Integrand size = 13, antiderivative size = 46 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {3 \arctan (\sinh (x))}{8 a}-\frac {3 \text {sech}(x) \tanh (x)}{8 a}-\frac {\text {sech}(x) \tanh ^3(x)}{4 a}-\frac {\tanh ^5(x)}{5 a} \] Output:
3/8*arctan(sinh(x))/a-3/8*sech(x)*tanh(x)/a-1/4*sech(x)*tanh(x)^3/a-1/5*ta nh(x)^5/a
Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {\cosh ^2\left (\frac {x}{2}\right ) \left (30 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\left (-8-25 \text {sech}(x)+16 \text {sech}^2(x)+10 \text {sech}^3(x)-8 \text {sech}^4(x)\right ) \tanh (x)\right )}{20 a (1+\cosh (x))} \] Input:
Integrate[Tanh[x]^6/(a + a*Cosh[x]),x]
Output:
(Cosh[x/2]^2*(30*ArcTan[Tanh[x/2]] + (-8 - 25*Sech[x] + 16*Sech[x]^2 + 10* Sech[x]^3 - 8*Sech[x]^4)*Tanh[x]))/(20*a*(1 + Cosh[x]))
Time = 0.48 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.077, Rules used = {3042, 25, 3185, 25, 3042, 3087, 15, 3091, 25, 3042, 25, 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.
\(\displaystyle \int \frac {\tanh ^6(x)}{a \cosh (x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\tan \left (-\frac {\pi }{2}+i x\right )^6 \left (a-a \sin \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\left (a-a \sin \left (i x-\frac {\pi }{2}\right )\right ) \tan \left (i x-\frac {\pi }{2}\right )^6}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle -\frac {\int \text {sech}^2(x) \tanh ^4(x)dx}{a}-\frac {\int -\text {sech}(x) \tanh ^4(x)dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \text {sech}(x) \tanh ^4(x)dx}{a}-\frac {\int \text {sech}^2(x) \tanh ^4(x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (i x) \tan (i x)^4dx}{a}-\frac {\int \sec (i x)^2 \tan (i x)^4dx}{a}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {i \int \tanh ^4(x)d(i \tanh (x))}{a}+\frac {\int \sec (i x) \tan (i x)^4dx}{a}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {\tanh ^5(x)}{5 a}+\frac {\int \sec (i x) \tan (i x)^4dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {-\frac {3}{4} \int -\text {sech}(x) \tanh ^2(x)dx-\frac {1}{4} \tanh ^3(x) \text {sech}(x)}{a}-\frac {\tanh ^5(x)}{5 a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {3}{4} \int \text {sech}(x) \tanh ^2(x)dx-\frac {1}{4} \tanh ^3(x) \text {sech}(x)}{a}-\frac {\tanh ^5(x)}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tanh ^5(x)}{5 a}+\frac {-\frac {1}{4} \tanh ^3(x) \text {sech}(x)+\frac {3}{4} \int -\sec (i x) \tan (i x)^2dx}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tanh ^5(x)}{5 a}+\frac {-\frac {1}{4} \tanh ^3(x) \text {sech}(x)-\frac {3}{4} \int \sec (i x) \tan (i x)^2dx}{a}\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {-\frac {3}{4} \left (\frac {1}{2} \tanh (x) \text {sech}(x)-\frac {\int \text {sech}(x)dx}{2}\right )-\frac {1}{4} \tanh ^3(x) \text {sech}(x)}{a}-\frac {\tanh ^5(x)}{5 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\tanh ^5(x)}{5 a}+\frac {-\frac {1}{4} \tanh ^3(x) \text {sech}(x)-\frac {3}{4} \left (\frac {1}{2} \tanh (x) \text {sech}(x)-\frac {1}{2} \int \csc \left (i x+\frac {\pi }{2}\right )dx\right )}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {-\frac {3}{4} \left (\frac {1}{2} \tanh (x) \text {sech}(x)-\frac {1}{2} \arctan (\sinh (x))\right )-\frac {1}{4} \tanh ^3(x) \text {sech}(x)}{a}-\frac {\tanh ^5(x)}{5 a}\) |
Input:
Int[Tanh[x]^6/(a + a*Cosh[x]),x]
Output:
-1/5*Tanh[x]^5/a + (-1/4*(Sech[x]*Tanh[x]^3) - (3*(-1/2*ArcTan[Sinh[x]] + (Sech[x]*Tanh[x])/2))/4)/a
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 = 1.26 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.39
method | result | size |
default | \(\frac {\frac {64 \left (\frac {3 \tanh \left (\frac {x}{2}\right )^{9}}{256}+\frac {7 \tanh \left (\frac {x}{2}\right )^{7}}{128}-\frac {\tanh \left (\frac {x}{2}\right )^{5}}{10}-\frac {7 \tanh \left (\frac {x}{2}\right )^{3}}{128}-\frac {3 \tanh \left (\frac {x}{2}\right )}{256}\right )}{\left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )^{5}}+\frac {3 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{4}}{a}\) | \(64\) |
risch | \(-\frac {25 \,{\mathrm e}^{9 x}-40 \,{\mathrm e}^{8 x}+10 \,{\mathrm e}^{7 x}-80 \,{\mathrm e}^{4 x}-10 \,{\mathrm e}^{3 x}-25 \,{\mathrm e}^{x}-8}{20 \left ({\mathrm e}^{2 x}+1\right )^{5} a}+\frac {3 i \ln \left ({\mathrm e}^{x}+i\right )}{8 a}-\frac {3 i \ln \left ({\mathrm e}^{x}-i\right )}{8 a}\) | \(75\) |
Input:
int(tanh(x)^6/(a+cosh(x)*a),x,method=_RETURNVERBOSE)
Output:
64/a*((3/256*tanh(1/2*x)^9+7/128*tanh(1/2*x)^7-1/10*tanh(1/2*x)^5-7/128*ta nh(1/2*x)^3-3/256*tanh(1/2*x))/(tanh(1/2*x)^2+1)^5+3/256*arctan(tanh(1/2*x )))
Leaf count of result is larger than twice the leaf count of optimal. 750 vs. \(2 (38) = 76\).
Time = 0.09 (sec) , antiderivative size = 750, normalized size of antiderivative = 16.30 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\text {Too large to display} \] Input:
integrate(tanh(x)^6/(a+a*cosh(x)),x, algorithm="fricas")
Output:
-1/20*(25*cosh(x)^9 + 5*(45*cosh(x) - 8)*sinh(x)^8 + 25*sinh(x)^9 - 40*cos h(x)^8 + 10*(90*cosh(x)^2 - 32*cosh(x) + 1)*sinh(x)^7 + 10*cosh(x)^7 + 70* (30*cosh(x)^3 - 16*cosh(x)^2 + cosh(x))*sinh(x)^6 + 70*(45*cosh(x)^4 - 32* cosh(x)^3 + 3*cosh(x)^2)*sinh(x)^5 + 10*(315*cosh(x)^5 - 280*cosh(x)^4 + 3 5*cosh(x)^3 - 8)*sinh(x)^4 - 80*cosh(x)^4 + 10*(210*cosh(x)^6 - 224*cosh(x )^5 + 35*cosh(x)^4 - 32*cosh(x) - 1)*sinh(x)^3 - 10*cosh(x)^3 + 10*(90*cos h(x)^7 - 112*cosh(x)^6 + 21*cosh(x)^5 - 48*cosh(x)^2 - 3*cosh(x))*sinh(x)^ 2 - 15*(cosh(x)^10 + 10*cosh(x)*sinh(x)^9 + sinh(x)^10 + 5*(9*cosh(x)^2 + 1)*sinh(x)^8 + 5*cosh(x)^8 + 40*(3*cosh(x)^3 + cosh(x))*sinh(x)^7 + 10*(21 *cosh(x)^4 + 14*cosh(x)^2 + 1)*sinh(x)^6 + 10*cosh(x)^6 + 4*(63*cosh(x)^5 + 70*cosh(x)^3 + 15*cosh(x))*sinh(x)^5 + 10*(21*cosh(x)^6 + 35*cosh(x)^4 + 15*cosh(x)^2 + 1)*sinh(x)^4 + 10*cosh(x)^4 + 40*(3*cosh(x)^7 + 7*cosh(x)^ 5 + 5*cosh(x)^3 + cosh(x))*sinh(x)^3 + 5*(9*cosh(x)^8 + 28*cosh(x)^6 + 30* cosh(x)^4 + 12*cosh(x)^2 + 1)*sinh(x)^2 + 5*cosh(x)^2 + 10*(cosh(x)^9 + 4* cosh(x)^7 + 6*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*sinh(x) + 1)*arctan(cosh( x) + sinh(x)) + 5*(45*cosh(x)^8 - 64*cosh(x)^7 + 14*cosh(x)^6 - 64*cosh(x) ^3 - 6*cosh(x)^2 - 5)*sinh(x) - 25*cosh(x) - 8)/(a*cosh(x)^10 + 10*a*cosh( x)*sinh(x)^9 + a*sinh(x)^10 + 5*a*cosh(x)^8 + 5*(9*a*cosh(x)^2 + a)*sinh(x )^8 + 40*(3*a*cosh(x)^3 + a*cosh(x))*sinh(x)^7 + 10*a*cosh(x)^6 + 10*(21*a *cosh(x)^4 + 14*a*cosh(x)^2 + a)*sinh(x)^6 + 4*(63*a*cosh(x)^5 + 70*a*c...
\[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {\int \frac {\tanh ^{6}{\left (x \right )}}{\cosh {\left (x \right )} + 1}\, dx}{a} \] Input:
integrate(tanh(x)**6/(a+a*cosh(x)),x)
Output:
Integral(tanh(x)**6/(cosh(x) + 1), x)/a
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (38) = 76\).
Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.93 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=-\frac {25 \, e^{\left (-x\right )} + 10 \, e^{\left (-3 \, x\right )} + 80 \, e^{\left (-4 \, x\right )} - 10 \, e^{\left (-7 \, x\right )} + 40 \, e^{\left (-8 \, x\right )} - 25 \, e^{\left (-9 \, x\right )} + 8}{20 \, {\left (5 \, a e^{\left (-2 \, x\right )} + 10 \, a e^{\left (-4 \, x\right )} + 10 \, a e^{\left (-6 \, x\right )} + 5 \, a e^{\left (-8 \, x\right )} + a e^{\left (-10 \, x\right )} + a\right )}} - \frac {3 \, \arctan \left (e^{\left (-x\right )}\right )}{4 \, a} \] Input:
integrate(tanh(x)^6/(a+a*cosh(x)),x, algorithm="maxima")
Output:
-1/20*(25*e^(-x) + 10*e^(-3*x) + 80*e^(-4*x) - 10*e^(-7*x) + 40*e^(-8*x) - 25*e^(-9*x) + 8)/(5*a*e^(-2*x) + 10*a*e^(-4*x) + 10*a*e^(-6*x) + 5*a*e^(- 8*x) + a*e^(-10*x) + a) - 3/4*arctan(e^(-x))/a
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {3 \, \arctan \left (e^{x}\right )}{4 \, a} - \frac {25 \, e^{\left (9 \, x\right )} - 40 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (7 \, x\right )} - 80 \, e^{\left (4 \, x\right )} - 10 \, e^{\left (3 \, x\right )} - 25 \, e^{x} - 8}{20 \, a {\left (e^{\left (2 \, x\right )} + 1\right )}^{5}} \] Input:
integrate(tanh(x)^6/(a+a*cosh(x)),x, algorithm="giac")
Output:
3/4*arctan(e^x)/a - 1/20*(25*e^(9*x) - 40*e^(8*x) + 10*e^(7*x) - 80*e^(4*x ) - 10*e^(3*x) - 25*e^x - 8)/(a*(e^(2*x) + 1)^5)
Time = 1.99 (sec) , antiderivative size = 183, normalized size of antiderivative = 3.98 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {\frac {16}{a}-\frac {6\,{\mathrm {e}}^x}{a}}{3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1}-\frac {\frac {8}{a}-\frac {9\,{\mathrm {e}}^x}{2\,a}}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {32}{5\,a\,\left (5\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{4\,x}+10\,{\mathrm {e}}^{6\,x}+5\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{10\,x}+1\right )}-\frac {\frac {16}{a}-\frac {4\,{\mathrm {e}}^x}{a}}{4\,{\mathrm {e}}^{2\,x}+6\,{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1}+\frac {\frac {2}{a}-\frac {5\,{\mathrm {e}}^x}{4\,a}}{{\mathrm {e}}^{2\,x}+1}+\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^2}}{a}\right )}{4\,\sqrt {a^2}} \] Input:
int(tanh(x)^6/(a + a*cosh(x)),x)
Output:
(16/a - (6*exp(x))/a)/(3*exp(2*x) + 3*exp(4*x) + exp(6*x) + 1) - (8/a - (9 *exp(x))/(2*a))/(2*exp(2*x) + exp(4*x) + 1) + 32/(5*a*(5*exp(2*x) + 10*exp (4*x) + 10*exp(6*x) + 5*exp(8*x) + exp(10*x) + 1)) - (16/a - (4*exp(x))/a) /(4*exp(2*x) + 6*exp(4*x) + 4*exp(6*x) + exp(8*x) + 1) + (2/a - (5*exp(x)) /(4*a))/(exp(2*x) + 1) + (3*atan((exp(x)*(a^2)^(1/2))/a))/(4*(a^2)^(1/2))
Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 3.28 \[ \int \frac {\tanh ^6(x)}{a+a \cosh (x)} \, dx=\frac {15 e^{10 x} \mathit {atan} \left (e^{x}\right )+75 e^{8 x} \mathit {atan} \left (e^{x}\right )+150 e^{6 x} \mathit {atan} \left (e^{x}\right )+150 e^{4 x} \mathit {atan} \left (e^{x}\right )+75 e^{2 x} \mathit {atan} \left (e^{x}\right )+15 \mathit {atan} \left (e^{x}\right )-8 e^{10 x}-25 e^{9 x}-10 e^{7 x}-80 e^{6 x}+10 e^{3 x}-40 e^{2 x}+25 e^{x}}{20 a \left (e^{10 x}+5 e^{8 x}+10 e^{6 x}+10 e^{4 x}+5 e^{2 x}+1\right )} \] Input:
int(tanh(x)^6/(a+a*cosh(x)),x)
Output:
(15*e**(10*x)*atan(e**x) + 75*e**(8*x)*atan(e**x) + 150*e**(6*x)*atan(e**x ) + 150*e**(4*x)*atan(e**x) + 75*e**(2*x)*atan(e**x) + 15*atan(e**x) - 8*e **(10*x) - 25*e**(9*x) - 10*e**(7*x) - 80*e**(6*x) + 10*e**(3*x) - 40*e**( 2*x) + 25*e**x)/(20*a*(e**(10*x) + 5*e**(8*x) + 10*e**(6*x) + 10*e**(4*x) + 5*e**(2*x) + 1))