Integrand size = 13, antiderivative size = 45 \[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-2 \sqrt {1+\tanh (x)}-\frac {2}{5} (1+\tanh (x))^{5/2} \]
Time = 0.67 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00 \[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\tanh (x)}}{\sqrt {2}}\right )-2 \sqrt {1+\tanh (x)}-\frac {2}{5} (1+\tanh (x))^{5/2} \]
2*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - 2*Sqrt[1 + Tanh[x]] - (2*(1 + Tanh[x])^(5/2))/5
Time = 0.33 (sec) , antiderivative size = 45, 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, 3959, 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.
\(\displaystyle \int \tanh ^2(x) (\tanh (x)+1)^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -(1-i \tan (i x))^{3/2} \tan (i x)^2dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int (1-i \tan (i x))^{3/2} \tan (i x)^2dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle -\int -(\tanh (x)+1)^{3/2}dx-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int (\tanh (x)+1)^{3/2}dx-\frac {2}{5} (\tanh (x)+1)^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2}{5} (\tanh (x)+1)^{5/2}+\int (1-i \tan (i x))^{3/2}dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle 2 \int \sqrt {\tanh (x)+1}dx-\frac {2}{5} (\tanh (x)+1)^{5/2}-2 \sqrt {\tanh (x)+1}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 \int \sqrt {1-i \tan (i x)}dx-\frac {2}{5} (\tanh (x)+1)^{5/2}-2 \sqrt {\tanh (x)+1}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle 4 \int \frac {1}{1-\tanh (x)}d\sqrt {\tanh (x)+1}-\frac {2}{5} (\tanh (x)+1)^{5/2}-2 \sqrt {\tanh (x)+1}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {\tanh (x)+1}}{\sqrt {2}}\right )-\frac {2}{5} (\tanh (x)+1)^{5/2}-2 \sqrt {\tanh (x)+1}\) |
2*Sqrt[2]*ArcTanh[Sqrt[1 + Tanh[x]]/Sqrt[2]] - 2*Sqrt[1 + Tanh[x]] - (2*(1 + Tanh[x])^(5/2))/5
3.2.29.3.1 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[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[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] && GtQ[n , 1]
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 = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78
method | result | size |
derivativedivides | \(2 \,\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}-2 \sqrt {1+\tanh \left (x \right )}-\frac {2 \left (1+\tanh \left (x \right )\right )^{\frac {5}{2}}}{5}\) | \(35\) |
default | \(2 \,\operatorname {arctanh}\left (\frac {\sqrt {1+\tanh \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}-2 \sqrt {1+\tanh \left (x \right )}-\frac {2 \left (1+\tanh \left (x \right )\right )^{\frac {5}{2}}}{5}\) | \(35\) |
Leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (34) = 68\).
Time = 0.27 (sec) , antiderivative size = 429, normalized size of antiderivative = 9.53 \[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=-\frac {2 \, \sqrt {2} {\left (9 \, \sqrt {2} \cosh \left (x\right )^{5} + 45 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{4} + 9 \, \sqrt {2} \sinh \left (x\right )^{5} + 10 \, {\left (9 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right )^{3} + 10 \, \sqrt {2} \cosh \left (x\right )^{3} + 30 \, {\left (3 \, \sqrt {2} \cosh \left (x\right )^{3} + \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + 5 \, {\left (9 \, \sqrt {2} \cosh \left (x\right )^{4} + 6 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right ) + 5 \, \sqrt {2} \cosh \left (x\right )\right )} \sqrt {\frac {\cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 5 \, {\left (\sqrt {2} \cosh \left (x\right )^{6} + 6 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sqrt {2} \sinh \left (x\right )^{6} + 3 \, {\left (5 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right )^{4} + 3 \, \sqrt {2} \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \sqrt {2} \cosh \left (x\right )^{3} + 3 \, \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 3 \, {\left (5 \, \sqrt {2} \cosh \left (x\right )^{4} + 6 \, \sqrt {2} \cosh \left (x\right )^{2} + \sqrt {2}\right )} \sinh \left (x\right )^{2} + 3 \, \sqrt {2} \cosh \left (x\right )^{2} + 6 \, {\left (\sqrt {2} \cosh \left (x\right )^{5} + 2 \, \sqrt {2} \cosh \left (x\right )^{3} + \sqrt {2} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt {2}\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 )}{5 \, {\left (\cosh \left (x\right )^{6} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} + 3 \, {\left (5 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{4} + 3 \, \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 3 \, {\left (5 \, \cosh \left (x\right )^{4} + 6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 3 \, \cosh \left (x\right )^{2} + 6 \, {\left (\cosh \left (x\right )^{5} + 2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )}} \]
-1/5*(2*sqrt(2)*(9*sqrt(2)*cosh(x)^5 + 45*sqrt(2)*cosh(x)*sinh(x)^4 + 9*sq rt(2)*sinh(x)^5 + 10*(9*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x)^3 + 10*sqrt(2 )*cosh(x)^3 + 30*(3*sqrt(2)*cosh(x)^3 + sqrt(2)*cosh(x))*sinh(x)^2 + 5*(9* sqrt(2)*cosh(x)^4 + 6*sqrt(2)*cosh(x)^2 + sqrt(2))*sinh(x) + 5*sqrt(2)*cos h(x))*sqrt(cosh(x)/(cosh(x) - sinh(x))) - 5*(sqrt(2)*cosh(x)^6 + 6*sqrt(2) *cosh(x)*sinh(x)^5 + sqrt(2)*sinh(x)^6 + 3*(5*sqrt(2)*cosh(x)^2 + sqrt(2)) *sinh(x)^4 + 3*sqrt(2)*cosh(x)^4 + 4*(5*sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh (x))*sinh(x)^3 + 3*(5*sqrt(2)*cosh(x)^4 + 6*sqrt(2)*cosh(x)^2 + sqrt(2))*s inh(x)^2 + 3*sqrt(2)*cosh(x)^2 + 6*(sqrt(2)*cosh(x)^5 + 2*sqrt(2)*cosh(x)^ 3 + sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*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*s inh(x)^2 - 1))/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x) ^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + 1)
Time = 5.29 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29 \[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=- \sqrt {2} \left (\log {\left (\sqrt {\tanh {\left (x \right )} + 1} - \sqrt {2} \right )} - \log {\left (\sqrt {\tanh {\left (x \right )} + 1} + \sqrt {2} \right )}\right ) - \frac {2 \left (\tanh {\left (x \right )} + 1\right )^{\frac {5}{2}}}{5} - 2 \sqrt {\tanh {\left (x \right )} + 1} \]
-sqrt(2)*(log(sqrt(tanh(x) + 1) - sqrt(2)) - log(sqrt(tanh(x) + 1) + sqrt( 2))) - 2*(tanh(x) + 1)**(5/2)/5 - 2*sqrt(tanh(x) + 1)
\[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=\int { {\left (\tanh \left (x\right ) + 1\right )}^{\frac {3}{2}} \tanh \left (x\right )^{2} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (34) = 68\).
Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 3.11 \[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=\frac {1}{5} \, \sqrt {2} {\left (\frac {2 \, {\left (25 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{4} - 60 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} + 70 \, {\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{2} - 40 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 40 \, e^{\left (2 \, x\right )} + 9\right )}}{{\left (\sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )} - 1\right )}^{5}} - 5 \, \log \left (-2 \, \sqrt {e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}} + 2 \, e^{\left (2 \, x\right )} + 1\right )\right )} \]
1/5*sqrt(2)*(2*(25*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^4 - 60*(sqrt(e^(4*x ) + e^(2*x)) - e^(2*x))^3 + 70*(sqrt(e^(4*x) + e^(2*x)) - e^(2*x))^2 - 40* sqrt(e^(4*x) + e^(2*x)) + 40*e^(2*x) + 9)/(sqrt(e^(4*x) + e^(2*x)) - e^(2* x) - 1)^5 - 5*log(-2*sqrt(e^(4*x) + e^(2*x)) + 2*e^(2*x) + 1))
Time = 1.76 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int \tanh ^2(x) (1+\tanh (x))^{3/2} \, dx=2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {tanh}\left (x\right )+1}}{2}\right )-2\,\sqrt {\mathrm {tanh}\left (x\right )+1}-\frac {2\,{\left (\mathrm {tanh}\left (x\right )+1\right )}^{5/2}}{5} \]