\(\int \frac {1}{(a \sinh ^4(x))^{3/2}} \, dx\) [156]

Optimal result

Integrand size = 10, antiderivative size = 68 \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=\frac {2 \cosh ^2(x) \coth (x)}{3 a \sqrt {a \sinh ^4(x)}}-\frac {\cosh ^2(x) \coth ^3(x)}{5 a \sqrt {a \sinh ^4(x)}}-\frac {\cosh (x) \sinh (x)}{a \sqrt {a \sinh ^4(x)}} \]

[Out]

2/3*cosh(x)^2*coth(x)/a/(a*sinh(x)^4)^(1/2)-1/5*cosh(x)^2*coth(x)^3/a/(a*sinh(x)^4)^(1/2)-cosh(x)*sinh(x)/a/(a
*sinh(x)^4)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3286, 3852} \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=-\frac {\sinh (x) \cosh (x)}{a \sqrt {a \sinh ^4(x)}}-\frac {\cosh ^2(x) \coth ^3(x)}{5 a \sqrt {a \sinh ^4(x)}}+\frac {2 \cosh ^2(x) \coth (x)}{3 a \sqrt {a \sinh ^4(x)}} \]

[In]

Int[(a*Sinh[x]^4)^(-3/2),x]

[Out]

(2*Cosh[x]^2*Coth[x])/(3*a*Sqrt[a*Sinh[x]^4]) - (Cosh[x]^2*Coth[x]^3)/(5*a*Sqrt[a*Sinh[x]^4]) - (Cosh[x]*Sinh[
x])/(a*Sqrt[a*Sinh[x]^4])

Rule 3286

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])), Int[ActivateTrig[u]
*(Sin[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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sinh ^2(x) \int \text {csch}^6(x) \, dx}{a \sqrt {a \sinh ^4(x)}} \\ & = -\frac {\left (i \sinh ^2(x)\right ) \text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \coth (x)\right )}{a \sqrt {a \sinh ^4(x)}} \\ & = \frac {2 \cosh ^2(x) \coth (x)}{3 a \sqrt {a \sinh ^4(x)}}-\frac {\cosh ^2(x) \coth ^3(x)}{5 a \sqrt {a \sinh ^4(x)}}-\frac {\cosh (x) \sinh (x)}{a \sqrt {a \sinh ^4(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=-\frac {\cosh (x) \left (8-4 \text {csch}^2(x)+3 \text {csch}^4(x)\right ) \sinh ^5(x)}{15 \left (a \sinh ^4(x)\right )^{3/2}} \]

[In]

Integrate[(a*Sinh[x]^4)^(-3/2),x]

[Out]

-1/15*(Cosh[x]*(8 - 4*Csch[x]^2 + 3*Csch[x]^4)*Sinh[x]^5)/(a*Sinh[x]^4)^(3/2)

Maple [A] (verified)

Time = 1.53 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {16 \,{\mathrm e}^{-2 x} \left (10 \,{\mathrm e}^{4 x}-5 \,{\mathrm e}^{2 x}+1\right )}{15 a \left ({\mathrm e}^{2 x}-1\right )^{3} \sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{4} {\mathrm e}^{-4 x}}}\)	\(48\)
default	\(-\frac {4 \left (2 \cosh \left (2 x \right )^{2}-6 \cosh \left (2 x \right )+7\right ) \sqrt {a \sinh \left (2 x \right )^{2}}\, \sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (1+\cosh \left (2 x \right )\right )}}{15 a^{2} \left (-1+\cosh \left (2 x \right )\right )^{2} \sinh \left (2 x \right ) \sqrt {\left (-1+\cosh \left (2 x \right )\right )^{2} a}}\)	\(74\)

[In]

int(1/(a*sinh(x)^4)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-16/15/a/(exp(2*x)-1)^3*exp(-2*x)/(a*(exp(2*x)-1)^4*exp(-4*x))^(1/2)*(10*exp(4*x)-5*exp(2*x)+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1163 vs. \(2 (58) = 116\).

Time = 0.30 (sec) , antiderivative size = 1163, normalized size of antiderivative = 17.10 \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(a*sinh(x)^4)^(3/2),x, algorithm="fricas")

[Out]

-16/15*(40*cosh(x)*e^(2*x)*sinh(x)^3 + 10*e^(2*x)*sinh(x)^4 + 5*(12*cosh(x)^2 - 1)*e^(2*x)*sinh(x)^2 + 10*(4*c
osh(x)^3 - cosh(x))*e^(2*x)*sinh(x) + (10*cosh(x)^4 - 5*cosh(x)^2 + 1)*e^(2*x))*sqrt(a*e^(8*x) - 4*a*e^(6*x) +
6*a*e^(4*x) - 4*a*e^(2*x) + a)*e^(-2*x)/(a^2*cosh(x)^10 + (a^2*e^(4*x) - 2*a^2*e^(2*x) + a^2)*sinh(x)^10 - 5*
a^2*cosh(x)^8 + 10*(a^2*cosh(x)*e^(4*x) - 2*a^2*cosh(x)*e^(2*x) + a^2*cosh(x))*sinh(x)^9 + 5*(9*a^2*cosh(x)^2
- a^2 + (9*a^2*cosh(x)^2 - a^2)*e^(4*x) - 2*(9*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^8 + 10*a^2*cosh(x)^6 + 40
*(3*a^2*cosh(x)^3 - a^2*cosh(x) + (3*a^2*cosh(x)^3 - a^2*cosh(x))*e^(4*x) - 2*(3*a^2*cosh(x)^3 - a^2*cosh(x))*
e^(2*x))*sinh(x)^7 + 10*(21*a^2*cosh(x)^4 - 14*a^2*cosh(x)^2 + a^2 + (21*a^2*cosh(x)^4 - 14*a^2*cosh(x)^2 + a^
2)*e^(4*x) - 2*(21*a^2*cosh(x)^4 - 14*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^6 - 10*a^2*cosh(x)^4 + 4*(63*a^2*c
osh(x)^5 - 70*a^2*cosh(x)^3 + 15*a^2*cosh(x) + (63*a^2*cosh(x)^5 - 70*a^2*cosh(x)^3 + 15*a^2*cosh(x))*e^(4*x)
- 2*(63*a^2*cosh(x)^5 - 70*a^2*cosh(x)^3 + 15*a^2*cosh(x))*e^(2*x))*sinh(x)^5 + 10*(21*a^2*cosh(x)^6 - 35*a^2*
cosh(x)^4 + 15*a^2*cosh(x)^2 - a^2 + (21*a^2*cosh(x)^6 - 35*a^2*cosh(x)^4 + 15*a^2*cosh(x)^2 - a^2)*e^(4*x) -
2*(21*a^2*cosh(x)^6 - 35*a^2*cosh(x)^4 + 15*a^2*cosh(x)^2 - a^2)*e^(2*x))*sinh(x)^4 + 5*a^2*cosh(x)^2 + 40*(3*
a^2*cosh(x)^7 - 7*a^2*cosh(x)^5 + 5*a^2*cosh(x)^3 - a^2*cosh(x) + (3*a^2*cosh(x)^7 - 7*a^2*cosh(x)^5 + 5*a^2*c
osh(x)^3 - a^2*cosh(x))*e^(4*x) - 2*(3*a^2*cosh(x)^7 - 7*a^2*cosh(x)^5 + 5*a^2*cosh(x)^3 - a^2*cosh(x))*e^(2*x
))*sinh(x)^3 + 5*(9*a^2*cosh(x)^8 - 28*a^2*cosh(x)^6 + 30*a^2*cosh(x)^4 - 12*a^2*cosh(x)^2 + a^2 + (9*a^2*cosh
(x)^8 - 28*a^2*cosh(x)^6 + 30*a^2*cosh(x)^4 - 12*a^2*cosh(x)^2 + a^2)*e^(4*x) - 2*(9*a^2*cosh(x)^8 - 28*a^2*co
sh(x)^6 + 30*a^2*cosh(x)^4 - 12*a^2*cosh(x)^2 + a^2)*e^(2*x))*sinh(x)^2 - a^2 + (a^2*cosh(x)^10 - 5*a^2*cosh(x
)^8 + 10*a^2*cosh(x)^6 - 10*a^2*cosh(x)^4 + 5*a^2*cosh(x)^2 - a^2)*e^(4*x) - 2*(a^2*cosh(x)^10 - 5*a^2*cosh(x)
^8 + 10*a^2*cosh(x)^6 - 10*a^2*cosh(x)^4 + 5*a^2*cosh(x)^2 - a^2)*e^(2*x) + 10*(a^2*cosh(x)^9 - 4*a^2*cosh(x)^
7 + 6*a^2*cosh(x)^5 - 4*a^2*cosh(x)^3 + a^2*cosh(x) + (a^2*cosh(x)^9 - 4*a^2*cosh(x)^7 + 6*a^2*cosh(x)^5 - 4*a
^2*cosh(x)^3 + a^2*cosh(x))*e^(4*x) - 2*(a^2*cosh(x)^9 - 4*a^2*cosh(x)^7 + 6*a^2*cosh(x)^5 - 4*a^2*cosh(x)^3 +
a^2*cosh(x))*e^(2*x))*sinh(x))

Sympy [F]

\[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=\int \frac {1}{\left (a \sinh ^{4}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(a*sinh(x)**4)**(3/2),x)

[Out]

Integral((a*sinh(x)**4)**(-3/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (58) = 116\).

Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.51 \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=-\frac {16 \, e^{\left (-2 \, x\right )}}{3 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} - 10 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} - 5 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + a^{\frac {3}{2}} e^{\left (-10 \, x\right )} - a^{\frac {3}{2}}\right )}} + \frac {32 \, e^{\left (-4 \, x\right )}}{3 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} - 10 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} - 5 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + a^{\frac {3}{2}} e^{\left (-10 \, x\right )} - a^{\frac {3}{2}}\right )}} + \frac {16}{15 \, {\left (5 \, a^{\frac {3}{2}} e^{\left (-2 \, x\right )} - 10 \, a^{\frac {3}{2}} e^{\left (-4 \, x\right )} + 10 \, a^{\frac {3}{2}} e^{\left (-6 \, x\right )} - 5 \, a^{\frac {3}{2}} e^{\left (-8 \, x\right )} + a^{\frac {3}{2}} e^{\left (-10 \, x\right )} - a^{\frac {3}{2}}\right )}} \]

[In]

integrate(1/(a*sinh(x)^4)^(3/2),x, algorithm="maxima")

[Out]

-16/3*e^(-2*x)/(5*a^(3/2)*e^(-2*x) - 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6*x) - 5*a^(3/2)*e^(-8*x) + a^(3/2)*
e^(-10*x) - a^(3/2)) + 32/3*e^(-4*x)/(5*a^(3/2)*e^(-2*x) - 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-6*x) - 5*a^(3/
2)*e^(-8*x) + a^(3/2)*e^(-10*x) - a^(3/2)) + 16/15/(5*a^(3/2)*e^(-2*x) - 10*a^(3/2)*e^(-4*x) + 10*a^(3/2)*e^(-
6*x) - 5*a^(3/2)*e^(-8*x) + a^(3/2)*e^(-10*x) - a^(3/2))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=-\frac {16 \, {\left (10 \, e^{\left (4 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + 1\right )}}{15 \, a^{\frac {3}{2}} {\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} \]

[In]

integrate(1/(a*sinh(x)^4)^(3/2),x, algorithm="giac")

[Out]

-16/15*(10*e^(4*x) - 5*e^(2*x) + 1)/(a^(3/2)*(e^(2*x) - 1)^5)

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a \sinh ^4(x)\right )^{3/2}} \, dx=-\frac {64\,{\mathrm {e}}^{2\,x}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x}}{2}-\frac {{\mathrm {e}}^x}{2}\right )}^4}\,\left (10\,{\mathrm {e}}^{4\,x}-5\,{\mathrm {e}}^{2\,x}+1\right )}{15\,a^2\,{\left ({\mathrm {e}}^{2\,x}-1\right )}^7} \]

[In]

int(1/(a*sinh(x)^4)^(3/2),x)

[Out]

-(64*exp(2*x)*(a*(exp(-x)/2 - exp(x)/2)^4)^(1/2)*(10*exp(4*x) - 5*exp(2*x) + 1))/(15*a^2*(exp(2*x) - 1)^7)