\(\int \frac {\cos ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [284]

Optimal result

Integrand size = 25, antiderivative size = 116 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {a E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{b f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(a+b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {\frac {a+b \sin ^2(e+f x)}{a}}}{b f \sqrt {a+b \sin ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \left (-a E\left (e+f x\left |-\frac {b}{a}\right .\right )+(a+b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )\right )}{b f \sqrt {2 a+b-b \cos (2 (e+f x))}} \] Input:

Integrate[Cos[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3671, 326, 323, 321, 330, 327}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3671

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\sqrt {1-\sin ^2(e+f x)}}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{f}\)

\(\Big \downarrow \) 326

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}-\frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{f}\)

\(\Big \downarrow \) 323

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{f}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )}{f}\)

\(\Big \downarrow \) 330

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )}{f}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {(a+b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )}{f}\)

Input:

Int[Cos[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]
 

Output:

(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*(-((EllipticE[ArcSin[Sin[e + f*x]], -(b 
/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*Sqrt[1 + (b*Sin[e + f*x]^2)/a])) + ((a 
 + b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[1 + (b*Sin[e + f*x]^2)/ 
a])/(b*Sqrt[a + b*Sin[e + f*x]^2])))/f
 

Defintions of rubi rules used

rule 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 323

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

rule 326

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

rule 327

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

rule 330

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

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 = 1.57 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \sin \left (f x +e \right )^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +\operatorname {EllipticF}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -\operatorname {EllipticE}\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \right )}{b \cos \left (f x +e \right ) \sqrt {a +b \sin \left (f x +e \right )^{2}}\, f}\)	\(111\)

Input:

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

Output:

(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*(EllipticF(sin(f*x+e),(- 
b/a)^(1/2))*a+EllipticF(sin(f*x+e),(-b/a)^(1/2))*b-EllipticE(sin(f*x+e),(- 
b/a)^(1/2))*a)/b/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
 

Fricas [F]

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

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

Output:

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)^2/(b*cos(f*x + e)^2 
 - a - b), x)
 

Sympy [F]

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

integrate(cos(f*x+e)**2/(a+b*sin(f*x+e)**2)**(1/2),x)
 

Output:

Integral(cos(e + f*x)**2/sqrt(a + b*sin(e + f*x)**2), x)
 

Maxima [F]

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

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

Output:

integrate(cos(f*x + e)^2/sqrt(b*sin(f*x + e)^2 + a), x)
 

Giac [F]

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

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

Output:

integrate(cos(f*x + e)^2/sqrt(b*sin(f*x + e)^2 + a), x)
 

Mupad [F(-1)]

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

int(cos(e + f*x)^2/(a + b*sin(e + f*x)^2)^(1/2),x)
 

Output:

int(cos(e + f*x)^2/(a + b*sin(e + f*x)^2)^(1/2), x)
 

Reduce [F]

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

int(cos(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x)
 

Output:

int((sqrt(sin(e + f*x)**2*b + a)*cos(e + f*x)**2)/(sin(e + f*x)**2*b + a), 
x)