Integrand size = 8, antiderivative size = 49 \[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {1}{3 (1+\tanh (x))^{3/2}}-\frac {1}{2 \sqrt {1+\tanh (x)}} \] Output:
1/4*2^(1/2)*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))-1/3/(1+tanh(x))^(3/2)-1 /2/(1+tanh(x))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {1}{2} (1+\tanh (x))\right )}{3 (1+\tanh (x))^{3/2}} \] Input:
Integrate[(1 + Tanh[x])^(-3/2),x]
Output:
-1/3*Hypergeometric2F1[-3/2, 1, -1/2, (1 + Tanh[x])/2]/(1 + Tanh[x])^(3/2)
Time = 0.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {3042, 3960, 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.
\(\displaystyle \int \frac {1}{(\tanh (x)+1)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(1-i \tan (i x))^{3/2}}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt {\tanh (x)+1}}dx-\frac {1}{3 (\tanh (x)+1)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{3 (\tanh (x)+1)^{3/2}}+\frac {1}{2} \int \frac {1}{\sqrt {1-i \tan (i x)}}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \sqrt {\tanh (x)+1}dx-\frac {1}{\sqrt {\tanh (x)+1}}\right )-\frac {1}{3 (\tanh (x)+1)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{3 (\tanh (x)+1)^{3/2}}+\frac {1}{2} \left (-\frac {1}{\sqrt {\tanh (x)+1}}+\frac {1}{2} \int \sqrt {1-i \tan (i x)}dx\right )\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {1}{2} \left (\int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)+1}-\frac {1}{\sqrt {\tanh (x)+1}}\right )-\frac {1}{3 (\tanh (x)+1)^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {\tanh (x)+1}}\right )-\frac {1}{3 (\tanh (x)+1)^{3/2}}\) |
Input:
Int[(1 + Tanh[x])^(-3/2),x]
Output:
-1/3*1/(1 + Tanh[x])^(3/2) + (ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/Sqrt[2] - 1/Sqrt[1 + Tanh[x]])/2
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]
Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right )}{4}-\frac {1}{3 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}-\frac {1}{2 \sqrt {1+\tanh \left (x \right )}}\) | \(35\) |
default | \(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right )}{4}-\frac {1}{3 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}-\frac {1}{2 \sqrt {1+\tanh \left (x \right )}}\) | \(35\) |
Input:
int(1/(1+tanh(x))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/4*2^(1/2)*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))-1/3/(1+tanh(x))^(3/2)-1 /2/(1+tanh(x))^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (34) = 68\).
Time = 0.11 (sec) , antiderivative size = 244, normalized size of antiderivative = 4.98 \[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=\frac {3 \, {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3}\right )} \log \left (-2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - \frac {\sqrt {2} {\left (\sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt {2} \sinh \left (x\right )^{3} + {\left (3 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right ) + \sqrt {2} \cosh \left (x\right )\right )}}{\sqrt {\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}} - 1\right ) - \frac {2 \, \sqrt {2} {\left (4 \, \cosh \left (x\right )^{4} + 16 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + 4 \, \sinh \left (x\right )^{4} + {\left (24 \, \cosh \left (x\right )^{2} + 5\right )} \sinh \left (x\right )^{2} + 5 \, \cosh \left (x\right )^{2} + 2 \, {\left (8 \, \cosh \left (x\right )^{3} + 5 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}}{\sqrt {\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1}}}{24 \, {\left (\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3}\right )}} \] Input:
integrate(1/(1+tanh(x))^(3/2),x, algorithm="fricas")
Output:
1/24*(3*(sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)^2*sinh(x) + 3*sqrt(2)*cosh( x)*sinh(x)^2 + sqrt(2)*sinh(x)^3)*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 + sq rt(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*sqrt(2)*(4 *cosh(x)^4 + 16*cosh(x)*sinh(x)^3 + 4*sinh(x)^4 + (24*cosh(x)^2 + 5)*sinh( x)^2 + 5*cosh(x)^2 + 2*(8*cosh(x)^3 + 5*cosh(x))*sinh(x) + 1)/sqrt(cosh(x) ^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1))/(cosh(x)^3 + 3*cosh(x)^2*sinh(x) + 3*cosh(x)*sinh(x)^2 + sinh(x)^3)
\[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=\int \frac {1}{\left (\tanh {\left (x \right )} + 1\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(1+tanh(x))**(3/2),x)
Output:
Integral((tanh(x) + 1)**(-3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=-\frac {1}{12} \, \sqrt {2} {\left (\frac {3}{e^{\left (-2 \, x\right )} + 1} + 1\right )} {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {3}{2}} - \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) \] Input:
integrate(1/(1+tanh(x))^(3/2),x, algorithm="maxima")
Output:
-1/12*sqrt(2)*(3/(e^(-2*x) + 1) + 1)*(e^(-2*x) + 1)^(3/2) - 1/8*sqrt(2)*lo g(-(sqrt(2) - sqrt(2)/sqrt(e^(-2*x) + 1))/(sqrt(2) + sqrt(2)/sqrt(e^(-2*x) + 1)))
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (34) = 68\).
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.94 \[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=-\frac {1}{24} \, \sqrt {2} {\left (\frac {2 \, {\left (6 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 3 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 3 \, e^{\left (2 \, x\right )} + 1\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3}} + 3 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \] Input:
integrate(1/(1+tanh(x))^(3/2),x, algorithm="giac")
Output:
-1/24*sqrt(2)*(2*(6*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^2 - 3*sqrt(e^(4*x) + e^(2*x)) + 3*e^(2*x) + 1)/(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^3 + 3*log (-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))
Time = 2.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )}{4}-\frac {\frac {\mathrm {tanh}\left (x\right )}{2}+\frac {5}{6}}{{\left (\mathrm {tanh}\left (x\right )+1\right )}^{3/2}} \] Input:
int(1/(tanh(x) + 1)^(3/2),x)
Output:
(2^(1/2)*atanh((2^(1/2)*(tanh(x) + 1)^(1/2))/2))/4 - (tanh(x)/2 + 5/6)/(ta nh(x) + 1)^(3/2)
\[ \int \frac {1}{(1+\tanh (x))^{3/2}} \, dx=\frac {-2 \sqrt {\tanh \left (x \right )+1}+\left (\int \frac {\sqrt {\tanh \left (x \right )+1}\, \tanh \left (x \right )^{2}}{\tanh \left (x \right )^{2}+2 \tanh \left (x \right )+1}d x \right ) \tanh \left (x \right )+\int \frac {\sqrt {\tanh \left (x \right )+1}\, \tanh \left (x \right )^{2}}{\tanh \left (x \right )^{2}+2 \tanh \left (x \right )+1}d x}{\tanh \left (x \right )+1} \] Input:
int(1/(1+tanh(x))^(3/2),x)
Output:
( - 2*sqrt(tanh(x) + 1) + int((sqrt(tanh(x) + 1)*tanh(x)**2)/(tanh(x)**2 + 2*tanh(x) + 1),x)*tanh(x) + int((sqrt(tanh(x) + 1)*tanh(x)**2)/(tanh(x)** 2 + 2*tanh(x) + 1),x))/(tanh(x) + 1)