3.92 \(\int \frac {\cosh ^5(x)}{a+b \text {csch}(x)} \, dx\)

Optimal. Leaf size=102 \[ -\frac {b \sinh ^4(x)}{4 a^2}-\frac {b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)}{a^6}+\frac {\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac {b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac {\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}+\frac {\sinh ^5(x)}{5 a} \]

[Out]

-b*(a^2+b^2)^2*ln(b+a*sinh(x))/a^6+(a^2+b^2)^2*sinh(x)/a^5-1/2*b*(2*a^2+b^2)*sinh(x)^2/a^4+1/3*(2*a^2+b^2)*sin
h(x)^3/a^3-1/4*b*sinh(x)^4/a^2+1/5*sinh(x)^5/a

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Rubi [A]  time = 0.20, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3872, 2837, 12, 772} \[ \frac {\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}-\frac {b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac {\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac {b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)}{a^6}-\frac {b \sinh ^4(x)}{4 a^2}+\frac {\sinh ^5(x)}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^5/(a + b*Csch[x]),x]

[Out]

-((b*(a^2 + b^2)^2*Log[b + a*Sinh[x]])/a^6) + ((a^2 + b^2)^2*Sinh[x])/a^5 - (b*(2*a^2 + b^2)*Sinh[x]^2)/(2*a^4
) + ((2*a^2 + b^2)*Sinh[x]^3)/(3*a^3) - (b*Sinh[x]^4)/(4*a^2) + Sinh[x]^5/(5*a)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\cosh ^5(x)}{a+b \text {csch}(x)} \, dx &=i \int \frac {\cosh ^5(x) \sinh (x)}{i b+i a \sinh (x)} \, dx\\ &=-\frac {i \operatorname {Subst}\left (\int \frac {x \left (a^2-x^2\right )^2}{a (i b+x)} \, dx,x,i a \sinh (x)\right )}{a^5}\\ &=-\frac {i \operatorname {Subst}\left (\int \frac {x \left (a^2-x^2\right )^2}{i b+x} \, dx,x,i a \sinh (x)\right )}{a^6}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (\left (a^2+b^2\right )^2-\frac {b \left (a^2+b^2\right )^2}{b-i x}+i b \left (2 a^2+b^2\right ) x-\left (2 a^2+b^2\right ) x^2-i b x^3+x^4\right ) \, dx,x,i a \sinh (x)\right )}{a^6}\\ &=-\frac {b \left (a^2+b^2\right )^2 \log (b+a \sinh (x))}{a^6}+\frac {\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac {b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac {\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}-\frac {b \sinh ^4(x)}{4 a^2}+\frac {\sinh ^5(x)}{5 a}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 97, normalized size = 0.95 \[ \frac {12 a^5 \sinh ^5(x)-15 a^4 b \sinh ^4(x)-30 a^2 b \left (2 a^2+b^2\right ) \sinh ^2(x)+60 a \left (a^2+b^2\right )^2 \sinh (x)-60 b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)+20 a^3 \left (2 a^2+b^2\right ) \sinh ^3(x)}{60 a^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*b*(a^2 + b^2)^2*Log[b + a*Sinh[x]] + 60*a*(a^2 + b^2)^2*Sinh[x] - 30*a^2*b*(2*a^2 + b^2)*Sinh[x]^2 + 20*a
^3*(2*a^2 + b^2)*Sinh[x]^3 - 15*a^4*b*Sinh[x]^4 + 12*a^5*Sinh[x]^5)/(60*a^6)

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fricas [B]  time = 0.56, size = 1398, normalized size = 13.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(6*a^5*cosh(x)^10 + 6*a^5*sinh(x)^10 - 15*a^4*b*cosh(x)^9 + 15*(4*a^5*cosh(x) - a^4*b)*sinh(x)^9 + 10*(5
*a^5 + 4*a^3*b^2)*cosh(x)^8 + 5*(54*a^5*cosh(x)^2 - 27*a^4*b*cosh(x) + 10*a^5 + 8*a^3*b^2)*sinh(x)^8 - 60*(3*a
^4*b + 2*a^2*b^3)*cosh(x)^7 + 20*(36*a^5*cosh(x)^3 - 27*a^4*b*cosh(x)^2 - 9*a^4*b - 6*a^2*b^3 + 4*(5*a^5 + 4*a
^3*b^2)*cosh(x))*sinh(x)^7 + 960*(a^4*b + 2*a^2*b^3 + b^5)*x*cosh(x)^5 + 60*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cos
h(x)^6 + 20*(63*a^5*cosh(x)^4 - 63*a^4*b*cosh(x)^3 + 15*a^5 + 42*a^3*b^2 + 24*a*b^4 + 14*(5*a^5 + 4*a^3*b^2)*c
osh(x)^2 - 21*(3*a^4*b + 2*a^2*b^3)*cosh(x))*sinh(x)^6 - 15*a^4*b*cosh(x) + 2*(756*a^5*cosh(x)^5 - 945*a^4*b*c
osh(x)^4 + 280*(5*a^5 + 4*a^3*b^2)*cosh(x)^3 - 630*(3*a^4*b + 2*a^2*b^3)*cosh(x)^2 + 480*(a^4*b + 2*a^2*b^3 +
b^5)*x + 180*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cosh(x))*sinh(x)^5 - 6*a^5 - 60*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cos
h(x)^4 + 10*(126*a^5*cosh(x)^6 - 189*a^4*b*cosh(x)^5 - 30*a^5 - 84*a^3*b^2 - 48*a*b^4 + 70*(5*a^5 + 4*a^3*b^2)
*cosh(x)^4 - 210*(3*a^4*b + 2*a^2*b^3)*cosh(x)^3 + 480*(a^4*b + 2*a^2*b^3 + b^5)*x*cosh(x) + 90*(5*a^5 + 14*a^
3*b^2 + 8*a*b^4)*cosh(x)^2)*sinh(x)^4 - 60*(3*a^4*b + 2*a^2*b^3)*cosh(x)^3 + 20*(36*a^5*cosh(x)^7 - 63*a^4*b*c
osh(x)^6 + 28*(5*a^5 + 4*a^3*b^2)*cosh(x)^5 - 9*a^4*b - 6*a^2*b^3 - 105*(3*a^4*b + 2*a^2*b^3)*cosh(x)^4 + 480*
(a^4*b + 2*a^2*b^3 + b^5)*x*cosh(x)^2 + 60*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cosh(x)^3 - 12*(5*a^5 + 14*a^3*b^2 +
8*a*b^4)*cosh(x))*sinh(x)^3 - 10*(5*a^5 + 4*a^3*b^2)*cosh(x)^2 + 10*(27*a^5*cosh(x)^8 - 54*a^4*b*cosh(x)^7 +
28*(5*a^5 + 4*a^3*b^2)*cosh(x)^6 - 126*(3*a^4*b + 2*a^2*b^3)*cosh(x)^5 - 5*a^5 - 4*a^3*b^2 + 960*(a^4*b + 2*a^
2*b^3 + b^5)*x*cosh(x)^3 + 90*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cosh(x)^4 - 36*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cos
h(x)^2 - 18*(3*a^4*b + 2*a^2*b^3)*cosh(x))*sinh(x)^2 - 960*((a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^5 + 5*(a^4*b + 2
*a^2*b^3 + b^5)*cosh(x)^4*sinh(x) + 10*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x)^3*sinh(x)^2 + 10*(a^4*b + 2*a^2*b^3 +
b^5)*cosh(x)^2*sinh(x)^3 + 5*(a^4*b + 2*a^2*b^3 + b^5)*cosh(x)*sinh(x)^4 + (a^4*b + 2*a^2*b^3 + b^5)*sinh(x)^
5)*log(2*(a*sinh(x) + b)/(cosh(x) - sinh(x))) + 5*(12*a^5*cosh(x)^9 - 27*a^4*b*cosh(x)^8 + 16*(5*a^5 + 4*a^3*b
^2)*cosh(x)^7 - 84*(3*a^4*b + 2*a^2*b^3)*cosh(x)^6 + 960*(a^4*b + 2*a^2*b^3 + b^5)*x*cosh(x)^4 + 72*(5*a^5 + 1
4*a^3*b^2 + 8*a*b^4)*cosh(x)^5 - 3*a^4*b - 48*(5*a^5 + 14*a^3*b^2 + 8*a*b^4)*cosh(x)^3 - 36*(3*a^4*b + 2*a^2*b
^3)*cosh(x)^2 - 4*(5*a^5 + 4*a^3*b^2)*cosh(x))*sinh(x))/(a^6*cosh(x)^5 + 5*a^6*cosh(x)^4*sinh(x) + 10*a^6*cosh
(x)^3*sinh(x)^2 + 10*a^6*cosh(x)^2*sinh(x)^3 + 5*a^6*cosh(x)*sinh(x)^4 + a^6*sinh(x)^5)

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giac [B]  time = 0.13, size = 194, normalized size = 1.90 \[ -\frac {6 \, a^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}^{5} + 15 \, a^{3} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{4} + 80 \, a^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 40 \, a^{2} b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )}^{3} + 240 \, a^{3} b {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 120 \, a b^{3} {\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 480 \, a^{4} {\left (e^{\left (-x\right )} - e^{x}\right )} + 960 \, a^{2} b^{2} {\left (e^{\left (-x\right )} - e^{x}\right )} + 480 \, b^{4} {\left (e^{\left (-x\right )} - e^{x}\right )}}{960 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | -a {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, b \right |}\right )}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(6*a^4*(e^(-x) - e^x)^5 + 15*a^3*b*(e^(-x) - e^x)^4 + 80*a^4*(e^(-x) - e^x)^3 + 40*a^2*b^2*(e^(-x) - e^
x)^3 + 240*a^3*b*(e^(-x) - e^x)^2 + 120*a*b^3*(e^(-x) - e^x)^2 + 480*a^4*(e^(-x) - e^x) + 960*a^2*b^2*(e^(-x)
- e^x) + 480*b^4*(e^(-x) - e^x))/a^5 - (a^4*b + 2*a^2*b^3 + b^5)*log(abs(-a*(e^(-x) - e^x) + 2*b))/a^6

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maple [B]  time = 0.14, size = 600, normalized size = 5.88 \[ -\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {7}{8 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{5 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{5}}-\frac {1}{5 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{5}}+\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {1}{2 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}-\frac {b}{4 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{4}}+\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {b^{4}}{a^{5} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b^{5} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{6}}+\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {b^{2}}{3 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {b^{5} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{6}}-\frac {b}{4 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}-\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b^{4}}{a^{5} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {b^{2}}{3 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {b^{2}}{2 a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b^{3}}{2 a^{4} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {b \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{a^{2}}-\frac {2 b^{3} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{a^{4}}-\frac {b^{5} \ln \left (\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -2 a \tanh \left (\frac {x}{2}\right )-b \right )}{a^{6}}-\frac {11}{12 a \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {11}{12 a \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {9 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {7 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {2 b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{2}}+\frac {2 b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a^{4}}-\frac {9 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {7 b}{8 a^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {2 b^{2}}{a^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{2}}+\frac {2 b^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/a/(tanh(1/2*x)-1)^2-1/a/(tanh(1/2*x)-1)+7/8/a/(tanh(1/2*x)+1)^2-1/a/(tanh(1/2*x)+1)-1/5/a/(tanh(1/2*x)-1)
^5-1/5/a/(tanh(1/2*x)+1)^5+1/2/a/(tanh(1/2*x)+1)^4-1/2/a/(tanh(1/2*x)-1)^4-1/4/a^2/(tanh(1/2*x)-1)^4*b+1/2/a^4
/(tanh(1/2*x)+1)*b^3-1/a^5/(tanh(1/2*x)+1)*b^4+b^5/a^6*ln(tanh(1/2*x)+1)-1/a^2*b*ln(tanh(1/2*x)^2*b-2*a*tanh(1
/2*x)-b)-2/a^4*b^3*ln(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)-1/a^6*b^5*ln(tanh(1/2*x)^2*b-2*a*tanh(1/2*x)-b)+1/2/a
^2/(tanh(1/2*x)+1)^3*b-1/3/a^3/(tanh(1/2*x)+1)^3*b^2+1/2/a^3/(tanh(1/2*x)+1)^2*b^2-1/2/a^4/(tanh(1/2*x)+1)^2*b
^3+b^5/a^6*ln(tanh(1/2*x)-1)-1/4/a^2/(tanh(1/2*x)+1)^4*b-1/2/a^4/(tanh(1/2*x)-1)*b^3-1/a^5/(tanh(1/2*x)-1)*b^4
-1/2/a^2/(tanh(1/2*x)-1)^3*b-1/3/a^3/(tanh(1/2*x)-1)^3*b^2-1/2/a^3/(tanh(1/2*x)-1)^2*b^2-1/2/a^4/(tanh(1/2*x)-
1)^2*b^3-11/12/a/(tanh(1/2*x)-1)^3-11/12/a/(tanh(1/2*x)+1)^3-9/8/a^2/(tanh(1/2*x)-1)^2*b-7/8/a^2/(tanh(1/2*x)-
1)*b-2/a^3/(tanh(1/2*x)-1)*b^2+b/a^2*ln(tanh(1/2*x)-1)+2*b^3/a^4*ln(tanh(1/2*x)-1)-9/8/a^2/(tanh(1/2*x)+1)^2*b
+7/8/a^2/(tanh(1/2*x)+1)*b-2/a^3/(tanh(1/2*x)+1)*b^2+b/a^2*ln(tanh(1/2*x)+1)+2*b^3/a^4*ln(tanh(1/2*x)+1)

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maxima [B]  time = 0.32, size = 242, normalized size = 2.37 \[ -\frac {{\left (15 \, a^{3} b e^{\left (-x\right )} - 6 \, a^{4} - 10 \, {\left (5 \, a^{4} + 4 \, a^{2} b^{2}\right )} e^{\left (-2 \, x\right )} + 60 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} e^{\left (-3 \, x\right )} - 60 \, {\left (5 \, a^{4} + 14 \, a^{2} b^{2} + 8 \, b^{4}\right )} e^{\left (-4 \, x\right )}\right )} e^{\left (5 \, x\right )}}{960 \, a^{5}} - \frac {15 \, a^{3} b e^{\left (-4 \, x\right )} + 6 \, a^{4} e^{\left (-5 \, x\right )} + 60 \, {\left (5 \, a^{4} + 14 \, a^{2} b^{2} + 8 \, b^{4}\right )} e^{\left (-x\right )} + 60 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} e^{\left (-2 \, x\right )} + 10 \, {\left (5 \, a^{4} + 4 \, a^{2} b^{2}\right )} e^{\left (-3 \, x\right )}}{960 \, a^{5}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} x}{a^{6}} - \frac {{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \log \left (-2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} - a\right )}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(15*a^3*b*e^(-x) - 6*a^4 - 10*(5*a^4 + 4*a^2*b^2)*e^(-2*x) + 60*(3*a^3*b + 2*a*b^3)*e^(-3*x) - 60*(5*a^
4 + 14*a^2*b^2 + 8*b^4)*e^(-4*x))*e^(5*x)/a^5 - 1/960*(15*a^3*b*e^(-4*x) + 6*a^4*e^(-5*x) + 60*(5*a^4 + 14*a^2
*b^2 + 8*b^4)*e^(-x) + 60*(3*a^3*b + 2*a*b^3)*e^(-2*x) + 10*(5*a^4 + 4*a^2*b^2)*e^(-3*x))/a^5 - (a^4*b + 2*a^2
*b^3 + b^5)*x/a^6 - (a^4*b + 2*a^2*b^3 + b^5)*log(-2*b*e^(-x) + a*e^(-2*x) - a)/a^6

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mupad [B]  time = 2.07, size = 228, normalized size = 2.24 \[ \frac {{\mathrm {e}}^{5\,x}}{160\,a}-\frac {{\mathrm {e}}^{-5\,x}}{160\,a}-\frac {{\mathrm {e}}^{-2\,x}\,\left (3\,a^2\,b+2\,b^3\right )}{16\,a^4}-\frac {{\mathrm {e}}^{2\,x}\,\left (3\,a^2\,b+2\,b^3\right )}{16\,a^4}+\frac {{\mathrm {e}}^x\,\left (5\,a^4+14\,a^2\,b^2+8\,b^4\right )}{16\,a^5}-\frac {b\,{\mathrm {e}}^{-4\,x}}{64\,a^2}-\frac {b\,{\mathrm {e}}^{4\,x}}{64\,a^2}-\frac {\ln \left (2\,b\,{\mathrm {e}}^x-a+a\,{\mathrm {e}}^{2\,x}\right )\,\left (a^4\,b+2\,a^2\,b^3+b^5\right )}{a^6}-\frac {{\mathrm {e}}^{-x}\,\left (5\,a^4+14\,a^2\,b^2+8\,b^4\right )}{16\,a^5}-\frac {{\mathrm {e}}^{-3\,x}\,\left (5\,a^2+4\,b^2\right )}{96\,a^3}+\frac {{\mathrm {e}}^{3\,x}\,\left (5\,a^2+4\,b^2\right )}{96\,a^3}+\frac {b\,x\,{\left (a^2+b^2\right )}^2}{a^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^5/(a + b/sinh(x)),x)

[Out]

exp(5*x)/(160*a) - exp(-5*x)/(160*a) - (exp(-2*x)*(3*a^2*b + 2*b^3))/(16*a^4) - (exp(2*x)*(3*a^2*b + 2*b^3))/(
16*a^4) + (exp(x)*(5*a^4 + 8*b^4 + 14*a^2*b^2))/(16*a^5) - (b*exp(-4*x))/(64*a^2) - (b*exp(4*x))/(64*a^2) - (l
og(2*b*exp(x) - a + a*exp(2*x))*(a^4*b + b^5 + 2*a^2*b^3))/a^6 - (exp(-x)*(5*a^4 + 8*b^4 + 14*a^2*b^2))/(16*a^
5) - (exp(-3*x)*(5*a^2 + 4*b^2))/(96*a^3) + (exp(3*x)*(5*a^2 + 4*b^2))/(96*a^3) + (b*x*(a^2 + b^2)^2)/a^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**5/(a+b*csch(x)),x)

[Out]

Integral(cosh(x)**5/(a + b*csch(x)), x)

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