Integrand size = 13, antiderivative size = 154 \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=-\frac {a \left (8 a^4-20 a^2 b^2+15 b^4\right ) x}{8 b^6}+\frac {2 (a-b)^{5/2} (a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^6}+\frac {\left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right ) \sinh (x)}{8 b^5}+\frac {\left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right ) \sinh ^3(x)}{12 b^3}+\frac {\sinh ^5(x)}{5 b} \]
-1/8*a*(8*a^4-20*a^2*b^2+15*b^4)*x/b^6+2*(a-b)^(5/2)*(a+b)^(5/2)*arctanh(( a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b^6+1/8*(8*(a^2-b^2)^2-a*b*(4*a^2-7*b^ 2)*cosh(x))*sinh(x)/b^5+1/12*(4*a^2-4*b^2-3*a*b*cosh(x))*sinh(x)^3/b^3+1/5 *sinh(x)^5/b
Time = 0.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=\frac {-60 a \left (8 a^4-20 a^2 b^2+15 b^4\right ) x+960 \left (-a^2+b^2\right )^{5/2} \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )+60 b \left (8 a^4-18 a^2 b^2+11 b^4\right ) \sinh (x)-120 a b^2 \left (a^2-2 b^2\right ) \sinh (2 x)-10 b^3 \left (-4 a^2+7 b^2\right ) \sinh (3 x)-15 a b^4 \sinh (4 x)+6 b^5 \sinh (5 x)}{480 b^6} \]
(-60*a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*x + 960*(-a^2 + b^2)^(5/2)*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]] + 60*b*(8*a^4 - 18*a^2*b^2 + 11*b^4)*Si nh[x] - 120*a*b^2*(a^2 - 2*b^2)*Sinh[2*x] - 10*b^3*(-4*a^2 + 7*b^2)*Sinh[3 *x] - 15*a*b^4*Sinh[4*x] + 6*b^5*Sinh[5*x])/(480*b^6)
Time = 0.96 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.19, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.154, Rules used = {3042, 25, 3174, 25, 3042, 3344, 3042, 25, 3344, 25, 3042, 3214, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cos \left (-\frac {\pi }{2}+i x\right )^6}{a-b \sin \left (-\frac {\pi }{2}+i x\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (i x-\frac {\pi }{2}\right )^6}{a-b \sin \left (i x-\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3174 |
| \(\displaystyle \frac {\int -\frac {(b+a \cosh (x)) \sinh ^4(x)}{a+b \cosh (x)}dx}{b}+\frac {\sinh ^5(x)}{5 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {\int \frac {(b+a \cosh (x)) \sinh ^4(x)}{a+b \cosh (x)}dx}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {\int \frac {\cos \left (i x+\frac {\pi }{2}\right )^4 \left (b+a \sin \left (i x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {\frac {\int \frac {\left (b \left (a^2-4 b^2\right )+a \left (4 a^2-7 b^2\right ) \cosh (x)\right ) \sinh ^2(x)}{a+b \cosh (x)}dx}{4 b^2}-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}+\frac {\int -\frac {\cos \left (i x+\frac {\pi }{2}\right )^2 \left (b \left (a^2-4 b^2\right )+a \left (4 a^2-7 b^2\right ) \sin \left (i x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{4 b^2}}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}-\frac {\int \frac {\cos \left (i x+\frac {\pi }{2}\right )^2 \left (b \left (a^2-4 b^2\right )+a \left (4 a^2-7 b^2\right ) \sin \left (i x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{4 b^2}}{b}\) |
| \(\Big \downarrow \) 3344 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\frac {\int -\frac {b \left (4 a^4-9 b^2 a^2+8 b^4\right )+a \left (8 a^4-20 b^2 a^2+15 b^4\right ) \cosh (x)}{a+b \cosh (x)}dx}{2 b^2}+\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}}{4 b^2}-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}-\frac {\int \frac {b \left (4 a^4-9 b^2 a^2+8 b^4\right )+a \left (8 a^4-20 b^2 a^2+15 b^4\right ) \cosh (x)}{a+b \cosh (x)}dx}{2 b^2}}{4 b^2}-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}-\frac {\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}-\frac {\int \frac {b \left (4 a^4-9 b^2 a^2+8 b^4\right )+a \left (8 a^4-20 b^2 a^2+15 b^4\right ) \sin \left (i x+\frac {\pi }{2}\right )}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{2 b^2}}{4 b^2}}{b}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (8 a^4-20 a^2 b^2+15 b^4\right )}{b}-\frac {8 \left (a^2-b^2\right )^3 \int \frac {1}{a+b \cosh (x)}dx}{b}}{2 b^2}}{4 b^2}-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}-\frac {\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (8 a^4-20 a^2 b^2+15 b^4\right )}{b}-\frac {8 \left (a^2-b^2\right )^3 \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{b}}{2 b^2}}{4 b^2}}{b}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (8 a^4-20 a^2 b^2+15 b^4\right )}{b}-\frac {16 \left (a^2-b^2\right )^3 \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{b}}{2 b^2}}{4 b^2}-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}}{b}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sinh ^5(x)}{5 b}-\frac {-\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^2}-\frac {\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{2 b^2}-\frac {\frac {a x \left (8 a^4-20 a^2 b^2+15 b^4\right )}{b}-\frac {16 \left (a^2-b^2\right )^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}}}{2 b^2}}{4 b^2}}{b}\) |
Sinh[x]^5/(5*b) - (-1/12*((4*(a^2 - b^2) - 3*a*b*Cosh[x])*Sinh[x]^3)/b^2 - (-1/2*((a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*x)/b - (16*(a^2 - b^2)^3*ArcTanh[ (Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]))/b^2 + ( (8*(a^2 - b^2)^2 - a*b*(4*a^2 - 7*b^2)*Cosh[x])*Sinh[x])/(2*b^2))/(4*b^2)) /b
3.2.66.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(b*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g*(g* Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c*(m + p + 1) - a*d* p + b*d*(m + p)*Sin[e + f*x])/(b^2*f*(m + p)*(m + p + 1))), x] + Simp[g^2*( (p - 1)/(b^2*(m + p)*(m + p + 1))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Si n[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1) - d*(a^ 2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1 , 0] && IntegerQ[2*m]
Leaf count of result is larger than twice the leaf count of optimal. \(408\) vs. \(2(135)=270\).
Time = 0.05 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.66
\[-\frac {1}{5 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{5}}-\frac {a +2 b}{4 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {4 a^{2}+6 a b -b^{2}}{12 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {4 a^{3}+4 a^{2} b -5 a \,b^{2}-5 b^{3}}{8 b^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {8 a^{4}+4 a^{3} b -16 a^{2} b^{2}-7 a \,b^{3}+8 b^{4}}{8 b^{5} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \left (8 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8 b^{6}}-\frac {2 \left (-a^{6}+3 a^{4} b^{2}-3 a^{2} b^{4}+b^{6}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{6} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{5 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}-\frac {-a -2 b}{4 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {4 a^{2}+6 a b -b^{2}}{12 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-4 a^{3}-4 a^{2} b +5 a \,b^{2}+5 b^{3}}{8 b^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {8 a^{4}+4 a^{3} b -16 a^{2} b^{2}-7 a \,b^{3}+8 b^{4}}{8 b^{5} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \left (8 a^{4}-20 a^{2} b^{2}+15 b^{4}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8 b^{6}}\]
-1/5/b/(tanh(1/2*x)-1)^5-1/4*(a+2*b)/b^2/(tanh(1/2*x)-1)^4-1/12*(4*a^2+6*a *b-b^2)/b^3/(tanh(1/2*x)-1)^3-1/8*(4*a^3+4*a^2*b-5*a*b^2-5*b^3)/b^4/(tanh( 1/2*x)-1)^2-1/8*(8*a^4+4*a^3*b-16*a^2*b^2-7*a*b^3+8*b^4)/b^5/(tanh(1/2*x)- 1)+1/8*a*(8*a^4-20*a^2*b^2+15*b^4)/b^6*ln(tanh(1/2*x)-1)-2/b^6*(-a^6+3*a^4 *b^2-3*a^2*b^4+b^6)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*( a-b))^(1/2))-1/5/b/(tanh(1/2*x)+1)^5-1/4*(-a-2*b)/b^2/(tanh(1/2*x)+1)^4-1/ 12*(4*a^2+6*a*b-b^2)/b^3/(tanh(1/2*x)+1)^3-1/8*(-4*a^3-4*a^2*b+5*a*b^2+5*b ^3)/b^4/(tanh(1/2*x)+1)^2-1/8*(8*a^4+4*a^3*b-16*a^2*b^2-7*a*b^3+8*b^4)/b^5 /(tanh(1/2*x)+1)-1/8*a*(8*a^4-20*a^2*b^2+15*b^4)/b^6*ln(tanh(1/2*x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 1422 vs. \(2 (134) = 268\).
Time = 0.30 (sec) , antiderivative size = 2913, normalized size of antiderivative = 18.92 \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \]
[1/960*(6*b^5*cosh(x)^10 + 6*b^5*sinh(x)^10 - 15*a*b^4*cosh(x)^9 + 15*(4*b ^5*cosh(x) - a*b^4)*sinh(x)^9 + 10*(4*a^2*b^3 - 7*b^5)*cosh(x)^8 + 5*(54*b ^5*cosh(x)^2 - 27*a*b^4*cosh(x) + 8*a^2*b^3 - 14*b^5)*sinh(x)^8 - 120*(a^3 *b^2 - 2*a*b^4)*cosh(x)^7 + 20*(36*b^5*cosh(x)^3 - 27*a*b^4*cosh(x)^2 - 6* a^3*b^2 + 12*a*b^4 + 4*(4*a^2*b^3 - 7*b^5)*cosh(x))*sinh(x)^7 - 120*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x*cosh(x)^5 + 60*(8*a^4*b - 18*a^2*b^3 + 11*b^5) *cosh(x)^6 + 20*(63*b^5*cosh(x)^4 - 63*a*b^4*cosh(x)^3 + 24*a^4*b - 54*a^2 *b^3 + 33*b^5 + 14*(4*a^2*b^3 - 7*b^5)*cosh(x)^2 - 42*(a^3*b^2 - 2*a*b^4)* cosh(x))*sinh(x)^6 + 15*a*b^4*cosh(x) + 2*(756*b^5*cosh(x)^5 - 945*a*b^4*c osh(x)^4 + 280*(4*a^2*b^3 - 7*b^5)*cosh(x)^3 - 1260*(a^3*b^2 - 2*a*b^4)*co sh(x)^2 - 60*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x + 180*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x))*sinh(x)^5 - 6*b^5 - 60*(8*a^4*b - 18*a^2*b^3 + 11*b^5) *cosh(x)^4 + 10*(126*b^5*cosh(x)^6 - 189*a*b^4*cosh(x)^5 - 48*a^4*b + 108* a^2*b^3 - 66*b^5 + 70*(4*a^2*b^3 - 7*b^5)*cosh(x)^4 - 420*(a^3*b^2 - 2*a*b ^4)*cosh(x)^3 - 60*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x*cosh(x) + 90*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x)^2)*sinh(x)^4 + 120*(a^3*b^2 - 2*a*b^4)*cos h(x)^3 + 20*(36*b^5*cosh(x)^7 - 63*a*b^4*cosh(x)^6 + 28*(4*a^2*b^3 - 7*b^5 )*cosh(x)^5 + 6*a^3*b^2 - 12*a*b^4 - 210*(a^3*b^2 - 2*a*b^4)*cosh(x)^4 - 6 0*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x*cosh(x)^2 + 60*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x)^3 - 12*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x))*sinh(x...
Timed out. \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.25 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.73 \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=\frac {6 \, b^{4} e^{\left (5 \, x\right )} - 15 \, a b^{3} e^{\left (4 \, x\right )} + 40 \, a^{2} b^{2} e^{\left (3 \, x\right )} - 70 \, b^{4} e^{\left (3 \, x\right )} - 120 \, a^{3} b e^{\left (2 \, x\right )} + 240 \, a b^{3} e^{\left (2 \, x\right )} + 480 \, a^{4} e^{x} - 1080 \, a^{2} b^{2} e^{x} + 660 \, b^{4} e^{x}}{960 \, b^{5}} - \frac {{\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x}{8 \, b^{6}} + \frac {{\left (15 \, a b^{4} e^{x} - 6 \, b^{5} - 60 \, {\left (8 \, a^{4} b - 18 \, a^{2} b^{3} + 11 \, b^{5}\right )} e^{\left (4 \, x\right )} + 120 \, {\left (a^{3} b^{2} - 2 \, a b^{4}\right )} e^{\left (3 \, x\right )} - 10 \, {\left (4 \, a^{2} b^{3} - 7 \, b^{5}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-5 \, x\right )}}{960 \, b^{6}} + \frac {2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{6}} \]
1/960*(6*b^4*e^(5*x) - 15*a*b^3*e^(4*x) + 40*a^2*b^2*e^(3*x) - 70*b^4*e^(3 *x) - 120*a^3*b*e^(2*x) + 240*a*b^3*e^(2*x) + 480*a^4*e^x - 1080*a^2*b^2*e ^x + 660*b^4*e^x)/b^5 - 1/8*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x/b^6 + 1/960* (15*a*b^4*e^x - 6*b^5 - 60*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*e^(4*x) + 120*( a^3*b^2 - 2*a*b^4)*e^(3*x) - 10*(4*a^2*b^3 - 7*b^5)*e^(2*x))*e^(-5*x)/b^6 + 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*arctan((b*e^x + a)/sqrt(-a^2 + b^2 ))/(sqrt(-a^2 + b^2)*b^6)
Time = 1.42 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.26 \[ \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^{5\,x}}{160\,b}-\frac {{\mathrm {e}}^{-5\,x}}{160\,b}-\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a\,b^2-a^3\right )}{8\,b^4}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a\,b^2-a^3\right )}{8\,b^4}-\frac {x\,\left (8\,a^5-20\,a^3\,b^2+15\,a\,b^4\right )}{8\,b^6}+\frac {{\mathrm {e}}^x\,\left (8\,a^4-18\,a^2\,b^2+11\,b^4\right )}{16\,b^5}+\frac {a\,{\mathrm {e}}^{-4\,x}}{64\,b^2}-\frac {a\,{\mathrm {e}}^{4\,x}}{64\,b^2}-\frac {{\mathrm {e}}^{-x}\,\left (8\,a^4-18\,a^2\,b^2+11\,b^4\right )}{16\,b^5}-\frac {{\mathrm {e}}^{-3\,x}\,\left (4\,a^2-7\,b^2\right )}{96\,b^3}+\frac {{\mathrm {e}}^{3\,x}\,\left (4\,a^2-7\,b^2\right )}{96\,b^3}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{b^7}-\frac {2\,{\left (a+b\right )}^{5/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{5/2}}{b^7}\right )\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}{b^6}-\frac {\ln \left (\frac {2\,{\left (a+b\right )}^{5/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{5/2}}{b^7}-\frac {2\,{\mathrm {e}}^x\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{b^7}\right )\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}{b^6} \]
exp(5*x)/(160*b) - exp(-5*x)/(160*b) - (exp(-2*x)*(2*a*b^2 - a^3))/(8*b^4) + (exp(2*x)*(2*a*b^2 - a^3))/(8*b^4) - (x*(15*a*b^4 + 8*a^5 - 20*a^3*b^2) )/(8*b^6) + (exp(x)*(8*a^4 + 11*b^4 - 18*a^2*b^2))/(16*b^5) + (a*exp(-4*x) )/(64*b^2) - (a*exp(4*x))/(64*b^2) - (exp(-x)*(8*a^4 + 11*b^4 - 18*a^2*b^2 ))/(16*b^5) - (exp(-3*x)*(4*a^2 - 7*b^2))/(96*b^3) + (exp(3*x)*(4*a^2 - 7* b^2))/(96*b^3) + (log(- (2*exp(x)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/b^7 - (2*(a + b)^(5/2)*(b + a*exp(x))*(a - b)^(5/2))/b^7)*(a + b)^(5/2)*(a - b)^(5/2))/b^6 - (log((2*(a + b)^(5/2)*(b + a*exp(x))*(a - b)^(5/2))/b^7 - (2*exp(x)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))/b^7)*(a + b)^(5/2)*(a - b)^ (5/2))/b^6