\(\int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx\) [155]

Optimal result

Integrand size = 10, antiderivative size = 16 \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=-\frac {\cosh (x) \sinh (x)}{\sqrt {a \sinh ^4(x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3286, 3852, 8} \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=-\frac {\sinh (x) \cosh (x)}{\sqrt {a \sinh ^4(x)}} \]

[In]

Int[1/Sqrt[a*Sinh[x]^4],x]

[Out]

-((Cosh[x]*Sinh[x])/Sqrt[a*Sinh[x]^4])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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}^2(x) \, dx}{\sqrt {a \sinh ^4(x)}} \\ & = -\frac {\left (i \sinh ^2(x)\right ) \text {Subst}(\int 1 \, dx,x,-i \coth (x))}{\sqrt {a \sinh ^4(x)}} \\ & = -\frac {\cosh (x) \sinh (x)}{\sqrt {a \sinh ^4(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=-\frac {\cosh (x) \sinh (x)}{\sqrt {a \sinh ^4(x)}} \]

[In]

Integrate[1/Sqrt[a*Sinh[x]^4],x]

[Out]

-((Cosh[x]*Sinh[x])/Sqrt[a*Sinh[x]^4])

Maple [A] (verified)

Time = 1.55 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.81

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{-2 x} \left ({\mathrm e}^{2 x}-1\right )}{\sqrt {a \left ({\mathrm e}^{2 x}-1\right )^{4} {\mathrm e}^{-4 x}}}\)	\(29\)
default	\(-\frac {\sqrt {a \left (-1+\cosh \left (2 x \right )\right ) \left (1+\cosh \left (2 x \right )\right )}\, \sqrt {a \sinh \left (2 x \right )^{2}}}{a \sinh \left (2 x \right ) \sqrt {\left (-1+\cosh \left (2 x \right )\right )^{2} a}}\)	\(50\)

[In]

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

[Out]

-2/(a*(exp(2*x)-1)^4*exp(-4*x))^(1/2)*exp(-2*x)*(exp(2*x)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (14) = 28\).

Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 7.62 \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=-\frac {2 \, \sqrt {a e^{\left (8 \, x\right )} - 4 \, a e^{\left (6 \, x\right )} + 6 \, a e^{\left (4 \, x\right )} - 4 \, a e^{\left (2 \, x\right )} + a}}{a \cosh \left (x\right )^{2} + {\left (a e^{\left (4 \, x\right )} - 2 \, a e^{\left (2 \, x\right )} + a\right )} \sinh \left (x\right )^{2} + {\left (a \cosh \left (x\right )^{2} - a\right )} e^{\left (4 \, x\right )} - 2 \, {\left (a \cosh \left (x\right )^{2} - a\right )} e^{\left (2 \, x\right )} + 2 \, {\left (a \cosh \left (x\right ) e^{\left (4 \, x\right )} - 2 \, a \cosh \left (x\right ) e^{\left (2 \, x\right )} + a \cosh \left (x\right )\right )} \sinh \left (x\right ) - a} \]

[In]

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

[Out]

-2*sqrt(a*e^(8*x) - 4*a*e^(6*x) + 6*a*e^(4*x) - 4*a*e^(2*x) + a)/(a*cosh(x)^2 + (a*e^(4*x) - 2*a*e^(2*x) + a)*
sinh(x)^2 + (a*cosh(x)^2 - a)*e^(4*x) - 2*(a*cosh(x)^2 - a)*e^(2*x) + 2*(a*cosh(x)*e^(4*x) - 2*a*cosh(x)*e^(2*
x) + a*cosh(x))*sinh(x) - a)

Sympy [F]

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

[In]

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

[Out]

Integral(1/sqrt(a*sinh(x)**4), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=\frac {2}{\sqrt {a} e^{\left (-2 \, x\right )} - \sqrt {a}} \]

[In]

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

[Out]

2/(sqrt(a)*e^(-2*x) - sqrt(a))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=-\frac {2}{\sqrt {a} {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

[In]

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

[Out]

-2/(sqrt(a)*(e^(2*x) - 1))

Mupad [B] (verification not implemented)

Time = 1.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.38 \[ \int \frac {1}{\sqrt {a \sinh ^4(x)}} \, dx=\frac {{\mathrm {e}}^{-x}\,\sqrt {a\,{\left (\frac {{\mathrm {e}}^{-x}}{2}-\frac {{\mathrm {e}}^x}{2}\right )}^4}}{a\,{\left (\frac {{\mathrm {e}}^{-x}}{2}-\frac {{\mathrm {e}}^x}{2}\right )}^3} \]

[In]

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

[Out]

(exp(-x)*(a*(exp(-x)/2 - exp(x)/2)^4)^(1/2))/(a*(exp(-x)/2 - exp(x)/2)^3)