\(\int \frac {\cosh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) [342]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \frac {\cosh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=-\frac {E\left (\arctan (\sinh (e+f x))\left |1-\frac {b}{a}\right .\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{b f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\operatorname {EllipticF}\left (\arctan (\sinh (e+f x)),1-\frac {b}{a}\right ) \text {sech}(e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{a f \sqrt {\frac {\text {sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac {\sqrt {a+b \sinh ^2(e+f x)} \tanh (e+f x)}{b f} \] Output:

-EllipticE(sinh(f*x+e)/(1+sinh(f*x+e)^2)^(1/2),(1-b/a)^(1/2))*sech(f*x+e)* 
(a+b*sinh(f*x+e)^2)^(1/2)/b/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+ 
InverseJacobiAM(arctan(sinh(f*x+e)),(1-b/a)^(1/2))*sech(f*x+e)*(a+b*sinh(f 
*x+e)^2)^(1/2)/a/f/(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)/a)^(1/2)+(a+b*sinh(f 
*x+e)^2)^(1/2)*tanh(f*x+e)/b/f
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.54 \[ \int \frac {\cosh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx=-\frac {i \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}} \left (a E\left (i (e+f x)\left |\frac {b}{a}\right .\right )+(-a+b) \operatorname {EllipticF}\left (i (e+f x),\frac {b}{a}\right )\right )}{b f \sqrt {2 a-b+b \cosh (2 (e+f x))}} \] Input:

Integrate[Cosh[e + f*x]^2/Sqrt[a + b*Sinh[e + f*x]^2],x]
 

Output:

((-I)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*(a*EllipticE[I*(e + f*x), b/ 
a] + (-a + b)*EllipticF[I*(e + f*x), b/a]))/(b*f*Sqrt[2*a - b + b*Cosh[2*( 
e + f*x)]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.29, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3671, 324, 320, 388, 313}

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.

\(\displaystyle \int \frac {\cosh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (i e+i f x)^2}{\sqrt {a-b \sin (i e+i f x)^2}}dx\)

\(\Big \downarrow \) 3671

\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \int \frac {\sqrt {\sinh ^2(e+f x)+1}}{\sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)}{f}\)

\(\Big \downarrow \) 324

\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \left (\int \frac {1}{\sqrt {\sinh ^2(e+f x)+1} \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)+\int \frac {\sinh ^2(e+f x)}{\sqrt {\sinh ^2(e+f x)+1} \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)\right )}{f}\)

\(\Big \downarrow \) 320

\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \left (\int \frac {\sinh ^2(e+f x)}{\sqrt {\sinh ^2(e+f x)+1} \sqrt {b \sinh ^2(e+f x)+a}}d\sinh (e+f x)+\frac {\sqrt {a+b \sinh ^2(e+f x)} \operatorname {EllipticF}\left (\arctan (\sinh (e+f x)),1-\frac {b}{a}\right )}{a \sqrt {\sinh ^2(e+f x)+1} \sqrt {\frac {a+b \sinh ^2(e+f x)}{a \left (\sinh ^2(e+f x)+1\right )}}}\right )}{f}\)

\(\Big \downarrow \) 388

\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \left (-\frac {\int \frac {\sqrt {b \sinh ^2(e+f x)+a}}{\left (\sinh ^2(e+f x)+1\right )^{3/2}}d\sinh (e+f x)}{b}+\frac {\sqrt {a+b \sinh ^2(e+f x)} \operatorname {EllipticF}\left (\arctan (\sinh (e+f x)),1-\frac {b}{a}\right )}{a \sqrt {\sinh ^2(e+f x)+1} \sqrt {\frac {a+b \sinh ^2(e+f x)}{a \left (\sinh ^2(e+f x)+1\right )}}}+\frac {\sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{b \sqrt {\sinh ^2(e+f x)+1}}\right )}{f}\)

\(\Big \downarrow \) 313

\(\displaystyle \frac {\sqrt {\cosh ^2(e+f x)} \text {sech}(e+f x) \left (\frac {\sqrt {a+b \sinh ^2(e+f x)} \operatorname {EllipticF}\left (\arctan (\sinh (e+f x)),1-\frac {b}{a}\right )}{a \sqrt {\sinh ^2(e+f x)+1} \sqrt {\frac {a+b \sinh ^2(e+f x)}{a \left (\sinh ^2(e+f x)+1\right )}}}-\frac {\sqrt {a+b \sinh ^2(e+f x)} E\left (\arctan (\sinh (e+f x))\left |1-\frac {b}{a}\right .\right )}{b \sqrt {\sinh ^2(e+f x)+1} \sqrt {\frac {a+b \sinh ^2(e+f x)}{a \left (\sinh ^2(e+f x)+1\right )}}}+\frac {\sinh (e+f x) \sqrt {a+b \sinh ^2(e+f x)}}{b \sqrt {\sinh ^2(e+f x)+1}}\right )}{f}\)

Input:

Int[Cosh[e + f*x]^2/Sqrt[a + b*Sinh[e + f*x]^2],x]
 

Output:

(Sqrt[Cosh[e + f*x]^2]*Sech[e + f*x]*((Sinh[e + f*x]*Sqrt[a + b*Sinh[e + f 
*x]^2])/(b*Sqrt[1 + Sinh[e + f*x]^2]) - (EllipticE[ArcTan[Sinh[e + f*x]], 
1 - b/a]*Sqrt[a + b*Sinh[e + f*x]^2])/(b*Sqrt[1 + Sinh[e + f*x]^2]*Sqrt[(a 
 + b*Sinh[e + f*x]^2)/(a*(1 + Sinh[e + f*x]^2))]) + (EllipticF[ArcTan[Sinh 
[e + f*x]], 1 - b/a]*Sqrt[a + b*Sinh[e + f*x]^2])/(a*Sqrt[1 + Sinh[e + f*x 
]^2]*Sqrt[(a + b*Sinh[e + f*x]^2)/(a*(1 + Sinh[e + f*x]^2))])))/f
 

Defintions of rubi rules used

rule 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c 
+ d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; FreeQ 
[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]
 

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 324

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
a   Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] + Simp[b   Int[x^2/(Sqr 
t[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[d/c 
] && PosQ[b/a]
 

rule 388

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt[c + d*x^2])), x] - Simp[c/b   Int[Sqrt[ 
a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 3042

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

rule 3671

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff*(Sqrt[ 
Cos[e + f*x]^2]/(f*Cos[e + f*x]))   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a 
 + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
 && IntegerQ[m/2] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {\sqrt {\frac {a +b \sinh \left (f x +e \right )^{2}}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \operatorname {EllipticE}\left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )}{\sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \sinh \left (f x +e \right )^{2}}\, f}\)	\(86\)

Input:

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

Output:

((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)* 
(-b/a)^(1/2),(1/b*a)^(1/2))/(-b/a)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^( 
1/2)/f
 

Fricas [F]

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

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

Output:

integral(cosh(f*x + e)^2/sqrt(b*sinh(f*x + e)^2 + a), x)
 

Sympy [F]

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

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

Output:

Integral(cosh(e + f*x)**2/sqrt(a + b*sinh(e + f*x)**2), x)
 

Maxima [F]

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

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

Output:

integrate(cosh(f*x + e)^2/sqrt(b*sinh(f*x + e)^2 + a), x)
                                                                                    
                                                                                    
 

Giac [F]

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

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

Output:

integrate(cosh(f*x + e)^2/sqrt(b*sinh(f*x + e)^2 + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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