

Integral(sech(c + d*x)**4/(a + b*sinh(c + d*x)**2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sech(d*x+c)^4/(a+b*sinh(d*x+c)^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 [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.57 \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {\frac {3 \, b^{2} \arctan \left (\frac {b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt {-a^{2} + a b}}\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-a^{2} + a b}} + \frac {2 \, {\left (3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a + 5 \, b\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{3 \, d} \] Input:

integrate(sech(d*x+c)^4/(a+b*sinh(d*x+c)^2),x, algorithm="giac")
 

Output:

1/3*(3*b^2*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/((a^ 
2 - 2*a*b + b^2)*sqrt(-a^2 + a*b)) + 2*(3*b*e^(4*d*x + 4*c) - 6*a*e^(2*d*x 
 + 2*c) + 12*b*e^(2*d*x + 2*c) - 2*a + 5*b)/((a^2 - 2*a*b + b^2)*(e^(2*d*x 
 + 2*c) + 1)^3))/d
 

Mupad [B] (verification not implemented)

Time = 3.26 (sec) , antiderivative size = 710, normalized size of antiderivative = 8.07 \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx=\frac {8}{3\,\left (a\,d-b\,d\right )\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {4}{\left (a\,d-b\,d\right )\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {\mathrm {atan}\left (\left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (\frac {4}{d\,{\left (a-b\right )}^2\,\sqrt {b^4}\,\left (a^2-2\,a\,b+b^2\right )}+\frac {\left (2\,a-b\right )\,\left (2\,a^3\,d\,\sqrt {b^4}-b^3\,d\,\sqrt {b^4}+4\,a\,b^2\,d\,\sqrt {b^4}-5\,a^2\,b\,d\,\sqrt {b^4}\right )}{b^4\,\left (a^2-2\,a\,b+b^2\right )\,\sqrt {-a\,d^2\,{\left (a-b\right )}^5}\,\sqrt {-a^6\,d^2+5\,a^5\,b\,d^2-10\,a^4\,b^2\,d^2+10\,a^3\,b^3\,d^2-5\,a^2\,b^4\,d^2+a\,b^5\,d^2}}\right )+\frac {\left (2\,a-b\right )\,\left (b^3\,d\,\sqrt {b^4}-2\,a\,b^2\,d\,\sqrt {b^4}+a^2\,b\,d\,\sqrt {b^4}\right )}{b^4\,\left (a^2-2\,a\,b+b^2\right )\,\sqrt {-a\,d^2\,{\left (a-b\right )}^5}\,\sqrt {-a^6\,d^2+5\,a^5\,b\,d^2-10\,a^4\,b^2\,d^2+10\,a^3\,b^3\,d^2-5\,a^2\,b^4\,d^2+a\,b^5\,d^2}}\right )\,\left (\frac {b^3\,\sqrt {-a^6\,d^2+5\,a^5\,b\,d^2-10\,a^4\,b^2\,d^2+10\,a^3\,b^3\,d^2-5\,a^2\,b^4\,d^2+a\,b^5\,d^2}}{2}-a\,b^2\,\sqrt {-a^6\,d^2+5\,a^5\,b\,d^2-10\,a^4\,b^2\,d^2+10\,a^3\,b^3\,d^2-5\,a^2\,b^4\,d^2+a\,b^5\,d^2}+\frac {a^2\,b\,\sqrt {-a^6\,d^2+5\,a^5\,b\,d^2-10\,a^4\,b^2\,d^2+10\,a^3\,b^3\,d^2-5\,a^2\,b^4\,d^2+a\,b^5\,d^2}}{2}\right )\right )\,\sqrt {b^4}}{\sqrt {-a^6\,d^2+5\,a^5\,b\,d^2-10\,a^4\,b^2\,d^2+10\,a^3\,b^3\,d^2-5\,a^2\,b^4\,d^2+a\,b^5\,d^2}}+\frac {2\,b}{\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )\,\left (a-b\right )\,\left (a\,d-b\,d\right )} \] Input:

int(1/(cosh(c + d*x)^4*(a + b*sinh(c + d*x)^2)),x)
 

Output:

8/(3*(a*d - b*d)*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d* 
x) + 1)) - 4/((a*d - b*d)*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) + ( 
atan((exp(2*c)*exp(2*d*x)*(4/(d*(a - b)^2*(b^4)^(1/2)*(a^2 - 2*a*b + b^2)) 
 + ((2*a - b)*(2*a^3*d*(b^4)^(1/2) - b^3*d*(b^4)^(1/2) + 4*a*b^2*d*(b^4)^( 
1/2) - 5*a^2*b*d*(b^4)^(1/2)))/(b^4*(a^2 - 2*a*b + b^2)*(-a*d^2*(a - b)^5) 
^(1/2)*(a*b^5*d^2 - a^6*d^2 + 5*a^5*b*d^2 - 5*a^2*b^4*d^2 + 10*a^3*b^3*d^2 
 - 10*a^4*b^2*d^2)^(1/2))) + ((2*a - b)*(b^3*d*(b^4)^(1/2) - 2*a*b^2*d*(b^ 
4)^(1/2) + a^2*b*d*(b^4)^(1/2)))/(b^4*(a^2 - 2*a*b + b^2)*(-a*d^2*(a - b)^ 
5)^(1/2)*(a*b^5*d^2 - a^6*d^2 + 5*a^5*b*d^2 - 5*a^2*b^4*d^2 + 10*a^3*b^3*d 
^2 - 10*a^4*b^2*d^2)^(1/2)))*((b^3*(a*b^5*d^2 - a^6*d^2 + 5*a^5*b*d^2 - 5* 
a^2*b^4*d^2 + 10*a^3*b^3*d^2 - 10*a^4*b^2*d^2)^(1/2))/2 - a*b^2*(a*b^5*d^2 
 - a^6*d^2 + 5*a^5*b*d^2 - 5*a^2*b^4*d^2 + 10*a^3*b^3*d^2 - 10*a^4*b^2*d^2 
)^(1/2) + (a^2*b*(a*b^5*d^2 - a^6*d^2 + 5*a^5*b*d^2 - 5*a^2*b^4*d^2 + 10*a 
^3*b^3*d^2 - 10*a^4*b^2*d^2)^(1/2))/2))*(b^4)^(1/2))/(a*b^5*d^2 - a^6*d^2 
+ 5*a^5*b*d^2 - 5*a^2*b^4*d^2 + 10*a^3*b^3*d^2 - 10*a^4*b^2*d^2)^(1/2) + ( 
2*b)/((exp(2*c + 2*d*x) + 1)*(a - b)*(a*d - b*d))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 911, normalized size of antiderivative = 10.35 \[ \int \frac {\text {sech}^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx =\text {Too large to display} \] Input:

int(sech(d*x+c)^4/(a+b*sinh(d*x+c)^2),x)
 

Output:

(3*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**2 + 3*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 
**2 - 3*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**2 + 9*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))*b**2 
 + 9*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))*b**2 - 9*e**(4*c + 4*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**2 + 9*e 
**(2*c + 2*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**2 + 9*e**(2*c + 2*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**2 
- 9*e**(2*c + 2*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**2 + 3*sqrt(a)*sqrt(a - b)*log( - sqrt(2*sqrt(a 
)*sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt(b))*b**2 + 3*sqrt(a)*sqrt(a - 
 b)*log(sqrt(2*sqrt(a)*sqrt(a - b) - 2*a + b) + e**(c + d*x)*sqrt(b))*b**2 
 - 3*sqrt(a)*sqrt(a - b)*log(2*sqrt(a)*sqrt(a - b) + e**(2*c + 2*d*x)*b + 
2*a - b)*b**2 - 4*e**(6*c + 6*d*x)*a**2*b + 4*e**(6*c + 6*d*x)*a*b**2 - 24 
*e**(2*c + 2*d*x)*a**3 + 60*e**(2*c + 2*d*x)*a**2*b - 36*e**(2*c + 2*d*x)* 
a*b**2 - 8*a**3 + 24*a**2*b - 16*a*b**2)/(6*a*d*(e**(6*c + 6*d*x)*a**3 ...