\(\int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx\) [110]

Optimal result

Integrand size = 10, antiderivative size = 197 \[ \int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx=-\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}-\frac {8 a b \cosh (x)}{3 \left (a^2+b^2\right )^2 \sqrt {a+b \sinh (x)}}+\frac {8 i a E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right ) \sqrt {a+b \sinh (x)}}{3 \left (a^2+b^2\right )^2 \sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}}}{3 \left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}} \] Output:

-2/3*b*cosh(x)/(a^2+b^2)/(a+b*sinh(x))^(3/2)-8/3*a*b*cosh(x)/(a^2+b^2)^2/( 
a+b*sinh(x))^(1/2)+8/3*I*a*EllipticE(cos(1/4*Pi+1/2*I*x),2^(1/2)*(b/(I*a+b 
))^(1/2))*(a+b*sinh(x))^(1/2)/(a^2+b^2)^2/((a+b*sinh(x))/(a-I*b))^(1/2)+2/ 
3*I*InverseJacobiAM(-1/4*Pi+1/2*I*x,2^(1/2)*(b/(I*a+b))^(1/2))*((a+b*sinh( 
x))/(a-I*b))^(1/2)/(a^2+b^2)/(a+b*sinh(x))^(1/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx=\frac {\frac {8 i a E\left (\frac {1}{4} (\pi -2 i x)|-\frac {2 i b}{a-i b}\right ) (a+b \sinh (x))^2}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-2 i \left (a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (\pi -2 i x),-\frac {2 i b}{a-i b}\right ) (a+b \sinh (x)) \sqrt {\frac {a+b \sinh (x)}{a-i b}}-2 b \cosh (x) \left (5 a^2+b^2+4 a b \sinh (x)\right )}{3 \left (a^2+b^2\right )^2 (a+b \sinh (x))^{3/2}} \] Input:

Integrate[(a + b*Sinh[x])^(-5/2),x]
 

Output:

(((8*I)*a*EllipticE[(Pi - (2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*(a + b*Sinh[x] 
)^2)/Sqrt[(a + b*Sinh[x])/(a - I*b)] - (2*I)*(a^2 + b^2)*EllipticF[(Pi - ( 
2*I)*x)/4, ((-2*I)*b)/(a - I*b)]*(a + b*Sinh[x])*Sqrt[(a + b*Sinh[x])/(a - 
 I*b)] - 2*b*Cosh[x]*(5*a^2 + b^2 + 4*a*b*Sinh[x]))/(3*(a^2 + b^2)^2*(a + 
b*Sinh[x])^(3/2))
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.500, Rules used = {3042, 3143, 27, 3042, 3233, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

\(\displaystyle \int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a-i b \sin (i x))^{5/2}}dx\)

\(\Big \downarrow \) 3143

\(\displaystyle -\frac {2 \int -\frac {3 a-b \sinh (x)}{2 (a+b \sinh (x))^{3/2}}dx}{3 \left (a^2+b^2\right )}-\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 a-b \sinh (x)}{(a+b \sinh (x))^{3/2}}dx}{3 \left (a^2+b^2\right )}-\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {\int \frac {3 a+i b \sin (i x)}{(a-i b \sin (i x))^{3/2}}dx}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {-\frac {2 \int -\frac {3 a^2+4 b \sinh (x) a-b^2}{2 \sqrt {a+b \sinh (x)}}dx}{a^2+b^2}-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}}{3 \left (a^2+b^2\right )}-\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {3 a^2+4 b \sinh (x) a-b^2}{\sqrt {a+b \sinh (x)}}dx}{a^2+b^2}-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}}{3 \left (a^2+b^2\right )}-\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\int \frac {3 a^2-4 i b \sin (i x) a-b^2}{\sqrt {a-i b \sin (i x)}}dx}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {\frac {4 a \int \sqrt {a+b \sinh (x)}dx-\left (a^2+b^2\right ) \int \frac {1}{\sqrt {a+b \sinh (x)}}dx}{a^2+b^2}-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}}{3 \left (a^2+b^2\right )}-\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {4 a \int \sqrt {a-i b \sin (i x)}dx-\left (a^2+b^2\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3134

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\frac {4 a \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}dx}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\left (a^2+b^2\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\frac {4 a \sqrt {a+b \sinh (x)} \int \sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}dx}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\left (a^2+b^2\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3132

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\frac {8 i a \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\left (a^2+b^2\right ) \int \frac {1}{\sqrt {a-i b \sin (i x)}}dx}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3142

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\frac {8 i a \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}+\frac {b \sinh (x)}{a-i b}}}dx}{\sqrt {a+b \sinh (x)}}}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\frac {8 i a \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {\left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \int \frac {1}{\sqrt {\frac {a}{a-i b}-\frac {i b \sin (i x)}{a-i b}}}dx}{\sqrt {a+b \sinh (x)}}}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3140

\(\displaystyle -\frac {2 b \cosh (x)}{3 \left (a^2+b^2\right ) (a+b \sinh (x))^{3/2}}+\frac {-\frac {8 a b \cosh (x)}{\left (a^2+b^2\right ) \sqrt {a+b \sinh (x)}}+\frac {\frac {8 i a \sqrt {a+b \sinh (x)} E\left (\frac {\pi }{4}-\frac {i x}{2}|\frac {2 b}{i a+b}\right )}{\sqrt {\frac {a+b \sinh (x)}{a-i b}}}-\frac {2 i \left (a^2+b^2\right ) \sqrt {\frac {a+b \sinh (x)}{a-i b}} \operatorname {EllipticF}\left (\frac {\pi }{4}-\frac {i x}{2},\frac {2 b}{i a+b}\right )}{\sqrt {a+b \sinh (x)}}}{a^2+b^2}}{3 \left (a^2+b^2\right )}\)

Input:

Int[(a + b*Sinh[x])^(-5/2),x]
 

Output:

(-2*b*Cosh[x])/(3*(a^2 + b^2)*(a + b*Sinh[x])^(3/2)) + ((-8*a*b*Cosh[x])/( 
(a^2 + b^2)*Sqrt[a + b*Sinh[x]]) + (((8*I)*a*EllipticE[Pi/4 - (I/2)*x, (2* 
b)/(I*a + b)]*Sqrt[a + b*Sinh[x]])/Sqrt[(a + b*Sinh[x])/(a - I*b)] - ((2*I 
)*(a^2 + b^2)*EllipticF[Pi/4 - (I/2)*x, (2*b)/(I*a + b)]*Sqrt[(a + b*Sinh[ 
x])/(a - I*b)])/Sqrt[a + b*Sinh[x]])/(a^2 + b^2))/(3*(a^2 + b^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3143

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

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 3233

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + 
 f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( 
m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (172 ) = 344\).

Time = 0.46 (sec) , antiderivative size = 438, normalized size of antiderivative = 2.22

method	result	size
default	\(\frac {\sqrt {\cosh \left (x \right )^{2} \left (a +b \sinh \left (x \right )\right )}\, \left (-\frac {2 \sqrt {\cosh \left (x \right )^{2} \left (a +b \sinh \left (x \right )\right )}}{3 b \left (a^{2}+b^{2}\right ) \left (\sinh \left (x \right )+\frac {a}{b}\right )^{2}}-\frac {8 b \cosh \left (x \right )^{2} a}{3 \left (a^{2}+b^{2}\right )^{2} \sqrt {\cosh \left (x \right )^{2} \left (a +b \sinh \left (x \right )\right )}}+\frac {2 \left (3 a^{2}-b^{2}\right ) \left (\frac {a}{b}-i\right ) \sqrt {\frac {-a -b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {b \left (i+\sinh \left (x \right )\right )}{i b -a}}\, \operatorname {EllipticF}\left (\sqrt {\frac {-a -b \sinh \left (x \right )}{i b -a}}, \sqrt {\frac {-i b +a}{i b +a}}\right )}{\left (3 a^{4}+6 a^{2} b^{2}+3 b^{4}\right ) \sqrt {\cosh \left (x \right )^{2} \left (a +b \sinh \left (x \right )\right )}}+\frac {8 a b \left (\frac {a}{b}-i\right ) \sqrt {\frac {-a -b \sinh \left (x \right )}{i b -a}}\, \sqrt {\frac {\left (i-\sinh \left (x \right )\right ) b}{i b +a}}\, \sqrt {\frac {b \left (i+\sinh \left (x \right )\right )}{i b -a}}\, \left (\left (-\frac {a}{b}-i\right ) \operatorname {EllipticE}\left (\sqrt {\frac {-a -b \sinh \left (x \right )}{i b -a}}, \sqrt {\frac {-i b +a}{i b +a}}\right )+i \operatorname {EllipticF}\left (\sqrt {\frac {-a -b \sinh \left (x \right )}{i b -a}}, \sqrt {\frac {-i b +a}{i b +a}}\right )\right )}{3 \left (a^{2}+b^{2}\right )^{2} \sqrt {\cosh \left (x \right )^{2} \left (a +b \sinh \left (x \right )\right )}}\right )}{\cosh \left (x \right ) \sqrt {a +b \sinh \left (x \right )}}\)	\(438\)

Input:

int(1/(a+b*sinh(x))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(cosh(x)^2*(a+b*sinh(x)))^(1/2)*(-2/3/b/(a^2+b^2)*(cosh(x)^2*(a+b*sinh(x)) 
)^(1/2)/(sinh(x)+1/b*a)^2-8/3*b*cosh(x)^2/(a^2+b^2)^2*a/(cosh(x)^2*(a+b*si 
nh(x)))^(1/2)+2*(3*a^2-b^2)/(3*a^4+6*a^2*b^2+3*b^4)*(1/b*a-I)*((-a-b*sinh( 
x))/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*(b*(I+sinh(x))/(I*b-a))^( 
1/2)/(cosh(x)^2*(a+b*sinh(x)))^(1/2)*EllipticF(((-a-b*sinh(x))/(I*b-a))^(1 
/2),((a-I*b)/(I*b+a))^(1/2))+8/3*a*b/(a^2+b^2)^2*(1/b*a-I)*((-a-b*sinh(x)) 
/(I*b-a))^(1/2)*((I-sinh(x))*b/(I*b+a))^(1/2)*(b*(I+sinh(x))/(I*b-a))^(1/2 
)/(cosh(x)^2*(a+b*sinh(x)))^(1/2)*((-1/b*a-I)*EllipticE(((-a-b*sinh(x))/(I 
*b-a))^(1/2),((a-I*b)/(I*b+a))^(1/2))+I*EllipticF(((-a-b*sinh(x))/(I*b-a)) 
^(1/2),((a-I*b)/(I*b+a))^(1/2))))/cosh(x)/(a+b*sinh(x))^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1198 vs. \(2 (164) = 328\).

Time = 0.12 (sec) , antiderivative size = 1198, normalized size of antiderivative = 6.08 \[ \int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sinh(x))^(5/2),x, algorithm="fricas")
 

Output:

4/9*(sqrt(1/2)*((a^2*b^2 - 3*b^4)*cosh(x)^4 + (a^2*b^2 - 3*b^4)*sinh(x)^4 
+ a^2*b^2 - 3*b^4 + 4*(a^3*b - 3*a*b^3)*cosh(x)^3 + 4*(a^3*b - 3*a*b^3 + ( 
a^2*b^2 - 3*b^4)*cosh(x))*sinh(x)^3 + 2*(2*a^4 - 7*a^2*b^2 + 3*b^4)*cosh(x 
)^2 + 2*(2*a^4 - 7*a^2*b^2 + 3*b^4 + 3*(a^2*b^2 - 3*b^4)*cosh(x)^2 + 6*(a^ 
3*b - 3*a*b^3)*cosh(x))*sinh(x)^2 - 4*(a^3*b - 3*a*b^3)*cosh(x) - 4*(a^3*b 
 - 3*a*b^3 - (a^2*b^2 - 3*b^4)*cosh(x)^3 - 3*(a^3*b - 3*a*b^3)*cosh(x)^2 - 
 (2*a^4 - 7*a^2*b^2 + 3*b^4)*cosh(x))*sinh(x))*sqrt(b)*weierstrassPInverse 
(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, 1/3*(3*b*cosh(x) + 
3*b*sinh(x) + 2*a)/b) - 12*sqrt(1/2)*(a*b^3*cosh(x)^4 + a*b^3*sinh(x)^4 + 
4*a^2*b^2*cosh(x)^3 - 4*a^2*b^2*cosh(x) + a*b^3 + 4*(a*b^3*cosh(x) + a^2*b 
^2)*sinh(x)^3 + 2*(2*a^3*b - a*b^3)*cosh(x)^2 + 2*(3*a*b^3*cosh(x)^2 + 6*a 
^2*b^2*cosh(x) + 2*a^3*b - a*b^3)*sinh(x)^2 + 4*(a*b^3*cosh(x)^3 + 3*a^2*b 
^2*cosh(x)^2 - a^2*b^2 + (2*a^3*b - a*b^3)*cosh(x))*sinh(x))*sqrt(b)*weier 
strassZeta(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, weierstra 
ssPInverse(4/3*(4*a^2 + 3*b^2)/b^2, -8/27*(8*a^3 + 9*a*b^2)/b^3, 1/3*(3*b* 
cosh(x) + 3*b*sinh(x) + 2*a)/b)) - 3*(4*a*b^3*cosh(x)^4 + 4*a*b^3*sinh(x)^ 
4 + (13*a^2*b^2 + b^4)*cosh(x)^3 + (16*a*b^3*cosh(x) + 13*a^2*b^2 + b^4)*s 
inh(x)^3 + 4*(2*a^3*b - a*b^3)*cosh(x)^2 + (24*a*b^3*cosh(x)^2 + 8*a^3*b - 
 4*a*b^3 + 3*(13*a^2*b^2 + b^4)*cosh(x))*sinh(x)^2 - (3*a^2*b^2 - b^4)*cos 
h(x) + (16*a*b^3*cosh(x)^3 - 3*a^2*b^2 + b^4 + 3*(13*a^2*b^2 + b^4)*cos...
 

Sympy [F]

\[ \int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx=\int \frac {1}{\left (a + b \sinh {\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a+b*sinh(x))**(5/2),x)
 

Output:

Integral((a + b*sinh(x))**(-5/2), x)
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate(1/(a+b*sinh(x))^(5/2),x, algorithm="maxima")
 

Output:

integrate((b*sinh(x) + a)^(-5/2), x)
 

Giac [F]

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

integrate(1/(a+b*sinh(x))^(5/2),x, algorithm="giac")
 

Output:

integrate((b*sinh(x) + a)^(-5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sinh (x))^{5/2}} \, dx=\int \frac {1}{{\left (a+b\,\mathrm {sinh}\left (x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(a + b*sinh(x))^(5/2),x)
 

Output:

int(1/(a + b*sinh(x))^(5/2), x)
 

Reduce [F]

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

int(1/(a+b*sinh(x))^(5/2),x)
 

Output:

int(sqrt(sinh(x)*b + a)/(sinh(x)**3*b**3 + 3*sinh(x)**2*a*b**2 + 3*sinh(x) 
*a**2*b + a**3),x)