\(\int \frac {e^x}{(a-\tanh (2 x))^2} \, dx\) [227]

Optimal result

Integrand size = 14, antiderivative size = 152 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (1+a) \left (1+a+(-1+a) e^{4 x}\right )}-\frac {(1+4 a) \arctan \left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{2 (1-a)^2 (1+a)^{3/2} \sqrt [4]{1-a^2}}-\frac {(1+4 a) \text {arctanh}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{2 (1-a)^2 (1+a)^{3/2} \sqrt [4]{1-a^2}} \]

[Out]

exp(x)/(1-a)^2+exp(x)/(1-a)^2/(1+a)/(1+a+(-1+a)*exp(4*x))-1/2*(1+4*a)*arctan((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)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2320, 398, 393, 218, 214, 211} \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=-\frac {(4 a+1) \arctan \left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}-\frac {(4 a+1) \text {arctanh}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{2 (1-a)^2 (a+1)^{3/2} \sqrt [4]{1-a^2}}+\frac {e^x}{(1-a)^2}+\frac {e^x}{(1-a)^2 (a+1) \left ((a-1) e^{4 x}+a+1\right )} \]

[In]

Int[E^x/(a - Tanh[2*x])^2,x]

[Out]

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))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 398

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] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\left (1+x^4\right )^2}{\left (1+a-(1-a) x^4\right )^2} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (\frac {1}{(-1+a)^2}-\frac {4 \left (a-(1-a) x^4\right )}{(-1+a)^2 \left (1+a+(-1+a) x^4\right )^2}\right ) \, dx,x,e^x\right ) \\ & = \frac {e^x}{(1-a)^2}-\frac {4 \text {Subst}\left (\int \frac {a-(1-a) x^4}{\left (1+a+(-1+a) x^4\right )^2} \, dx,x,e^x\right )}{(1-a)^2} \\ & = \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) \text {Subst}\left (\int \frac {1}{1+a+(-1+a) x^4} \, dx,x,e^x\right )}{(1-a)^2 (1+a)} \\ & = \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) \text {Subst}\left (\int \frac {1}{\sqrt {1+a}-\sqrt {1-a} x^2} \, dx,x,e^x\right )}{2 (1-a)^2 (1+a)^{3/2}}-\frac {(1+4 a) \text {Subst}\left (\int \frac {1}{\sqrt {1+a}+\sqrt {1-a} x^2} \, dx,x,e^x\right )}{2 (1-a)^2 (1+a)^{3/2}} \\ & = \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}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.70 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\frac {\frac {4 (-1+a) e^x \left (2+2 a-e^{4 x}+a^2 \left (1+e^{4 x}\right )\right )}{1+a-e^{4 x}+a e^{4 x}}+(1+4 a) \text {RootSum}\left [1+a-\text {$\#$1}^4+a \text {$\#$1}^4\&,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ]}{4 (-1+a)^3 (1+a)} \]

[In]

Integrate[E^x/(a - Tanh[2*x])^2,x]

[Out]

((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))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.40 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.26

method	result	size
default	\(-\frac {2}{\left (-1+a \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2 \left (\frac {-\frac {\left (a -2\right ) \tanh \left (\frac {x}{2}\right )^{3}}{2 a \left (a +1\right )}-\frac {3 \tanh \left (\frac {x}{2}\right )^{2}}{2 \left (a +1\right )}+\frac {\left (a +2\right ) \tanh \left (\frac {x}{2}\right )}{2 a \left (a +1\right )}-\frac {1}{2 \left (a +1\right )}}{\tanh \left (\frac {x}{2}\right )^{4} a +6 \tanh \left (\frac {x}{2}\right )^{2} a -4 \tanh \left (\frac {x}{2}\right )^{3}+a -4 \tanh \left (\frac {x}{2}\right )}+\frac {\left (1+4 a \right ) \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{4}-4 \textit {\_Z}^{3}+6 a \,\textit {\_Z}^{2}-4 \textit {\_Z} +a \right )}{\sum }\frac {\left (\textit {\_R}^{2}-2 \textit {\_R} +1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{3} a -3 \textit {\_R}^{2}+3 \textit {\_R} a -1}\right )}{8 a +8}\right )}{\left (-1+a \right )^{2}}\)	\(191\)
risch	\(\frac {{\mathrm e}^{x}}{a^{2}-2 a +1}+\frac {{\mathrm e}^{x}}{\left (a +1\right ) \left (a^{2}-2 a +1\right ) \left (a \,{\mathrm e}^{4 x}-{\mathrm e}^{4 x}+a +1\right )}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a^{16}-512 a^{15}-1536 a^{14}+3584 a^{13}+3584 a^{12}-10752 a^{11}-3584 a^{10}+17920 a^{9}-17920 a^{7}+3584 a^{6}+10752 a^{5}-3584 a^{4}-3584 a^{3}+1536 a^{2}+512 a -256\right ) \textit {\_Z}^{4}+256 a^{4}+256 a^{3}+96 a^{2}+16 a +1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\frac {\left (-\frac {4 a^{4}}{1+4 a}+\frac {8 a^{2}}{1+4 a}-\frac {4}{1+4 a}\right ) \textit {\_R}}{\frac {4 a}{1+4 a}+\frac {1}{1+4 a}}\right )\right )\)	\(215\)

[In]

int(exp(x)/(a-tanh(2*x))^2,x,method=_RETURNVERBOSE)

[Out]

-2/(-1+a)^2/(tanh(1/2*x)-1)-2/(-1+a)^2*((-1/2*(a-2)/a/(a+1)*tanh(1/2*x)^3-3/2/(a+1)*tanh(1/2*x)^2+1/2*(a+2)/a/
(a+1)*tanh(1/2*x)-1/2/(a+1))/(tanh(1/2*x)^4*a+6*tanh(1/2*x)^2*a-4*tanh(1/2*x)^3+a-4*tanh(1/2*x))+1/8*(1+4*a)/(
a+1)*sum((_R^2-2*_R+1)/(_R^3*a-3*_R^2+3*_R*a-1)*ln(tanh(1/2*x)-_R),_R=RootOf(_Z^4*a-4*_Z^3+6*_Z^2*a-4*_Z+a)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 1604, normalized size of antiderivative = 10.55 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\text {Too large to display} \]

[In]

integrate(exp(x)/(a-tanh(2*x))^2,x, algorithm="fricas")

[Out]

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)*c
osh(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 + 4
2*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)*cosh(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)*si
nh(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)) + ((I*a^4 - 2*I*a^3 + 2*I*a - I)*cosh(x)^4 - 4*(-I*a
^4 + 2*I*a^3 - 2*I*a + I)*cosh(x)^3*sinh(x) - 6*(-I*a^4 + 2*I*a^3 - 2*I*a + I)*cosh(x)^2*sinh(x)^2 - 4*(-I*a^4
+ 2*I*a^3 - 2*I*a + I)*cosh(x)*sinh(x)^3 + (I*a^4 - 2*I*a^3 + 2*I*a - I)*sinh(x)^4 + I*a^4 - 2*I*a^2 + I)*(-(
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)*sin
h(x) - (I*a^4 - 2*I*a^2 + I)*(-(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)) + ((
-I*a^4 + 2*I*a^3 - 2*I*a + I)*cosh(x)^4 - 4*(I*a^4 - 2*I*a^3 + 2*I*a - I)*cosh(x)^3*sinh(x) - 6*(I*a^4 - 2*I*a
^3 + 2*I*a - I)*cosh(x)^2*sinh(x)^2 - 4*(I*a^4 - 2*I*a^3 + 2*I*a - I)*cosh(x)*sinh(x)^3 + (-I*a^4 + 2*I*a^3 -
2*I*a + I)*sinh(x)^4 - I*a^4 + 2*I*a^2 - I)*(-(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) - (-I*a^4 + 2*I*a^2 - I)*(-(256*a^4 + 256*a^3 + 96*a^2 + 1
6*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)) + 4*(a^2 + 2*a + 2)*cosh(x) + 4*(5*(a^2 - 1)*cosh(x)^4 + a^2 + 2*a
+ 2)*sinh(x))/((a^4 - 2*a^3 + 2*a - 1)*cosh(x)^4 + 4*(a^4 - 2*a^3 + 2*a - 1)*cosh(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)

Sympy [F]

\[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\int \frac {e^{x}}{\left (a - \tanh {\left (2 x \right )}\right )^{2}}\, dx \]

[In]

integrate(exp(x)/(a-tanh(2*x))**2,x)

[Out]

Integral(exp(x)/(a - tanh(2*x))**2, x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(exp(x)/(a-tanh(2*x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a-1>0)', see `assume?` for mor
e details)Is

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 456 vs. \(2 (119) = 238\).

Time = 0.27 (sec) , antiderivative size = 456, normalized size of antiderivative = 3.00 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=-\frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} + 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} - 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \log \left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{4 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} + \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} {\left (4 \, a + 1\right )} \log \left (-\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{4 \, {\left (\sqrt {2} a^{5} - \sqrt {2} a^{4} - 2 \, \sqrt {2} a^{3} + 2 \, \sqrt {2} a^{2} + \sqrt {2} a - \sqrt {2}\right )}} + \frac {e^{x}}{a^{2} - 2 \, a + 1} + \frac {e^{x}}{{\left (a^{3} - a^{2} - a + 1\right )} {\left (a e^{\left (4 \, x\right )} + a - e^{\left (4 \, x\right )} + 1\right )}} \]

[In]

integrate(exp(x)/(a-tanh(2*x))^2,x, algorithm="giac")

[Out]

-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)) - 1/2*(
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 + 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)) + 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 - sq
rt(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))

Mupad [B] (verification not implemented)

Time = 23.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.84 \[ \int \frac {e^x}{(a-\tanh (2 x))^2} \, dx=\frac {{\mathrm {e}}^x}{{\left (a-1\right )}^2}+\frac {\ln \left (\frac {4\,a+1}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}+\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{a^4-2\,a^3+2\,a-1}\right )\,\left (4\,a+1\right )}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}-\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{a^4-2\,a^3+2\,a-1}-\frac {4\,a+1}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}+\frac {{\mathrm {e}}^x}{{\left (a-1\right )}^2\,\left (a+1\right )\,\left (a+{\mathrm {e}}^{4\,x}\,\left (a-1\right )+1\right )}-\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{{\left (a-1\right )}^3\,\left (a+1\right )}-\frac {\left (4\,a+1\right )\,1{}\mathrm {i}}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )\,1{}\mathrm {i}}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}}+\frac {\ln \left (\frac {{\mathrm {e}}^x\,\left (4\,a+1\right )}{{\left (a-1\right )}^3\,\left (a+1\right )}+\frac {\left (4\,a+1\right )\,1{}\mathrm {i}}{{\left (a-1\right )}^{13/4}\,{\left (-a-1\right )}^{3/4}}\right )\,\left (4\,a+1\right )\,1{}\mathrm {i}}{4\,{\left (a-1\right )}^{9/4}\,{\left (-a-1\right )}^{7/4}} \]

[In]

int(exp(x)/(a - tanh(2*x))^2,x)

[Out]

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)) + (
exp(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)) + exp(x)/((a - 1)^2*(a + 1)*(a + exp(4*x)*(a - 1) + 1))