\(\int \frac {1}{(a \sinh ^2(x))^{5/2}} \, dx\) [145]

Optimal result

Integrand size = 10, antiderivative size = 61 \[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}+\frac {3 \coth (x)}{8 a^2 \sqrt {a \sinh ^2(x)}}-\frac {3 \text {arctanh}(\cosh (x)) \sinh (x)}{8 a^2 \sqrt {a \sinh ^2(x)}} \] Output:

-1/4*coth(x)/a/(a*sinh(x)^2)^(3/2)+3/8*coth(x)/a^2/(a*sinh(x)^2)^(1/2)-3/8 
*arctanh(cosh(x))*sinh(x)/a^2/(a*sinh(x)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=-\frac {\text {csch}(x) \left (-6 \text {csch}^2\left (\frac {x}{2}\right )+\text {csch}^4\left (\frac {x}{2}\right )+24 \left (\log \left (\cosh \left (\frac {x}{2}\right )\right )-\log \left (\sinh \left (\frac {x}{2}\right )\right )\right )-6 \text {sech}^2\left (\frac {x}{2}\right )-\text {sech}^4\left (\frac {x}{2}\right )\right ) \sqrt {a \sinh ^2(x)}}{64 a^3} \] Input:

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

Output:

-1/64*(Csch[x]*(-6*Csch[x/2]^2 + Csch[x/2]^4 + 24*(Log[Cosh[x/2]] - Log[Si 
nh[x/2]]) - 6*Sech[x/2]^2 - Sech[x/2]^4)*Sqrt[a*Sinh[x]^2])/a^3
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3683, 3042, 3683, 3042, 3686, 3042, 26, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (-a \sin (i x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3683

\(\displaystyle -\frac {3 \int \frac {1}{\left (a \sinh ^2(x)\right )^{3/2}}dx}{4 a}-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}-\frac {3 \int \frac {1}{\left (-a \sin (i x)^2\right )^{3/2}}dx}{4 a}\)

\(\Big \downarrow \) 3683

\(\displaystyle -\frac {3 \left (-\frac {\int \frac {1}{\sqrt {a \sinh ^2(x)}}dx}{2 a}-\frac {\coth (x)}{2 a \sqrt {a \sinh ^2(x)}}\right )}{4 a}-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}-\frac {3 \left (-\frac {\coth (x)}{2 a \sqrt {a \sinh ^2(x)}}-\frac {\int \frac {1}{\sqrt {-a \sin (i x)^2}}dx}{2 a}\right )}{4 a}\)

\(\Big \downarrow \) 3686

\(\displaystyle -\frac {3 \left (-\frac {\sinh (x) \int \text {csch}(x)dx}{2 a \sqrt {a \sinh ^2(x)}}-\frac {\coth (x)}{2 a \sqrt {a \sinh ^2(x)}}\right )}{4 a}-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}-\frac {3 \left (-\frac {\coth (x)}{2 a \sqrt {a \sinh ^2(x)}}-\frac {\sinh (x) \int i \csc (i x)dx}{2 a \sqrt {a \sinh ^2(x)}}\right )}{4 a}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}-\frac {3 \left (-\frac {\coth (x)}{2 a \sqrt {a \sinh ^2(x)}}-\frac {i \sinh (x) \int \csc (i x)dx}{2 a \sqrt {a \sinh ^2(x)}}\right )}{4 a}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {3 \left (\frac {\sinh (x) \text {arctanh}(\cosh (x))}{2 a \sqrt {a \sinh ^2(x)}}-\frac {\coth (x)}{2 a \sqrt {a \sinh ^2(x)}}\right )}{4 a}-\frac {\coth (x)}{4 a \left (a \sinh ^2(x)\right )^{3/2}}\)

Input:

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

Output:

-1/4*Coth[x]/(a*(a*Sinh[x]^2)^(3/2)) - (3*(-1/2*Coth[x]/(a*Sqrt[a*Sinh[x]^ 
2]) + (ArcTanh[Cosh[x]]*Sinh[x])/(2*a*Sqrt[a*Sinh[x]^2])))/(4*a)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3683

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[Cot[e + f*x]* 
((b*Sin[e + f*x]^2)^(p + 1)/(b*f*(2*p + 1))), x] + Simp[2*((p + 1)/(b*(2*p 
+ 1)))   Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x] && 
!IntegerQ[p] && LtQ[p, -1]
 

rule 3686

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ 
n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Si 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {\sqrt {a \cosh \left (x \right )^{2}}\, \left (-3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \cosh \left (x \right )^{2}}+2 a}{\sinh \left (x \right )}\right ) a \sinh \left (x \right )^{4}+3 \sinh \left (x \right )^{2} \sqrt {a \cosh \left (x \right )^{2}}\, \sqrt {a}-2 \sqrt {a}\, \sqrt {a \cosh \left (x \right )^{2}}\right )}{8 a^{\frac {7}{2}} \sinh \left (x \right )^{3} \cosh \left (x \right ) \sqrt {a \sinh \left (x \right )^{2}}}\)	\(89\)
risch	\(\frac {3 \,{\mathrm e}^{6 x}-11 \,{\mathrm e}^{4 x}-11 \,{\mathrm e}^{2 x}+3}{4 a^{2} \left ({\mathrm e}^{2 x}-1\right )^{3} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}-\frac {3 \left ({\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-x} \ln \left ({\mathrm e}^{x}+1\right )}{8 a^{2} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}+\frac {3 \left ({\mathrm e}^{2 x}-1\right ) {\mathrm e}^{-x} \ln \left ({\mathrm e}^{x}-1\right )}{8 a^{2} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{2} {\mathrm e}^{-2 x}}}\)	\(123\)

Input:

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

Output:

1/8*(a*cosh(x)^2)^(1/2)*(-3*ln(2*(a^(1/2)*(a*cosh(x)^2)^(1/2)+a)/sinh(x))* 
a*sinh(x)^4+3*sinh(x)^2*(a*cosh(x)^2)^(1/2)*a^(1/2)-2*a^(1/2)*(a*cosh(x)^2 
)^(1/2))/a^(7/2)/sinh(x)^3/cosh(x)/(a*sinh(x)^2)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 875 vs. \(2 (49) = 98\).

Time = 0.10 (sec) , antiderivative size = 875, normalized size of antiderivative = 14.34 \[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/8*(42*cosh(x)*e^x*sinh(x)^6 + 6*e^x*sinh(x)^7 + 2*(63*cosh(x)^2 - 11)*e 
^x*sinh(x)^5 + 10*(21*cosh(x)^3 - 11*cosh(x))*e^x*sinh(x)^4 + 2*(105*cosh( 
x)^4 - 110*cosh(x)^2 - 11)*e^x*sinh(x)^3 + 2*(63*cosh(x)^5 - 110*cosh(x)^3 
 - 33*cosh(x))*e^x*sinh(x)^2 + 2*(21*cosh(x)^6 - 55*cosh(x)^4 - 33*cosh(x) 
^2 + 3)*e^x*sinh(x) + 2*(3*cosh(x)^7 - 11*cosh(x)^5 - 11*cosh(x)^3 + 3*cos 
h(x))*e^x + 3*(8*cosh(x)*e^x*sinh(x)^7 + e^x*sinh(x)^8 + 4*(7*cosh(x)^2 - 
1)*e^x*sinh(x)^6 + 8*(7*cosh(x)^3 - 3*cosh(x))*e^x*sinh(x)^5 + 2*(35*cosh( 
x)^4 - 30*cosh(x)^2 + 3)*e^x*sinh(x)^4 + 8*(7*cosh(x)^5 - 10*cosh(x)^3 + 3 
*cosh(x))*e^x*sinh(x)^3 + 4*(7*cosh(x)^6 - 15*cosh(x)^4 + 9*cosh(x)^2 - 1) 
*e^x*sinh(x)^2 + 8*(cosh(x)^7 - 3*cosh(x)^5 + 3*cosh(x)^3 - cosh(x))*e^x*s 
inh(x) + (cosh(x)^8 - 4*cosh(x)^6 + 6*cosh(x)^4 - 4*cosh(x)^2 + 1)*e^x)*lo 
g((cosh(x) + sinh(x) - 1)/(cosh(x) + sinh(x) + 1)))*sqrt(a*e^(4*x) - 2*a*e 
^(2*x) + a)*e^(-x)/(a^3*cosh(x)^8 - 4*a^3*cosh(x)^6 - (a^3*e^(2*x) - a^3)* 
sinh(x)^8 - 8*(a^3*cosh(x)*e^(2*x) - a^3*cosh(x))*sinh(x)^7 + 6*a^3*cosh(x 
)^4 + 4*(7*a^3*cosh(x)^2 - a^3 - (7*a^3*cosh(x)^2 - a^3)*e^(2*x))*sinh(x)^ 
6 + 8*(7*a^3*cosh(x)^3 - 3*a^3*cosh(x) - (7*a^3*cosh(x)^3 - 3*a^3*cosh(x)) 
*e^(2*x))*sinh(x)^5 - 4*a^3*cosh(x)^2 + 2*(35*a^3*cosh(x)^4 - 30*a^3*cosh( 
x)^2 + 3*a^3 - (35*a^3*cosh(x)^4 - 30*a^3*cosh(x)^2 + 3*a^3)*e^(2*x))*sinh 
(x)^4 + 8*(7*a^3*cosh(x)^5 - 10*a^3*cosh(x)^3 + 3*a^3*cosh(x) - (7*a^3*cos 
h(x)^5 - 10*a^3*cosh(x)^3 + 3*a^3*cosh(x))*e^(2*x))*sinh(x)^3 + a^3 + 4...
 

Sympy [F]

\[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (a \sinh ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral((a*sinh(x)**2)**(-5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=\frac {3 \, e^{\left (-x\right )} - 11 \, e^{\left (-3 \, x\right )} - 11 \, e^{\left (-5 \, x\right )} + 3 \, e^{\left (-7 \, x\right )}}{4 \, {\left (4 \, a^{\frac {5}{2}} e^{\left (-2 \, x\right )} - 6 \, a^{\frac {5}{2}} e^{\left (-4 \, x\right )} + 4 \, a^{\frac {5}{2}} e^{\left (-6 \, x\right )} - a^{\frac {5}{2}} e^{\left (-8 \, x\right )} - a^{\frac {5}{2}}\right )}} + \frac {3 \, \log \left (e^{\left (-x\right )} + 1\right )}{8 \, a^{\frac {5}{2}}} - \frac {3 \, \log \left (e^{\left (-x\right )} - 1\right )}{8 \, a^{\frac {5}{2}}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=\frac {3 \, {\left (e^{\left (-x\right )} + e^{x}\right )}^{3} - 20 \, e^{\left (-x\right )} - 20 \, e^{x}}{4 \, {\left ({\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} \] Input:

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

Output:

1/4*(3*(e^(-x) + e^x)^3 - 20*e^(-x) - 20*e^x)/(((e^(-x) + e^x)^2 - 4)^2*a^ 
(5/2)*sgn(e^(3*x) - e^x))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 184, normalized size of antiderivative = 3.02 \[ \int \frac {1}{\left (a \sinh ^2(x)\right )^{5/2}} \, dx=\frac {\sqrt {a}\, \left (3 e^{8 x} \mathrm {log}\left (e^{x}-1\right )-3 e^{8 x} \mathrm {log}\left (e^{x}+1\right )+6 e^{7 x}-12 e^{6 x} \mathrm {log}\left (e^{x}-1\right )+12 e^{6 x} \mathrm {log}\left (e^{x}+1\right )-22 e^{5 x}+18 e^{4 x} \mathrm {log}\left (e^{x}-1\right )-18 e^{4 x} \mathrm {log}\left (e^{x}+1\right )-22 e^{3 x}-12 e^{2 x} \mathrm {log}\left (e^{x}-1\right )+12 e^{2 x} \mathrm {log}\left (e^{x}+1\right )+6 e^{x}+3 \,\mathrm {log}\left (e^{x}-1\right )-3 \,\mathrm {log}\left (e^{x}+1\right )\right )}{8 a^{3} \left (e^{8 x}-4 e^{6 x}+6 e^{4 x}-4 e^{2 x}+1\right )} \] Input:

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

Output:

(sqrt(a)*(3*e**(8*x)*log(e**x - 1) - 3*e**(8*x)*log(e**x + 1) + 6*e**(7*x) 
 - 12*e**(6*x)*log(e**x - 1) + 12*e**(6*x)*log(e**x + 1) - 22*e**(5*x) + 1 
8*e**(4*x)*log(e**x - 1) - 18*e**(4*x)*log(e**x + 1) - 22*e**(3*x) - 12*e* 
*(2*x)*log(e**x - 1) + 12*e**(2*x)*log(e**x + 1) + 6*e**x + 3*log(e**x - 1 
) - 3*log(e**x + 1)))/(8*a**3*(e**(8*x) - 4*e**(6*x) + 6*e**(4*x) - 4*e**( 
2*x) + 1))