3.166 \(\int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=154 \[ \frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{8 b^5}+\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^3}-\frac {a x \left (8 a^4-20 a^2 b^2+15 b^4\right )}{8 b^6}+\frac {2 (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^6}+\frac {\sinh ^5(x)}{5 b} \]

[Out]

-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

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Rubi [A]  time = 0.43, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2695, 2865, 2735, 2659, 208} \[ -\frac {a x \left (-20 a^2 b^2+8 a^4+15 b^4\right )}{8 b^6}+\frac {\sinh ^3(x) \left (4 \left (a^2-b^2\right )-3 a b \cosh (x)\right )}{12 b^3}+\frac {\sinh (x) \left (8 \left (a^2-b^2\right )^2-a b \left (4 a^2-7 b^2\right ) \cosh (x)\right )}{8 b^5}+\frac {2 (a-b)^{5/2} (a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^6}+\frac {\sinh ^5(x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^6/(a + b*Cosh[x]),x]

[Out]

-(a*(8*a^4 - 20*a^2*b^2 + 15*b^4)*x)/(8*b^6) + (2*(a - b)^(5/2)*(a + b)^(5/2)*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/
Sqrt[a + b]])/b^6 + ((8*(a^2 - b^2)^2 - a*b*(4*a^2 - 7*b^2)*Cosh[x])*Sinh[x])/(8*b^5) + ((4*(a^2 - b^2) - 3*a*
b*Cosh[x])*Sinh[x]^3)/(12*b^3) + Sinh[x]^5/(5*b)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(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]

Rule 2695

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] + Dist[(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] /; FreeQ[{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]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(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]

Rule 2865

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] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[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]

Rubi steps

\begin {align*} \int \frac {\sinh ^6(x)}{a+b \cosh (x)} \, dx &=\frac {\sinh ^5(x)}{5 b}+\frac {\int \frac {(-b-a \cosh (x)) \sinh ^4(x)}{a+b \cosh (x)} \, dx}{b}\\ &=\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}-\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^3}\\ &=\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}+\frac {\int \frac {-b \left (4 a^4-9 a^2 b^2+8 b^4\right )-a \left (8 a^4-20 a^2 b^2+15 b^4\right ) \cosh (x)}{a+b \cosh (x)} \, dx}{8 b^5}\\ &=-\frac {a \left (8 a^4-20 a^2 b^2+15 b^4\right ) x}{8 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}+\frac {\left (a^2-b^2\right )^3 \int \frac {1}{a+b \cosh (x)} \, dx}{b^6}\\ &=-\frac {a \left (8 a^4-20 a^2 b^2+15 b^4\right ) x}{8 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}+\frac {\left (2 \left (a^2-b^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^6}\\ &=-\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} \tanh ^{-1}\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}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 154, normalized size = 1.00 \[ \frac {-120 a b^2 \left (a^2-2 b^2\right ) \sinh (2 x)+960 \left (b^2-a^2\right )^{5/2} \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )-10 b^3 \left (7 b^2-4 a^2\right ) \sinh (3 x)-60 a x \left (8 a^4-20 a^2 b^2+15 b^4\right )+60 b \left (8 a^4-18 a^2 b^2+11 b^4\right ) \sinh (x)-15 a b^4 \sinh (4 x)+6 b^5 \sinh (5 x)}{480 b^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^6/(a + b*Cosh[x]),x]

[Out]

(-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)*Sinh[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)

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fricas [B]  time = 1.51, size = 2913, normalized size = 18.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^6/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[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*cosh(x)^4 + 280*(4*a^2*b^3 - 7*b^5)*cosh(x)^3 - 1260*(a^3*b^2 - 2*a*b^4)*cosh(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) + 9
0*(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)*cosh(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)*c
osh(x)^4 - 60*(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)^3 - 10*(4*a^2*b^3 - 7*b^5)*cosh(x)^2 + 10*(27*b^5*cosh(x)^8 -
54*a*b^4*cosh(x)^7 + 28*(4*a^2*b^3 - 7*b^5)*cosh(x)^6 - 252*(a^3*b^2 - 2*a*b^4)*cosh(x)^5 - 4*a^2*b^3 + 7*b^5
- 120*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x*cosh(x)^3 + 90*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x)^4 - 36*(8*a^4*
b - 18*a^2*b^3 + 11*b^5)*cosh(x)^2 + 36*(a^3*b^2 - 2*a*b^4)*cosh(x))*sinh(x)^2 + 960*((a^4 - 2*a^2*b^2 + b^4)*
cosh(x)^5 + 5*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^4*sinh(x) + 10*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3*sinh(x)^2 + 10*
(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^2*sinh(x)^3 + 5*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^4 + (a^4 - 2*a^2*b^2 +
b^4)*sinh(x)^5)*sqrt(a^2 - b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 - b^2 + 2*(b^2*cos
h(x) + a*b)*sinh(x) - 2*sqrt(a^2 - b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x)
+ 2*(b*cosh(x) + a)*sinh(x) + b)) + 5*(12*b^5*cosh(x)^9 - 27*a*b^4*cosh(x)^8 + 16*(4*a^2*b^3 - 7*b^5)*cosh(x)^
7 - 168*(a^3*b^2 - 2*a*b^4)*cosh(x)^6 - 120*(8*a^5 - 20*a^3*b^2 + 15*a*b^4)*x*cosh(x)^4 + 72*(8*a^4*b - 18*a^2
*b^3 + 11*b^5)*cosh(x)^5 + 3*a*b^4 - 48*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x)^3 + 72*(a^3*b^2 - 2*a*b^4)*cos
h(x)^2 - 4*(4*a^2*b^3 - 7*b^5)*cosh(x))*sinh(x))/(b^6*cosh(x)^5 + 5*b^6*cosh(x)^4*sinh(x) + 10*b^6*cosh(x)^3*s
inh(x)^2 + 10*b^6*cosh(x)^2*sinh(x)^3 + 5*b^6*cosh(x)*sinh(x)^4 + b^6*sinh(x)^5), 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*c
osh(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*cosh(x)^4 + 280*(4*a^2*b
^3 - 7*b^5)*cosh(x)^3 - 1260*(a^3*b^2 - 2*a*b^4)*cosh(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)*cosh(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 - 60*(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)^3 - 10*(4*a^2*b^3 - 7*b^5)*cosh(x)^2 + 10*(27*b^5*cosh(x)^8 - 54*a*b^4*cosh(x)^7 + 28*(4*a
^2*b^3 - 7*b^5)*cosh(x)^6 - 252*(a^3*b^2 - 2*a*b^4)*cosh(x)^5 - 4*a^2*b^3 + 7*b^5 - 120*(8*a^5 - 20*a^3*b^2 +
15*a*b^4)*x*cosh(x)^3 + 90*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x)^4 - 36*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh
(x)^2 + 36*(a^3*b^2 - 2*a*b^4)*cosh(x))*sinh(x)^2 - 1920*((a^4 - 2*a^2*b^2 + b^4)*cosh(x)^5 + 5*(a^4 - 2*a^2*b
^2 + b^4)*cosh(x)^4*sinh(x) + 10*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)^3*sinh(x)^2 + 10*(a^4 - 2*a^2*b^2 + b^4)*cosh
(x)^2*sinh(x)^3 + 5*(a^4 - 2*a^2*b^2 + b^4)*cosh(x)*sinh(x)^4 + (a^4 - 2*a^2*b^2 + b^4)*sinh(x)^5)*sqrt(-a^2 +
b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 5*(12*b^5*cosh(x)^9 - 27*a*b^4*cosh(
x)^8 + 16*(4*a^2*b^3 - 7*b^5)*cosh(x)^7 - 168*(a^3*b^2 - 2*a*b^4)*cosh(x)^6 - 120*(8*a^5 - 20*a^3*b^2 + 15*a*b
^4)*x*cosh(x)^4 + 72*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*cosh(x)^5 + 3*a*b^4 - 48*(8*a^4*b - 18*a^2*b^3 + 11*b^5)*
cosh(x)^3 + 72*(a^3*b^2 - 2*a*b^4)*cosh(x)^2 - 4*(4*a^2*b^3 - 7*b^5)*cosh(x))*sinh(x))/(b^6*cosh(x)^5 + 5*b^6*
cosh(x)^4*sinh(x) + 10*b^6*cosh(x)^3*sinh(x)^2 + 10*b^6*cosh(x)^2*sinh(x)^3 + 5*b^6*cosh(x)*sinh(x)^4 + b^6*si
nh(x)^5)]

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giac [A]  time = 0.16, size = 266, normalized size = 1.73 \[ \frac {6 \, b^{4} e^{\left (5 \, x\right )} - 15 \, a b^{3} e^{\left (4 \, x\right )} + 40 \, a^{2} b^{2} e^{\left (3 \, x\right )} - 70 \, b^{4} e^{\left (3 \, x\right )} - 120 \, a^{3} b e^{\left (2 \, x\right )} + 240 \, a b^{3} e^{\left (2 \, x\right )} + 480 \, a^{4} e^{x} - 1080 \, a^{2} b^{2} e^{x} + 660 \, b^{4} e^{x}}{960 \, b^{5}} - \frac {{\left (8 \, a^{5} - 20 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x}{8 \, b^{6}} + \frac {{\left (15 \, a b^{4} e^{x} - 6 \, b^{5} - 60 \, {\left (8 \, a^{4} b - 18 \, a^{2} b^{3} + 11 \, b^{5}\right )} e^{\left (4 \, x\right )} + 120 \, {\left (a^{3} b^{2} - 2 \, a b^{4}\right )} e^{\left (3 \, x\right )} - 10 \, {\left (4 \, a^{2} b^{3} - 7 \, b^{5}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-5 \, x\right )}}{960 \, b^{6}} + \frac {2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^6/(a+b*cosh(x)),x, algorithm="giac")

[Out]

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 + 1/96
0*(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)

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maple [B]  time = 0.08, size = 679, normalized size = 4.41 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^6/(a+b*cosh(x)),x)

[Out]

-1/5/b/(tanh(1/2*x)-1)^5-1/2/b/(tanh(1/2*x)-1)^4-1/5/b/(tanh(1/2*x)+1)^5+1/2/b/(tanh(1/2*x)+1)^4+1/12/b/(tanh(
1/2*x)-1)^3+5/8/b/(tanh(1/2*x)-1)^2-1/b/(tanh(1/2*x)-1)+1/12/b/(tanh(1/2*x)+1)^3-5/8/b/(tanh(1/2*x)+1)^2-1/b/(
tanh(1/2*x)+1)+2/b^6/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*a^6-6/b^4/((a+b)*(a-b)
)^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))*a^4+6/b^2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)
/((a+b)*(a-b))^(1/2))*a^2-1/b^5/(tanh(1/2*x)+1)*a^4-1/2/b^4/(tanh(1/2*x)+1)*a^3+1/4/b^2/(tanh(1/2*x)+1)^4*a-1/
3/b^3/(tanh(1/2*x)+1)^3*a^2-1/2/b^2/(tanh(1/2*x)+1)^3*a+1/2/b^4/(tanh(1/2*x)+1)^2*a^3+1/2/b^3/(tanh(1/2*x)+1)^
2*a^2-1/3/b^3/(tanh(1/2*x)-1)^3*a^2-1/2/b^2/(tanh(1/2*x)-1)^3*a-1/2/b^4/(tanh(1/2*x)-1)^2*a^3-1/2/b^3/(tanh(1/
2*x)-1)^2*a^2-1/b^5/(tanh(1/2*x)-1)*a^4-1/2/b^4/(tanh(1/2*x)-1)*a^3-1/4/b^2/(tanh(1/2*x)-1)^4*a+7/8/b^2/(tanh(
1/2*x)+1)*a+5/2*a^3/b^4*ln(tanh(1/2*x)+1)-15/8*a/b^2*ln(tanh(1/2*x)+1)+5/8/b^2/(tanh(1/2*x)-1)^2*a+2/b^3/(tanh
(1/2*x)-1)*a^2+7/8/b^2/(tanh(1/2*x)-1)*a-5/2*a^3/b^4*ln(tanh(1/2*x)-1)+15/8*a/b^2*ln(tanh(1/2*x)-1)-5/8/b^2/(t
anh(1/2*x)+1)^2*a+2/b^3/(tanh(1/2*x)+1)*a^2-2/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2
))-a^5/b^6*ln(tanh(1/2*x)+1)+a^5/b^6*ln(tanh(1/2*x)-1)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^6/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.70, size = 348, normalized size = 2.26 \[ \frac {{\mathrm {e}}^{5\,x}}{160\,b}-\frac {{\mathrm {e}}^{-5\,x}}{160\,b}-\frac {{\mathrm {e}}^{-2\,x}\,\left (2\,a\,b^2-a^3\right )}{8\,b^4}+\frac {{\mathrm {e}}^{2\,x}\,\left (2\,a\,b^2-a^3\right )}{8\,b^4}-\frac {x\,\left (8\,a^5-20\,a^3\,b^2+15\,a\,b^4\right )}{8\,b^6}+\frac {{\mathrm {e}}^x\,\left (8\,a^4-18\,a^2\,b^2+11\,b^4\right )}{16\,b^5}+\frac {a\,{\mathrm {e}}^{-4\,x}}{64\,b^2}-\frac {a\,{\mathrm {e}}^{4\,x}}{64\,b^2}-\frac {{\mathrm {e}}^{-x}\,\left (8\,a^4-18\,a^2\,b^2+11\,b^4\right )}{16\,b^5}-\frac {{\mathrm {e}}^{-3\,x}\,\left (4\,a^2-7\,b^2\right )}{96\,b^3}+\frac {{\mathrm {e}}^{3\,x}\,\left (4\,a^2-7\,b^2\right )}{96\,b^3}+\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{b^7}-\frac {2\,{\left (a+b\right )}^{5/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{5/2}}{b^7}\right )\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}{b^6}-\frac {\ln \left (\frac {2\,{\left (a+b\right )}^{5/2}\,\left (b+a\,{\mathrm {e}}^x\right )\,{\left (a-b\right )}^{5/2}}{b^7}-\frac {2\,{\mathrm {e}}^x\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}{b^7}\right )\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}{b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^6/(a + b*cosh(x)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**6/(a+b*cosh(x)),x)

[Out]

Timed out

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