Integrand size = 14, antiderivative size = 155 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (1+a) \left (1+a-(1-a) e^{4 x}\right )}-\frac {(1+4 a) \arctan \left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{2 (1-a)^2 (1+a)^{3/2} \sqrt [4]{1-a^2}}-\frac {(1+4 a) \text {arctanh}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{2 (1-a)^2 (1+a)^{3/2} \sqrt [4]{1-a^2}} \] Output:
exp(x)/(1-a)^2+exp(x)/(1-a)^2/(1+a)/(1+a-(1-a)*exp(4*x))-1/2*(1+4*a)*arcta n((1-a)^(1/4)*exp(x)/(1+a)^(1/4))/(1-a)^2/(1+a)^(3/2)/(-a^2+1)^(1/4)-1/2*( 1+4*a)*arctanh((1-a)^(1/4)*exp(x)/(1+a)^(1/4))/(1-a)^2/(1+a)^(3/2)/(-a^2+1 )^(1/4)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\frac {\frac {4 (-1+a) e^x \left (2+2 a-e^{4 x}+a^2 \left (1+e^{4 x}\right )\right )}{1+a-e^{4 x}+a e^{4 x}}+(1+4 a) \text {RootSum}\left [1+a-\text {$\#$1}^4+a \text {$\#$1}^4\&,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ]}{4 (-1+a)^3 (1+a)} \] Input:
Integrate[E^x/(a - Tanh[2*x])^2,x]
Output:
((4*(-1 + a)*E^x*(2 + 2*a - E^(4*x) + a^2*(1 + E^(4*x))))/(1 + a - E^(4*x) + a*E^(4*x)) + (1 + 4*a)*RootSum[1 + a - #1^4 + a*#1^4 & , (x - Log[E^x - #1])/#1^3 & ])/(4*(-1 + a)^3*(1 + a))
Time = 0.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {2720, 915, 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.
\(\displaystyle \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int \frac {\left (e^{4 x}+1\right )^2}{\left ((1-a) \left (-e^{4 x}\right )+a+1\right )^2}de^x\) |
\(\Big \downarrow \) 915 |
\(\displaystyle \int \left (\frac {1}{(a-1)^2}-\frac {4 \left (a-(1-a) e^{4 x}\right )}{(a-1)^2 \left ((a-1) e^{4 x}+a+1\right )^2}\right )de^x\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(4 a+1) \arctan \left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}-\frac {(4 a+1) \text {arctanh}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}+\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (a+1) \left ((1-a) \left (-e^{4 x}\right )+a+1\right )}\) |
Input:
Int[E^x/(a - Tanh[2*x])^2,x]
Output:
E^x/(1 - a)^2 + E^x/((1 - a)^2*(1 + a)*(1 + a - (1 - a)*E^(4*x))) - ((1 + 4*a)*ArcTan[((1 - a)^(1/4)*E^x)/(1 + a)^(1/4)])/(2*(1 - a)^2*(1 + a)^(3/2) *(1 - a^2)^(1/4)) - ((1 + 4*a)*ArcTanh[((1 - a)^(1/4)*E^x)/(1 + a)^(1/4)]) /(2*(1 - a)^2*(1 + a)^(3/2)*(1 - a^2)^(1/4))
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a , b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.08 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.39
method | result | size |
risch | \(\frac {{\mathrm e}^{x}}{a^{2}-2 a +1}+\frac {{\mathrm e}^{x}}{\left (a +1\right ) \left (a^{2}-2 a +1\right ) \left ({\mathrm e}^{4 x} a -{\mathrm e}^{4 x}+a +1\right )}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a^{16}-512 a^{15}-1536 a^{14}+3584 a^{13}+3584 a^{12}-10752 a^{11}-3584 a^{10}+17920 a^{9}-17920 a^{7}+3584 a^{6}+10752 a^{5}-3584 a^{4}-3584 a^{3}+1536 a^{2}+512 a -256\right ) \textit {\_Z}^{4}+256 a^{4}+256 a^{3}+96 a^{2}+16 a +1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\frac {\left (-\frac {4 a^{4}}{1+4 a}+\frac {8 a^{2}}{1+4 a}-\frac {4}{1+4 a}\right ) \textit {\_R}}{\frac {4 a}{1+4 a}+\frac {1}{1+4 a}}\right )\right )\) | \(215\) |
Input:
int(exp(x)/(a-tanh(2*x))^2,x,method=_RETURNVERBOSE)
Output:
exp(x)/(a^2-2*a+1)+1/(a+1)*exp(x)/(a^2-2*a+1)/(a*exp(x)^4-exp(x)^4+a+1)+su m(_R*ln(exp(x)+(-4/(1+4*a)*a^4+8/(1+4*a)*a^2-4/(1+4*a))/(4/(1+4*a)*a+1/(1+ 4*a))*_R),_R=RootOf((256*a^16-512*a^15-1536*a^14+3584*a^13+3584*a^12-10752 *a^11-3584*a^10+17920*a^9-17920*a^7+3584*a^6+10752*a^5-3584*a^4-3584*a^3+1 536*a^2+512*a-256)*_Z^4+256*a^4+256*a^3+96*a^2+16*a+1))
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 1604, normalized size of antiderivative = 10.35 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\text {Too large to display} \] Input:
integrate(exp(x)/(a-tanh(2*x))^2,x, algorithm="fricas")
Output:
1/4*(4*(a^2 - 1)*cosh(x)^5 + 40*(a^2 - 1)*cosh(x)^3*sinh(x)^2 + 40*(a^2 - 1)*cosh(x)^2*sinh(x)^3 + 20*(a^2 - 1)*cosh(x)*sinh(x)^4 + 4*(a^2 - 1)*sinh (x)^5 - ((a^4 - 2*a^3 + 2*a - 1)*cosh(x)^4 + 4*(a^4 - 2*a^3 + 2*a - 1)*cos h(x)^3*sinh(x) + 6*(a^4 - 2*a^3 + 2*a - 1)*cosh(x)^2*sinh(x)^2 + 4*(a^4 - 2*a^3 + 2*a - 1)*cosh(x)*sinh(x)^3 + (a^4 - 2*a^3 + 2*a - 1)*sinh(x)^4 + a ^4 - 2*a^2 + 1)*(-(256*a^4 + 256*a^3 + 96*a^2 + 16*a + 1)/(a^16 - 2*a^15 - 6*a^14 + 14*a^13 + 14*a^12 - 42*a^11 - 14*a^10 + 70*a^9 - 70*a^7 + 14*a^6 + 42*a^5 - 14*a^4 - 14*a^3 + 6*a^2 + 2*a - 1))^(1/4)*log((4*a + 1)*cosh(x ) + (4*a + 1)*sinh(x) + (a^4 - 2*a^2 + 1)*(-(256*a^4 + 256*a^3 + 96*a^2 + 16*a + 1)/(a^16 - 2*a^15 - 6*a^14 + 14*a^13 + 14*a^12 - 42*a^11 - 14*a^10 + 70*a^9 - 70*a^7 + 14*a^6 + 42*a^5 - 14*a^4 - 14*a^3 + 6*a^2 + 2*a - 1))^ (1/4)) + ((a^4 - 2*a^3 + 2*a - 1)*cosh(x)^4 + 4*(a^4 - 2*a^3 + 2*a - 1)*co sh(x)^3*sinh(x) + 6*(a^4 - 2*a^3 + 2*a - 1)*cosh(x)^2*sinh(x)^2 + 4*(a^4 - 2*a^3 + 2*a - 1)*cosh(x)*sinh(x)^3 + (a^4 - 2*a^3 + 2*a - 1)*sinh(x)^4 + a^4 - 2*a^2 + 1)*(-(256*a^4 + 256*a^3 + 96*a^2 + 16*a + 1)/(a^16 - 2*a^15 - 6*a^14 + 14*a^13 + 14*a^12 - 42*a^11 - 14*a^10 + 70*a^9 - 70*a^7 + 14*a^ 6 + 42*a^5 - 14*a^4 - 14*a^3 + 6*a^2 + 2*a - 1))^(1/4)*log((4*a + 1)*cosh( x) + (4*a + 1)*sinh(x) - (a^4 - 2*a^2 + 1)*(-(256*a^4 + 256*a^3 + 96*a^2 + 16*a + 1)/(a^16 - 2*a^15 - 6*a^14 + 14*a^13 + 14*a^12 - 42*a^11 - 14*a^10 + 70*a^9 - 70*a^7 + 14*a^6 + 42*a^5 - 14*a^4 - 14*a^3 + 6*a^2 + 2*a - ...
\[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\int \frac {e^{x}}{\left (a - \tanh {\left (2 x \right )}\right )^{2}}\, dx \] Input:
integrate(exp(x)/(a-tanh(2*x))**2,x)
Output:
Integral(exp(x)/(a - tanh(2*x))**2, x)
Exception generated. \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(exp(x)/(a-tanh(2*x))^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(a-1>0)', see `assume?` for more details)Is
Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (119) = 238\).
Time = 0.11 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.94 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=-\frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} + 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} - 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \log \left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{4 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} + \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \log \left (-\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{4 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} + \frac {e^{x}}{a^{2} - 2 \, a + 1} + \frac {e^{x}}{{\left (a^{3} - a^{2} - a + 1\right )} {\left (a e^{\left (4 \, x\right )} + a - e^{\left (4 \, x\right )} + 1\right )}} \] Input:
integrate(exp(x)/(a-tanh(2*x))^2,x, algorithm="giac")
Output:
-1/2*(a^4 - 2*a^3 + 2*a - 1)^(1/4)*(4*a + 1)*arctan(1/2*sqrt(2)*(sqrt(2)*( (a + 1)/(a - 1))^(1/4) + 2*e^x)/((a + 1)/(a - 1))^(1/4))/(sqrt(2)*a^5 - sq rt(2)*a^4 - 2*sqrt(2)*a^3 + 2*sqrt(2)*a^2 + sqrt(2)*a - sqrt(2)) - 1/2*(a^ 4 - 2*a^3 + 2*a - 1)^(1/4)*(4*a + 1)*arctan(-1/2*sqrt(2)*(sqrt(2)*((a + 1) /(a - 1))^(1/4) - 2*e^x)/((a + 1)/(a - 1))^(1/4))/(sqrt(2)*a^5 - sqrt(2)*a ^4 - 2*sqrt(2)*a^3 + 2*sqrt(2)*a^2 + sqrt(2)*a - sqrt(2)) - 1/4*(a^4 - 2*a ^3 + 2*a - 1)^(1/4)*(4*a + 1)*log(sqrt(2)*((a + 1)/(a - 1))^(1/4)*e^x + sq rt((a + 1)/(a - 1)) + e^(2*x))/(sqrt(2)*a^5 - sqrt(2)*a^4 - 2*sqrt(2)*a^3 + 2*sqrt(2)*a^2 + sqrt(2)*a - sqrt(2)) + 1/4*(a^4 - 2*a^3 + 2*a - 1)^(1/4) *(4*a + 1)*log(-sqrt(2)*((a + 1)/(a - 1))^(1/4)*e^x + sqrt((a + 1)/(a - 1) ) + e^(2*x))/(sqrt(2)*a^5 - sqrt(2)*a^4 - 2*sqrt(2)*a^3 + 2*sqrt(2)*a^2 + sqrt(2)*a - sqrt(2)) + e^x/(a^2 - 2*a + 1) + e^x/((a^3 - a^2 - a + 1)*(a*e ^(4*x) + a - e^(4*x) + 1))
Time = 24.35 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.81 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\frac {{\mathrm {e}}^x}{{\left (a-1\right )}^2}+\frac {\ln \left (\frac {4\,a+1}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}+\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{a^4-2\,a^3+2\,a-1}\right )\,\left (4\,a+1\right )}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}-\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{a^4-2\,a^3+2\,a-1}-\frac {4\,a+1}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}+\frac {{\mathrm {e}}^x}{{\left (a-1\right )}^2\,\left (a+1\right )\,\left (a+{\mathrm {e}}^{4\,x}\,\left (a-1\right )+1\right )}-\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{{\left (a-1\right )}^3\,\left (a+1\right )}-\frac {\left (4\,a+1\right )\,1{}\mathrm {i}}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )\,1{}\mathrm {i}}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}+\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{{\left (a-1\right )}^3\,\left (a+1\right )}+\frac {\left (4\,a+1\right )\,1{}\mathrm {i}}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )\,1{}\mathrm {i}}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}} \] Input:
int(exp(x)/(a - tanh(2*x))^2,x)
Output:
exp(x)/(a - 1)^2 - (log((exp(x)*(4*a + 1))/((a - 1)^3*(a + 1)) - ((4*a + 1 )*1i)/((a - 1)^(13/4)*(- a - 1)^(3/4)))*(4*a + 1)*1i)/(4*(a - 1)^(9/4)*(- a - 1)^(7/4)) + (log(((4*a + 1)*1i)/((a - 1)^(13/4)*(- a - 1)^(3/4)) + (ex p(x)*(4*a + 1))/((a - 1)^3*(a + 1)))*(4*a + 1)*1i)/(4*(a - 1)^(9/4)*(- a - 1)^(7/4)) + (log((4*a + 1)/((a - 1)^(13/4)*(- a - 1)^(3/4)) + (exp(x)*(4* a + 1))/(2*a - 2*a^3 + a^4 - 1))*(4*a + 1))/(4*(a - 1)^(9/4)*(- a - 1)^(7/ 4)) - (log((exp(x)*(4*a + 1))/(2*a - 2*a^3 + a^4 - 1) - (4*a + 1)/((a - 1) ^(13/4)*(- a - 1)^(3/4)))*(4*a + 1))/(4*(a - 1)^(9/4)*(- a - 1)^(7/4)) + e xp(x)/((a - 1)^2*(a + 1)*(a + exp(4*x)*(a - 1) + 1))
Time = 0.27 (sec) , antiderivative size = 5067, normalized size of antiderivative = 32.69 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx =\text {Too large to display} \] Input:
int(exp(x)/(a-tanh(2*x))^2,x)
Output:
(16*e**(4*x)*(a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2)*atan(((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2) - 2*e**x*sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4)* sqrt(2)))*tanh(2*x)*a**4 - 4*e**(4*x)*(a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2 )*atan(((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2) - 2*e**x*sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2)))*tanh(2*x)*a**3 + 6*e**(4*x)*(a + 1)**(1 /4)*(a - 1)**(3/4)*sqrt(2)*atan(((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2) - 2 *e**x*sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2)))*tanh(2*x)*a**2 - 14*e**(4*x)*(a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2)*atan(((a + 1)**(1/4)* (a - 1)**(1/4)*sqrt(2) - 2*e**x*sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4 )*sqrt(2)))*tanh(2*x)*a - 4*e**(4*x)*(a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2) *atan(((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2) - 2*e**x*sqrt(a - 1))/((a + 1 )**(1/4)*(a - 1)**(1/4)*sqrt(2)))*tanh(2*x) - 16*e**(4*x)*(a + 1)**(1/4)*( a - 1)**(3/4)*sqrt(2)*atan(((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2) - 2*e**x *sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2)))*a**5 + 4*e**(4*x)*( a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2)*atan(((a + 1)**(1/4)*(a - 1)**(1/4)*s qrt(2) - 2*e**x*sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2)))*a**4 - 6*e**(4*x)*(a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2)*atan(((a + 1)**(1/4)*( a - 1)**(1/4)*sqrt(2) - 2*e**x*sqrt(a - 1))/((a + 1)**(1/4)*(a - 1)**(1/4) *sqrt(2)))*a**3 + 14*e**(4*x)*(a + 1)**(1/4)*(a - 1)**(3/4)*sqrt(2)*atan(( (a + 1)**(1/4)*(a - 1)**(1/4)*sqrt(2) - 2*e**x*sqrt(a - 1))/((a + 1)**(...