Integrand size = 15, antiderivative size = 17 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=\frac {\tanh ^6(x)}{6}-\frac {\tanh ^8(x)}{8} \]
Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=-\frac {1}{4} \text {sech}^4(x)+\frac {\text {sech}^6(x)}{3}-\frac {\text {sech}^8(x)}{8} \]
Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 26, 4608, 26, 3042, 26, 3087, 244, 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.
\(\displaystyle \int \tanh (x) \text {sech}^4(x) \left (\text {sech}^2(x)-1\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -i \tan (i x) \sec (i x)^4 \left (-1+\sec (i x)^2\right )^2dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \sec (i x)^4 \left (1-\sec (i x)^2\right )^2 \tan (i x)dx\) |
\(\Big \downarrow \) 4608 |
\(\displaystyle -i \int i \text {sech}^4(x) \tanh ^5(x)dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \tanh ^5(x) \text {sech}^4(x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -i \tan (i x)^5 \sec (i x)^4dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \sec (i x)^4 \tan (i x)^5dx\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle -\int i \tanh ^5(x) \left (1-\tanh ^2(x)\right )d(i \tanh (x))\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\int \left (i \tanh ^5(x)-i \tanh ^7(x)\right )d(i \tanh (x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\tanh ^6(x)}{6}-\frac {\tanh ^8(x)}{8}\) |
3.11.3.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ b^p Int[ActivateTrig[u*tan[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82
\[\frac {\tanh \left (x \right )^{6}}{6}-\frac {\tanh \left (x \right )^{8}}{8}\]
Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (13) = 26\).
Time = 0.25 (sec) , antiderivative size = 340, normalized size of antiderivative = 20.00 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=-\frac {4 \, {\left (3 \, \cosh \left (x\right )^{6} + 18 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + 3 \, \sinh \left (x\right )^{6} + {\left (45 \, \cosh \left (x\right )^{2} - 4\right )} \sinh \left (x\right )^{4} - 4 \, \cosh \left (x\right )^{4} + 4 \, {\left (15 \, \cosh \left (x\right )^{3} - 4 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + {\left (45 \, \cosh \left (x\right )^{4} - 24 \, \cosh \left (x\right )^{2} + 13\right )} \sinh \left (x\right )^{2} + 13 \, \cosh \left (x\right )^{2} + 2 \, {\left (9 \, \cosh \left (x\right )^{5} - 8 \, \cosh \left (x\right )^{3} + 7 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) - 4\right )}}{3 \, {\left (\cosh \left (x\right )^{10} + 10 \, \cosh \left (x\right ) \sinh \left (x\right )^{9} + \sinh \left (x\right )^{10} + {\left (45 \, \cosh \left (x\right )^{2} + 8\right )} \sinh \left (x\right )^{8} + 8 \, \cosh \left (x\right )^{8} + 8 \, {\left (15 \, \cosh \left (x\right )^{3} + 8 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{7} + {\left (210 \, \cosh \left (x\right )^{4} + 224 \, \cosh \left (x\right )^{2} + 29\right )} \sinh \left (x\right )^{6} + 29 \, \cosh \left (x\right )^{6} + 2 \, {\left (126 \, \cosh \left (x\right )^{5} + 224 \, \cosh \left (x\right )^{3} + 81 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} + {\left (210 \, \cosh \left (x\right )^{6} + 560 \, \cosh \left (x\right )^{4} + 435 \, \cosh \left (x\right )^{2} + 64\right )} \sinh \left (x\right )^{4} + 64 \, \cosh \left (x\right )^{4} + 4 \, {\left (30 \, \cosh \left (x\right )^{7} + 112 \, \cosh \left (x\right )^{5} + 135 \, \cosh \left (x\right )^{3} + 48 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + {\left (45 \, \cosh \left (x\right )^{8} + 224 \, \cosh \left (x\right )^{6} + 435 \, \cosh \left (x\right )^{4} + 384 \, \cosh \left (x\right )^{2} + 98\right )} \sinh \left (x\right )^{2} + 98 \, \cosh \left (x\right )^{2} + 2 \, {\left (5 \, \cosh \left (x\right )^{9} + 32 \, \cosh \left (x\right )^{7} + 81 \, \cosh \left (x\right )^{5} + 96 \, \cosh \left (x\right )^{3} + 42 \, \cosh \left (x\right )\right )} \sinh \left (x\right ) + 56\right )}} \]
-4/3*(3*cosh(x)^6 + 18*cosh(x)*sinh(x)^5 + 3*sinh(x)^6 + (45*cosh(x)^2 - 4 )*sinh(x)^4 - 4*cosh(x)^4 + 4*(15*cosh(x)^3 - 4*cosh(x))*sinh(x)^3 + (45*c osh(x)^4 - 24*cosh(x)^2 + 13)*sinh(x)^2 + 13*cosh(x)^2 + 2*(9*cosh(x)^5 - 8*cosh(x)^3 + 7*cosh(x))*sinh(x) - 4)/(cosh(x)^10 + 10*cosh(x)*sinh(x)^9 + sinh(x)^10 + (45*cosh(x)^2 + 8)*sinh(x)^8 + 8*cosh(x)^8 + 8*(15*cosh(x)^3 + 8*cosh(x))*sinh(x)^7 + (210*cosh(x)^4 + 224*cosh(x)^2 + 29)*sinh(x)^6 + 29*cosh(x)^6 + 2*(126*cosh(x)^5 + 224*cosh(x)^3 + 81*cosh(x))*sinh(x)^5 + (210*cosh(x)^6 + 560*cosh(x)^4 + 435*cosh(x)^2 + 64)*sinh(x)^4 + 64*cosh( x)^4 + 4*(30*cosh(x)^7 + 112*cosh(x)^5 + 135*cosh(x)^3 + 48*cosh(x))*sinh( x)^3 + (45*cosh(x)^8 + 224*cosh(x)^6 + 435*cosh(x)^4 + 384*cosh(x)^2 + 98) *sinh(x)^2 + 98*cosh(x)^2 + 2*(5*cosh(x)^9 + 32*cosh(x)^7 + 81*cosh(x)^5 + 96*cosh(x)^3 + 42*cosh(x))*sinh(x) + 56)
Time = 0.92 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=- \frac {\operatorname {sech}^{8}{\left (x \right )}}{8} + \frac {\operatorname {sech}^{6}{\left (x \right )}}{3} - \frac {\operatorname {sech}^{4}{\left (x \right )}}{4} \]
Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (13) = 26\).
Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.00 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=-\frac {4}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{4}} + \frac {64}{3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{6}} - \frac {32}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{8}} \]
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (13) = 26\).
Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.41 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=-\frac {4 \, {\left (3 \, e^{\left (12 \, x\right )} - 4 \, e^{\left (10 \, x\right )} + 10 \, e^{\left (8 \, x\right )} - 4 \, e^{\left (6 \, x\right )} + 3 \, e^{\left (4 \, x\right )}\right )}}{3 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{8}} \]
Time = 2.48 (sec) , antiderivative size = 375, normalized size of antiderivative = 22.06 \[ \int \text {sech}^4(x) \left (-1+\text {sech}^2(x)\right )^2 \tanh (x) \, dx=\frac {{\mathrm {e}}^{2\,x}-5\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}-10\,{\mathrm {e}}^{8\,x}+5\,{\mathrm {e}}^{10\,x}-{\mathrm {e}}^{12\,x}}{8\,{\mathrm {e}}^{2\,x}+28\,{\mathrm {e}}^{4\,x}+56\,{\mathrm {e}}^{6\,x}+70\,{\mathrm {e}}^{8\,x}+56\,{\mathrm {e}}^{10\,x}+28\,{\mathrm {e}}^{12\,x}+8\,{\mathrm {e}}^{14\,x}+{\mathrm {e}}^{16\,x}+1}-\frac {\frac {20\,{\mathrm {e}}^{4\,x}}{7}-\frac {10\,{\mathrm {e}}^{2\,x}}{7}-\frac {50\,{\mathrm {e}}^{6\,x}}{21}+\frac {5\,{\mathrm {e}}^{8\,x}}{7}+\frac {5}{21}}{6\,{\mathrm {e}}^{2\,x}+15\,{\mathrm {e}}^{4\,x}+20\,{\mathrm {e}}^{6\,x}+15\,{\mathrm {e}}^{8\,x}+6\,{\mathrm {e}}^{10\,x}+{\mathrm {e}}^{12\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^{2\,x}}{7}-\frac {10\,{\mathrm {e}}^{4\,x}}{7}+\frac {4\,{\mathrm {e}}^{6\,x}}{7}-\frac {2}{7}}{5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}+5\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{10\,x}+1}-\frac {\frac {10\,{\mathrm {e}}^{2\,x}}{7}-\frac {30\,{\mathrm {e}}^{4\,x}}{7}+\frac {40\,{\mathrm {e}}^{6\,x}}{7}-\frac {25\,{\mathrm {e}}^{8\,x}}{7}+\frac {6\,{\mathrm {e}}^{10\,x}}{7}-\frac {1}{7}}{7\,{\mathrm {e}}^{2\,x}+21\,{\mathrm {e}}^{4\,x}+35\,{\mathrm {e}}^{6\,x}+35\,{\mathrm {e}}^{8\,x}+21\,{\mathrm {e}}^{10\,x}+7\,{\mathrm {e}}^{12\,x}+{\mathrm {e}}^{14\,x}+1}-\frac {\frac {2\,{\mathrm {e}}^{2\,x}}{7}-\frac {5}{21}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {1}{7\,\left (2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1\right )}-\frac {\frac {3\,{\mathrm {e}}^{4\,x}}{7}-\frac {5\,{\mathrm {e}}^{2\,x}}{7}+\frac {2}{7}}{4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1} \]
(exp(2*x) - 5*exp(4*x) + 10*exp(6*x) - 10*exp(8*x) + 5*exp(10*x) - exp(12* x))/(8*exp(2*x) + 28*exp(4*x) + 56*exp(6*x) + 70*exp(8*x) + 56*exp(10*x) + 28*exp(12*x) + 8*exp(14*x) + exp(16*x) + 1) - ((20*exp(4*x))/7 - (10*exp( 2*x))/7 - (50*exp(6*x))/21 + (5*exp(8*x))/7 + 5/21)/(6*exp(2*x) + 15*exp(4 *x) + 20*exp(6*x) + 15*exp(8*x) + 6*exp(10*x) + exp(12*x) + 1) - ((8*exp(2 *x))/7 - (10*exp(4*x))/7 + (4*exp(6*x))/7 - 2/7)/(5*exp(2*x) + 10*exp(4*x) + 10*exp(6*x) + 5*exp(8*x) + exp(10*x) + 1) - ((10*exp(2*x))/7 - (30*exp( 4*x))/7 + (40*exp(6*x))/7 - (25*exp(8*x))/7 + (6*exp(10*x))/7 - 1/7)/(7*ex p(2*x) + 21*exp(4*x) + 35*exp(6*x) + 35*exp(8*x) + 21*exp(10*x) + 7*exp(12 *x) + exp(14*x) + 1) - ((2*exp(2*x))/7 - 5/21)/(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1) - 1/(7*(2*exp(2*x) + exp(4*x) + 1)) - ((3*exp(4*x))/7 - (5*e xp(2*x))/7 + 2/7)/(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1)