Integrand size = 13, antiderivative size = 42 \[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {1+\tanh (x)}}-2 \sqrt {1+\tanh (x)} \] Output:
1/2*2^(1/2)*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))-1/(1+tanh(x))^(1/2)-2*( 1+tanh(x))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.48 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.81 \[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {-\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+\tanh (x))\right )-2 (1+\tanh (x))}{\sqrt {1+\tanh (x)}} \] Input:
Integrate[Tanh[x]^2/Sqrt[1 + Tanh[x]],x]
Output:
(-Hypergeometric2F1[-1/2, 1, 1/2, (1 + Tanh[x])/2] - 2*(1 + Tanh[x]))/Sqrt [1 + Tanh[x]]
Time = 0.34 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3042, 25, 4026, 25, 3042, 3960, 3042, 3961, 219}
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.
\(\displaystyle \int \frac {\tanh ^2(x)}{\sqrt {\tanh (x)+1}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan (i x)^2}{\sqrt {1-i \tan (i x)}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan (i x)^2}{\sqrt {1-i \tan (i x)}}dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle -\int -\frac {1}{\sqrt {\tanh (x)+1}}dx-2 \sqrt {\tanh (x)+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {1}{\sqrt {\tanh (x)+1}}dx-2 \sqrt {\tanh (x)+1}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 \sqrt {\tanh (x)+1}+\int \frac {1}{\sqrt {1-i \tan (i x)}}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {1}{2} \int \sqrt {\tanh (x)+1}dx-2 \sqrt {\tanh (x)+1}-\frac {1}{\sqrt {\tanh (x)+1}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \sqrt {1-i \tan (i x)}dx-2 \sqrt {\tanh (x)+1}-\frac {1}{\sqrt {\tanh (x)+1}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)+1}-2 \sqrt {\tanh (x)+1}-\frac {1}{\sqrt {\tanh (x)+1}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-2 \sqrt {\tanh (x)+1}-\frac {1}{\sqrt {\tanh (x)+1}}\) |
Input:
Int[Tanh[x]^2/Sqrt[1 + Tanh[x]],x]
Output:
ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/Sqrt[2] - 1/Sqrt[1 + Tanh[x]] - 2*Sqrt[ 1 + Tanh[x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a) Int[(a + b*Tan[c + d*x])^ (n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83
method | result | size |
derivativedivides | \(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right )}{2}-\frac {1}{\sqrt {1+\tanh \left (x \right )}}-2 \sqrt {1+\tanh \left (x \right )}\) | \(35\) |
default | \(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right )}{2}-\frac {1}{\sqrt {1+\tanh \left (x \right )}}-2 \sqrt {1+\tanh \left (x \right )}\) | \(35\) |
Input:
int(tanh(x)^2/(1+tanh(x))^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*2^(1/2)*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))-1/(1+tanh(x))^(1/2)-2*( 1+tanh(x))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 161 vs. \(2 (34) = 68\).
Time = 0.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 3.83 \[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {{\left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right )\right )} \log \left (-2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - \frac {\sqrt {2} {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3} + {\left (3 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right ) + \sqrt {2} \cosh \left (x\right )\right )}}{\sqrt {\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}} - 1\right ) - \frac {2 \, \sqrt {2} {\left (5 \, \cosh \left (x\right )^{2} + 10 \, \cosh \left (x\right ) \sinh \left (x\right ) + 5 \, \sinh \left (x\right )^{2} + 1\right )}}{\sqrt {\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}}}{4 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \] Input:
integrate(tanh(x)^2/(1+tanh(x))^(1/2),x, algorithm="fricas")
Output:
1/4*((sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*log(-2*cosh(x)^2 - 4*cosh(x)*sinh (x) - 2*sinh(x)^2 - sqrt(2)*(sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)*sinh(x) ^2 + sqrt(2)*sinh(x)^3 + (3*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x) + sqrt(2) *cosh(x))/sqrt(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1) - 1) - 2*sqr t(2)*(5*cosh(x)^2 + 10*cosh(x)*sinh(x) + 5*sinh(x)^2 + 1)/sqrt(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1))/(cosh(x) + sinh(x))
Time = 1.38 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.38 \[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=- \frac {\sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right )}{4} - 2 \sqrt {\tanh {\left (x \right )} + 1} - \frac {1}{\sqrt {\tanh {\left (x \right )} + 1}} \] Input:
integrate(tanh(x)**2/(1+tanh(x))**(1/2),x)
Output:
-sqrt(2)*(log(sqrt(tanh(x) + 1) - sqrt(2)) - log(sqrt(tanh(x) + 1) + sqrt( 2)))/4 - 2*sqrt(tanh(x) + 1) - 1/sqrt(tanh(x) + 1)
\[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=\int { \frac {\tanh \left (x\right )^{2}}{\sqrt {\tanh \left (x\right ) + 1}} \,d x } \] Input:
integrate(tanh(x)^2/(1+tanh(x))^(1/2),x, algorithm="maxima")
Output:
integrate(tanh(x)^2/sqrt(tanh(x) + 1), x)
Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29 \[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=-\frac {1}{4} \, \sqrt {2} \log \left (-4 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 4 \, e^{\left (2 \, x\right )} + 2\right ) - \frac {5 \, \sqrt {2} e^{\left (2 \, x\right )} + \sqrt {2}}{2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}}} \] Input:
integrate(tanh(x)^2/(1+tanh(x))^(1/2),x, algorithm="giac")
Output:
-1/4*sqrt(2)*log(-4*sqrt(e^(4*x) + e^(2*x)) + 4*e^(2*x) + 2) - 1/2*(5*sqrt (2)*e^(2*x) + sqrt(2))/sqrt(e^(4*x) + e^(2*x))
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )}{2}-\frac {3}{\sqrt {\mathrm {tanh}\left (x\right )+1}}-\frac {2\,\mathrm {tanh}\left (x\right )}{\sqrt {\mathrm {tanh}\left (x\right )+1}} \] Input:
int(tanh(x)^2/(tanh(x) + 1)^(1/2),x)
Output:
(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/2 - 3/(tanh(x) + 1)^(1/2) - (2*tanh(x))/(tanh(x) + 1)^(1/2)
\[ \int \frac {\tanh ^2(x)}{\sqrt {1+\tanh (x)}} \, dx=\int \frac {\sqrt {\tanh \left (x \right )+1}\, \tanh \left (x \right )^{2}}{\tanh \left (x \right )+1}d x \] Input:
int(tanh(x)^2/(1+tanh(x))^(1/2),x)
Output:
int((sqrt(tanh(x) + 1)*tanh(x)**2)/(tanh(x) + 1),x)