Integrand size = 13, antiderivative size = 94 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=-\frac {5 \text {arctanh}(\cosh (x))}{16 a}-\frac {1}{8 (a-a \cosh (x))}+\frac {3}{16 (a+a \cosh (x))}+\frac {a^5}{24 \left (a^2+a^2 \cosh (x)\right )^3}-\frac {a^5}{32 \left (a^3-a^3 \cosh (x)\right )^2}+\frac {3 a^5}{32 \left (a^3+a^3 \cosh (x)\right )^2} \] Output:
-5/16*arctanh(cosh(x))/a-1/(8*a-8*a*cosh(x))+3/(16*a+16*a*cosh(x))+1/24*a^ 5/(a^2+a^2*cosh(x))^3-1/32*a^5/(a^3-a^3*cosh(x))^2+3/32*a^5/(a^3+a^3*cosh( x))^2
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.95 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=\frac {\cosh ^2\left (\frac {x}{2}\right ) \left (24 \text {csch}^2\left (\frac {x}{2}\right )-3 \text {csch}^4\left (\frac {x}{2}\right )-120 \log \left (\cosh \left (\frac {x}{2}\right )\right )+120 \log \left (\sinh \left (\frac {x}{2}\right )\right )+36 \text {sech}^2\left (\frac {x}{2}\right )+9 \text {sech}^4\left (\frac {x}{2}\right )+2 \text {sech}^6\left (\frac {x}{2}\right )\right )}{192 (a+a \cosh (x))} \] Input:
Integrate[Csch[x]^5/(a + a*Cosh[x]),x]
Output:
(Cosh[x/2]^2*(24*Csch[x/2]^2 - 3*Csch[x/2]^4 - 120*Log[Cosh[x/2]] + 120*Lo g[Sinh[x/2]] + 36*Sech[x/2]^2 + 9*Sech[x/2]^4 + 2*Sech[x/2]^6))/(192*(a + a*Cosh[x]))
Time = 0.35 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 26, 3146, 54, 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.
| \(\displaystyle \int \frac {\text {csch}^5(x)}{a \cosh (x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i}{\cos \left (-\frac {\pi }{2}+i x\right )^5 \left (a-a \sin \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {1}{\cos \left (i x-\frac {\pi }{2}\right )^5 \left (a-a \sin \left (i x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle -a^5 \int \frac {1}{(a-a \cosh (x))^3 (\cosh (x) a+a)^4}d(a \cosh (x))\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle -a^5 \int \left (\frac {1}{8 a^5 (a-a \cosh (x))^2}+\frac {3}{16 a^5 (\cosh (x) a+a)^2}+\frac {1}{16 a^4 (a-a \cosh (x))^3}+\frac {3}{16 a^4 (\cosh (x) a+a)^3}+\frac {1}{8 a^3 (\cosh (x) a+a)^4}+\frac {5}{16 a^5 \left (a^2-a^2 \cosh ^2(x)\right )}\right )d(a \cosh (x))\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -a^5 \left (\frac {5 \text {arctanh}(\cosh (x))}{16 a^6}+\frac {1}{8 a^5 (a-a \cosh (x))}-\frac {3}{16 a^5 (a \cosh (x)+a)}+\frac {1}{32 a^4 (a-a \cosh (x))^2}-\frac {3}{32 a^4 (a \cosh (x)+a)^2}-\frac {1}{24 a^3 (a \cosh (x)+a)^3}\right )\) |
Input:
Int[Csch[x]^5/(a + a*Cosh[x]),x]
Output:
-(a^5*((5*ArcTanh[Cosh[x]])/(16*a^6) + 1/(32*a^4*(a - a*Cosh[x])^2) + 1/(8 *a^5*(a - a*Cosh[x])) - 1/(24*a^3*(a + a*Cosh[x])^3) - 3/(32*a^4*(a + a*Co sh[x])^2) - 3/(16*a^5*(a + a*Cosh[x]))))
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[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 10.96 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.57
| method | result | size |
| default | \(\frac {-\frac {\tanh \left (\frac {x}{2}\right )^{6}}{6}+\frac {5 \tanh \left (\frac {x}{2}\right )^{4}}{4}-5 \tanh \left (\frac {x}{2}\right )^{2}+\frac {5}{2 \tanh \left (\frac {x}{2}\right )^{2}}-\frac {1}{4 \tanh \left (\frac {x}{2}\right )^{4}}+10 \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{32 a}\) | \(54\) |
| risch | \(\frac {{\mathrm e}^{x} \left (15 \,{\mathrm e}^{8 x}+30 \,{\mathrm e}^{7 x}-40 \,{\mathrm e}^{6 x}-110 \,{\mathrm e}^{5 x}+18 \,{\mathrm e}^{4 x}-110 \,{\mathrm e}^{3 x}-40 \,{\mathrm e}^{2 x}+30 \,{\mathrm e}^{x}+15\right )}{24 \left ({\mathrm e}^{x}+1\right )^{6} a \left ({\mathrm e}^{x}-1\right )^{4}}+\frac {5 \ln \left ({\mathrm e}^{x}-1\right )}{16 a}-\frac {5 \ln \left ({\mathrm e}^{x}+1\right )}{16 a}\) | \(89\) |
Input:
int(csch(x)^5/(a+cosh(x)*a),x,method=_RETURNVERBOSE)
Output:
1/32/a*(-1/6*tanh(1/2*x)^6+5/4*tanh(1/2*x)^4-5*tanh(1/2*x)^2+5/2/tanh(1/2* x)^2-1/4/tanh(1/2*x)^4+10*ln(tanh(1/2*x)))
Leaf count of result is larger than twice the leaf count of optimal. 1551 vs. \(2 (84) = 168\).
Time = 0.10 (sec) , antiderivative size = 1551, normalized size of antiderivative = 16.50 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=\text {Too large to display} \] Input:
integrate(csch(x)^5/(a+a*cosh(x)),x, algorithm="fricas")
Output:
1/48*(30*cosh(x)^9 + 30*(9*cosh(x) + 2)*sinh(x)^8 + 30*sinh(x)^9 + 60*cosh (x)^8 + 40*(27*cosh(x)^2 + 12*cosh(x) - 2)*sinh(x)^7 - 80*cosh(x)^7 + 20*( 126*cosh(x)^3 + 84*cosh(x)^2 - 28*cosh(x) - 11)*sinh(x)^6 - 220*cosh(x)^6 + 12*(315*cosh(x)^4 + 280*cosh(x)^3 - 140*cosh(x)^2 - 110*cosh(x) + 3)*sin h(x)^5 + 36*cosh(x)^5 + 20*(189*cosh(x)^5 + 210*cosh(x)^4 - 140*cosh(x)^3 - 165*cosh(x)^2 + 9*cosh(x) - 11)*sinh(x)^4 - 220*cosh(x)^4 + 40*(63*cosh( x)^6 + 84*cosh(x)^5 - 70*cosh(x)^4 - 110*cosh(x)^3 + 9*cosh(x)^2 - 22*cosh (x) - 2)*sinh(x)^3 - 80*cosh(x)^3 + 60*(18*cosh(x)^7 + 28*cosh(x)^6 - 28*c osh(x)^5 - 55*cosh(x)^4 + 6*cosh(x)^3 - 22*cosh(x)^2 - 4*cosh(x) + 1)*sinh (x)^2 + 60*cosh(x)^2 - 15*(cosh(x)^10 + 2*(5*cosh(x) + 1)*sinh(x)^9 + sinh (x)^10 + 2*cosh(x)^9 + 3*(15*cosh(x)^2 + 6*cosh(x) - 1)*sinh(x)^8 - 3*cosh (x)^8 + 8*(15*cosh(x)^3 + 9*cosh(x)^2 - 3*cosh(x) - 1)*sinh(x)^7 - 8*cosh( x)^7 + 2*(105*cosh(x)^4 + 84*cosh(x)^3 - 42*cosh(x)^2 - 28*cosh(x) + 1)*si nh(x)^6 + 2*cosh(x)^6 + 12*(21*cosh(x)^5 + 21*cosh(x)^4 - 14*cosh(x)^3 - 1 4*cosh(x)^2 + cosh(x) + 1)*sinh(x)^5 + 12*cosh(x)^5 + 2*(105*cosh(x)^6 + 1 26*cosh(x)^5 - 105*cosh(x)^4 - 140*cosh(x)^3 + 15*cosh(x)^2 + 30*cosh(x) + 1)*sinh(x)^4 + 2*cosh(x)^4 + 8*(15*cosh(x)^7 + 21*cosh(x)^6 - 21*cosh(x)^ 5 - 35*cosh(x)^4 + 5*cosh(x)^3 + 15*cosh(x)^2 + cosh(x) - 1)*sinh(x)^3 - 8 *cosh(x)^3 + 3*(15*cosh(x)^8 + 24*cosh(x)^7 - 28*cosh(x)^6 - 56*cosh(x)^5 + 10*cosh(x)^4 + 40*cosh(x)^3 + 4*cosh(x)^2 - 8*cosh(x) - 1)*sinh(x)^2 ...
\[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=\frac {\int \frac {\operatorname {csch}^{5}{\left (x \right )}}{\cosh {\left (x \right )} + 1}\, dx}{a} \] Input:
integrate(csch(x)**5/(a+a*cosh(x)),x)
Output:
Integral(csch(x)**5/(cosh(x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.65 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=\frac {15 \, e^{\left (-x\right )} + 30 \, e^{\left (-2 \, x\right )} - 40 \, e^{\left (-3 \, x\right )} - 110 \, e^{\left (-4 \, x\right )} + 18 \, e^{\left (-5 \, x\right )} - 110 \, e^{\left (-6 \, x\right )} - 40 \, e^{\left (-7 \, x\right )} + 30 \, e^{\left (-8 \, x\right )} + 15 \, e^{\left (-9 \, x\right )}}{24 \, {\left (2 \, a e^{\left (-x\right )} - 3 \, a e^{\left (-2 \, x\right )} - 8 \, a e^{\left (-3 \, x\right )} + 2 \, a e^{\left (-4 \, x\right )} + 12 \, a e^{\left (-5 \, x\right )} + 2 \, a e^{\left (-6 \, x\right )} - 8 \, a e^{\left (-7 \, x\right )} - 3 \, a e^{\left (-8 \, x\right )} + 2 \, a e^{\left (-9 \, x\right )} + a e^{\left (-10 \, x\right )} + a\right )}} - \frac {5 \, \log \left (e^{\left (-x\right )} + 1\right )}{16 \, a} + \frac {5 \, \log \left (e^{\left (-x\right )} - 1\right )}{16 \, a} \] Input:
integrate(csch(x)^5/(a+a*cosh(x)),x, algorithm="maxima")
Output:
1/24*(15*e^(-x) + 30*e^(-2*x) - 40*e^(-3*x) - 110*e^(-4*x) + 18*e^(-5*x) - 110*e^(-6*x) - 40*e^(-7*x) + 30*e^(-8*x) + 15*e^(-9*x))/(2*a*e^(-x) - 3*a *e^(-2*x) - 8*a*e^(-3*x) + 2*a*e^(-4*x) + 12*a*e^(-5*x) + 2*a*e^(-6*x) - 8 *a*e^(-7*x) - 3*a*e^(-8*x) + 2*a*e^(-9*x) + a*e^(-10*x) + a) - 5/16*log(e^ (-x) + 1)/a + 5/16*log(e^(-x) - 1)/a
Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=-\frac {5 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{32 \, a} + \frac {5 \, \log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{32 \, a} - \frac {15 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 76 \, e^{\left (-x\right )} - 76 \, e^{x} + 100}{64 \, a {\left (e^{\left (-x\right )} + e^{x} - 2\right )}^{2}} + \frac {55 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 402 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 1020 \, e^{\left (-x\right )} + 1020 \, e^{x} + 936}{192 \, a {\left (e^{\left (-x\right )} + e^{x} + 2\right )}^{3}} \] Input:
integrate(csch(x)^5/(a+a*cosh(x)),x, algorithm="giac")
Output:
-5/32*log(e^(-x) + e^x + 2)/a + 5/32*log(e^(-x) + e^x - 2)/a - 1/64*(15*(e ^(-x) + e^x)^2 - 76*e^(-x) - 76*e^x + 100)/(a*(e^(-x) + e^x - 2)^2) + 1/19 2*(55*(e^(-x) + e^x)^3 + 402*(e^(-x) + e^x)^2 + 1020*e^(-x) + 1020*e^x + 9 36)/(a*(e^(-x) + e^x + 2)^3)
Time = 2.07 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.60 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=\frac {1}{a\,\left (10\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{5\,x}+5\,{\mathrm {e}}^x+1\right )}+\frac {1}{4\,a\,\left (3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x+1\right )}+\frac {1}{8\,a\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )}-\frac {1}{8\,a\,\left (6\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^x+1\right )}-\frac {5}{8\,a\,\left (6\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^x+1\right )}+\frac {1}{4\,a\,\left ({\mathrm {e}}^x-1\right )}+\frac {3}{8\,a\,\left ({\mathrm {e}}^x+1\right )}-\frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {-a^2}}{a}\right )}{8\,\sqrt {-a^2}}-\frac {1}{3\,a\,\left (15\,{\mathrm {e}}^{2\,x}+20\,{\mathrm {e}}^{3\,x}+15\,{\mathrm {e}}^{4\,x}+6\,{\mathrm {e}}^{5\,x}+{\mathrm {e}}^{6\,x}+6\,{\mathrm {e}}^x+1\right )}-\frac {5}{12\,a\,\left (3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1\right )} \] Input:
int(1/(sinh(x)^5*(a + a*cosh(x))),x)
Output:
1/(a*(10*exp(2*x) + 10*exp(3*x) + 5*exp(4*x) + exp(5*x) + 5*exp(x) + 1)) + 1/(4*a*(3*exp(2*x) - exp(3*x) - 3*exp(x) + 1)) + 1/(8*a*(exp(2*x) - 2*exp (x) + 1)) - 1/(8*a*(6*exp(2*x) - 4*exp(3*x) + exp(4*x) - 4*exp(x) + 1)) - 5/(8*a*(6*exp(2*x) + 4*exp(3*x) + exp(4*x) + 4*exp(x) + 1)) + 1/(4*a*(exp( x) - 1)) + 3/(8*a*(exp(x) + 1)) - (5*atan((exp(x)*(-a^2)^(1/2))/a))/(8*(-a ^2)^(1/2)) - 1/(3*a*(15*exp(2*x) + 20*exp(3*x) + 15*exp(4*x) + 6*exp(5*x) + exp(6*x) + 6*exp(x) + 1)) - 5/(12*a*(3*exp(2*x) + exp(3*x) + 3*exp(x) + 1))
Time = 0.27 (sec) , antiderivative size = 405, normalized size of antiderivative = 4.31 \[ \int \frac {\text {csch}^5(x)}{a+a \cosh (x)} \, dx=\frac {-15-250 e^{4 x}+15 \,\mathrm {log}\left (e^{x}-1\right )-15 \,\mathrm {log}\left (e^{x}+1\right )+105 e^{8 x}-250 e^{6 x}+40 e^{3 x}-144 e^{5 x}+30 e^{6 x} \mathrm {log}\left (e^{x}-1\right )-30 e^{6 x} \mathrm {log}\left (e^{x}+1\right )+180 e^{5 x} \mathrm {log}\left (e^{x}-1\right )-180 e^{5 x} \mathrm {log}\left (e^{x}+1\right )+30 e^{4 x} \mathrm {log}\left (e^{x}-1\right )-30 e^{4 x} \mathrm {log}\left (e^{x}+1\right )-120 e^{3 x} \mathrm {log}\left (e^{x}-1\right )+120 e^{3 x} \mathrm {log}\left (e^{x}+1\right )+40 e^{7 x}-15 e^{10 x}+105 e^{2 x}-45 e^{2 x} \mathrm {log}\left (e^{x}-1\right )+45 e^{2 x} \mathrm {log}\left (e^{x}+1\right )+15 e^{10 x} \mathrm {log}\left (e^{x}-1\right )-15 e^{10 x} \mathrm {log}\left (e^{x}+1\right )+30 e^{9 x} \mathrm {log}\left (e^{x}-1\right )-30 e^{9 x} \mathrm {log}\left (e^{x}+1\right )-45 e^{8 x} \mathrm {log}\left (e^{x}-1\right )+45 e^{8 x} \mathrm {log}\left (e^{x}+1\right )-120 e^{7 x} \mathrm {log}\left (e^{x}-1\right )+120 e^{7 x} \mathrm {log}\left (e^{x}+1\right )+30 e^{x} \mathrm {log}\left (e^{x}-1\right )-30 e^{x} \mathrm {log}\left (e^{x}+1\right )}{48 a \left (e^{10 x}+2 e^{9 x}-3 e^{8 x}-8 e^{7 x}+2 e^{6 x}+12 e^{5 x}+2 e^{4 x}-8 e^{3 x}-3 e^{2 x}+2 e^{x}+1\right )} \] Input:
int(csch(x)^5/(a+a*cosh(x)),x)
Output:
(15*e**(10*x)*log(e**x - 1) - 15*e**(10*x)*log(e**x + 1) - 15*e**(10*x) + 30*e**(9*x)*log(e**x - 1) - 30*e**(9*x)*log(e**x + 1) - 45*e**(8*x)*log(e* *x - 1) + 45*e**(8*x)*log(e**x + 1) + 105*e**(8*x) - 120*e**(7*x)*log(e**x - 1) + 120*e**(7*x)*log(e**x + 1) + 40*e**(7*x) + 30*e**(6*x)*log(e**x - 1) - 30*e**(6*x)*log(e**x + 1) - 250*e**(6*x) + 180*e**(5*x)*log(e**x - 1) - 180*e**(5*x)*log(e**x + 1) - 144*e**(5*x) + 30*e**(4*x)*log(e**x - 1) - 30*e**(4*x)*log(e**x + 1) - 250*e**(4*x) - 120*e**(3*x)*log(e**x - 1) + 1 20*e**(3*x)*log(e**x + 1) + 40*e**(3*x) - 45*e**(2*x)*log(e**x - 1) + 45*e **(2*x)*log(e**x + 1) + 105*e**(2*x) + 30*e**x*log(e**x - 1) - 30*e**x*log (e**x + 1) + 15*log(e**x - 1) - 15*log(e**x + 1) - 15)/(48*a*(e**(10*x) + 2*e**(9*x) - 3*e**(8*x) - 8*e**(7*x) + 2*e**(6*x) + 12*e**(5*x) + 2*e**(4* x) - 8*e**(3*x) - 3*e**(2*x) + 2*e**x + 1))