3.644 \(\int (a \coth (x)+b \text {csch}(x))^5 \, dx\)

Optimal. Leaf size=124 \[ a^5 \log (\sinh (x))+\frac {1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac {1}{8} \text {csch}^2(x) (a \cosh (x)+b)^2 \left (b \left (5 a^2-3 b^2\right ) \cosh (x)+2 a \left (2 a^2-b^2\right )\right )-\frac {1}{8} b \left (15 a^4-10 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\cosh (x))-\frac {1}{4} \text {csch}^4(x) (a \cosh (x)+b)^4 (a+b \cosh (x)) \]

[Out]

-1/8*b*(15*a^4-10*a^2*b^2+3*b^4)*arctanh(cosh(x))+1/8*a^2*b*(7*a^2-3*b^2)*cosh(x)-1/8*(b+a*cosh(x))^2*(2*a*(2*
a^2-b^2)+b*(5*a^2-3*b^2)*cosh(x))*csch(x)^2-1/4*(b+a*cosh(x))^4*(a+b*cosh(x))*csch(x)^4+a^5*ln(sinh(x))

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Rubi [A]  time = 0.24, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {4392, 2668, 739, 819, 774, 635, 204, 260} \[ \frac {1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac {1}{8} b \left (-10 a^2 b^2+15 a^4+3 b^4\right ) \tanh ^{-1}(\cosh (x))-\frac {1}{8} \text {csch}^2(x) (a \cosh (x)+b)^2 \left (b \left (5 a^2-3 b^2\right ) \cosh (x)+2 a \left (2 a^2-b^2\right )\right )+a^5 \log (\sinh (x))-\frac {1}{4} \text {csch}^4(x) (a \cosh (x)+b)^4 (a+b \cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(a*Coth[x] + b*Csch[x])^5,x]

[Out]

-(b*(15*a^4 - 10*a^2*b^2 + 3*b^4)*ArcTanh[Cosh[x]])/8 + (a^2*b*(7*a^2 - 3*b^2)*Cosh[x])/8 - ((b + a*Cosh[x])^2
*(2*a*(2*a^2 - b^2) + b*(5*a^2 - 3*b^2)*Cosh[x])*Csch[x]^2)/8 - ((b + a*Cosh[x])^4*(a + b*Cosh[x])*Csch[x]^4)/
4 + a^5*Log[Sinh[x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int (a \coth (x)+b \text {csch}(x))^5 \, dx &=-\left (i \int (i b+i a \cosh (x))^5 \text {csch}^5(x) \, dx\right )\\ &=-\left (a^5 \operatorname {Subst}\left (\int \frac {(i b+x)^5}{\left (-a^2-x^2\right )^3} \, dx,x,i a \cosh (x)\right )\right )\\ &=-\frac {1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text {csch}^4(x)-\frac {1}{4} a^3 \operatorname {Subst}\left (\int \frac {(i b+x)^3 \left (-4 a^2+3 b^2+i b x\right )}{\left (-a^2-x^2\right )^2} \, dx,x,i a \cosh (x)\right )\\ &=-\frac {1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text {csch}^2(x)-\frac {1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text {csch}^4(x)-\frac {1}{8} a \operatorname {Subst}\left (\int \frac {(i b+x) \left (8 a^4-7 a^2 b^2+3 b^4-i b \left (7 a^2-3 b^2\right ) x\right )}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )\\ &=\frac {1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac {1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text {csch}^2(x)-\frac {1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text {csch}^4(x)+\frac {1}{8} a \operatorname {Subst}\left (\int \frac {-i a^2 b \left (7 a^2-3 b^2\right )-i b \left (8 a^4-7 a^2 b^2+3 b^4\right )-\left (8 a^4-7 a^2 b^2+3 b^4+b^2 \left (7 a^2-3 b^2\right )\right ) x}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )\\ &=\frac {1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac {1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text {csch}^2(x)-\frac {1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text {csch}^4(x)-a^5 \operatorname {Subst}\left (\int \frac {x}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )-\frac {1}{8} \left (i a b \left (15 a^4-10 a^2 b^2+3 b^4\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-a^2-x^2} \, dx,x,i a \cosh (x)\right )\\ &=-\frac {1}{8} b \left (15 a^4-10 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\cosh (x))+\frac {1}{8} a^2 b \left (7 a^2-3 b^2\right ) \cosh (x)-\frac {1}{8} (b+a \cosh (x))^2 \left (2 a \left (2 a^2-b^2\right )+b \left (5 a^2-3 b^2\right ) \cosh (x)\right ) \text {csch}^2(x)-\frac {1}{4} (b+a \cosh (x))^4 (a+b \cosh (x)) \text {csch}^4(x)+a^5 \log (\sinh (x))\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 244, normalized size = 1.97 \[ -\frac {1}{64} \text {csch}^4(x) \left (-24 a^5 \log (\sinh (x))-8 a^5 \cosh (4 x) \log (\sinh (x))-16 a^5+50 a^4 b \cosh (3 x)-45 a^4 b \log \left (\tanh \left (\frac {x}{2}\right )\right )-15 a^4 b \cosh (4 x) \log \left (\tanh \left (\frac {x}{2}\right )\right )+20 a^2 b^3 \cosh (3 x)+30 a^2 b^3 \log \left (\tanh \left (\frac {x}{2}\right )\right )+10 a^2 b^3 \cosh (4 x) \log \left (\tanh \left (\frac {x}{2}\right )\right )+2 b \left (15 a^4+70 a^2 b^2+11 b^4\right ) \cosh (x)+4 \cosh (2 x) \left (8 a^5 \log (\sinh (x))+8 \left (a^5+5 a^3 b^2\right )+b \left (15 a^4-10 a^2 b^2+3 b^4\right ) \log \left (\tanh \left (\frac {x}{2}\right )\right )\right )+80 a b^4-6 b^5 \cosh (3 x)-9 b^5 \log \left (\tanh \left (\frac {x}{2}\right )\right )-3 b^5 \cosh (4 x) \log \left (\tanh \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Coth[x] + b*Csch[x])^5,x]

[Out]

-1/64*(Csch[x]^4*(-16*a^5 + 80*a*b^4 + 2*b*(15*a^4 + 70*a^2*b^2 + 11*b^4)*Cosh[x] + 50*a^4*b*Cosh[3*x] + 20*a^
2*b^3*Cosh[3*x] - 6*b^5*Cosh[3*x] - 24*a^5*Log[Sinh[x]] - 8*a^5*Cosh[4*x]*Log[Sinh[x]] - 45*a^4*b*Log[Tanh[x/2
]] + 30*a^2*b^3*Log[Tanh[x/2]] - 9*b^5*Log[Tanh[x/2]] - 15*a^4*b*Cosh[4*x]*Log[Tanh[x/2]] + 10*a^2*b^3*Cosh[4*
x]*Log[Tanh[x/2]] - 3*b^5*Cosh[4*x]*Log[Tanh[x/2]] + 4*Cosh[2*x]*(8*(a^5 + 5*a^3*b^2) + 8*a^5*Log[Sinh[x]] + b
*(15*a^4 - 10*a^2*b^2 + 3*b^4)*Log[Tanh[x/2]])))

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fricas [B]  time = 0.47, size = 2716, normalized size = 21.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))^5,x, algorithm="fricas")

[Out]

-1/8*(8*a^5*x*cosh(x)^8 + 8*a^5*x*sinh(x)^8 + 2*(25*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^7 + 2*(32*a^5*x*cosh(x
) + 25*a^4*b + 10*a^2*b^3 - 3*b^5)*sinh(x)^7 - 32*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x)^6 + 2*(112*a^5*x*cosh(x)^2
- 16*a^5*x + 16*a^5 + 80*a^3*b^2 + 7*(25*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x))*sinh(x)^6 + 8*a^5*x + 2*(15*a^4
*b + 70*a^2*b^3 + 11*b^5)*cosh(x)^5 + 2*(224*a^5*x*cosh(x)^3 + 15*a^4*b + 70*a^2*b^3 + 11*b^5 + 21*(25*a^4*b +
10*a^2*b^3 - 3*b^5)*cosh(x)^2 - 96*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x))*sinh(x)^5 + 16*(3*a^5*x - 2*a^5 + 10*a*
b^4)*cosh(x)^4 + 2*(280*a^5*x*cosh(x)^4 + 24*a^5*x - 16*a^5 + 80*a*b^4 + 35*(25*a^4*b + 10*a^2*b^3 - 3*b^5)*co
sh(x)^3 - 240*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x)^2 + 5*(15*a^4*b + 70*a^2*b^3 + 11*b^5)*cosh(x))*sinh(x)^4 + 2*
(15*a^4*b + 70*a^2*b^3 + 11*b^5)*cosh(x)^3 + 2*(224*a^5*x*cosh(x)^5 + 15*a^4*b + 70*a^2*b^3 + 11*b^5 + 35*(25*
a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^4 - 320*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x)^3 + 10*(15*a^4*b + 70*a^2*b^3 +
11*b^5)*cosh(x)^2 + 32*(3*a^5*x - 2*a^5 + 10*a*b^4)*cosh(x))*sinh(x)^3 - 32*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x)^
2 + 2*(112*a^5*x*cosh(x)^6 - 16*a^5*x + 21*(25*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^5 + 16*a^5 + 80*a^3*b^2 - 2
40*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x)^4 + 10*(15*a^4*b + 70*a^2*b^3 + 11*b^5)*cosh(x)^3 + 48*(3*a^5*x - 2*a^5 +
10*a*b^4)*cosh(x)^2 + 3*(15*a^4*b + 70*a^2*b^3 + 11*b^5)*cosh(x))*sinh(x)^2 + 2*(25*a^4*b + 10*a^2*b^3 - 3*b^
5)*cosh(x) - ((8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^8 + 8*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*co
sh(x)*sinh(x)^7 + (8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*sinh(x)^8 - 4*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5
)*cosh(x)^6 - 4*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5 - 7*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^2)*
sinh(x)^6 + 8*(7*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^3 - 3*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)
*cosh(x))*sinh(x)^5 + 8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5 + 6*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x
)^4 + 2*(24*a^5 - 45*a^4*b + 30*a^2*b^3 - 9*b^5 + 35*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^4 - 30*(8
*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^2)*sinh(x)^4 + 8*(7*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh
(x)^5 - 10*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^3 + 3*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(
x))*sinh(x)^3 - 4*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^2 + 4*(7*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*
b^5)*cosh(x)^6 - 8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5 - 15*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^4
+ 9*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^2)*sinh(x)^2 + 8*((8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*
cosh(x)^7 - 3*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x)^5 + 3*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*co
sh(x)^3 - (8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) + 1) - ((8*a^5 + 15*
a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^8 + 8*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)*sinh(x)^7 + (8*a^5 +
15*a^4*b - 10*a^2*b^3 + 3*b^5)*sinh(x)^8 - 4*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^6 - 4*(8*a^5 + 1
5*a^4*b - 10*a^2*b^3 + 3*b^5 - 7*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^2)*sinh(x)^6 + 8*(7*(8*a^5 +
15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^3 - 3*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x))*sinh(x)^5 + 8*a^
5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5 + 6*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^4 + 2*(24*a^5 + 45*a^4*b
- 30*a^2*b^3 + 9*b^5 + 35*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^4 - 30*(8*a^5 + 15*a^4*b - 10*a^2*b
^3 + 3*b^5)*cosh(x)^2)*sinh(x)^4 + 8*(7*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^5 - 10*(8*a^5 + 15*a^4
*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^3 + 3*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x))*sinh(x)^3 - 4*(8*a^5 +
15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^2 + 4*(7*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^6 - 8*a^5 - 1
5*a^4*b + 10*a^2*b^3 - 3*b^5 - 15*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^4 + 9*(8*a^5 + 15*a^4*b - 10
*a^2*b^3 + 3*b^5)*cosh(x)^2)*sinh(x)^2 + 8*((8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^7 - 3*(8*a^5 + 15*
a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^5 + 3*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b^5)*cosh(x)^3 - (8*a^5 + 15*a^4*
b - 10*a^2*b^3 + 3*b^5)*cosh(x))*sinh(x))*log(cosh(x) + sinh(x) - 1) + 2*(32*a^5*x*cosh(x)^7 + 7*(25*a^4*b + 1
0*a^2*b^3 - 3*b^5)*cosh(x)^6 - 96*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x)^5 + 25*a^4*b + 10*a^2*b^3 - 3*b^5 + 5*(15*
a^4*b + 70*a^2*b^3 + 11*b^5)*cosh(x)^4 + 32*(3*a^5*x - 2*a^5 + 10*a*b^4)*cosh(x)^3 + 3*(15*a^4*b + 70*a^2*b^3
+ 11*b^5)*cosh(x)^2 - 32*(a^5*x - a^5 - 5*a^3*b^2)*cosh(x))*sinh(x))/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x
)^8 + 4*(7*cosh(x)^2 - 1)*sinh(x)^6 - 4*cosh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*sinh(x)^5 + 2*(35*cosh(x)^4 -
30*cosh(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x
)^6 - 15*cosh(x)^4 + 9*cosh(x)^2 - 1)*sinh(x)^2 - 4*cosh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cos
h(x))*sinh(x) + 1)

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giac [B]  time = 0.14, size = 234, normalized size = 1.89 \[ \frac {1}{16} \, {\left (8 \, a^{5} - 15 \, a^{4} b + 10 \, a^{2} b^{3} - 3 \, b^{5}\right )} \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + \frac {1}{16} \, {\left (8 \, a^{5} + 15 \, a^{4} b - 10 \, a^{2} b^{3} + 3 \, b^{5}\right )} \log \left (e^{\left (-x\right )} + e^{x} - 2\right ) - \frac {3 \, a^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{4} + 25 \, a^{4} b {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} + 10 \, a^{2} b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 3 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 8 \, a^{5} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} + 80 \, a^{3} b^{2} {\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 60 \, a^{4} b {\left (e^{\left (-x\right )} + e^{x}\right )} + 40 \, a^{2} b^{3} {\left (e^{\left (-x\right )} + e^{x}\right )} + 20 \, b^{5} {\left (e^{\left (-x\right )} + e^{x}\right )} - 160 \, a^{3} b^{2} + 80 \, a b^{4}}{4 \, {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))^5,x, algorithm="giac")

[Out]

1/16*(8*a^5 - 15*a^4*b + 10*a^2*b^3 - 3*b^5)*log(e^(-x) + e^x + 2) + 1/16*(8*a^5 + 15*a^4*b - 10*a^2*b^3 + 3*b
^5)*log(e^(-x) + e^x - 2) - 1/4*(3*a^5*(e^(-x) + e^x)^4 + 25*a^4*b*(e^(-x) + e^x)^3 + 10*a^2*b^3*(e^(-x) + e^x
)^3 - 3*b^5*(e^(-x) + e^x)^3 - 8*a^5*(e^(-x) + e^x)^2 + 80*a^3*b^2*(e^(-x) + e^x)^2 - 60*a^4*b*(e^(-x) + e^x)
+ 40*a^2*b^3*(e^(-x) + e^x) + 20*b^5*(e^(-x) + e^x) - 160*a^3*b^2 + 80*a*b^4)/((e^(-x) + e^x)^2 - 4)^2

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maple [A]  time = 0.38, size = 201, normalized size = 1.62 \[ a^{5} \ln \left (\sinh \relax (x )\right )-\frac {a^{5} \left (\coth ^{2}\relax (x )\right )}{2}-\frac {a^{5} \left (\coth ^{4}\relax (x )\right )}{4}-\frac {5 a^{4} b \left (\cosh ^{3}\relax (x )\right )}{\sinh \relax (x )^{4}}+\frac {5 a^{4} b \cosh \relax (x )}{\sinh \relax (x )^{4}}-\frac {5 a^{4} b \coth \relax (x ) \mathrm {csch}\relax (x )^{3}}{4}+\frac {15 a^{4} b \,\mathrm {csch}\relax (x ) \coth \relax (x )}{8}-\frac {15 a^{4} b \arctanh \left ({\mathrm e}^{x}\right )}{4}-\frac {5 a^{3} b^{2} \left (\cosh ^{2}\relax (x )\right )}{\sinh \relax (x )^{4}}+\frac {5 a^{3} b^{2}}{2 \sinh \relax (x )^{4}}-\frac {10 a^{2} b^{3} \cosh \relax (x )}{3 \sinh \relax (x )^{4}}+\frac {5 a^{2} b^{3} \coth \relax (x ) \mathrm {csch}\relax (x )^{3}}{6}-\frac {5 a^{2} b^{3} \mathrm {csch}\relax (x ) \coth \relax (x )}{4}+\frac {5 a^{2} b^{3} \arctanh \left ({\mathrm e}^{x}\right )}{2}-\frac {5 a \,b^{4}}{4 \sinh \relax (x )^{4}}-\frac {b^{5} \coth \relax (x ) \mathrm {csch}\relax (x )^{3}}{4}+\frac {3 b^{5} \mathrm {csch}\relax (x ) \coth \relax (x )}{8}-\frac {3 b^{5} \arctanh \left ({\mathrm e}^{x}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*coth(x)+b*csch(x))^5,x)

[Out]

a^5*ln(sinh(x))-1/2*a^5*coth(x)^2-1/4*a^5*coth(x)^4-5*a^4*b/sinh(x)^4*cosh(x)^3+5*a^4*b/sinh(x)^4*cosh(x)-5/4*
a^4*b*coth(x)*csch(x)^3+15/8*a^4*b*csch(x)*coth(x)-15/4*a^4*b*arctanh(exp(x))-5*a^3*b^2/sinh(x)^4*cosh(x)^2+5/
2*a^3*b^2/sinh(x)^4-10/3*a^2*b^3/sinh(x)^4*cosh(x)+5/6*a^2*b^3*coth(x)*csch(x)^3-5/4*a^2*b^3*csch(x)*coth(x)+5
/2*a^2*b^3*arctanh(exp(x))-5/4*a*b^4/sinh(x)^4-1/4*b^5*coth(x)*csch(x)^3+3/8*b^5*csch(x)*coth(x)-3/4*b^5*arcta
nh(exp(x))

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maxima [B]  time = 0.39, size = 330, normalized size = 2.66 \[ -\frac {5}{2} \, a^{3} b^{2} \coth \relax (x)^{4} + a^{5} {\left (x + \frac {4 \, {\left (e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + \log \left (e^{\left (-x\right )} + 1\right ) + \log \left (e^{\left (-x\right )} - 1\right )\right )} + \frac {5}{8} \, a^{4} b {\left (\frac {2 \, {\left (5 \, e^{\left (-x\right )} + 3 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-5 \, x\right )} + 5 \, e^{\left (-7 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} - 3 \, \log \left (e^{\left (-x\right )} + 1\right ) + 3 \, \log \left (e^{\left (-x\right )} - 1\right )\right )} - \frac {1}{8} \, b^{5} {\left (\frac {2 \, {\left (3 \, e^{\left (-x\right )} - 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} + 3 \, e^{\left (-7 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + 3 \, \log \left (e^{\left (-x\right )} + 1\right ) - 3 \, \log \left (e^{\left (-x\right )} - 1\right )\right )} + \frac {5}{4} \, a^{2} b^{3} {\left (\frac {2 \, {\left (e^{\left (-x\right )} + 7 \, e^{\left (-3 \, x\right )} + 7 \, e^{\left (-5 \, x\right )} + e^{\left (-7 \, x\right )}\right )}}{4 \, e^{\left (-2 \, x\right )} - 6 \, e^{\left (-4 \, x\right )} + 4 \, e^{\left (-6 \, x\right )} - e^{\left (-8 \, x\right )} - 1} + \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right )\right )} - \frac {20 \, a b^{4}}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))^5,x, algorithm="maxima")

[Out]

-5/2*a^3*b^2*coth(x)^4 + a^5*(x + 4*(e^(-2*x) - e^(-4*x) + e^(-6*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e
^(-8*x) - 1) + log(e^(-x) + 1) + log(e^(-x) - 1)) + 5/8*a^4*b*(2*(5*e^(-x) + 3*e^(-3*x) + 3*e^(-5*x) + 5*e^(-7
*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1) - 3*log(e^(-x) + 1) + 3*log(e^(-x) - 1)) - 1/8*b^5*
(2*(3*e^(-x) - 11*e^(-3*x) - 11*e^(-5*x) + 3*e^(-7*x))/(4*e^(-2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1) +
3*log(e^(-x) + 1) - 3*log(e^(-x) - 1)) + 5/4*a^2*b^3*(2*(e^(-x) + 7*e^(-3*x) + 7*e^(-5*x) + e^(-7*x))/(4*e^(-
2*x) - 6*e^(-4*x) + 4*e^(-6*x) - e^(-8*x) - 1) + log(e^(-x) + 1) - log(e^(-x) - 1)) - 20*a*b^4/(e^(-x) - e^x)^
4

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mupad [B]  time = 0.20, size = 392, normalized size = 3.16 \[ \ln \left (\frac {15\,a^4\,b}{4}+\frac {3\,b^5}{4}-\frac {5\,a^2\,b^3}{2}-\frac {3\,b^5\,{\mathrm {e}}^x}{4}-\frac {15\,a^4\,b\,{\mathrm {e}}^x}{4}+\frac {5\,a^2\,b^3\,{\mathrm {e}}^x}{2}\right )\,\left (a^5+\frac {15\,a^4\,b}{8}-\frac {5\,a^2\,b^3}{4}+\frac {3\,b^5}{8}\right )-\frac {{\mathrm {e}}^x\,\left (20\,a^4\,b+40\,a^2\,b^3+4\,b^5\right )+20\,a\,b^4+4\,a^5+40\,a^3\,b^2}{6\,{\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^{6\,x}+{\mathrm {e}}^{8\,x}+1}-\frac {{\mathrm {e}}^x\,\left (30\,a^4\,b+60\,a^2\,b^3+6\,b^5\right )+40\,a\,b^4+8\,a^5+80\,a^3\,b^2}{3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1}-a^5\,x-\ln \left (\frac {5\,a^2\,b^3}{2}-\frac {3\,b^5}{4}-\frac {15\,a^4\,b}{4}-\frac {3\,b^5\,{\mathrm {e}}^x}{4}-\frac {15\,a^4\,b\,{\mathrm {e}}^x}{4}+\frac {5\,a^2\,b^3\,{\mathrm {e}}^x}{2}\right )\,\left (-a^5+\frac {15\,a^4\,b}{8}-\frac {5\,a^2\,b^3}{4}+\frac {3\,b^5}{8}\right )-\frac {{\mathrm {e}}^x\,\left (\frac {25\,a^4\,b}{4}+\frac {5\,a^2\,b^3}{2}-\frac {3\,b^5}{4}\right )+4\,a^5+20\,a^3\,b^2}{{\mathrm {e}}^{2\,x}-1}-\frac {{\mathrm {e}}^x\,\left (\frac {45\,a^4\,b}{2}+25\,a^2\,b^3+\frac {b^5}{2}\right )+20\,a\,b^4+8\,a^5+60\,a^3\,b^2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/sinh(x) + a*coth(x))^5,x)

[Out]

log((15*a^4*b)/4 + (3*b^5)/4 - (5*a^2*b^3)/2 - (3*b^5*exp(x))/4 - (15*a^4*b*exp(x))/4 + (5*a^2*b^3*exp(x))/2)*
((15*a^4*b)/8 + a^5 + (3*b^5)/8 - (5*a^2*b^3)/4) - (exp(x)*(20*a^4*b + 4*b^5 + 40*a^2*b^3) + 20*a*b^4 + 4*a^5
+ 40*a^3*b^2)/(6*exp(4*x) - 4*exp(2*x) - 4*exp(6*x) + exp(8*x) + 1) - (exp(x)*(30*a^4*b + 6*b^5 + 60*a^2*b^3)
+ 40*a*b^4 + 8*a^5 + 80*a^3*b^2)/(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1) - a^5*x - log((5*a^2*b^3)/2 - (3*b^5
)/4 - (15*a^4*b)/4 - (3*b^5*exp(x))/4 - (15*a^4*b*exp(x))/4 + (5*a^2*b^3*exp(x))/2)*((15*a^4*b)/8 - a^5 + (3*b
^5)/8 - (5*a^2*b^3)/4) - (exp(x)*((25*a^4*b)/4 - (3*b^5)/4 + (5*a^2*b^3)/2) + 4*a^5 + 20*a^3*b^2)/(exp(2*x) -
1) - (exp(x)*((45*a^4*b)/2 + b^5/2 + 25*a^2*b^3) + 20*a*b^4 + 8*a^5 + 60*a^3*b^2)/(exp(4*x) - 2*exp(2*x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \coth {\relax (x )} + b \operatorname {csch}{\relax (x )}\right )^{5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*coth(x)+b*csch(x))**5,x)

[Out]

Integral((a*coth(x) + b*csch(x))**5, x)

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