Integrand size = 16, antiderivative size = 110 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=2 e^{-x} \sqrt {e^x+e^{2 x}}-\frac {\arctan \left (\frac {i-(1-2 i) e^x}{2 \sqrt {1+i} \sqrt {e^x+e^{2 x}}}\right )}{\sqrt {1+i}}+\frac {\arctan \left (\frac {i+(1+2 i) e^x}{2 \sqrt {1-i} \sqrt {e^x+e^{2 x}}}\right )}{\sqrt {1-i}} \]
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)
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.10 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\frac {2+2 e^x-(1-i)^{3/2} e^{x/2} \sqrt {1+e^x} \text {arctanh}\left (\frac {\sqrt {1-i} e^{x/2}}{\sqrt {1+e^x}}\right )-(1+i)^{3/2} e^{x/2} \sqrt {1+e^x} \text {arctanh}\left (\frac {\sqrt {1+i} e^{x/2}}{\sqrt {1+e^x}}\right )}{\sqrt {e^x \left (1+e^x\right )}} \]
(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)]
Time = 0.52 (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 |
\(\displaystyle -\frac {2 \sqrt {e^x} \sqrt {e^x+1} \left (\frac {1}{2} (1-i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1-i} \sqrt {e^x}}{\sqrt {e^{2 x}+1}}\right )+\frac {1}{2} (1+i)^{3/2} \text {arctanh}\left (\frac {\sqrt {1+i} \sqrt {e^x}}{\sqrt {e^{2 x}+1}}\right )-e^{-x} \sqrt {e^{2 x}+1}\right )}{\sqrt {e^x+e^{2 x}}}\) |
(-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)]
3.1.24.3.1 Defintions of rubi rules used
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]))
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]
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]
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]
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]]]
Leaf count of result is larger than twice the leaf count of optimal. \(365\) vs. \(2(81)=162\).
Time = 0.24 (sec) , antiderivative size = 366, normalized size of antiderivative = 3.33
method | result | size |
default | \(-\frac {\sqrt {2}\, \left (\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right ) \sqrt {2 \sqrt {2}+2}-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right ) \sqrt {2 \sqrt {2}+2}+4 \sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \arctan \left (\frac {2 \sqrt {\tanh \left (\frac {x}{2}\right )+1}-\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )+4 \sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \arctan \left (\frac {2 \sqrt {\tanh \left (\frac {x}{2}\right )+1}+\sqrt {2 \sqrt {2}+2}}{\sqrt {-2+2 \sqrt {2}}}\right )-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1-\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right ) \sqrt {2 \sqrt {2}+2}+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\tanh \left (\frac {x}{2}\right )+1+\sqrt {\tanh \left (\frac {x}{2}\right )+1}\, \sqrt {2 \sqrt {2}+2}+\sqrt {2}\right ) \sqrt {2 \sqrt {2}+2}+8 \sqrt {-2+2 \sqrt {2}}\right )}{4 \sqrt {-2+2 \sqrt {2}}\, \left (\tanh \left (\frac {x}{2}\right )-1\right ) \sqrt {\frac {\tanh \left (\frac {x}{2}\right )+1}{\left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}}}\) | \(366\) |
-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^(1/2)+2)^(1/2)+2^(1/2))*(2*2^(1/2)+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^(1/2)+2)^(1/2)+2^(1/2))*(2*2^(1/2)+2)^(1/2)+4*(t anh(1/2*x)+1)^(1/2)*arctan((2*(tanh(1/2*x)+1)^(1/2)-(2*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^(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^(1/2)+2)^(1/2)+ 2^(1/2))*(2*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^(1/2)+2)^(1/2)+2^(1/2))*(2*2^(1/ 2)+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)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (67) = 134\).
Time = 0.26 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.00 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\frac {\sqrt {2 i - 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \log \left (\left (i + 1\right ) \, \sqrt {2 i - 2} + 2 \, \sqrt {\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 2 \, \cosh \left (x\right ) - 2 \, \sinh \left (x\right ) - 2 i\right ) - \sqrt {2 i - 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \log \left (-\left (i + 1\right ) \, \sqrt {2 i - 2} + 2 \, \sqrt {\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 2 \, \cosh \left (x\right ) - 2 \, \sinh \left (x\right ) - 2 i\right ) + \sqrt {-2 i - 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \log \left (-\left (i - 1\right ) \, \sqrt {-2 i - 2} + 2 \, \sqrt {\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 2 \, \cosh \left (x\right ) - 2 \, \sinh \left (x\right ) + 2 i\right ) - \sqrt {-2 i - 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \log \left (\left (i - 1\right ) \, \sqrt {-2 i - 2} + 2 \, \sqrt {\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 2 \, \cosh \left (x\right ) - 2 \, \sinh \left (x\right ) + 2 i\right ) + 4 \, \sqrt {\frac {\cosh \left (x\right ) + \sinh \left (x\right ) + 1}{\cosh \left (x\right ) - \sinh \left (x\right )}} + 4 \, \cosh \left (x\right ) + 4 \, \sinh \left (x\right )}{2 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]
1/2*(sqrt(2*I - 2)*(cosh(x) + sinh(x))*log((I + 1)*sqrt(2*I - 2) + 2*sqrt( (cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) - 2*cosh(x) - 2*sinh(x) - 2*I ) - sqrt(2*I - 2)*(cosh(x) + sinh(x))*log(-(I + 1)*sqrt(2*I - 2) + 2*sqrt( (cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) - 2*cosh(x) - 2*sinh(x) - 2*I ) + sqrt(-2*I - 2)*(cosh(x) + sinh(x))*log(-(I - 1)*sqrt(-2*I - 2) + 2*sqr t((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) - 2*cosh(x) - 2*sinh(x) + 2 *I) - sqrt(-2*I - 2)*(cosh(x) + sinh(x))*log((I - 1)*sqrt(-2*I - 2) + 2*sq rt((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) - 2*cosh(x) - 2*sinh(x) + 2*I) + 4*sqrt((cosh(x) + sinh(x) + 1)/(cosh(x) - sinh(x))) + 4*cosh(x) + 4 *sinh(x))/(cosh(x) + sinh(x))
\[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int \frac {\tanh {\left (x \right )}}{\sqrt {\left (e^{x} + 1\right ) e^{x}}}\, dx \]
\[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int { \frac {\tanh \left (x\right )}{\sqrt {e^{\left (2 \, x\right )} + e^{x}}} \,d x } \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 379 vs. \(2 (67) = 134\).
Time = 0.40 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.45 \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=-\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 ) - \left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2 \, \sqrt {2} - 2} {\left (\frac {i}{\sqrt {2} - 1} + 1\right )} \log \left (-20 \, \sqrt {2} {\left (-\left (i + 2\right ) \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + \left (i + 2\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 (-20 \, \sqrt {2} {\left (-\left (i + 2\right ) \, \sqrt {e^{\left (2 \, x\right )} + e^{x}} + \left (i + 2\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 ) + \frac {2}{\sqrt {e^{\left (2 \, x\right )} + e^{x}} - e^{x}} \]
-(1/4*I + 1/4)*sqrt(2*sqrt(2) + 2)*(I/(sqrt(2) + 1) + 1)*log(4*sqrt(2)*(sq rt(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*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(-20*sqrt(2)*(-(I + 2)*s qrt(e^(2*x) + e^x) + (I + 2)*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(-20*sqrt(2)*(-(I + 2)*sqrt(e^(2*x) + e^x) + (I + 2)*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) + 2/(sqrt(e^(2*x) + e^x) - e^x)
Timed out. \[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int \frac {\mathrm {tanh}\left (x\right )}{\sqrt {{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x}} \,d x \]
\[ \int \frac {\tanh (x)}{\sqrt {e^x+e^{2 x}}} \, dx=\int \frac {\sqrt {e^{x}+1}\, \tanh \left (x \right )}{e^{\frac {3 x}{2}}+e^{\frac {x}{2}}}d x \]