\(\int \frac {1}{(a \cosh ^3(x))^{3/2}} \, dx\) [132]

Optimal result

Integrand size = 10, antiderivative size = 75 \[ \int \frac {1}{\left (a \cosh ^3(x)\right )^{3/2}} \, dx=-\frac {10 i \cosh ^{\frac {3}{2}}(x) \operatorname {EllipticF}\left (\frac {i x}{2},2\right )}{21 a \sqrt {a \cosh ^3(x)}}+\frac {10 \sinh (x)}{21 a \sqrt {a \cosh ^3(x)}}+\frac {2 \text {sech}(x) \tanh (x)}{7 a \sqrt {a \cosh ^3(x)}} \] Output:

-10/21*I*cosh(x)^(3/2)*InverseJacobiAM(1/2*I*x,2^(1/2))/a/(a*cosh(x)^3)^(1 
/2)+10/21*sinh(x)/a/(a*cosh(x)^3)^(1/2)+2/7*sech(x)*tanh(x)/a/(a*cosh(x)^3 
)^(1/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\left (a \cosh ^3(x)\right )^{3/2}} \, dx=\frac {2 \cosh ^2(x) \left (-5 i \cosh ^{\frac {5}{2}}(x) \operatorname {EllipticF}\left (\frac {i x}{2},2\right )+5 \cosh (x) \sinh (x)+3 \tanh (x)\right )}{21 \left (a \cosh ^3(x)\right )^{3/2}} \] Input:

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

Output:

(2*Cosh[x]^2*((-5*I)*Cosh[x]^(5/2)*EllipticF[(I/2)*x, 2] + 5*Cosh[x]*Sinh[ 
x] + 3*Tanh[x]))/(21*(a*Cosh[x]^3)^(3/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3686, 3042, 3116, 3042, 3116, 3042, 3120}

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.

\(\displaystyle \int \frac {1}{\left (a \cosh ^3(x)\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a \sin \left (\frac {\pi }{2}+i x\right )^3\right )^{3/2}}dx\)

\(\Big \downarrow \) 3686

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \int \frac {1}{\cosh ^{\frac {9}{2}}(x)}dx}{a \sqrt {a \cosh ^3(x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \int \frac {1}{\sin \left (i x+\frac {\pi }{2}\right )^{9/2}}dx}{a \sqrt {a \cosh ^3(x)}}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \left (\frac {5}{7} \int \frac {1}{\cosh ^{\frac {5}{2}}(x)}dx+\frac {2 \sinh (x)}{7 \cosh ^{\frac {7}{2}}(x)}\right )}{a \sqrt {a \cosh ^3(x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \left (\frac {2 \sinh (x)}{7 \cosh ^{\frac {7}{2}}(x)}+\frac {5}{7} \int \frac {1}{\sin \left (i x+\frac {\pi }{2}\right )^{5/2}}dx\right )}{a \sqrt {a \cosh ^3(x)}}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\cosh (x)}}dx+\frac {2 \sinh (x)}{3 \cosh ^{\frac {3}{2}}(x)}\right )+\frac {2 \sinh (x)}{7 \cosh ^{\frac {7}{2}}(x)}\right )}{a \sqrt {a \cosh ^3(x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \left (\frac {2 \sinh (x)}{7 \cosh ^{\frac {7}{2}}(x)}+\frac {5}{7} \left (\frac {2 \sinh (x)}{3 \cosh ^{\frac {3}{2}}(x)}+\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (i x+\frac {\pi }{2}\right )}}dx\right )\right )}{a \sqrt {a \cosh ^3(x)}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\cosh ^{\frac {3}{2}}(x) \left (\frac {2 \sinh (x)}{7 \cosh ^{\frac {7}{2}}(x)}+\frac {5}{7} \left (\frac {2 \sinh (x)}{3 \cosh ^{\frac {3}{2}}(x)}-\frac {2}{3} i \operatorname {EllipticF}\left (\frac {i x}{2},2\right )\right )\right )}{a \sqrt {a \cosh ^3(x)}}\)

Input:

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

Output:

(Cosh[x]^(3/2)*((2*Sinh[x])/(7*Cosh[x]^(7/2)) + (5*(((-2*I)/3)*EllipticF[( 
I/2)*x, 2] + (2*Sinh[x])/(3*Cosh[x]^(3/2))))/7))/(a*Sqrt[a*Cosh[x]^3])
 

Defintions of rubi rules used

rule 3042

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

rule 3116

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] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3120

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 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]])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 554 vs. \(2 (57) = 114\).

Time = 0.13 (sec) , antiderivative size = 554, normalized size of antiderivative = 7.39 \[ \int \frac {1}{\left (a \cosh ^3(x)\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

4/21*(5*sqrt(1/2)*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 4*(7*cosh 
(x)^2 + 1)*sinh(x)^6 + 4*cosh(x)^6 + 8*(7*cosh(x)^3 + 3*cosh(x))*sinh(x)^5 
 + 2*(35*cosh(x)^4 + 30*cosh(x)^2 + 3)*sinh(x)^4 + 6*cosh(x)^4 + 8*(7*cosh 
(x)^5 + 10*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 + 15*cosh(x)^ 
4 + 9*cosh(x)^2 + 1)*sinh(x)^2 + 4*cosh(x)^2 + 8*(cosh(x)^7 + 3*cosh(x)^5 
+ 3*cosh(x)^3 + cosh(x))*sinh(x) + 1)*sqrt(a)*weierstrassPInverse(-4, 0, c 
osh(x) + sinh(x)) + (5*cosh(x)^7 + 35*cosh(x)*sinh(x)^6 + 5*sinh(x)^7 + (1 
05*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*cosh(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*cosh(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(-1)]

Timed out. \[ \int \frac {1}{\left (a \cosh ^3(x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((a*cosh(x)^3)^(-3/2), x)
 

Giac [F]

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

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

Output:

integrate((a*cosh(x)^3)^(-3/2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\left (a \cosh ^3(x)\right )^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cosh \left (x \right )}}{\cosh \left (x \right )^{5}}d x \right )}{a^{2}} \] Input:

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

Output:

(sqrt(a)*int(sqrt(cosh(x))/cosh(x)**5,x))/a**2