\(\int \frac {1}{(a \sinh ^3(x))^{3/2}} \, dx\) [150]

Optimal result

Integrand size = 10, antiderivative size = 87 \[ \int \frac {1}{\left (a \sinh ^3(x)\right )^{3/2}} \, dx=\frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {10 i \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right ) \sqrt {i \sinh (x)} \sinh (x)}{21 a \sqrt {a \sinh ^3(x)}} \]

[Out]

10/21*cosh(x)/a/(a*sinh(x)^3)^(1/2)-2/7*coth(x)*csch(x)/a/(a*sinh(x)^3)^(1/2)+10/21*I*(sin(1/4*Pi+1/2*I*x)^2)^
(1/2)/sin(1/4*Pi+1/2*I*x)*EllipticF(cos(1/4*Pi+1/2*I*x),2^(1/2))*sinh(x)*(I*sinh(x))^(1/2)/a/(a*sinh(x)^3)^(1/
2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3286, 2716, 2721, 2720} \[ \int \frac {1}{\left (a \sinh ^3(x)\right )^{3/2}} \, dx=\frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}+\frac {10 i \sqrt {i \sinh (x)} \sinh (x) \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right )}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}} \]

[In]

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

[Out]

(10*Cosh[x])/(21*a*Sqrt[a*Sinh[x]^3]) - (2*Coth[x]*Csch[x])/(7*a*Sqrt[a*Sinh[x]^3]) + (((10*I)/21)*EllipticF[P
i/4 - (I/2)*x, 2]*Sqrt[I*Sinh[x]]*Sinh[x])/(a*Sqrt[a*Sinh[x]^3])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

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
]])

Rubi steps \begin{align*} \text {integral}& = \frac {\sinh ^{\frac {3}{2}}(x) \int \frac {1}{\sinh ^{\frac {9}{2}}(x)} \, dx}{a \sqrt {a \sinh ^3(x)}} \\ & = -\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}-\frac {\left (5 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sinh ^{\frac {5}{2}}(x)} \, dx}{7 a \sqrt {a \sinh ^3(x)}} \\ & = \frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {\left (5 \sinh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\sqrt {\sinh (x)}} \, dx}{21 a \sqrt {a \sinh ^3(x)}} \\ & = \frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {\left (5 \sqrt {i \sinh (x)} \sinh (x)\right ) \int \frac {1}{\sqrt {i \sinh (x)}} \, dx}{21 a \sqrt {a \sinh ^3(x)}} \\ & = \frac {10 \cosh (x)}{21 a \sqrt {a \sinh ^3(x)}}-\frac {2 \coth (x) \text {csch}(x)}{7 a \sqrt {a \sinh ^3(x)}}+\frac {10 i \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},2\right ) \sqrt {i \sinh (x)} \sinh (x)}{21 a \sqrt {a \sinh ^3(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.61 \[ \int \frac {1}{\left (a \sinh ^3(x)\right )^{3/2}} \, dx=\frac {2 \left (5 \cosh (x)-3 \coth (x) \text {csch}(x)+5 \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),2\right ) (i \sinh (x))^{3/2}\right )}{21 a \sqrt {a \sinh ^3(x)}} \]

[In]

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

[Out]

(2*(5*Cosh[x] - 3*Coth[x]*Csch[x] + 5*EllipticF[(Pi - (2*I)*x)/4, 2]*(I*Sinh[x])^(3/2)))/(21*a*Sqrt[a*Sinh[x]^
3])

Maple [F]

\[\int \frac {1}{\left (a \sinh \left (x \right )^{3}\right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 639, normalized size of antiderivative = 7.34 \[ \int \frac {1}{\left (a \sinh ^3(x)\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

2/21*(5*(sqrt(2)*cosh(x)^8 + 8*sqrt(2)*cosh(x)*sinh(x)^7 + sqrt(2)*sinh(x)^8 + 4*(7*sqrt(2)*cosh(x)^2 - sqrt(2
))*sinh(x)^6 - 4*sqrt(2)*cosh(x)^6 + 8*(7*sqrt(2)*cosh(x)^3 - 3*sqrt(2)*cosh(x))*sinh(x)^5 + 2*(35*sqrt(2)*cos
h(x)^4 - 30*sqrt(2)*cosh(x)^2 + 3*sqrt(2))*sinh(x)^4 + 6*sqrt(2)*cosh(x)^4 + 8*(7*sqrt(2)*cosh(x)^5 - 10*sqrt(
2)*cosh(x)^3 + 3*sqrt(2)*cosh(x))*sinh(x)^3 + 4*(7*sqrt(2)*cosh(x)^6 - 15*sqrt(2)*cosh(x)^4 + 9*sqrt(2)*cosh(x
)^2 - sqrt(2))*sinh(x)^2 - 4*sqrt(2)*cosh(x)^2 + 8*(sqrt(2)*cosh(x)^7 - 3*sqrt(2)*cosh(x)^5 + 3*sqrt(2)*cosh(x
)^3 - sqrt(2)*cosh(x))*sinh(x) + sqrt(2))*sqrt(a)*weierstrassPInverse(4, 0, cosh(x) + sinh(x)) + 2*(5*cosh(x)^
7 + 35*cosh(x)*sinh(x)^6 + 5*sinh(x)^7 + (105*cosh(x)^2 - 17)*sinh(x)^5 - 17*cosh(x)^5 + 5*(35*cosh(x)^3 - 17*
cosh(x))*sinh(x)^4 + (175*cosh(x)^4 - 170*cosh(x)^2 - 17)*sinh(x)^3 - 17*cosh(x)^3 + (105*cosh(x)^5 - 170*cosh
(x)^3 - 51*cosh(x))*sinh(x)^2 + (35*cosh(x)^6 - 85*cosh(x)^4 - 51*cosh(x)^2 + 5)*sinh(x) + 5*cosh(x))*sqrt(a*s
inh(x)))/(a^2*cosh(x)^8 + 8*a^2*cosh(x)*sinh(x)^7 + a^2*sinh(x)^8 - 4*a^2*cosh(x)^6 + 4*(7*a^2*cosh(x)^2 - a^2
)*sinh(x)^6 + 6*a^2*cosh(x)^4 + 8*(7*a^2*cosh(x)^3 - 3*a^2*cosh(x))*sinh(x)^5 + 2*(35*a^2*cosh(x)^4 - 30*a^2*c
osh(x)^2 + 3*a^2)*sinh(x)^4 - 4*a^2*cosh(x)^2 + 8*(7*a^2*cosh(x)^5 - 10*a^2*cosh(x)^3 + 3*a^2*cosh(x))*sinh(x)
^3 + 4*(7*a^2*cosh(x)^6 - 15*a^2*cosh(x)^4 + 9*a^2*cosh(x)^2 - a^2)*sinh(x)^2 + a^2 + 8*(a^2*cosh(x)^7 - 3*a^2
*cosh(x)^5 + 3*a^2*cosh(x)^3 - a^2*cosh(x))*sinh(x))

Sympy [F]

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

[In]

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

[Out]

Integral((a*sinh(x)**3)**(-3/2), x)

Maxima [F]

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

[In]

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

[Out]

integrate((a*sinh(x)^3)^(-3/2), x)

Giac [F]

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

[In]

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

[Out]

integrate((a*sinh(x)^3)^(-3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a \sinh ^3(x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (a\,{\mathrm {sinh}\left (x\right )}^3\right )}^{3/2}} \,d x \]

[In]

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

[Out]

int(1/(a*sinh(x)^3)^(3/2), x)