Integrand size = 15, antiderivative size = 94 \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=-\frac {b^{5/2} \arctan \left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)^3}-\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}(\cosh (x))}{8 (a+b)^3}+\frac {(3 a+7 b) \coth (x) \text {csch}(x)}{8 (a+b)^2}-\frac {\coth (x) \text {csch}^3(x)}{4 (a+b)} \] Output:
-b^(5/2)*arctan(b^(1/2)*cosh(x)/a^(1/2))/a^(1/2)/(a+b)^3-1/8*(3*a^2+10*a*b +15*b^2)*arctanh(cosh(x))/(a+b)^3+1/8*(3*a+7*b)*coth(x)*csch(x)/(a+b)^2-co th(x)*csch(x)^3/(4*a+4*b)
Result contains complex when optimal does not.
Time = 0.51 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.44 \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=\frac {2 \sqrt {a} \left (3 a^2+10 a b+7 b^2\right ) \text {csch}^2\left (\frac {x}{2}\right )-\sqrt {a} (a+b)^2 \text {csch}^4\left (\frac {x}{2}\right )-8 \left (8 b^{5/2} \arctan \left (\frac {\sqrt {b}-i \sqrt {a+b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a}}\right )+8 b^{5/2} \arctan \left (\frac {\sqrt {b}+i \sqrt {a+b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a}}\right )+\sqrt {a} \left (3 a^2+10 a b+15 b^2\right ) \left (\log \left (\cosh \left (\frac {x}{2}\right )\right )-\log \left (\sinh \left (\frac {x}{2}\right )\right )\right )\right )+2 \sqrt {a} \left (3 a^2+10 a b+7 b^2\right ) \text {sech}^2\left (\frac {x}{2}\right )+\sqrt {a} (a+b)^2 \text {sech}^4\left (\frac {x}{2}\right )}{64 \sqrt {a} (a+b)^3} \] Input:
Integrate[Csch[x]^5/(a + b*Cosh[x]^2),x]
Output:
(2*Sqrt[a]*(3*a^2 + 10*a*b + 7*b^2)*Csch[x/2]^2 - Sqrt[a]*(a + b)^2*Csch[x /2]^4 - 8*(8*b^(5/2)*ArcTan[(Sqrt[b] - I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]] + 8*b^(5/2)*ArcTan[(Sqrt[b] + I*Sqrt[a + b]*Tanh[x/2])/Sqrt[a]] + Sqrt[a]*( 3*a^2 + 10*a*b + 15*b^2)*(Log[Cosh[x/2]] - Log[Sinh[x/2]])) + 2*Sqrt[a]*(3 *a^2 + 10*a*b + 7*b^2)*Sech[x/2]^2 + Sqrt[a]*(a + b)^2*Sech[x/2]^4)/(64*Sq rt[a]*(a + b)^3)
Time = 0.38 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.33, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 26, 3669, 316, 402, 397, 218, 219}
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.
\(\displaystyle \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i}{\cos \left (\frac {\pi }{2}+i x\right )^5 \left (a+b \sin \left (\frac {\pi }{2}+i x\right )^2\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {1}{\cos \left (i x+\frac {\pi }{2}\right )^5 \left (b \sin \left (i x+\frac {\pi }{2}\right )^2+a\right )}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle -\int \frac {1}{\left (1-\cosh ^2(x)\right )^3 \left (b \cosh ^2(x)+a\right )}d\cosh (x)\) |
\(\Big \downarrow \) 316 |
\(\displaystyle -\frac {\int \frac {3 b \cosh ^2(x)+3 a+4 b}{\left (1-\cosh ^2(x)\right )^2 \left (b \cosh ^2(x)+a\right )}d\cosh (x)}{4 (a+b)}-\frac {\cosh (x)}{4 (a+b) \left (1-\cosh ^2(x)\right )^2}\) |
\(\Big \downarrow \) 402 |
\(\displaystyle -\frac {\frac {\int \frac {3 a^2+7 b a+8 b^2+b (3 a+7 b) \cosh ^2(x)}{\left (1-\cosh ^2(x)\right ) \left (b \cosh ^2(x)+a\right )}d\cosh (x)}{2 (a+b)}+\frac {(3 a+7 b) \cosh (x)}{2 (a+b) \left (1-\cosh ^2(x)\right )}}{4 (a+b)}-\frac {\cosh (x)}{4 (a+b) \left (1-\cosh ^2(x)\right )^2}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle -\frac {\frac {\frac {\left (3 a^2+10 a b+15 b^2\right ) \int \frac {1}{1-\cosh ^2(x)}d\cosh (x)}{a+b}+\frac {8 b^3 \int \frac {1}{b \cosh ^2(x)+a}d\cosh (x)}{a+b}}{2 (a+b)}+\frac {(3 a+7 b) \cosh (x)}{2 (a+b) \left (1-\cosh ^2(x)\right )}}{4 (a+b)}-\frac {\cosh (x)}{4 (a+b) \left (1-\cosh ^2(x)\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {\frac {\frac {\left (3 a^2+10 a b+15 b^2\right ) \int \frac {1}{1-\cosh ^2(x)}d\cosh (x)}{a+b}+\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 (a+b)}+\frac {(3 a+7 b) \cosh (x)}{2 (a+b) \left (1-\cosh ^2(x)\right )}}{4 (a+b)}-\frac {\cosh (x)}{4 (a+b) \left (1-\cosh ^2(x)\right )^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\frac {\left (3 a^2+10 a b+15 b^2\right ) \text {arctanh}(\cosh (x))}{a+b}+\frac {8 b^{5/2} \arctan \left (\frac {\sqrt {b} \cosh (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 (a+b)}+\frac {(3 a+7 b) \cosh (x)}{2 (a+b) \left (1-\cosh ^2(x)\right )}}{4 (a+b)}-\frac {\cosh (x)}{4 (a+b) \left (1-\cosh ^2(x)\right )^2}\) |
Input:
Int[Csch[x]^5/(a + b*Cosh[x]^2),x]
Output:
-1/4*Cosh[x]/((a + b)*(1 - Cosh[x]^2)^2) - (((8*b^(5/2)*ArcTan[(Sqrt[b]*Co sh[x])/Sqrt[a]])/(Sqrt[a]*(a + b)) + ((3*a^2 + 10*a*b + 15*b^2)*ArcTanh[Co sh[x]])/(a + b))/(2*(a + b)) + ((3*a + 7*b)*Cosh[x])/(2*(a + b)*(1 - Cosh[ x]^2)))/(4*(a + b))
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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)^(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]
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[((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]
Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 52.70 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.44
method | result | size |
default | \(\frac {\left (\tanh \left (\frac {x}{2}\right )^{2} a +b \tanh \left (\frac {x}{2}\right )^{2}-4 a -8 b \right )^{2}}{64 \left (a +b \right )^{3}}-\frac {b^{3} \arctan \left (\frac {2 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{2}-2 a +2 b}{4 \sqrt {a b}}\right )}{\left (a +b \right )^{3} \sqrt {a b}}-\frac {1}{64 \left (a +b \right ) \tanh \left (\frac {x}{2}\right )^{4}}-\frac {-4 a -8 b}{32 \left (a +b \right )^{2} \tanh \left (\frac {x}{2}\right )^{2}}+\frac {\left (6 a^{2}+20 a b +30 b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{16 \left (a +b \right )^{3}}\) | \(135\) |
risch | \(\frac {{\mathrm e}^{x} \left (3 \,{\mathrm e}^{6 x} a +7 \,{\mathrm e}^{6 x} b -11 \,{\mathrm e}^{4 x} a -15 \,{\mathrm e}^{4 x} b -11 \,{\mathrm e}^{2 x} a -15 \,{\mathrm e}^{2 x} b +3 a +7 b \right )}{4 \left ({\mathrm e}^{2 x}-1\right )^{4} \left (a +b \right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{x}-1\right ) a^{2}}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {5 \ln \left ({\mathrm e}^{x}-1\right ) a b}{4 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {15 \ln \left ({\mathrm e}^{x}-1\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {3 \ln \left ({\mathrm e}^{x}+1\right ) a^{2}}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {5 \ln \left ({\mathrm e}^{x}+1\right ) a b}{4 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}-\frac {15 \ln \left ({\mathrm e}^{x}+1\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 b^{2} a +b^{3}\right )}+\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 x}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{x}}{b}+1\right )}{2 a \left (a +b \right )^{3}}-\frac {\sqrt {-a b}\, b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{x}}{b}+1\right )}{2 a \left (a +b \right )^{3}}\) | \(330\) |
Input:
int(csch(x)^5/(a+b*cosh(x)^2),x,method=_RETURNVERBOSE)
Output:
1/64*(tanh(1/2*x)^2*a+b*tanh(1/2*x)^2-4*a-8*b)^2/(a+b)^3-1/(a+b)^3*b^3/(a* b)^(1/2)*arctan(1/4*(2*(a+b)*tanh(1/2*x)^2-2*a+2*b)/(a*b)^(1/2))-1/64/(a+b )/tanh(1/2*x)^4-1/32*(-4*a-8*b)/(a+b)^2/tanh(1/2*x)^2+1/16/(a+b)^3*(6*a^2+ 20*a*b+30*b^2)*ln(tanh(1/2*x))
Leaf count of result is larger than twice the leaf count of optimal. 2724 vs. \(2 (80) = 160\).
Time = 0.19 (sec) , antiderivative size = 5326, normalized size of antiderivative = 56.66 \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Too large to display} \] Input:
integrate(csch(x)^5/(a+b*cosh(x)^2),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Timed out} \] Input:
integrate(csch(x)**5/(a+b*cosh(x)**2),x)
Output:
Timed out
\[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=\int { \frac {\operatorname {csch}\left (x\right )^{5}}{b \cosh \left (x\right )^{2} + a} \,d x } \] Input:
integrate(csch(x)^5/(a+b*cosh(x)^2),x, algorithm="maxima")
Output:
-1/8*(3*a^2 + 10*a*b + 15*b^2)*log(e^x + 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3 ) + 1/8*(3*a^2 + 10*a*b + 15*b^2)*log(e^x - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 1/4*((3*a + 7*b)*e^(7*x) - (11*a + 15*b)*e^(5*x) - (11*a + 15*b)*e^ (3*x) + (3*a + 7*b)*e^x)/(a^2 + 2*a*b + b^2 + (a^2 + 2*a*b + b^2)*e^(8*x) - 4*(a^2 + 2*a*b + b^2)*e^(6*x) + 6*(a^2 + 2*a*b + b^2)*e^(4*x) - 4*(a^2 + 2*a*b + b^2)*e^(2*x)) - 32*integrate(1/16*(b^3*e^(3*x) - b^3*e^x)/(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4 + (a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*e^(4*x) + 2*(2*a^4 + 7*a^3*b + 9*a^2*b^2 + 5*a*b^3 + b^4)*e^(2*x)), x)
Exception generated. \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(csch(x)^5/(a+b*cosh(x)^2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 17.45 (sec) , antiderivative size = 5056, normalized size of antiderivative = 53.79 \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx=\text {Too large to display} \] Input:
int(1/(sinh(x)^5*(a + b*cosh(x)^2)),x)
Output:
(atan((exp(x)*(243*a^12*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20 *a^3*b^3 - 15*a^4*b^2)^(3/2) + 3840*b^12*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2) + 110560*a*b^11*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2) + 4050* a^11*b*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4 *b^2)^(3/2) + 976143*a^2*b^10*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^ 4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2) + 2740050*a^3*b^9*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2) + 4252775*a^4*b^ 8*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2) ^(3/2) + 4316760*a^5*b^7*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 2 0*a^3*b^3 - 15*a^4*b^2)^(3/2) + 3087390*a^6*b^6*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2) + 1608364*a^7*b^5*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2 ) + 615750*a^8*b^4*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3* b^3 - 15*a^4*b^2)^(3/2) + 171000*a^9*b^3*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2) + 33075*a^10*b^2*(- 6*a*b^5 - 6*a^5*b - a^6 - b^6 - 15*a^2*b^4 - 20*a^3*b^3 - 15*a^4*b^2)^(3/2)))/(81* a^19*(300*a*b^3 + 60*a^3*b + 9*a^4 + 225*b^4 + 190*a^2*b^2)^(1/2) + 256*b^ 19*(300*a*b^3 + 60*a^3*b + 9*a^4 + 225*b^4 + 190*a^2*b^2)^(1/2) + 9504*a*b ^18*(300*a*b^3 + 60*a^3*b + 9*a^4 + 225*b^4 + 190*a^2*b^2)^(1/2) + 1809...
Time = 0.52 (sec) , antiderivative size = 2380, normalized size of antiderivative = 25.32 \[ \int \frac {\text {csch}^5(x)}{a+b \cosh ^2(x)} \, dx =\text {Too large to display} \] Input:
int(csch(x)^5/(a+b*cosh(x)^2),x)
Output:
( - 8*e**(8*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2* a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 32*e**(6*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b - 48* e**(4*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b) *atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 32*e** (2*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*at an((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b - 8*sqrt(b) *sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/( sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*b + 8*e**(8*x)*sqrt(b)*sqr t(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*s qrt(a + b) + 2*a + b)))*a*b - 32*e**(6*x)*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b) ))*a*b + 48*e**(4*x)*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e **x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a*b - 32*e**(2*x)* sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt( 2*sqrt(a)*sqrt(a + b) + 2*a + b)))*a*b + 8*sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b)*atan((e**x*b)/(sqrt(b)*sqrt(2*sqrt(a)*sqrt(a + b) + 2*a + b )))*a*b - 4*e**(8*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(a)*sqrt(a + b ) - 2*a - b)*log( - sqrt(2*sqrt(a)*sqrt(a + b) - 2*a - b) + e**x*sqrt(b...