\(\int \frac {1}{(a \text {csch}^2(x))^{5/2}} \, dx\) [34]

Optimal result

Integrand size = 10, antiderivative size = 55 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {\coth (x)}{5 \left (a \text {csch}^2(x)\right )^{5/2}}-\frac {4 \coth (x)}{15 a \left (a \text {csch}^2(x)\right )^{3/2}}+\frac {8 \coth (x)}{15 a^2 \sqrt {a \text {csch}^2(x)}} \]

[Out]

1/5*coth(x)/(a*csch(x)^2)^(5/2)-4/15*coth(x)/a/(a*csch(x)^2)^(3/2)+8/15*coth(x)/a^2/(a*csch(x)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4207, 198, 197} \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {8 \coth (x)}{15 a^2 \sqrt {a \text {csch}^2(x)}}-\frac {4 \coth (x)}{15 a \left (a \text {csch}^2(x)\right )^{3/2}}+\frac {\coth (x)}{5 \left (a \text {csch}^2(x)\right )^{5/2}} \]

[In]

Int[(a*Csch[x]^2)^(-5/2),x]

[Out]

Coth[x]/(5*(a*Csch[x]^2)^(5/2)) - (4*Coth[x])/(15*a*(a*Csch[x]^2)^(3/2)) + (8*Coth[x])/(15*a^2*Sqrt[a*Csch[x]^
2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 4207

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[b*(ff/
f), Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps \begin{align*} \text {integral}& = -\left (a \text {Subst}\left (\int \frac {1}{\left (-a+a x^2\right )^{7/2}} \, dx,x,\coth (x)\right )\right ) \\ & = \frac {\coth (x)}{5 \left (a \text {csch}^2(x)\right )^{5/2}}+\frac {4}{5} \text {Subst}\left (\int \frac {1}{\left (-a+a x^2\right )^{5/2}} \, dx,x,\coth (x)\right ) \\ & = \frac {\coth (x)}{5 \left (a \text {csch}^2(x)\right )^{5/2}}-\frac {4 \coth (x)}{15 a \left (a \text {csch}^2(x)\right )^{3/2}}-\frac {8 \text {Subst}\left (\int \frac {1}{\left (-a+a x^2\right )^{3/2}} \, dx,x,\coth (x)\right )}{15 a} \\ & = \frac {\coth (x)}{5 \left (a \text {csch}^2(x)\right )^{5/2}}-\frac {4 \coth (x)}{15 a \left (a \text {csch}^2(x)\right )^{3/2}}+\frac {8 \coth (x)}{15 a^2 \sqrt {a \text {csch}^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {(150 \cosh (x)-25 \cosh (3 x)+3 \cosh (5 x)) \sqrt {a \text {csch}^2(x)} \sinh (x)}{240 a^3} \]

[In]

Integrate[(a*Csch[x]^2)^(-5/2),x]

[Out]

((150*Cosh[x] - 25*Cosh[3*x] + 3*Cosh[5*x])*Sqrt[a*Csch[x]^2]*Sinh[x])/(240*a^3)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(195\) vs. \(2(43)=86\).

Time = 0.16 (sec) , antiderivative size = 196, normalized size of antiderivative = 3.56

method	result	size
risch	\(\frac {{\mathrm e}^{6 x}}{160 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}-\frac {5 \,{\mathrm e}^{4 x}}{96 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}+\frac {5 \,{\mathrm e}^{2 x}}{16 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}+\frac {5}{16 \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \left ({\mathrm e}^{2 x}-1\right ) a^{2}}-\frac {5 \,{\mathrm e}^{-2 x}}{96 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}+\frac {{\mathrm e}^{-4 x}}{160 a^{2} \left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 x}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}}\)	\(196\)

[In]

int(1/(a*csch(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/160/a^2*exp(6*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)-5/96/a^2*exp(4*x)/(exp(2*x)-1)/(a*exp(2*x)/(
exp(2*x)-1)^2)^(1/2)+5/16/a^2*exp(2*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)+5/16/(a*exp(2*x)/(exp(2*
x)-1)^2)^(1/2)/(exp(2*x)-1)/a^2-5/96/a^2*exp(-2*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)+1/160/a^2*ex
p(-4*x)/(exp(2*x)-1)/(a*exp(2*x)/(exp(2*x)-1)^2)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (43) = 86\).

Time = 0.27 (sec) , antiderivative size = 590, normalized size of antiderivative = 10.73 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a*csch(x)^2)^(5/2),x, algorithm="fricas")

[Out]

1/480*(3*(e^(2*x) - 1)*sinh(x)^10 - 3*cosh(x)^10 + 30*(cosh(x)*e^(2*x) - cosh(x))*sinh(x)^9 - 5*(27*cosh(x)^2
- (27*cosh(x)^2 - 5)*e^(2*x) - 5)*sinh(x)^8 + 25*cosh(x)^8 - 40*(9*cosh(x)^3 - (9*cosh(x)^3 - 5*cosh(x))*e^(2*
x) - 5*cosh(x))*sinh(x)^7 - 10*(63*cosh(x)^4 - 70*cosh(x)^2 - (63*cosh(x)^4 - 70*cosh(x)^2 + 15)*e^(2*x) + 15)
*sinh(x)^6 - 150*cosh(x)^6 - 4*(189*cosh(x)^5 - 350*cosh(x)^3 - (189*cosh(x)^5 - 350*cosh(x)^3 + 225*cosh(x))*
e^(2*x) + 225*cosh(x))*sinh(x)^5 - 10*(63*cosh(x)^6 - 175*cosh(x)^4 + 225*cosh(x)^2 - (63*cosh(x)^6 - 175*cosh
(x)^4 + 225*cosh(x)^2 + 15)*e^(2*x) + 15)*sinh(x)^4 - 150*cosh(x)^4 - 40*(9*cosh(x)^7 - 35*cosh(x)^5 + 75*cosh
(x)^3 - (9*cosh(x)^7 - 35*cosh(x)^5 + 75*cosh(x)^3 + 15*cosh(x))*e^(2*x) + 15*cosh(x))*sinh(x)^3 - 5*(27*cosh(
x)^8 - 140*cosh(x)^6 + 450*cosh(x)^4 + 180*cosh(x)^2 - (27*cosh(x)^8 - 140*cosh(x)^6 + 450*cosh(x)^4 + 180*cos
h(x)^2 - 5)*e^(2*x) - 5)*sinh(x)^2 + 25*cosh(x)^2 + (3*cosh(x)^10 - 25*cosh(x)^8 + 150*cosh(x)^6 + 150*cosh(x)
^4 - 25*cosh(x)^2 + 3)*e^(2*x) - 10*(3*cosh(x)^9 - 20*cosh(x)^7 + 90*cosh(x)^5 + 60*cosh(x)^3 - (3*cosh(x)^9 -
20*cosh(x)^7 + 90*cosh(x)^5 + 60*cosh(x)^3 - 5*cosh(x))*e^(2*x) - 5*cosh(x))*sinh(x) - 3)*sqrt(a/(e^(4*x) - 2
*e^(2*x) + 1))*e^x/(a^3*cosh(x)^5*e^x + 5*a^3*cosh(x)^4*e^x*sinh(x) + 10*a^3*cosh(x)^3*e^x*sinh(x)^2 + 10*a^3*
cosh(x)^2*e^x*sinh(x)^3 + 5*a^3*cosh(x)*e^x*sinh(x)^4 + a^3*e^x*sinh(x)^5)

Sympy [F]

\[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \operatorname {csch}^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(a*csch(x)**2)**(5/2),x)

[Out]

Integral((a*csch(x)**2)**(-5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=-\frac {e^{\left (5 \, x\right )}}{160 \, a^{\frac {5}{2}}} + \frac {5 \, e^{\left (3 \, x\right )}}{96 \, a^{\frac {5}{2}}} - \frac {5 \, e^{\left (-x\right )}}{16 \, a^{\frac {5}{2}}} + \frac {5 \, e^{\left (-3 \, x\right )}}{96 \, a^{\frac {5}{2}}} - \frac {e^{\left (-5 \, x\right )}}{160 \, a^{\frac {5}{2}}} - \frac {5 \, e^{x}}{16 \, a^{\frac {5}{2}}} \]

[In]

integrate(1/(a*csch(x)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/160*e^(5*x)/a^(5/2) + 5/96*e^(3*x)/a^(5/2) - 5/16*e^(-x)/a^(5/2) + 5/96*e^(-3*x)/a^(5/2) - 1/160*e^(-5*x)/a
^(5/2) - 5/16*e^x/a^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\frac {{\left (150 \, e^{\left (4 \, x\right )} - 25 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-5 \, x\right )} + 3 \, e^{\left (5 \, x\right )} - 25 \, e^{\left (3 \, x\right )} + 150 \, e^{x}}{480 \, a^{\frac {5}{2}} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} \]

[In]

integrate(1/(a*csch(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/480*((150*e^(4*x) - 25*e^(2*x) + 3)*e^(-5*x) + 3*e^(5*x) - 25*e^(3*x) + 150*e^x)/(a^(5/2)*sgn(e^(3*x) - e^x)
)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a \text {csch}^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (\frac {a}{{\mathrm {sinh}\left (x\right )}^2}\right )}^{5/2}} \,d x \]

[In]

int(1/(a/sinh(x)^2)^(5/2),x)

[Out]

int(1/(a/sinh(x)^2)^(5/2), x)