3.24 \(\int \frac{\tanh (x)}{\sqrt{e^x+e^{2 x}}} \, dx\)

Optimal. Leaf size=110 \[ 2 e^{-x} \sqrt{e^x+e^{2 x}}-\frac{\tan ^{-1}\left (\frac{i-(1-2 i) e^x}{2 \sqrt{1+i} \sqrt{e^x+e^{2 x}}}\right )}{\sqrt{1+i}}+\frac{\tan ^{-1}\left (\frac{(1+2 i) e^x+i}{2 \sqrt{1-i} \sqrt{e^x+e^{2 x}}}\right )}{\sqrt{1-i}} \]

[Out]

(2*Sqrt[E^x + E^(2*x)])/E^x - ArcTan[(I - (1 - 2*I)*E^x)/(2*Sqrt[1 + I]*Sqrt[E^x + E^(2*x)])]/Sqrt[1 + I] + Ar
cTan[(I + (1 + 2*I)*E^x)/(2*Sqrt[1 - I]*Sqrt[E^x + E^(2*x)])]/Sqrt[1 - I]

________________________________________________________________________________________

Rubi [A]  time = 0.606929, antiderivative size = 147, normalized size of antiderivative = 1.34, number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {2282, 6724, 1586, 6725, 94, 93, 208} \[ \frac{2 \left (e^x+1\right )}{\sqrt{e^x+e^{2 x}}}-\frac{(1-i)^{3/2} \sqrt{e^x} \sqrt{e^x+1} \tanh ^{-1}\left (\frac{\sqrt{1-i} \sqrt{e^x}}{\sqrt{e^x+1}}\right )}{\sqrt{e^x+e^{2 x}}}-\frac{(1+i)^{3/2} \sqrt{e^x} \sqrt{e^x+1} \tanh ^{-1}\left (\frac{\sqrt{1+i} \sqrt{e^x}}{\sqrt{e^x+1}}\right )}{\sqrt{e^x+e^{2 x}}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/Sqrt[E^x + E^(2*x)],x]

[Out]

(2*(1 + E^x))/Sqrt[E^x + E^(2*x)] - ((1 - I)^(3/2)*Sqrt[E^x]*Sqrt[1 + E^x]*ArcTanh[(Sqrt[1 - I]*Sqrt[E^x])/Sqr
t[1 + E^x]])/Sqrt[E^x + E^(2*x)] - ((1 + I)^(3/2)*Sqrt[E^x]*Sqrt[1 + E^x]*ArcTanh[(Sqrt[1 + I]*Sqrt[E^x])/Sqrt
[1 + E^x]])/Sqrt[E^x + E^(2*x)]

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 6724

Int[(u_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(m_), x_Symbol] :> With[{v = (a*x^r + b*x^s)^FracPart[m]/(x^(r
*FracPart[m])*(a + b*x^(s - r))^FracPart[m])}, Dist[v, Int[u*x^(m*r)*(a + b*x^(s - r))^m, x], x] /; NeQ[Simpli
fy[v], 1]] /; FreeQ[{a, b, m, r, s}, x] &&  !IntegerQ[m] && PosQ[s - r]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

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]

Rubi steps

\begin{align*} \int \frac{\tanh (x)}{\sqrt{e^x+e^{2 x}}} \, dx &=\operatorname{Subst}\left (\int \frac{-1+x^2}{x \left (1+x^2\right ) \sqrt{x+x^2}} \, dx,x,e^x\right )\\ &=\frac{\left (\sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{-1+x^2}{x^{3/2} \sqrt{1+x} \left (1+x^2\right )} \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}\\ &=\frac{\left (\sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{(-1+x) \sqrt{1+x}}{x^{3/2} \left (1+x^2\right )} \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}\\ &=\frac{\left (\sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \left (-\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{1+x}}{(i-x) x^{3/2}}+\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \sqrt{1+x}}{x^{3/2} (i+x)}\right ) \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}\\ &=-\frac{\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{(i-x) x^{3/2}} \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}+\frac{\left (\left (\frac{1}{2}-\frac{i}{2}\right ) \sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{x^{3/2} (i+x)} \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}\\ &=\frac{2 \left (1+e^x\right )}{\sqrt{e^x+e^{2 x}}}-\frac{\left (\sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{1+x}} \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}+\frac{\left (\sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{1+x}} \, dx,x,e^x\right )}{\sqrt{e^x+e^{2 x}}}\\ &=\frac{2 \left (1+e^x\right )}{\sqrt{e^x+e^{2 x}}}-\frac{\left (2 \sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(1+i) x^2} \, dx,x,\frac{\sqrt{e^x}}{\sqrt{1+e^x}}\right )}{\sqrt{e^x+e^{2 x}}}+\frac{\left (2 \sqrt{e^x} \sqrt{1+e^x}\right ) \operatorname{Subst}\left (\int \frac{1}{i+(1-i) x^2} \, dx,x,\frac{\sqrt{e^x}}{\sqrt{1+e^x}}\right )}{\sqrt{e^x+e^{2 x}}}\\ &=\frac{2 \left (1+e^x\right )}{\sqrt{e^x+e^{2 x}}}-\frac{(1-i)^{3/2} \sqrt{e^x} \sqrt{1+e^x} \tanh ^{-1}\left (\frac{\sqrt{1-i} \sqrt{e^x}}{\sqrt{1+e^x}}\right )}{\sqrt{e^x+e^{2 x}}}-\frac{(1+i)^{3/2} \sqrt{e^x} \sqrt{1+e^x} \tanh ^{-1}\left (\frac{\sqrt{1+i} \sqrt{e^x}}{\sqrt{1+e^x}}\right )}{\sqrt{e^x+e^{2 x}}}\\ \end{align*}

Mathematica [A]  time = 0.118181, size = 121, normalized size = 1.1 \[ \frac{2 e^x-(1-i)^{3/2} e^{x/2} \sqrt{e^x+1} \tanh ^{-1}\left (\frac{\sqrt{1-i} e^{x/2}}{\sqrt{e^x+1}}\right )-(1+i)^{3/2} e^{x/2} \sqrt{e^x+1} \tanh ^{-1}\left (\frac{\sqrt{1+i} e^{x/2}}{\sqrt{e^x+1}}\right )+2}{\sqrt{e^x \left (e^x+1\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/Sqrt[E^x + E^(2*x)],x]

[Out]

(2 + 2*E^x - (1 - I)^(3/2)*E^(x/2)*Sqrt[1 + E^x]*ArcTanh[(Sqrt[1 - I]*E^(x/2))/Sqrt[1 + E^x]] - (1 + I)^(3/2)*
E^(x/2)*Sqrt[1 + E^x]*ArcTanh[(Sqrt[1 + I]*E^(x/2))/Sqrt[1 + E^x]])/Sqrt[E^x*(1 + E^x)]

________________________________________________________________________________________

Maple [B]  time = 0.113, size = 366, normalized size = 3.3 \begin{align*} -{\frac{\sqrt{2}}{4\,\sqrt{-2+2\,\sqrt{2}}} \left ( \sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1-\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}+\sqrt{2} \right ) \sqrt{-2+2\,\sqrt{2}}\sqrt{2}-\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1+\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}+\sqrt{2} \right ) \sqrt{-2+2\,\sqrt{2}}\sqrt{2}-\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1-\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}+\sqrt{2} \right ) \sqrt{-2+2\,\sqrt{2}}+\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1+\sqrt{\tanh \left ({\frac{x}{2}} \right ) +1}\sqrt{2+2\,\sqrt{2}}+\sqrt{2} \right ) \sqrt{-2+2\,\sqrt{2}}+4\,\sqrt{\tanh \left ( x/2 \right ) +1}\arctan \left ({\frac{2\,\sqrt{\tanh \left ( x/2 \right ) +1}-\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) +4\,\sqrt{\tanh \left ( x/2 \right ) +1}\arctan \left ({\frac{2\,\sqrt{\tanh \left ( x/2 \right ) +1}+\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) +8\,\sqrt{-2+2\,\sqrt{2}} \right ){\frac{1}{\sqrt{{1 \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}}}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)/(exp(x)+exp(2*x))^(1/2),x)

[Out]

-1/4*2^(1/2)*((tanh(1/2*x)+1)^(1/2)*(2+2*2^(1/2))^(1/2)*ln(tanh(1/2*x)+1-(tanh(1/2*x)+1)^(1/2)*(2+2*2^(1/2))^(
1/2)+2^(1/2))*(-2+2*2^(1/2))^(1/2)*2^(1/2)-(tanh(1/2*x)+1)^(1/2)*(2+2*2^(1/2))^(1/2)*ln(tanh(1/2*x)+1+(tanh(1/
2*x)+1)^(1/2)*(2+2*2^(1/2))^(1/2)+2^(1/2))*(-2+2*2^(1/2))^(1/2)*2^(1/2)-(tanh(1/2*x)+1)^(1/2)*(2+2*2^(1/2))^(1
/2)*ln(tanh(1/2*x)+1-(tanh(1/2*x)+1)^(1/2)*(2+2*2^(1/2))^(1/2)+2^(1/2))*(-2+2*2^(1/2))^(1/2)+(tanh(1/2*x)+1)^(
1/2)*(2+2*2^(1/2))^(1/2)*ln(tanh(1/2*x)+1+(tanh(1/2*x)+1)^(1/2)*(2+2*2^(1/2))^(1/2)+2^(1/2))*(-2+2*2^(1/2))^(1
/2)+4*(tanh(1/2*x)+1)^(1/2)*arctan((2*(tanh(1/2*x)+1)^(1/2)-(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))+4*(tanh
(1/2*x)+1)^(1/2)*arctan((2*(tanh(1/2*x)+1)^(1/2)+(2+2*2^(1/2))^(1/2))/(-2+2*2^(1/2))^(1/2))+8*(-2+2*2^(1/2))^(
1/2))/((tanh(1/2*x)+1)/(tanh(1/2*x)-1)^2)^(1/2)/(tanh(1/2*x)-1)/(-2+2*2^(1/2))^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (x\right )}{\sqrt{e^{\left (2 \, x\right )} + e^{x}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(exp(x)+exp(2*x))^(1/2),x, algorithm="maxima")

[Out]

integrate(tanh(x)/sqrt(e^(2*x) + e^x), x)

________________________________________________________________________________________

Fricas [B]  time = 2.66269, size = 2684, normalized size = 24.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(exp(x)+exp(2*x))^(1/2),x, algorithm="fricas")

[Out]

-1/8*(8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*e^x*log(2*(8^(1/4)*(sqrt(2) - 1)*e^x - 8^(1/4))*sqrt(2*sqrt(2)
+ 4) - 2*(8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1) + 4*e^x)*sqrt(e^(2*x) + e^x) + 4*sqrt(2) + 8*e^(2*x) + 4*
e^x + 4) - 8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*e^x*log(-2*(8^(1/4)*(sqrt(2) - 1)*e^x - 8^(1/4))*sqrt(2*s
qrt(2) + 4) + 2*(8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1) - 4*e^x)*sqrt(e^(2*x) + e^x) + 4*sqrt(2) + 8*e^(2*x
) + 4*e^x + 4) + 4*8^(1/4)*sqrt(2)*sqrt(2*sqrt(2) + 4)*arctan(1/7*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)*e^
x + 1/112*(8*sqrt(2)*(5*sqrt(2) + 6) + (8^(3/4)*(5*sqrt(2) + 6) + 8*8^(1/4)*(2*sqrt(2) + 1))*sqrt(2*sqrt(2) +
4) + 64*sqrt(2) + 32)*sqrt(2*(8^(1/4)*(sqrt(2) - 1)*e^x - 8^(1/4))*sqrt(2*sqrt(2) + 4) - 2*(8^(1/4)*sqrt(2*sqr
t(2) + 4)*(sqrt(2) - 1) + 4*e^x)*sqrt(e^(2*x) + e^x) + 4*sqrt(2) + 8*e^(2*x) + 4*e^x + 4) + 1/56*((8^(3/4)*(5*
sqrt(2) + 6) + 8*8^(1/4)*(2*sqrt(2) + 1))*e^x + 8^(3/4)*(sqrt(2) + 4) - 8*8^(1/4)*(sqrt(2) - 3))*sqrt(2*sqrt(2
) + 4) + 1/7*sqrt(2)*(sqrt(2) + 4) - 1/56*(8*sqrt(2)*(5*sqrt(2) + 6) + (8^(3/4)*(5*sqrt(2) + 6) + 8*8^(1/4)*(2
*sqrt(2) + 1))*sqrt(2*sqrt(2) + 4) + 64*sqrt(2) + 32)*sqrt(e^(2*x) + e^x) + 3/7*sqrt(2) + 5/7)*e^x + 4*8^(1/4)
*sqrt(2)*sqrt(2*sqrt(2) + 4)*arctan(-1/7*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)*e^x - 1/112*(8*sqrt(2)*(5*s
qrt(2) + 6) - (8^(3/4)*(5*sqrt(2) + 6) + 8*8^(1/4)*(2*sqrt(2) + 1))*sqrt(2*sqrt(2) + 4) + 64*sqrt(2) + 32)*sqr
t(-2*(8^(1/4)*(sqrt(2) - 1)*e^x - 8^(1/4))*sqrt(2*sqrt(2) + 4) + 2*(8^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1)
- 4*e^x)*sqrt(e^(2*x) + e^x) + 4*sqrt(2) + 8*e^(2*x) + 4*e^x + 4) + 1/56*((8^(3/4)*(5*sqrt(2) + 6) + 8*8^(1/4)
*(2*sqrt(2) + 1))*e^x + 8^(3/4)*(sqrt(2) + 4) - 8*8^(1/4)*(sqrt(2) - 3))*sqrt(2*sqrt(2) + 4) - 1/7*sqrt(2)*(sq
rt(2) + 4) + 1/56*(8*sqrt(2)*(5*sqrt(2) + 6) - (8^(3/4)*(5*sqrt(2) + 6) + 8*8^(1/4)*(2*sqrt(2) + 1))*sqrt(2*sq
rt(2) + 4) + 64*sqrt(2) + 32)*sqrt(e^(2*x) + e^x) - 3/7*sqrt(2) - 5/7)*e^x - 16*sqrt(e^(2*x) + e^x) - 16*e^x)*
e^(-x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh{\left (x \right )}}{\sqrt{\left (e^{x} + 1\right ) e^{x}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(exp(x)+exp(2*x))**(1/2),x)

[Out]

Integral(tanh(x)/sqrt((exp(x) + 1)*exp(x)), x)

________________________________________________________________________________________

Giac [B]  time = 1.36504, size = 366, normalized size = 3.33 \begin{align*} -\left (\frac{1}{4} i + \frac{1}{4}\right ) \, \sqrt{2 \, \sqrt{2} - 2}{\left (\frac{i}{\sqrt{2} - 1} + 1\right )} \log \left (2 \, \sqrt{10 \, \sqrt{2} - 14}{\left (-\frac{i}{5 \, \sqrt{2} - 7} + 1\right )} + \left (4 i + 8\right ) \, \sqrt{e^{\left (2 \, x\right )} + e^{x}} - \left (4 i + 8\right ) \, e^{x} + 8 i - 4\right ) + \left (\frac{1}{4} i + \frac{1}{4}\right ) \, \sqrt{2 \, \sqrt{2} - 2}{\left (\frac{i}{\sqrt{2} - 1} + 1\right )} \log \left (-2 \, \sqrt{10 \, \sqrt{2} - 14}{\left (-\frac{i}{5 \, \sqrt{2} - 7} + 1\right )} + \left (4 i + 8\right ) \, \sqrt{e^{\left (2 \, x\right )} + e^{x}} - \left (4 i + 8\right ) \, e^{x} + 8 i - 4\right ) - \left (\frac{1}{4} i + \frac{1}{4}\right ) \, \sqrt{2 \, \sqrt{2} + 2}{\left (\frac{i}{\sqrt{2} + 1} + 1\right )} \log \left (2 \, \sqrt{2 \, \sqrt{2} - 2}{\left (-\frac{i}{\sqrt{2} - 1} + 1\right )} + 4 \, \sqrt{e^{\left (2 \, x\right )} + e^{x}} - 4 \, e^{x} - 4 i\right ) + \left (\frac{1}{4} i + \frac{1}{4}\right ) \, \sqrt{2 \, \sqrt{2} + 2}{\left (\frac{i}{\sqrt{2} + 1} + 1\right )} \log \left (-2 \, \sqrt{2 \, \sqrt{2} - 2}{\left (-\frac{i}{\sqrt{2} - 1} + 1\right )} + 4 \, \sqrt{e^{\left (2 \, x\right )} + e^{x}} - 4 \, e^{x} - 4 i\right ) + \frac{2}{\sqrt{e^{\left (2 \, x\right )} + e^{x}} - e^{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)/(exp(x)+exp(2*x))^(1/2),x, algorithm="giac")

[Out]

-(1/4*I + 1/4)*sqrt(2*sqrt(2) - 2)*(I/(sqrt(2) - 1) + 1)*log(2*sqrt(10*sqrt(2) - 14)*(-I/(5*sqrt(2) - 7) + 1)
+ (4*I + 8)*sqrt(e^(2*x) + e^x) - (4*I + 8)*e^x + 8*I - 4) + (1/4*I + 1/4)*sqrt(2*sqrt(2) - 2)*(I/(sqrt(2) - 1
) + 1)*log(-2*sqrt(10*sqrt(2) - 14)*(-I/(5*sqrt(2) - 7) + 1) + (4*I + 8)*sqrt(e^(2*x) + e^x) - (4*I + 8)*e^x +
8*I - 4) - (1/4*I + 1/4)*sqrt(2*sqrt(2) + 2)*(I/(sqrt(2) + 1) + 1)*log(2*sqrt(2*sqrt(2) - 2)*(-I/(sqrt(2) - 1
) + 1) + 4*sqrt(e^(2*x) + e^x) - 4*e^x - 4*I) + (1/4*I + 1/4)*sqrt(2*sqrt(2) + 2)*(I/(sqrt(2) + 1) + 1)*log(-2
*sqrt(2*sqrt(2) - 2)*(-I/(sqrt(2) - 1) + 1) + 4*sqrt(e^(2*x) + e^x) - 4*e^x - 4*I) + 2/(sqrt(e^(2*x) + e^x) -
e^x)