\(\int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx\) [354]

Optimal result

Integrand size = 13, antiderivative size = 112 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {2 (b \cosh (x)+a \sinh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {2 i E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \] Output:

2*(b*cosh(x)+a*sinh(x))/(a^2-b^2)/(a*cosh(x)+b*sinh(x))^(1/2)+2*I*Elliptic 
E(sin(1/2*I*x-1/2*arctan(-I*b,a)),2^(1/2))*(a*cosh(x)+b*sinh(x))^(1/2)/(a^ 
2-b^2)/((a*cosh(x)+b*sinh(x))/(a^2-b^2)^(1/2))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.32 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {b \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cosh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )\right ) \sinh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )-\sqrt {-\sinh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )} \left (2 a \sqrt {1-\frac {b^2}{a^2}} \cosh (x)-2 a \cosh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )+b \sinh \left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )\right )}{a b \sqrt {1-\frac {b^2}{a^2}} \sqrt {a \cosh (x)+b \sinh (x)} \sqrt {-\sinh ^2\left (x+\text {arctanh}\left (\frac {b}{a}\right )\right )}} \] Input:

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

Output:

(b*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cosh[x + ArcTanh[b/a]]^2]*Sinh[x 
 + ArcTanh[b/a]] - Sqrt[-Sinh[x + ArcTanh[b/a]]^2]*(2*a*Sqrt[1 - b^2/a^2]* 
Cosh[x] - 2*a*Cosh[x + ArcTanh[b/a]] + b*Sinh[x + ArcTanh[b/a]]))/(a*b*Sqr 
t[1 - b^2/a^2]*Sqrt[a*Cosh[x] + b*Sinh[x]]*Sqrt[-Sinh[x + ArcTanh[b/a]]^2] 
)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3555, 3042, 3557, 3042, 3119}

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.

\(\displaystyle \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \cos (i x)-i b \sin (i x))^{3/2}}dx\)

\(\Big \downarrow \) 3555

\(\displaystyle \frac {2 (a \sinh (x)+b \cosh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}-\frac {\int \sqrt {a \cosh (x)+b \sinh (x)}dx}{a^2-b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (a \sinh (x)+b \cosh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}-\frac {\int \sqrt {a \cos (i x)-i b \sin (i x)}dx}{a^2-b^2}\)

\(\Big \downarrow \) 3557

\(\displaystyle \frac {2 (a \sinh (x)+b \cosh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}-\frac {\sqrt {a \cosh (x)+b \sinh (x)} \int \sqrt {\cosh \left (x+i \tan ^{-1}(a,-i b)\right )}dx}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 (a \sinh (x)+b \cosh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}-\frac {\sqrt {a \cosh (x)+b \sinh (x)} \int \sqrt {\sin \left (i x-\tan ^{-1}(a,-i b)+\frac {\pi }{2}\right )}dx}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 (a \sinh (x)+b \cosh (x))}{\left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}+\frac {2 i \sqrt {a \cosh (x)+b \sinh (x)} E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{\left (a^2-b^2\right ) \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\)

Input:

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

Output:

(2*(b*Cosh[x] + a*Sinh[x]))/((a^2 - b^2)*Sqrt[a*Cosh[x] + b*Sinh[x]]) + (( 
2*I)*EllipticE[(I*x - ArcTan[a, (-I)*b])/2, 2]*Sqrt[a*Cosh[x] + b*Sinh[x]] 
)/((a^2 - b^2)*Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])
 

Defintions of rubi rules used

rule 3042

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

rule 3119

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

rule 3555

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x 
_Symbol] :> Simp[(b*Cos[c + d*x] - a*Sin[c + d*x])*((a*Cos[c + d*x] + b*Sin 
[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[(n + 2)/((n + 1)*(a^ 
2 + b^2))   Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1] && NeQ[n, -2]
 

rule 3557

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x 
_Symbol] :> Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*S 
in[c + d*x])/Sqrt[a^2 + b^2])^n   Int[Cos[c + d*x - ArcTan[a, b]]^n, x], x] 
 /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 
+ b^2, 0] || EqQ[a^2 + b^2, 0])
 

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.29

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\cosh \left (x \right )\right )}{\sqrt {a^{2}-b^{2}}\, \sqrt {-\sinh \left (x \right ) \sqrt {a^{2}-b^{2}}}}\)	\(33\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (100) = 200\).

Time = 0.10 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(a \cosh (x)+b \sinh (x))^{3/2}} \, dx=\frac {4 \, {\left ({\left ({\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a + b\right )} \sinh \left (x\right )^{2} + a - b\right )} \sqrt {\frac {1}{2} \, a + \frac {1}{2} \, b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) + {\left ({\left (a + b\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a + b\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a + b\right )} \sinh \left (x\right )^{2}\right )} \sqrt {a \cosh \left (x\right ) + b \sinh \left (x\right )}\right )}}{a^{3} - a^{2} b - a b^{2} + b^{3} + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} \sinh \left (x\right )^{2}} \] Input:

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

Output:

4*(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + a 
- b)*sqrt(1/2*a + 1/2*b)*weierstrassZeta(-4*(a - b)/(a + b), 0, weierstras 
sPInverse(-4*(a - b)/(a + b), 0, cosh(x) + sinh(x))) + ((a + b)*cosh(x)^2 
+ 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2)*sqrt(a*cosh(x) + b*sinh(x 
)))/(a^3 - a^2*b - a*b^2 + b^3 + (a^3 + a^2*b - a*b^2 - b^3)*cosh(x)^2 + 2 
*(a^3 + a^2*b - a*b^2 - b^3)*cosh(x)*sinh(x) + (a^3 + a^2*b - a*b^2 - b^3) 
*sinh(x)^2)
 

Sympy [F]

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

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

Output:

Integral((a*cosh(x) + b*sinh(x))**(-3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cosh(x)*a + sinh(x)*b)/(cosh(x)**2*a**2 + 2*cosh(x)*sinh(x)*a*b + 
 sinh(x)**2*b**2),x)