Integrand size = 13, antiderivative size = 187 \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx=\frac {x}{a}-\frac {3 \arctan (\sinh (x))}{8 b}-\frac {\left (a^2-3 b^2\right ) \arctan (\sinh (x))}{2 b^3}-\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \arctan (\sinh (x))}{b^5}+\frac {2 (a-b)^{5/2} (a+b)^{5/2} \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b^5}+\frac {a \tanh (x)}{b^2}+\frac {a \left (a^2-3 b^2\right ) \tanh (x)}{b^4}-\frac {3 \text {sech}(x) \tanh (x)}{8 b}-\frac {\left (a^2-3 b^2\right ) \text {sech}(x) \tanh (x)}{2 b^3}-\frac {\text {sech}^3(x) \tanh (x)}{4 b}-\frac {a \tanh ^3(x)}{3 b^2} \] Output:
x/a-3/8*arctan(sinh(x))/b-1/2*(a^2-3*b^2)*arctan(sinh(x))/b^3-(a^4-3*a^2*b ^2+3*b^4)*arctan(sinh(x))/b^5+2*(a-b)^(5/2)*(a+b)^(5/2)*arctan((a-b)^(1/2) *tanh(1/2*x)/(a+b)^(1/2))/a/b^5+a*tanh(x)/b^2+a*(a^2-3*b^2)*tanh(x)/b^4-3/ 8*sech(x)*tanh(x)/b-1/2*(a^2-3*b^2)*sech(x)*tanh(x)/b^3-1/4*sech(x)^3*tanh (x)/b-1/3*a*tanh(x)^3/b^2
Time = 0.50 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.99 \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx=\frac {-12 \left (8 a^4-20 a^2 b^2+15 b^4\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {48 \left (b^5 \sqrt {a^2-b^2} x-2 \left (a^2-b^2\right )^3 \arctan \left (\frac {(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )\right )}{a \sqrt {a^2-b^2}}+b \left (-12 a^2 b+15 b^3+4 a \left (9 a^2-17 b^2\right ) \cosh (x)+3 b \left (-4 a^2+9 b^2\right ) \cosh (2 x)+12 a^3 \cosh (3 x)-28 a b^2 \cosh (3 x)\right ) \text {sech}^3(x) \tanh (x)}{48 b^5} \] Input:
Integrate[Tanh[x]^6/(a + b*Sech[x]),x]
Output:
(-12*(8*a^4 - 20*a^2*b^2 + 15*b^4)*ArcTan[Tanh[x/2]] + (48*(b^5*Sqrt[a^2 - b^2]*x - 2*(a^2 - b^2)^3*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]]))/( a*Sqrt[a^2 - b^2]) + b*(-12*a^2*b + 15*b^3 + 4*a*(9*a^2 - 17*b^2)*Cosh[x] + 3*b*(-4*a^2 + 9*b^2)*Cosh[2*x] + 12*a^3*Cosh[3*x] - 28*a*b^2*Cosh[3*x])* Sech[x]^3*Tanh[x])/(48*b^5)
Time = 0.64 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 25, 4386, 25, 3042, 25, 3376, 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.
| \(\displaystyle \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cot \left (\frac {\pi }{2}+i x\right )^6}{a+b \csc \left (\frac {\pi }{2}+i x\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cot \left (i x+\frac {\pi }{2}\right )^6}{a+b \csc \left (i x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4386 |
| \(\displaystyle -\int -\frac {\sinh (x) \tanh ^5(x)}{b+a \cosh (x)}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\sinh (x) \tanh ^5(x)}{a \cosh (x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cos \left (\frac {\pi }{2}+i x\right )^6}{\sin \left (\frac {\pi }{2}+i x\right )^5 \left (b+a \sin \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (i x+\frac {\pi }{2}\right )^6}{\sin \left (i x+\frac {\pi }{2}\right )^5 \left (b+a \sin \left (i x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3376 |
| \(\displaystyle -\int \left (\frac {\text {sech}^5(x)}{b}-\frac {a \text {sech}^4(x)}{b^2}+\frac {\left (a^2-3 b^2\right ) \text {sech}^3(x)}{b^3}+\frac {\left (3 a b^2-a^3\right ) \text {sech}^2(x)}{b^4}+\frac {\left (a^4-3 b^2 a^2+3 b^4\right ) \text {sech}(x)}{b^5}-\frac {1}{a}-\frac {\left (a^2-b^2\right )^3}{a b^5 (b+a \cosh (x))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\left (a^2-3 b^2\right ) \arctan (\sinh (x))}{2 b^3}+\frac {a \left (a^2-3 b^2\right ) \tanh (x)}{b^4}-\frac {\left (a^2-3 b^2\right ) \tanh (x) \text {sech}(x)}{2 b^3}-\frac {\left (a^4-3 a^2 b^2+3 b^4\right ) \arctan (\sinh (x))}{b^5}+\frac {2 (a-b)^{5/2} (a+b)^{5/2} \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a b^5}-\frac {a \tanh ^3(x)}{3 b^2}+\frac {a \tanh (x)}{b^2}+\frac {x}{a}-\frac {3 \arctan (\sinh (x))}{8 b}-\frac {\tanh (x) \text {sech}^3(x)}{4 b}-\frac {3 \tanh (x) \text {sech}(x)}{8 b}\) |
Input:
Int[Tanh[x]^6/(a + b*Sech[x]),x]
Output:
x/a - (3*ArcTan[Sinh[x]])/(8*b) - ((a^2 - 3*b^2)*ArcTan[Sinh[x]])/(2*b^3) - ((a^4 - 3*a^2*b^2 + 3*b^4)*ArcTan[Sinh[x]])/b^5 + (2*(a - b)^(5/2)*(a + b)^(5/2)*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a*b^5) + (a*Tanh[x] )/b^2 + (a*(a^2 - 3*b^2)*Tanh[x])/b^4 - (3*Sech[x]*Tanh[x])/(8*b) - ((a^2 - 3*b^2)*Sech[x]*Tanh[x])/(2*b^3) - (Sech[x]^3*Tanh[x])/(4*b) - (a*Tanh[x] ^3)/(3*b^2)
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Int[Cos[c + d*x]^m*((b + a*Sin[c + d*x])^n/Sin[c + d*x]^(m + n)), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])
Time = 1.75 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.42
| method | result | size |
| default | \(-\frac {2 \left (\frac {\left (-a^{3} b -\frac {1}{2} a^{2} b^{2}+2 a \,b^{3}+\frac {7}{8} b^{4}\right ) \tanh \left (\frac {x}{2}\right )^{7}+\left (-3 a^{3} b -\frac {1}{2} a^{2} b^{2}+\frac {15}{8} b^{4}+\frac {22}{3} a \,b^{3}\right ) \tanh \left (\frac {x}{2}\right )^{5}+\left (\frac {1}{2} a^{2} b^{2}-\frac {15}{8} b^{4}-3 a^{3} b +\frac {22}{3} a \,b^{3}\right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (-a^{3} b +2 a \,b^{3}+\frac {1}{2} a^{2} b^{2}-\frac {7}{8} b^{4}\right ) \tanh \left (\frac {x}{2}\right )}{\left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )^{4}}+\frac {\left (8 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{8}\right )}{b^{5}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}+\frac {2 \left (a -b \right )^{3} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a \,b^{5} \sqrt {\left (a -b \right ) \left (a +b \right )}}\) | \(266\) |
| risch | \(\frac {x}{a}-\frac {12 a^{2} b \,{\mathrm e}^{7 x}-27 b^{3} {\mathrm e}^{7 x}+24 a^{3} {\mathrm e}^{6 x}-72 a \,b^{2} {\mathrm e}^{6 x}+12 a^{2} b \,{\mathrm e}^{5 x}-3 \,{\mathrm e}^{5 x} b^{3}+72 a^{3} {\mathrm e}^{4 x}-168 a \,b^{2} {\mathrm e}^{4 x}-12 a^{2} b \,{\mathrm e}^{3 x}+3 b^{3} {\mathrm e}^{3 x}+72 a^{3} {\mathrm e}^{2 x}-152 a \,b^{2} {\mathrm e}^{2 x}-12 b \,{\mathrm e}^{x} a^{2}+27 b^{3} {\mathrm e}^{x}+24 a^{3}-56 a \,b^{2}}{12 b^{4} \left ({\mathrm e}^{2 x}+1\right )^{4}}+\frac {i \ln \left ({\mathrm e}^{x}-i\right ) a^{4}}{b^{5}}-\frac {5 i \ln \left ({\mathrm e}^{x}-i\right ) a^{2}}{2 b^{3}}+\frac {15 i \ln \left ({\mathrm e}^{x}-i\right )}{8 b}-\frac {i \ln \left ({\mathrm e}^{x}+i\right ) a^{4}}{b^{5}}+\frac {5 i \ln \left ({\mathrm e}^{x}+i\right ) a^{2}}{2 b^{3}}-\frac {15 i \ln \left ({\mathrm e}^{x}+i\right )}{8 b}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{3} \ln \left ({\mathrm e}^{x}+\frac {b +\sqrt {-a^{2}+b^{2}}}{a}\right )}{b^{5}}-\frac {2 \sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{x}+\frac {b +\sqrt {-a^{2}+b^{2}}}{a}\right )}{b^{3}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{x}+\frac {b +\sqrt {-a^{2}+b^{2}}}{a}\right )}{b a}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{3} \ln \left ({\mathrm e}^{x}-\frac {\sqrt {-a^{2}+b^{2}}-b}{a}\right )}{b^{5}}+\frac {2 \sqrt {-a^{2}+b^{2}}\, a \ln \left ({\mathrm e}^{x}-\frac {\sqrt {-a^{2}+b^{2}}-b}{a}\right )}{b^{3}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{x}-\frac {\sqrt {-a^{2}+b^{2}}-b}{a}\right )}{b a}\) | \(488\) |
Input:
int(tanh(x)^6/(a+b*sech(x)),x,method=_RETURNVERBOSE)
Output:
-2/b^5*(((-a^3*b-1/2*a^2*b^2+2*a*b^3+7/8*b^4)*tanh(1/2*x)^7+(-3*a^3*b-1/2* a^2*b^2+15/8*b^4+22/3*a*b^3)*tanh(1/2*x)^5+(1/2*a^2*b^2-15/8*b^4-3*a^3*b+2 2/3*a*b^3)*tanh(1/2*x)^3+(-a^3*b+2*a*b^3+1/2*a^2*b^2-7/8*b^4)*tanh(1/2*x)) /(tanh(1/2*x)^2+1)^4+1/8*(8*a^4-20*a^2*b^2+15*b^4)*arctan(tanh(1/2*x)))-1/ a*ln(tanh(1/2*x)-1)+1/a*ln(tanh(1/2*x)+1)+2/a*(a-b)^3*(a^3+3*a^2*b+3*a*b^2 +b^3)/b^5/((a-b)*(a+b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a-b)*(a+b))^(1/2) )
Leaf count of result is larger than twice the leaf count of optimal. 2417 vs. \(2 (165) = 330\).
Time = 0.39 (sec) , antiderivative size = 4914, normalized size of antiderivative = 26.28 \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)^6/(a+b*sech(x)),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\tanh ^{6}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \] Input:
integrate(tanh(x)**6/(a+b*sech(x)),x)
Output:
Integral(tanh(x)**6/(a + b*sech(x)), x)
Exception generated. \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(tanh(x)^6/(a+b*sech(x)),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(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.12 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.34 \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx=\frac {x}{a} - \frac {{\left (8 \, a^{4} - 20 \, a^{2} b^{2} + 15 \, b^{4}\right )} \arctan \left (e^{x}\right )}{4 \, b^{5}} + \frac {2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} a b^{5}} - \frac {12 \, a^{2} b e^{\left (7 \, x\right )} - 27 \, b^{3} e^{\left (7 \, x\right )} + 24 \, a^{3} e^{\left (6 \, x\right )} - 72 \, a b^{2} e^{\left (6 \, x\right )} + 12 \, a^{2} b e^{\left (5 \, x\right )} - 3 \, b^{3} e^{\left (5 \, x\right )} + 72 \, a^{3} e^{\left (4 \, x\right )} - 168 \, a b^{2} e^{\left (4 \, x\right )} - 12 \, a^{2} b e^{\left (3 \, x\right )} + 3 \, b^{3} e^{\left (3 \, x\right )} + 72 \, a^{3} e^{\left (2 \, x\right )} - 152 \, a b^{2} e^{\left (2 \, x\right )} - 12 \, a^{2} b e^{x} + 27 \, b^{3} e^{x} + 24 \, a^{3} - 56 \, a b^{2}}{12 \, b^{4} {\left (e^{\left (2 \, x\right )} + 1\right )}^{4}} \] Input:
integrate(tanh(x)^6/(a+b*sech(x)),x, algorithm="giac")
Output:
x/a - 1/4*(8*a^4 - 20*a^2*b^2 + 15*b^4)*arctan(e^x)/b^5 + 2*(a^6 - 3*a^4*b ^2 + 3*a^2*b^4 - b^6)*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2) *a*b^5) - 1/12*(12*a^2*b*e^(7*x) - 27*b^3*e^(7*x) + 24*a^3*e^(6*x) - 72*a* b^2*e^(6*x) + 12*a^2*b*e^(5*x) - 3*b^3*e^(5*x) + 72*a^3*e^(4*x) - 168*a*b^ 2*e^(4*x) - 12*a^2*b*e^(3*x) + 3*b^3*e^(3*x) + 72*a^3*e^(2*x) - 152*a*b^2* e^(2*x) - 12*a^2*b*e^x + 27*b^3*e^x + 24*a^3 - 56*a*b^2)/(b^4*(e^(2*x) + 1 )^4)
Time = 8.73 (sec) , antiderivative size = 1001, normalized size of antiderivative = 5.35 \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx =\text {Too large to display} \] Input:
int(tanh(x)^6/(a + b/cosh(x)),x)
Output:
((8*a)/(3*b^2) + (6*exp(x))/b)/(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1) - ((exp(x)*(4*a^2 - 9*b^2))/(4*b^3) + (2*(a^4 - 3*a^2*b^2))/(a*b^4))/(exp(2* x) + 1) - ((4*a)/b^2 - (exp(x)*(4*a^2 - 13*b^2))/(2*b^3))/(2*exp(2*x) + ex p(4*x) + 1) + x/a + (log(exp(x) - 1i)*(a^4*8i + b^4*15i - a^2*b^2*20i))/(8 *b^5) - (log(exp(x) + 1i)*(a^4*8i + b^4*15i - a^2*b^2*20i))/(8*b^5) - (4*e xp(x))/(b*(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1)) + (log((( -(a + b)^5*(a - b)^5)^(1/2)*((128*a^12 + 64*b^12 - 834*a^2*b^10 + 2385*a^4 *b^8 - 3160*a^6*b^6 + 2240*a^8*b^4 - 832*a^10*b^2 - 900*a*b^11*exp(x) + 19 2*a^11*b*exp(x) + 3075*a^3*b^9*exp(x) - 4360*a^5*b^7*exp(x) + 3200*a^7*b^5 *exp(x) - 1216*a^9*b^3*exp(x))/(2*a^6*b^8) - ((-(a + b)^5*(a - b)^5)^(1/2) *((4*(a^2 - b^2)*(16*a*b^4 + 16*a^5 - 32*a^3*b^2 + 32*b^5*exp(x) + 28*a^4* b*exp(x) - 57*a^2*b^3*exp(x)))/(a^6*b^2) + (32*(-(a + b)^5*(a - b)^5)^(1/2 )*(3*a*b^2 - 2*a^3 + 4*b^3*exp(x) - 3*a^2*b*exp(x)))/(a^6*b^3)))/(a*b^5))) /(a*b^5) - ((a^2 - b^2)^3*(8*a^4 + 15*b^4 - 20*a^2*b^2)*(30*a*b^4 + 16*a^5 - 40*a^3*b^2 + 52*b^5*exp(x) + 28*a^4*b*exp(x) - 71*a^2*b^3*exp(x)))/(2*a ^6*b^12))*(-(a + b)^5*(a - b)^5)^(1/2))/(a*b^5) - (log(- ((-(a + b)^5*(a - b)^5)^(1/2)*((128*a^12 + 64*b^12 - 834*a^2*b^10 + 2385*a^4*b^8 - 3160*a^6 *b^6 + 2240*a^8*b^4 - 832*a^10*b^2 - 900*a*b^11*exp(x) + 192*a^11*b*exp(x) + 3075*a^3*b^9*exp(x) - 4360*a^5*b^7*exp(x) + 3200*a^7*b^5*exp(x) - 1216* a^9*b^3*exp(x))/(2*a^6*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((4*(a^2 - ...
Time = 0.19 (sec) , antiderivative size = 1097, normalized size of antiderivative = 5.87 \[ \int \frac {\tanh ^6(x)}{a+b \text {sech}(x)} \, dx =\text {Too large to display} \] Input:
int(tanh(x)^6/(a+b*sech(x)),x)
Output:
( - 24*e**(8*x)*atan(e**x)*a**5 + 60*e**(8*x)*atan(e**x)*a**3*b**2 - 45*e* *(8*x)*atan(e**x)*a*b**4 - 96*e**(6*x)*atan(e**x)*a**5 + 240*e**(6*x)*atan (e**x)*a**3*b**2 - 180*e**(6*x)*atan(e**x)*a*b**4 - 144*e**(4*x)*atan(e**x )*a**5 + 360*e**(4*x)*atan(e**x)*a**3*b**2 - 270*e**(4*x)*atan(e**x)*a*b** 4 - 96*e**(2*x)*atan(e**x)*a**5 + 240*e**(2*x)*atan(e**x)*a**3*b**2 - 180* e**(2*x)*atan(e**x)*a*b**4 - 24*atan(e**x)*a**5 + 60*atan(e**x)*a**3*b**2 - 45*atan(e**x)*a*b**4 + 24*e**(8*x)*sqrt(a**2 - b**2)*atan((e**x*a + b)/s qrt(a**2 - b**2))*a**4 - 48*e**(8*x)*sqrt(a**2 - b**2)*atan((e**x*a + b)/s qrt(a**2 - b**2))*a**2*b**2 + 24*e**(8*x)*sqrt(a**2 - b**2)*atan((e**x*a + b)/sqrt(a**2 - b**2))*b**4 + 96*e**(6*x)*sqrt(a**2 - b**2)*atan((e**x*a + b)/sqrt(a**2 - b**2))*a**4 - 192*e**(6*x)*sqrt(a**2 - b**2)*atan((e**x*a + b)/sqrt(a**2 - b**2))*a**2*b**2 + 96*e**(6*x)*sqrt(a**2 - b**2)*atan((e* *x*a + b)/sqrt(a**2 - b**2))*b**4 + 144*e**(4*x)*sqrt(a**2 - b**2)*atan((e **x*a + b)/sqrt(a**2 - b**2))*a**4 - 288*e**(4*x)*sqrt(a**2 - b**2)*atan(( e**x*a + b)/sqrt(a**2 - b**2))*a**2*b**2 + 144*e**(4*x)*sqrt(a**2 - b**2)* atan((e**x*a + b)/sqrt(a**2 - b**2))*b**4 + 96*e**(2*x)*sqrt(a**2 - b**2)* atan((e**x*a + b)/sqrt(a**2 - b**2))*a**4 - 192*e**(2*x)*sqrt(a**2 - b**2) *atan((e**x*a + b)/sqrt(a**2 - b**2))*a**2*b**2 + 96*e**(2*x)*sqrt(a**2 - b**2)*atan((e**x*a + b)/sqrt(a**2 - b**2))*b**4 + 24*sqrt(a**2 - b**2)*ata n((e**x*a + b)/sqrt(a**2 - b**2))*a**4 - 48*sqrt(a**2 - b**2)*atan((e**...