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]

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)

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Rubi [A]  time = 0.61, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.13, size = 121, normalized size = 1.10 \[ \frac {2 e^x-(1-i)^{3/2} e^{x/2} \sqrt {e^x+1} \tanh ^{-1}\left (\frac {\sqrt {1-i} e^{x/2}}{\sqrt {e^x+1}}\right )-(1+i)^{3/2} e^{x/2} \sqrt {e^x+1} \tanh ^{-1}\left (\frac {\sqrt {1+i} e^{x/2}}{\sqrt {e^x+1}}\right )+2}{\sqrt {e^x \left (e^x+1\right )}} \]

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)]

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fricas [B]  time = 0.46, size = 859, normalized size = 7.81 \[ \text {result too large to display} \]

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)

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giac [B]  time = 0.59, size = 377, normalized size = 3.43 \[ -\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2 \, \sqrt {2} - 2} {\left (\frac {i}{\sqrt {2} - 1} + 1\right )} \log \left (\sqrt {2} {\left (\left (20 i + 40\right ) \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} - \left (20 i + 40\right ) \, e^{x}\right )} + 10 \, \sqrt {2} \sqrt {10 \, \sqrt {2} - 14} + \left (40 i - 20\right ) \, \sqrt {2} - \left (2 i + 14\right ) \, \sqrt {10 \, \sqrt {2} - 14} - \left (28 i + 56\right ) \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + \left (28 i + 56\right ) \, e^{x} - 56 i + 28\right ) + \left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2 \, \sqrt {2} - 2} {\left (\frac {i}{\sqrt {2} - 1} + 1\right )} \log \left (\sqrt {2} {\left (\left (20 i + 40\right ) \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} - \left (20 i + 40\right ) \, e^{x}\right )} - 10 \, \sqrt {2} \sqrt {10 \, \sqrt {2} - 14} + \left (40 i - 20\right ) \, \sqrt {2} + \left (2 i + 14\right ) \, \sqrt {10 \, \sqrt {2} - 14} - \left (28 i + 56\right ) \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + \left (28 i + 56\right ) \, e^{x} - 56 i + 28\right ) - \left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2 \, \sqrt {2} + 2} {\left (\frac {i}{\sqrt {2} + 1} + 1\right )} \log \left (4 \, \sqrt {2} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} + 2 \, \sqrt {2} \sqrt {2 \, \sqrt {2} - 2} - 4 i \, \sqrt {2} - \left (2 i + 2\right ) \, \sqrt {2 \, \sqrt {2} - 2} - 4 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + 4 \, e^{x} + 4 i\right ) + \left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2 \, \sqrt {2} + 2} {\left (\frac {i}{\sqrt {2} + 1} + 1\right )} \log \left (4 \, \sqrt {2} {\left (\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}\right )} - 2 \, \sqrt {2} \sqrt {2 \, \sqrt {2} - 2} - 4 i \, \sqrt {2} + \left (2 i + 2\right ) \, \sqrt {2 \, \sqrt {2} - 2} - 4 \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + 4 \, e^{x} + 4 i\right ) + \frac {2}{\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}} \]

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(sqrt(2)*((20*I + 40)*sqrt(e^(2*x) + e^x) - (20*I
+ 40)*e^x) + 10*sqrt(2)*sqrt(10*sqrt(2) - 14) + (40*I - 20)*sqrt(2) - (2*I + 14)*sqrt(10*sqrt(2) - 14) - (28*I
+ 56)*sqrt(e^(2*x) + e^x) + (28*I + 56)*e^x - 56*I + 28) + (1/4*I + 1/4)*sqrt(2*sqrt(2) - 2)*(I/(sqrt(2) - 1)
+ 1)*log(sqrt(2)*((20*I + 40)*sqrt(e^(2*x) + e^x) - (20*I + 40)*e^x) - 10*sqrt(2)*sqrt(10*sqrt(2) - 14) + (40
*I - 20)*sqrt(2) + (2*I + 14)*sqrt(10*sqrt(2) - 14) - (28*I + 56)*sqrt(e^(2*x) + e^x) + (28*I + 56)*e^x - 56*I
+ 28) - (1/4*I + 1/4)*sqrt(2*sqrt(2) + 2)*(I/(sqrt(2) + 1) + 1)*log(4*sqrt(2)*(sqrt(e^(2*x) + e^x) - e^x) + 2
*sqrt(2)*sqrt(2*sqrt(2) - 2) - 4*I*sqrt(2) - (2*I + 2)*sqrt(2*sqrt(2) - 2) - 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(4*sqrt(2)*(sqrt(e^(2*x) + e^x) - e^x) - 2*sq
rt(2)*sqrt(2*sqrt(2) - 2) - 4*I*sqrt(2) + (2*I + 2)*sqrt(2*sqrt(2) - 2) - 4*sqrt(e^(2*x) + e^x) + 4*e^x + 4*I)
+ 2/(sqrt(e^(2*x) + e^x) - e^x)

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maple [B]  time = 0.18, size = 366, normalized size = 3.33 \[ -\frac {\sqrt {2}\, \left (4 \sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \arctan \left (\frac {2 \sqrt {\tanh \left (\frac {x}{2}\right )+1}-\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )+4 \sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \arctan \left (\frac {2 \sqrt {\tanh \left (\frac {x}{2}\right )+1}+\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \sqrt {2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {-2+2 \sqrt {2}}\, \sqrt {2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \sqrt {2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {-2+2 \sqrt {2}}\, \sqrt {2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2+2 \sqrt {2}}+\sqrt {2}\right )+8 \sqrt {-2+2 \sqrt {2}}\right )}{4 \sqrt {-2+2 \sqrt {2}}\, \left (\tanh \left (\frac {x}{2}\right )-1\right ) \sqrt {\frac {\tanh \left (\frac {x}{2}\right )+1}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}}} \]

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^(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^(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*(tan
h(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))-(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)+8*(-2+2*2^(1/2))^(
1/2))/(-2+2*2^(1/2))^(1/2)/(tanh(1/2*x)-1)/((tanh(1/2*x)+1)/(tanh(1/2*x)-1)^2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh \relax (x)}{\sqrt {e^{\left (2 \, x\right )} + e^{x}}}\,{d x} \]

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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {tanh}\relax (x)}{\sqrt {{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tanh {\relax (x )}}{\sqrt {\left (e^{x} + 1\right ) e^{x}}}\, dx \]

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)

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