Integrand size = 8, antiderivative size = 61 \[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )}{4 \sqrt {2}}-\frac {1}{5 (1+\tanh (x))^{5/2}}-\frac {1}{6 (1+\tanh (x))^{3/2}}-\frac {1}{4 \sqrt {1+\tanh (x)}} \]
1/8*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-1/4/(1+tanh(x))^(1/2)-1 /5/(1+tanh(x))^(5/2)-1/6/(1+tanh(x))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {1}{2} (1+\tanh (x))\right )}{5 (1+\tanh (x))^{5/2}} \]
Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {3042, 3960, 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.
\(\displaystyle \int \frac {1}{(\tanh (x)+1)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(1-i \tan (i x))^{5/2}}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {1}{2} \int \frac {1}{(\tanh (x)+1)^{3/2}}dx-\frac {1}{5 (\tanh (x)+1)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{5 (\tanh (x)+1)^{5/2}}+\frac {1}{2} \int \frac {1}{(1-i \tan (i x))^{3/2}}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\sqrt {\tanh (x)+1}}dx-\frac {1}{3 (\tanh (x)+1)^{3/2}}\right )-\frac {1}{5 (\tanh (x)+1)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{5 (\tanh (x)+1)^{5/2}}+\frac {1}{2} \left (-\frac {1}{3 (\tanh (x)+1)^{3/2}}+\frac {1}{2} \int \frac {1}{\sqrt {1-i \tan (i x)}}dx\right )\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \sqrt {\tanh (x)+1}dx-\frac {1}{\sqrt {\tanh (x)+1}}\right )-\frac {1}{3 (\tanh (x)+1)^{3/2}}\right )-\frac {1}{5 (\tanh (x)+1)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{5 (\tanh (x)+1)^{5/2}}+\frac {1}{2} \left (-\frac {1}{3 (\tanh (x)+1)^{3/2}}+\frac {1}{2} \left (-\frac {1}{\sqrt {\tanh (x)+1}}+\frac {1}{2} \int \sqrt {1-i \tan (i x)}dx\right )\right )\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)+1}-\frac {1}{\sqrt {\tanh (x)+1}}\right )-\frac {1}{3 (\tanh (x)+1)^{3/2}}\right )-\frac {1}{5 (\tanh (x)+1)^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {\text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {\tanh (x)+1}}\right )-\frac {1}{3 (\tanh (x)+1)^{3/2}}\right )-\frac {1}{5 (\tanh (x)+1)^{5/2}}\) |
-1/5*1/(1 + Tanh[x])^(5/2) + (-1/3*1/(1 + Tanh[x])^(3/2) + (ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]]/Sqrt[2] - 1/Sqrt[1 + Tanh[x]])/2)/2
3.1.56.3.1 Defintions of rubi rules used
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 = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.70
method | result | size |
derivativedivides | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{8}-\frac {1}{4 \sqrt {1+\tanh \left (x \right )}}-\frac {1}{5 \left (1+\tanh \left (x \right )\right )^{\frac {5}{2}}}-\frac {1}{6 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}\) | \(43\) |
default | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{8}-\frac {1}{4 \sqrt {1+\tanh \left (x \right )}}-\frac {1}{5 \left (1+\tanh \left (x \right )\right )^{\frac {5}{2}}}-\frac {1}{6 \left (1+\tanh \left (x \right )\right )^{\frac {3}{2}}}\) | \(43\) |
1/8*arctanh(1/2*(1+tanh(x))^(1/2)*2^(1/2))*2^(1/2)-1/4/(1+tanh(x))^(1/2)-1 /5/(1+tanh(x))^(5/2)-1/6/(1+tanh(x))^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (42) = 84\).
Time = 0.26 (sec) , antiderivative size = 266, normalized size of antiderivative = 4.36 \[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=-\frac {2 \, \sqrt {2} {\left (23 \, \sqrt {2} \cosh \left (x\right )^{4} + 92 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + 23 \, \sqrt {2} \sinh \left (x\right )^{4} + {\left (138 \, \sqrt {2} \cosh \left (x\right )^{2} + 11 \, \sqrt {2}\right )} \sinh \left (x\right )^{2} + 11 \, \sqrt {2} \cosh \left (x\right )^{2} + 2 \, {\left (46 \, \sqrt {2} \cosh \left (x\right )^{3} + 11 \, \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 3 \, \sqrt {2}\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 15 \, {\left (\sqrt {2} \cosh \left (x\right )^{5} + 5 \, \sqrt {2} \cosh \left (x\right )^{4} \sinh \left (x\right ) + 10 \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right )^{2} + 10 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{3} + 5 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{4} + \sqrt {2} \sinh \left (x\right )^{5}\right )} \log \left (-2 \, \sqrt {2} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} - 2 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right ) \sinh \left (x\right ) - 2 \, \sinh \left (x\right )^{2} - 1\right )}{240 \, {\left (\cosh \left (x\right )^{5} + 5 \, \cosh \left (x\right )^{4} \sinh \left (x\right ) + 10 \, \cosh \left (x\right )^{3} \sinh \left (x\right )^{2} + 10 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{3} + 5 \, \cosh \left (x\right ) \sinh \left (x\right )^{4} + \sinh \left (x\right )^{5}\right )}} \]
-1/240*(2*sqrt(2)*(23*sqrt(2)*cosh(x)^4 + 92*sqrt(2)*cosh(x)*sinh(x)^3 + 2 3*sqrt(2)*sinh(x)^4 + (138*sqrt(2)*cosh(x)^2 + 11*sqrt(2))*sinh(x)^2 + 11* sqrt(2)*cosh(x)^2 + 2*(46*sqrt(2)*cosh(x)^3 + 11*sqrt(2)*cosh(x))*sinh(x) + 3*sqrt(2))*sqrt(cosh(x)/(cosh(x) - sinh(x))) - 15*(sqrt(2)*cosh(x)^5 + 5 *sqrt(2)*cosh(x)^4*sinh(x) + 10*sqrt(2)*cosh(x)^3*sinh(x)^2 + 10*sqrt(2)*c osh(x)^2*sinh(x)^3 + 5*sqrt(2)*cosh(x)*sinh(x)^4 + sqrt(2)*sinh(x)^5)*log( -2*sqrt(2)*sqrt(cosh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x)) - 2*cosh( x)^2 - 4*cosh(x)*sinh(x) - 2*sinh(x)^2 - 1))/(cosh(x)^5 + 5*cosh(x)^4*sinh (x) + 10*cosh(x)^3*sinh(x)^2 + 10*cosh(x)^2*sinh(x)^3 + 5*cosh(x)*sinh(x)^ 4 + sinh(x)^5)
\[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=\int \frac {1}{\left (\tanh {\left (x \right )} + 1\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=-\frac {1}{120} \, \sqrt {2} {\left (\frac {5}{e^{\left (-2 \, x\right )} + 1} + \frac {15}{{\left (e^{\left (-2 \, x\right )} + 1\right )}^{2}} + 3\right )} {\left (e^{\left (-2 \, x\right )} + 1\right )}^{\frac {5}{2}} - \frac {1}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}{\sqrt {2} + \frac {\sqrt {2}}{\sqrt {e^{\left (-2 \, x\right )} + 1}}}\right ) \]
-1/120*sqrt(2)*(5/(e^(-2*x) + 1) + 15/(e^(-2*x) + 1)^2 + 3)*(e^(-2*x) + 1) ^(5/2) - 1/16*sqrt(2)*log(-(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. 139 vs. \(2 (42) = 84\).
Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.28 \[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=-\frac {1}{240} \, \sqrt {2} {\left (\frac {2 \, {\left (45 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} - 45 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 35 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 15 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 15 \, e^{\left (2 \, x\right )} + 3\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{5}} + 15 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]
-1/240*sqrt(2)*(2*(45*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^4 - 45*(sqrt(e^( 4*x) + e^(2*x)) - e^(2*x))^3 + 35*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^2 - 15*sqrt(e^(4*x) + e^(2*x)) + 15*e^(2*x) + 3)/(sqrt(e^(4*x) + e^(2*x)) - e^ (2*x))^5 + 15*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))
Time = 1.67 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(1+\tanh (x))^{5/2}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )}{8}-\frac {\frac {\mathrm {tanh}\left (x\right )}{6}+\frac {{\left (\mathrm {tanh}\left (x\right )+1\right )}^2}{4}+\frac {11}{30}}{{\left (\mathrm {tanh}\left (x\right )+1\right )}^{5/2}} \]