3.133 \(\int \frac {1}{(a \cosh ^3(x))^{5/2}} \, dx\)

Optimal. Leaf size=121 \[ \frac {154 \sinh (x) \cosh (x)}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 \tanh (x)}{585 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 i \cosh ^{\frac {3}{2}}(x) E\left (\left .\frac {i x}{2}\right |2\right )}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {2 \tanh (x) \text {sech}^4(x)}{13 a^2 \sqrt {a \cosh ^3(x)}}+\frac {22 \tanh (x) \text {sech}^2(x)}{117 a^2 \sqrt {a \cosh ^3(x)}} \]

[Out]

154/195*I*cosh(x)^(3/2)*(cosh(1/2*x)^2)^(1/2)/cosh(1/2*x)*EllipticE(I*sinh(1/2*x),2^(1/2))/a^2/(a*cosh(x)^3)^(
1/2)+154/195*cosh(x)*sinh(x)/a^2/(a*cosh(x)^3)^(1/2)+154/585*tanh(x)/a^2/(a*cosh(x)^3)^(1/2)+22/117*sech(x)^2*
tanh(x)/a^2/(a*cosh(x)^3)^(1/2)+2/13*sech(x)^4*tanh(x)/a^2/(a*cosh(x)^3)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3207, 2636, 2639} \[ \frac {154 \sinh (x) \cosh (x)}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 \tanh (x)}{585 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 i \cosh ^{\frac {3}{2}}(x) E\left (\left .\frac {i x}{2}\right |2\right )}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {2 \tanh (x) \text {sech}^4(x)}{13 a^2 \sqrt {a \cosh ^3(x)}}+\frac {22 \tanh (x) \text {sech}^2(x)}{117 a^2 \sqrt {a \cosh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x]^3)^(-5/2),x]

[Out]

(((154*I)/195)*Cosh[x]^(3/2)*EllipticE[(I/2)*x, 2])/(a^2*Sqrt[a*Cosh[x]^3]) + (154*Cosh[x]*Sinh[x])/(195*a^2*S
qrt[a*Cosh[x]^3]) + (154*Tanh[x])/(585*a^2*Sqrt[a*Cosh[x]^3]) + (22*Sech[x]^2*Tanh[x])/(117*a^2*Sqrt[a*Cosh[x]
^3]) + (2*Sech[x]^4*Tanh[x])/(13*a^2*Sqrt[a*Cosh[x]^3])

Rule 2636

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 2639

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

Rule 3207

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*} \int \frac {1}{\left (a \cosh ^3(x)\right )^{5/2}} \, dx &=\frac {\cosh ^{\frac {3}{2}}(x) \int \frac {1}{\cosh ^{\frac {15}{2}}(x)} \, dx}{a^2 \sqrt {a \cosh ^3(x)}}\\ &=\frac {2 \text {sech}^4(x) \tanh (x)}{13 a^2 \sqrt {a \cosh ^3(x)}}+\frac {\left (11 \cosh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\cosh ^{\frac {11}{2}}(x)} \, dx}{13 a^2 \sqrt {a \cosh ^3(x)}}\\ &=\frac {22 \text {sech}^2(x) \tanh (x)}{117 a^2 \sqrt {a \cosh ^3(x)}}+\frac {2 \text {sech}^4(x) \tanh (x)}{13 a^2 \sqrt {a \cosh ^3(x)}}+\frac {\left (77 \cosh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\cosh ^{\frac {7}{2}}(x)} \, dx}{117 a^2 \sqrt {a \cosh ^3(x)}}\\ &=\frac {154 \tanh (x)}{585 a^2 \sqrt {a \cosh ^3(x)}}+\frac {22 \text {sech}^2(x) \tanh (x)}{117 a^2 \sqrt {a \cosh ^3(x)}}+\frac {2 \text {sech}^4(x) \tanh (x)}{13 a^2 \sqrt {a \cosh ^3(x)}}+\frac {\left (77 \cosh ^{\frac {3}{2}}(x)\right ) \int \frac {1}{\cosh ^{\frac {3}{2}}(x)} \, dx}{195 a^2 \sqrt {a \cosh ^3(x)}}\\ &=\frac {154 \cosh (x) \sinh (x)}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 \tanh (x)}{585 a^2 \sqrt {a \cosh ^3(x)}}+\frac {22 \text {sech}^2(x) \tanh (x)}{117 a^2 \sqrt {a \cosh ^3(x)}}+\frac {2 \text {sech}^4(x) \tanh (x)}{13 a^2 \sqrt {a \cosh ^3(x)}}-\frac {\left (77 \cosh ^{\frac {3}{2}}(x)\right ) \int \sqrt {\cosh (x)} \, dx}{195 a^2 \sqrt {a \cosh ^3(x)}}\\ &=\frac {154 i \cosh ^{\frac {3}{2}}(x) E\left (\left .\frac {i x}{2}\right |2\right )}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 \cosh (x) \sinh (x)}{195 a^2 \sqrt {a \cosh ^3(x)}}+\frac {154 \tanh (x)}{585 a^2 \sqrt {a \cosh ^3(x)}}+\frac {22 \text {sech}^2(x) \tanh (x)}{117 a^2 \sqrt {a \cosh ^3(x)}}+\frac {2 \text {sech}^4(x) \tanh (x)}{13 a^2 \sqrt {a \cosh ^3(x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 61, normalized size = 0.50 \[ \frac {462 i \cosh ^{\frac {3}{2}}(x) E\left (\left .\frac {i x}{2}\right |2\right )+462 \sinh (x) \cosh (x)+2 \tanh (x) \left (45 \text {sech}^4(x)+55 \text {sech}^2(x)+77\right )}{585 a^2 \sqrt {a \cosh ^3(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x]^3)^(-5/2),x]

[Out]

((462*I)*Cosh[x]^(3/2)*EllipticE[(I/2)*x, 2] + 462*Cosh[x]*Sinh[x] + 2*(77 + 55*Sech[x]^2 + 45*Sech[x]^4)*Tanh
[x])/(585*a^2*Sqrt[a*Cosh[x]^3])

________________________________________________________________________________________

fricas [F]  time = 1.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \cosh \relax (x)^{3}}}{a^{3} \cosh \relax (x)^{9}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^3)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*cosh(x)^3)/(a^3*cosh(x)^9), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cosh \relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^3)^(5/2),x, algorithm="giac")

[Out]

integrate((a*cosh(x)^3)^(-5/2), x)

________________________________________________________________________________________

maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \left (\cosh ^{3}\relax (x )\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)^3)^(5/2),x)

[Out]

int(1/(a*cosh(x)^3)^(5/2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cosh \relax (x)^{3}\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)^3)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*cosh(x)^3)^(-5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a\,{\mathrm {cosh}\relax (x)}^3\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)^3)^(5/2),x)

[Out]

int(1/(a*cosh(x)^3)^(5/2), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)**3)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________