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]

-(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))

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Rubi [A]  time = 0.125877, antiderivative size = 107, normalized size of antiderivative = 1., 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 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]]]

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 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 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 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]

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 = 0.076932, 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))

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Maple [C]  time = 0.17, size = 70, normalized size = 0.7 \begin{align*}{\frac{{{\rm e}^{x}}}{-1+a}}+\sum _{{\it \_R}={\it RootOf} \left ( 1+ \left ( 16\,{a}^{8}-32\,{a}^{7}-32\,{a}^{6}+96\,{a}^{5}-96\,{a}^{3}+32\,{a}^{2}+32\,a-16 \right ){{\it \_Z}}^{4} \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}+ \left ( -2\,{a}^{2}+2 \right ){\it \_R} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/(-1+a)+sum(_R*ln(exp(x)+(-2*a^2+2)*_R),_R=RootOf(1+(16*a^8-32*a^7-32*a^6+96*a^5-96*a^3+32*a^2+32*a-16)*
_Z^4))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.25756, size = 1037, normalized size = 9.69 \begin{align*} -\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 )}} \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{e^{x}}{a - \tanh{\left (2 x \right )}}\, dx \end{align*}

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)

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Giac [B]  time = 1.29012, size = 443, normalized size = 4.14 \begin{align*} -\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} \end{align*}

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)