\(\int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx\) [24]

Optimal result

Integrand size = 16, antiderivative size = 110 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=2 e^{-x} \sqrt {e^x+e^{2 x}}-\frac {\arctan \left (\frac {i-(1-2 i) e^x}{2 \sqrt {1+i} \sqrt {e^x+e^{2 x}}}\right )}{\sqrt {1+i}}+\frac {\arctan \left (\frac {i+(1+2 i) e^x}{2 \sqrt {1-i} \sqrt {e^x+e^{2 x}}}\right )}{\sqrt {1-i}} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.10 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\frac {2+2 e^x-(1-i)^{3/2} e^{x/2} \sqrt {1+e^x} \text {arctanh}\left (\frac {\sqrt {1-i} e^{x/2}}{\sqrt {1+e^x}}\right )-(1+i)^{3/2} e^{x/2} \sqrt {1+e^x} \text {arctanh}\left (\frac {\sqrt {1+i} e^{x/2}}{\sqrt {1+e^x}}\right )}{\sqrt {e^x \left (1+e^x\right )}} \] Input:

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

Output:

(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)]
 

Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2720, 25, 2467, 2003, 2035, 2247, 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.

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

\(\Big \downarrow \) 2720

\(\displaystyle \int -\frac {e^{-x} \left (1-e^{2 x}\right )}{\left (e^{2 x}+1\right ) \sqrt {e^x+e^{2 x}}}de^x\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \frac {e^{-x} \left (1-e^{2 x}\right )}{\left (1+e^{2 x}\right ) \sqrt {e^x+e^{2 x}}}de^x\)

\(\Big \downarrow \) 2467

\(\displaystyle -\frac {\sqrt {e^x} \sqrt {e^x+1} \int \frac {1-e^{2 x}}{\left (e^x\right )^{3/2} \sqrt {1+e^x} \left (1+e^{2 x}\right )}de^x}{\sqrt {e^x+e^{2 x}}}\)

\(\Big \downarrow \) 2003

\(\displaystyle -\frac {\sqrt {e^x} \sqrt {e^x+1} \int \frac {\left (1-e^x\right ) \sqrt {1+e^x}}{\left (e^x\right )^{3/2} \left (1+e^{2 x}\right )}de^x}{\sqrt {e^x+e^{2 x}}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {2 \sqrt {e^x} \sqrt {e^x+1} \int \frac {e^{-2 x} \left (1-e^{2 x}\right ) \sqrt {1+e^{2 x}}}{1+e^{4 x}}d\sqrt {e^x}}{\sqrt {e^x+e^{2 x}}}\)

\(\Big \downarrow \) 2247

\(\displaystyle -\frac {2 \sqrt {e^x} \sqrt {e^x+1} \int \left (\frac {\sqrt {1+e^{2 x}} \left (-1-e^{2 x}\right )}{1+e^{4 x}}+e^{-2 x} \sqrt {1+e^{2 x}}\right )d\sqrt {e^x}}{\sqrt {e^x+e^{2 x}}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \sqrt {e^x} \sqrt {e^x+1} \left (\frac {1}{2} (1-i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1-i} \sqrt {e^x}}{\sqrt {e^{2 x}+1}}\right )+\frac {1}{2} (1+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+i} \sqrt {e^x}}{\sqrt {e^{2 x}+1}}\right )-e^{-x} \sqrt {e^{2 x}+1}\right )}{\sqrt {e^x+e^{2 x}}}\)

Input:

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

Output:

(-2*Sqrt[E^x]*Sqrt[1 + E^x]*(-(Sqrt[1 + E^(2*x)]/E^x) + ((1 - I)^(3/2)*Arc 
Tanh[(Sqrt[1 - I]*Sqrt[E^x])/Sqrt[1 + E^(2*x)]])/2 + ((1 + I)^(3/2)*ArcTan 
h[(Sqrt[1 + I]*Sqrt[E^x])/Sqrt[1 + E^(2*x)]])/2))/Sqrt[E^x + E^(2*x)]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2003

Int[(u_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] : 
> Int[u*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p} 
, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && 
  !IntegerQ[n]))
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2247

Int[(Px_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_) 
^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Px*(f*x)^m*(d + e*x^2)^q*(a + c 
*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && PolyQ[Px, x] && Integ 
erQ[p]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 2720

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]]]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(81)=162\).

Time = 0.41 (sec) , antiderivative size = 366, normalized size of antiderivative = 3.33

method	result	size
default	\(-\frac {\sqrt {2}\, \left (\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \sqrt {2 \sqrt {2}-2}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}\, \sqrt {2}\, \sqrt {2 \sqrt {2}-2}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}\, \sqrt {2 \sqrt {2}-2}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}\, \sqrt {2 \sqrt {2}-2}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right )+4 \sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \arctan \left (\frac {2 \sqrt {\tanh \left (\frac {x}{2}\right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {2 \sqrt {2}-2}}\right )+4 \sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \arctan \left (\frac {2 \sqrt {\tanh \left (\frac {x}{2}\right )+1}+\sqrt {2 \sqrt {2}+2}}{\sqrt {2 \sqrt {2}-2}}\right )+8 \sqrt {2 \sqrt {2}-2}\right )}{4 \sqrt {2 \sqrt {2}-2}\, \left (\tanh \left (\frac {x}{2}\right )-1\right ) \sqrt {\frac {\tanh \left (\frac {x}{2}\right )+1}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}}}\)	\(366\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (67) = 134\).

Time = 0.09 (sec) , antiderivative size = 501, normalized size of antiderivative = 4.55 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*(2*sqrt(sqrt(2) + 1)*(cosh(x) + sinh(x))*arctan((2*(sqrt(2) + 1)*sqrt( 
sqrt(2) - 1) + sqrt(2) + 2)*sqrt(sqrt(2) + 1)*sqrt((cosh(x) + sinh(x) + 1) 
/(cosh(x) - sinh(x))) - ((sqrt(2) + 2)*cosh(x) + (sqrt(2) + 2)*sinh(x) + ( 
2*(sqrt(2) + 1)*cosh(x) + 2*(sqrt(2) + 1)*sinh(x) + sqrt(2) + 1)*sqrt(sqrt 
(2) - 1) + sqrt(2))*sqrt(sqrt(2) + 1)) - 2*sqrt(sqrt(2) + 1)*(cosh(x) + si 
nh(x))*arctan((2*(sqrt(2) + 1)*sqrt(sqrt(2) - 1) - sqrt(2) - 2)*sqrt(sqrt( 
2) + 1)*sqrt((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) + ((sqrt(2) + 2) 
*cosh(x) + (sqrt(2) + 2)*sinh(x) - (2*(sqrt(2) + 1)*cosh(x) + 2*(sqrt(2) + 
 1)*sinh(x) + sqrt(2) + 1)*sqrt(sqrt(2) - 1) + sqrt(2))*sqrt(sqrt(2) + 1)) 
 + sqrt(sqrt(2) - 1)*(cosh(x) + sinh(x))*log(2*cosh(x)^2 + (4*cosh(x) + 1) 
*sinh(x) + 2*sinh(x)^2 + (sqrt(2)*cosh(x) + sqrt(2)*sinh(x) - sqrt(2) - 2) 
*sqrt(sqrt(2) - 1) - (sqrt(2)*sqrt(sqrt(2) - 1) + 2*cosh(x) + 2*sinh(x))*s 
qrt((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) + sqrt(2) + cosh(x) + 1) 
- sqrt(sqrt(2) - 1)*(cosh(x) + sinh(x))*log(2*cosh(x)^2 + (4*cosh(x) + 1)* 
sinh(x) + 2*sinh(x)^2 - (sqrt(2)*cosh(x) + sqrt(2)*sinh(x) - sqrt(2) - 2)* 
sqrt(sqrt(2) - 1) + (sqrt(2)*sqrt(sqrt(2) - 1) - 2*cosh(x) - 2*sinh(x))*sq 
rt((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) + sqrt(2) + cosh(x) + 1) + 
 4*sqrt((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) + 4*cosh(x) + 4*sinh( 
x))/(cosh(x) + sinh(x))
 

Sympy [F]

\[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int \frac {\tanh {\left (x \right )}}{\sqrt {\left (e^{x} + 1\right ) e^{x}}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int { \frac {\tanh \left (x\right )}{\sqrt {e^{\left (2 \, x\right )} + e^{x}}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (67) = 134\).

Time = 0.24 (sec) , antiderivative size = 470, normalized size of antiderivative = 4.27 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=-\frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left ({\left (65 \, \sqrt {2} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} + 13 \, \sqrt {2} \sqrt {13 \, \sqrt {2} - 17} - 13 \, \sqrt {2} - 17 \, \sqrt {13 \, \sqrt {2} - 17} - 85 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + 85 \, e^{x} + 17\right )}^{2} + {\left (13 \, \sqrt {2} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} + 65 \, \sqrt {2} + 7 \, \sqrt {13 \, \sqrt {2} - 17} - 17 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + 17 \, e^{x} - 85\right )}^{2}\right ) + \frac {1}{2} \, \sqrt {\sqrt {2} - 1} \log \left ({\left (65 \, \sqrt {2} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} - 13 \, \sqrt {2} \sqrt {13 \, \sqrt {2} - 17} - 13 \, \sqrt {2} + 17 \, \sqrt {13 \, \sqrt {2} - 17} - 85 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + 85 \, e^{x} + 17\right )}^{2} + {\left (13 \, \sqrt {2} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} + 65 \, \sqrt {2} - 7 \, \sqrt {13 \, \sqrt {2} - 17} - 17 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + 17 \, e^{x} - 85\right )}^{2}\right ) + \frac {\arctan \left (\frac {1}{5}\right ) + \arctan \left (\frac {1}{7} \, {\left (11 \, \sqrt {2} \sqrt {13 \, \sqrt {2} - 17} + 14 \, \sqrt {2} + 16 \, \sqrt {13 \, \sqrt {2} - 17} + 14\right )} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} - \frac {5}{7} \, \sqrt {2} \sqrt {13 \, \sqrt {2} - 17} - \sqrt {2} - \frac {6}{7} \, \sqrt {13 \, \sqrt {2} - 17} - 1\right )}{\sqrt {\sqrt {2} - 1}} - \frac {\arctan \left (\frac {1}{5}\right ) + \arctan \left (-\frac {1}{7} \, {\left (11 \, \sqrt {2} \sqrt {13 \, \sqrt {2} - 17} - 14 \, \sqrt {2} + 16 \, \sqrt {13 \, \sqrt {2} - 17} - 14\right )} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} + \frac {5}{7} \, \sqrt {2} \sqrt {13 \, \sqrt {2} - 17} - \sqrt {2} + \frac {6}{7} \, \sqrt {13 \, \sqrt {2} - 17} - 1\right )}{\sqrt {\sqrt {2} - 1}} + \frac {2}{\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}} \] Input:

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

Output:

-1/2*sqrt(sqrt(2) - 1)*log((65*sqrt(2)*(sqrt(e^(2*x) + e^x) - e^x) + 13*sq 
rt(2)*sqrt(13*sqrt(2) - 17) - 13*sqrt(2) - 17*sqrt(13*sqrt(2) - 17) - 85*s 
qrt(e^(2*x) + e^x) + 85*e^x + 17)^2 + (13*sqrt(2)*(sqrt(e^(2*x) + e^x) - e 
^x) + 65*sqrt(2) + 7*sqrt(13*sqrt(2) - 17) - 17*sqrt(e^(2*x) + e^x) + 17*e 
^x - 85)^2) + 1/2*sqrt(sqrt(2) - 1)*log((65*sqrt(2)*(sqrt(e^(2*x) + e^x) - 
 e^x) - 13*sqrt(2)*sqrt(13*sqrt(2) - 17) - 13*sqrt(2) + 17*sqrt(13*sqrt(2) 
 - 17) - 85*sqrt(e^(2*x) + e^x) + 85*e^x + 17)^2 + (13*sqrt(2)*(sqrt(e^(2* 
x) + e^x) - e^x) + 65*sqrt(2) - 7*sqrt(13*sqrt(2) - 17) - 17*sqrt(e^(2*x) 
+ e^x) + 17*e^x - 85)^2) + (arctan(1/5) + arctan(1/7*(11*sqrt(2)*sqrt(13*s 
qrt(2) - 17) + 14*sqrt(2) + 16*sqrt(13*sqrt(2) - 17) + 14)*(sqrt(e^(2*x) + 
 e^x) - e^x) - 5/7*sqrt(2)*sqrt(13*sqrt(2) - 17) - sqrt(2) - 6/7*sqrt(13*s 
qrt(2) - 17) - 1))/sqrt(sqrt(2) - 1) - (arctan(1/5) + arctan(-1/7*(11*sqrt 
(2)*sqrt(13*sqrt(2) - 17) - 14*sqrt(2) + 16*sqrt(13*sqrt(2) - 17) - 14)*(s 
qrt(e^(2*x) + e^x) - e^x) + 5/7*sqrt(2)*sqrt(13*sqrt(2) - 17) - sqrt(2) + 
6/7*sqrt(13*sqrt(2) - 17) - 1))/sqrt(sqrt(2) - 1) + 2/(sqrt(e^(2*x) + e^x) 
 - e^x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int \frac {\mathrm {tanh}\left (x\right )}{\sqrt {{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int \frac {\sqrt {e^{x}+1}\, \tanh \left (x \right )}{e^{\frac {3 x}{2}}+e^{\frac {x}{2}}}d x \] Input:

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

Output:

int((sqrt(e**x + 1)*tanh(x))/(e**((3*x)/2) + e**(x/2)),x)