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

Optimal. Leaf size=107 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{(1-a) \sqrt {a+1} \sqrt [4]{1-a^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{(1-a) \sqrt {a+1} \sqrt [4]{1-a^2}}-\frac {e^x}{1-a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2282, 388, 212, 208, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{(1-a) \sqrt {a+1} \sqrt [4]{1-a^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{(1-a) \sqrt {a+1} \sqrt [4]{1-a^2}}-\frac {e^x}{1-a} \]

Antiderivative was successfully verified.

[In]

Int[E^x/(a - Tanh[2*x]),x]

[Out]

-(E^x/(1 - a)) + ArcTan[((1 - a)^(1/4)*E^x)/(1 + a)^(1/4)]/((1 - a)*Sqrt[1 + a]*(1 - a^2)^(1/4)) + ArcTanh[((1
- a)^(1/4)*E^x)/(1 + a)^(1/4)]/((1 - a)*Sqrt[1 + a]*(1 - a^2)^(1/4))

Rule 205

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

Rule 208

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

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 388

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

Rule 2282

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*} \int \frac {e^x}{a-\tanh (2 x)} \, dx &=\operatorname {Subst}\left (\int \frac {1+x^4}{1+a-(1-a) x^4} \, dx,x,e^x\right )\\ &=-\frac {e^x}{1-a}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+a+(-1+a) x^4} \, dx,x,e^x\right )}{1-a}\\ &=-\frac {e^x}{1-a}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1+a}-\sqrt {1-a} x^2} \, dx,x,e^x\right )}{(1-a) \sqrt {1+a}}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1+a}+\sqrt {1-a} x^2} \, dx,x,e^x\right )}{(1-a) \sqrt {1+a}}\\ &=-\frac {e^x}{1-a}+\frac {\tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{(1-a) \sqrt {1+a} \sqrt [4]{1-a^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{1+a}}\right )}{(1-a) \sqrt {1+a} \sqrt [4]{1-a^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.09, size = 81, normalized size = 0.76 \[ \frac {-\sqrt [4]{1-a} (a+1)^{3/4} e^x+\tan ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )+\tanh ^{-1}\left (\frac {\sqrt [4]{1-a} e^x}{\sqrt [4]{a+1}}\right )}{(1-a)^{5/4} (a+1)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(a - Tanh[2*x]),x]

[Out]

(-((1 - a)^(1/4)*(1 + a)^(3/4)*E^x) + ArcTan[((1 - a)^(1/4)*E^x)/(1 + a)^(1/4)] + ArcTanh[((1 - a)^(1/4)*E^x)/
(1 + a)^(1/4)])/((1 - a)^(5/4)*(1 + a)^(3/4))

________________________________________________________________________________________

fricas [B]  time = 0.50, size = 440, normalized size = 4.11 \[ -\frac {4 \, {\left (a - 1\right )} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {1}{4}} \arctan \left (-{\left (a^{6} - 2 \, a^{5} - a^{4} + 4 \, a^{3} - a^{2} - 2 \, a + 1\right )} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {3}{4}} e^{x} + {\left (a^{6} - 2 \, a^{5} - a^{4} + 4 \, a^{3} - a^{2} - 2 \, a + 1\right )} \sqrt {{\left (a^{4} - 2 \, a^{2} + 1\right )} \sqrt {-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}} + e^{\left (2 \, x\right )}} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {3}{4}}\right ) + {\left (a - 1\right )} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {1}{4}} \log \left ({\left (a^{2} - 1\right )} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {1}{4}} + e^{x}\right ) - {\left (a - 1\right )} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {1}{4}} \log \left (-{\left (a^{2} - 1\right )} \left (-\frac {1}{a^{8} - 2 \, a^{7} - 2 \, a^{6} + 6 \, a^{5} - 6 \, a^{3} + 2 \, a^{2} + 2 \, a - 1}\right )^{\frac {1}{4}} + e^{x}\right ) - 2 \, e^{x}}{2 \, {\left (a - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(a - 1)*(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 + 2*a - 1))^(1/4)*arctan(-(a^6 - 2*a^5 - a^4
+ 4*a^3 - a^2 - 2*a + 1)*(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 + 2*a - 1))^(3/4)*e^x + (a^6 - 2*a^5
- a^4 + 4*a^3 - a^2 - 2*a + 1)*sqrt((a^4 - 2*a^2 + 1)*sqrt(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 +
2*a - 1)) + e^(2*x))*(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 + 2*a - 1))^(3/4)) + (a - 1)*(-1/(a^8 -
2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 + 2*a - 1))^(1/4)*log((a^2 - 1)*(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3
+ 2*a^2 + 2*a - 1))^(1/4) + e^x) - (a - 1)*(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 + 2*a - 1))^(1/4)
*log(-(a^2 - 1)*(-1/(a^8 - 2*a^7 - 2*a^6 + 6*a^5 - 6*a^3 + 2*a^2 + 2*a - 1))^(1/4) + e^x) - 2*e^x)/(a - 1)

________________________________________________________________________________________

giac [B]  time = 0.12, size = 328, normalized size = 3.07 \[ -\frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} + 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{\sqrt {2} a^{3} - \sqrt {2} a^{2} - \sqrt {2} a + \sqrt {2}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} - 2 \, e^{x}\right )}}{2 \, \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}}}\right )}{\sqrt {2} a^{3} - \sqrt {2} a^{2} - \sqrt {2} a + \sqrt {2}} - \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} \log \left (\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{2 \, {\left (\sqrt {2} a^{3} - \sqrt {2} a^{2} - \sqrt {2} a + \sqrt {2}\right )}} + \frac {{\left (a^{4} - 2 \, a^{3} + 2 \, a - 1\right )}^{\frac {1}{4}} \log \left (-\sqrt {2} \left (\frac {a + 1}{a - 1}\right )^{\frac {1}{4}} e^{x} + \sqrt {\frac {a + 1}{a - 1}} + e^{\left (2 \, x\right )}\right )}{2 \, {\left (\sqrt {2} a^{3} - \sqrt {2} a^{2} - \sqrt {2} a + \sqrt {2}\right )}} + \frac {e^{x}}{a - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^4 - 2*a^3 + 2*a - 1)^(1/4)*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^3 - sqrt(2)*a^2 - sqrt(2)*a + sqrt(2)) - (a^4 - 2*a^3 + 2*a - 1)^(1/4)*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^3 - sqrt(2)*a^2 - sqrt(2)*a + sqr
t(2)) - 1/2*(a^4 - 2*a^3 + 2*a - 1)^(1/4)*log(sqrt(2)*((a + 1)/(a - 1))^(1/4)*e^x + sqrt((a + 1)/(a - 1)) + e^
(2*x))/(sqrt(2)*a^3 - sqrt(2)*a^2 - sqrt(2)*a + sqrt(2)) + 1/2*(a^4 - 2*a^3 + 2*a - 1)^(1/4)*log(-sqrt(2)*((a
+ 1)/(a - 1))^(1/4)*e^x + sqrt((a + 1)/(a - 1)) + e^(2*x))/(sqrt(2)*a^3 - sqrt(2)*a^2 - sqrt(2)*a + sqrt(2)) +
e^x/(a - 1)

________________________________________________________________________________________

maple [C]  time = 0.29, size = 87, normalized size = 0.81 \[ \frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{4}-4 \textit {\_Z}^{3}+6 a \,\textit {\_Z}^{2}-4 \textit {\_Z} +a \right )}{\sum }\frac {\left (-\textit {\_R}^{2}+2 \textit {\_R} -1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{3} a -3 \textit {\_R}^{2}+3 \textit {\_R} a -1}}{-2+2 a}-\frac {2}{\left (-1+a \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(-1+a)*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))-2/(-1+a)/(tanh(1/2*x)-1)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(a-tanh(2*x)),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 a-1 negative or zero?

________________________________________________________________________________________

mupad [B]  time = 2.45, size = 163, normalized size = 1.52 \[ \frac {\ln \left (8\,a\,{\left (-a-1\right )}^{1/4}+8\,{\mathrm {e}}^x\,{\left (a-1\right )}^{5/4}-8\,{\left (-a-1\right )}^{1/4}\right )-\ln \left (8\,{\mathrm {e}}^x\,{\left (a-1\right )}^{5/4}-8\,a\,{\left (-a-1\right )}^{1/4}+8\,{\left (-a-1\right )}^{1/4}\right )+2\,{\mathrm {e}}^x\,{\left (a-1\right )}^{1/4}\,{\left (-a-1\right )}^{3/4}-\ln \left (8\,{\mathrm {e}}^x\,{\left (a-1\right )}^{5/4}-a\,{\left (-a-1\right )}^{1/4}\,8{}\mathrm {i}+{\left (-a-1\right )}^{1/4}\,8{}\mathrm {i}\right )\,1{}\mathrm {i}+\ln \left (a\,{\left (-a-1\right )}^{1/4}\,8{}\mathrm {i}+8\,{\mathrm {e}}^x\,{\left (a-1\right )}^{5/4}-{\left (-a-1\right )}^{1/4}\,8{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,{\left (a-1\right )}^{5/4}\,{\left (-a-1\right )}^{3/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(a - tanh(2*x)),x)

[Out]

(log(8*a*(- a - 1)^(1/4) + 8*exp(x)*(a - 1)^(5/4) - 8*(- a - 1)^(1/4)) - log(8*exp(x)*(a - 1)^(5/4) - 8*a*(- a
- 1)^(1/4) + 8*(- a - 1)^(1/4)) - log(8*exp(x)*(a - 1)^(5/4) - a*(- a - 1)^(1/4)*8i + (- a - 1)^(1/4)*8i)*1i
+ log(a*(- a - 1)^(1/4)*8i + 8*exp(x)*(a - 1)^(5/4) - (- a - 1)^(1/4)*8i)*1i + 2*exp(x)*(a - 1)^(1/4)*(- a - 1
)^(3/4))/(2*(a - 1)^(5/4)*(- a - 1)^(3/4))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{x}}{a - \tanh {\left (2 x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)/(a - tanh(2*x)), x)

________________________________________________________________________________________