Integrand size = 13, antiderivative size = 33 \[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=\frac {\arctan (\sinh (x))}{2 a}-\frac {\text {sech}(x) \tanh (x)}{2 a}-\frac {\tanh ^3(x)}{3 a} \]
Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=\frac {\cosh ^2\left (\frac {x}{2}\right ) \left (6 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\left (-2-3 \text {sech}(x)+2 \text {sech}^2(x)\right ) \tanh (x)\right )}{3 a (1+\cosh (x))} \]
(Cosh[x/2]^2*(6*ArcTan[Tanh[x/2]] + (-2 - 3*Sech[x] + 2*Sech[x]^2)*Tanh[x] ))/(3*a*(1 + Cosh[x]))
Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 3185, 25, 3042, 25, 3087, 15, 3091, 3042, 4257}
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.
\(\displaystyle \int \frac {\tanh ^4(x)}{a \cosh (x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan \left (-\frac {\pi }{2}+i x\right )^4 \left (a-a \sin \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle \frac {\int -\text {sech}^2(x) \tanh ^2(x)dx}{a}+\frac {\int \text {sech}(x) \tanh ^2(x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \text {sech}(x) \tanh ^2(x)dx}{a}-\frac {\int \text {sech}^2(x) \tanh ^2(x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int -\sec (i x) \tan (i x)^2dx}{a}-\frac {\int -\sec (i x)^2 \tan (i x)^2dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec (i x)^2 \tan (i x)^2dx}{a}-\frac {\int \sec (i x) \tan (i x)^2dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle -\frac {i \int -\tanh ^2(x)d(i \tanh (x))}{a}-\frac {\int \sec (i x) \tan (i x)^2dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\tanh ^3(x)}{3 a}-\frac {\int \sec (i x) \tan (i x)^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle -\frac {\frac {1}{2} \tanh (x) \text {sech}(x)-\frac {\int \text {sech}(x)dx}{2}}{a}-\frac {\tanh ^3(x)}{3 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tanh ^3(x)}{3 a}-\frac {\frac {1}{2} \tanh (x) \text {sech}(x)-\frac {1}{2} \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {\frac {1}{2} \tanh (x) \text {sech}(x)-\frac {1}{2} \arctan (\sinh (x))}{a}-\frac {\tanh ^3(x)}{3 a}\) |
3.2.89.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
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[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45
method | result | size |
default | \(\frac {\frac {16 \left (\frac {\tanh \left (\frac {x}{2}\right )^{5}}{16}-\frac {\tanh \left (\frac {x}{2}\right )^{3}}{6}-\frac {\tanh \left (\frac {x}{2}\right )}{16}\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{3}}+\arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a}\) | \(48\) |
risch | \(-\frac {3 \,{\mathrm e}^{5 x}-6 \,{\mathrm e}^{4 x}-3 \,{\mathrm e}^{x}-2}{3 \left (1+{\mathrm e}^{2 x}\right )^{3} a}+\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2 a}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2 a}\) | \(57\) |
16/a*((1/16*tanh(1/2*x)^5-1/6*tanh(1/2*x)^3-1/16*tanh(1/2*x))/(1+tanh(1/2* x)^2)^3+1/16*arctan(tanh(1/2*x)))
Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (27) = 54\).
Time = 0.29 (sec) , antiderivative size = 315, normalized size of antiderivative = 9.55 \[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=-\frac {3 \, \cosh \left (x\right )^{5} + 3 \, {\left (5 \, \cosh \left (x\right ) - 2\right )} \sinh \left (x\right )^{4} + 3 \, \sinh \left (x\right )^{5} - 6 \, \cosh \left (x\right )^{4} + 6 \, {\left (5 \, \cosh \left (x\right )^{2} - 4 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 6 \, {\left (5 \, \cosh \left (x\right )^{3} - 6 \, \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} - 3 \, {\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 )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 3 \, {\left (5 \, \cosh \left (x\right )^{4} - 8 \, \cosh \left (x\right )^{3} - 1\right )} \sinh \left (x\right ) - 3 \, \cosh \left (x\right ) - 2}{3 \, {\left (a \cosh \left (x\right )^{6} + 6 \, a \cosh \left (x\right ) \sinh \left (x\right )^{5} + a \sinh \left (x\right )^{6} + 3 \, a \cosh \left (x\right )^{4} + 3 \, {\left (5 \, a \cosh \left (x\right )^{2} + a\right )} \sinh \left (x\right )^{4} + 4 \, {\left (5 \, a \cosh \left (x\right )^{3} + 3 \, a \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 3 \, a \cosh \left (x\right )^{2} + 3 \, {\left (5 \, a \cosh \left (x\right )^{4} + 6 \, a \cosh \left (x\right )^{2} + a\right )} \sinh \left (x\right )^{2} + 6 \, {\left (a \cosh \left (x\right )^{5} + 2 \, a \cosh \left (x\right )^{3} + a \cosh \left (x\right )\right )} \sinh \left (x\right ) + a\right )}} \]
-1/3*(3*cosh(x)^5 + 3*(5*cosh(x) - 2)*sinh(x)^4 + 3*sinh(x)^5 - 6*cosh(x)^ 4 + 6*(5*cosh(x)^2 - 4*cosh(x))*sinh(x)^3 + 6*(5*cosh(x)^3 - 6*cosh(x)^2)* sinh(x)^2 - 3*(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)*arctan(cosh(x) + sinh(x)) + 3*(5*cosh( x)^4 - 8*cosh(x)^3 - 1)*sinh(x) - 3*cosh(x) - 2)/(a*cosh(x)^6 + 6*a*cosh(x )*sinh(x)^5 + a*sinh(x)^6 + 3*a*cosh(x)^4 + 3*(5*a*cosh(x)^2 + a)*sinh(x)^ 4 + 4*(5*a*cosh(x)^3 + 3*a*cosh(x))*sinh(x)^3 + 3*a*cosh(x)^2 + 3*(5*a*cos h(x)^4 + 6*a*cosh(x)^2 + a)*sinh(x)^2 + 6*(a*cosh(x)^5 + 2*a*cosh(x)^3 + a *cosh(x))*sinh(x) + a)
\[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=\frac {\int \frac {\tanh ^{4}{\left (x \right )}}{\cosh {\left (x \right )} + 1}\, dx}{a} \]
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (27) = 54\).
Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=-\frac {3 \, e^{\left (-x\right )} + 6 \, e^{\left (-4 \, x\right )} - 3 \, e^{\left (-5 \, x\right )} + 2}{3 \, {\left (3 \, a e^{\left (-2 \, x\right )} + 3 \, a e^{\left (-4 \, x\right )} + a e^{\left (-6 \, x\right )} + a\right )}} - \frac {\arctan \left (e^{\left (-x\right )}\right )}{a} \]
-1/3*(3*e^(-x) + 6*e^(-4*x) - 3*e^(-5*x) + 2)/(3*a*e^(-2*x) + 3*a*e^(-4*x) + a*e^(-6*x) + a) - arctan(e^(-x))/a
Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=\frac {\arctan \left (e^{x}\right )}{a} - \frac {3 \, e^{\left (5 \, x\right )} - 6 \, e^{\left (4 \, x\right )} - 3 \, e^{x} - 2}{3 \, a {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]
Time = 1.68 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.88 \[ \int \frac {\tanh ^4(x)}{a+a \cosh (x)} \, dx=\frac {8}{3\,a\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {\frac {4}{a}-\frac {2\,{\mathrm {e}}^x}{a}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {\frac {2}{a}-\frac {{\mathrm {e}}^x}{a}}{{\mathrm {e}}^{2\,x}+1}+\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^2}}{a}\right )}{\sqrt {a^2}} \]