3.92 \(\int \frac{\cosh ^5(x)}{a+b \text{csch}(x)} \, dx\)
Optimal. Leaf size=102 \[ \frac{\left (2 a^2+b^2\right ) \sinh ^3(x)}{3 a^3}-\frac{b \left (2 a^2+b^2\right ) \sinh ^2(x)}{2 a^4}+\frac{\left (a^2+b^2\right )^2 \sinh (x)}{a^5}-\frac{b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)}{a^6}-\frac{b \sinh ^4(x)}{4 a^2}+\frac{\sinh ^5(x)}{5 a} \]
[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)
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Rubi [A] time = 0.198158, antiderivative size = 102, normalized size of antiderivative = 1.,
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 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]
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 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]
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*}
Mathematica [A] time = 0.277388, size = 97, normalized size = 0.95 \[ \frac{20 a^3 \left (2 a^2+b^2\right ) \sinh ^3(x)-30 a^2 b \left (2 a^2+b^2\right ) \sinh ^2(x)+60 a \left (a^2+b^2\right )^2 \sinh (x)-60 b \left (a^2+b^2\right )^2 \log (a \sinh (x)+b)-15 a^4 b \sinh ^4(x)+12 a^5 \sinh ^5(x)}{60 a^6} \]
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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Maple [B] time = 0.041, size = 600, normalized size = 5.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cosh(x)^5/(a+b*csch(x)),x)
[Out]
-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)^4-11/12/a/(tanh(1/2*x)+1)^3+7/8/a/(tanh
(1/2*x)+1)^2-1/a/(tanh(1/2*x)+1)-1/2/a/(tanh(1/2*x)-1)^4-11/12/a/(tanh(1/2*x)-1)^3-7/8/a/(tanh(1/2*x)-1)^2-1/a
/(tanh(1/2*x)-1)+b^5/a^6*ln(tanh(1/2*x)-1)-1/2/a^4/(tanh(1/2*x)+1)^2*b^3-1/a^5/(tanh(1/2*x)+1)*b^4+b^5/a^6*ln(
tanh(1/2*x)+1)-1/4/a^2/(tanh(1/2*x)+1)^4*b-1/3/a^3/(tanh(1/2*x)+1)^3*b^2-1/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)-1/4/a
^2/(tanh(1/2*x)-1)^4*b-1/3/a^3/(tanh(1/2*x)-1)^3*b^2-1/2/a^4/(tanh(1/2*x)-1)^2*b^3-1/a^5/(tanh(1/2*x)-1)*b^4+b
/a^2*ln(tanh(1/2*x)+1)+2*b^3/a^4*ln(tanh(1/2*x)+1)+b/a^2*ln(tanh(1/2*x)-1)+2*b^3/a^4*ln(tanh(1/2*x)-1)-1/2/a^3
/(tanh(1/2*x)-1)^2*b^2-7/8/a^2/(tanh(1/2*x)-1)*b-2/a^3/(tanh(1/2*x)-1)*b^2-1/2/a^4/(tanh(1/2*x)-1)*b^3+7/8/a^2
/(tanh(1/2*x)+1)*b-2/a^3/(tanh(1/2*x)+1)*b^2+1/2/a^4/(tanh(1/2*x)+1)*b^3-1/2/a^2/(tanh(1/2*x)-1)^3*b-9/8/a^2/(
tanh(1/2*x)-1)^2*b+1/2/a^2/(tanh(1/2*x)+1)^3*b-9/8/a^2/(tanh(1/2*x)+1)^2*b+1/2/a^3/(tanh(1/2*x)+1)^2*b^2
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Maxima [B] time = 1.01022, size = 327, normalized size = 3.21 \begin{align*} -\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}} \end{align*}
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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Fricas [B] time = 1.75196, size = 3536, normalized size = 34.67 \begin{align*} \text{result too large to display} \end{align*}
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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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cosh(x)**5/(a+b*csch(x)),x)
[Out]
Timed out
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Giac [B] time = 1.15935, size = 262, normalized size = 2.57 \begin{align*} -\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}} \end{align*}
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