\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [653]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 214 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{a d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {(A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{a d \sqrt {a+b \cos (c+d x)}}+\frac {A \sqrt {a+b \cos (c+d x)} \tan (c+d x)}{a d} \] Output:

-A*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^( 
1/2))/a/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+(A+2*C)*((a+b*cos(d*x+c))/(a+b))^ 
(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/d/(a+b*cos(d* 
x+c))^(1/2)-A*b*((a+b*cos(d*x+c))/(a+b))^(1/2)*EllipticPi(sin(1/2*d*x+1/2* 
c),2,2^(1/2)*(b/(a+b))^(1/2))/a/d/(a+b*cos(d*x+c))^(1/2)+A*(a+b*cos(d*x+c) 
)^(1/2)*tan(d*x+c)/a/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 20.01 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.61 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 A \cos (c+d x) \sqrt {a+b \cos (c+d x)} \left (C+A \sec ^2(c+d x)\right ) \sin (c+d x)}{a d (2 A+C+C \cos (2 c+2 d x))}+\frac {\cos ^2(c+d x) \left (C+A \sec ^2(c+d x)\right ) \left (\frac {8 a C \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}-\frac {6 A b \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{\sqrt {a+b \cos (c+d x)}}+\frac {2 i A b \sqrt {\frac {b-b \cos (c+d x)}{a+b}} \sqrt {-\frac {b+b \cos (c+d x)}{a-b}} \cos (2 (c+d x)) \left (2 a (a-b) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right )|\frac {a+b}{a-b}\right )+b \left (2 a \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )-b \operatorname {EllipticPi}\left (\frac {a+b}{a},i \text {arcsinh}\left (\sqrt {-\frac {1}{a+b}} \sqrt {a+b \cos (c+d x)}\right ),\frac {a+b}{a-b}\right )\right )\right ) \sin (c+d x)}{a \sqrt {-\frac {1}{a+b}} \sqrt {1-\cos ^2(c+d x)} \sqrt {-\frac {a^2-b^2-2 a (a+b \cos (c+d x))+(a+b \cos (c+d x))^2}{b^2}} \left (2 a^2-b^2-4 a (a+b \cos (c+d x))+2 (a+b \cos (c+d x))^2\right )}\right )}{2 a d (2 A+C+C \cos (2 c+2 d x))} \] Input:

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/Sqrt[a + b*Cos[c + d*x]] 
,x]
 

Output:

(2*A*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(C + A*Sec[c + d*x]^2)*Sin[c + 
d*x])/(a*d*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^2*(C + A*Sec[c 
+ d*x]^2)*((8*a*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2 
, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + d*x]] - (6*A*b*Sqrt[(a + b*Cos[c + d* 
x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*b)/(a + b)])/Sqrt[a + b*Cos[c + 
 d*x]] + ((2*I)*A*b*Sqrt[(b - b*Cos[c + d*x])/(a + b)]*Sqrt[-((b + b*Cos[c 
 + d*x])/(a - b))]*Cos[2*(c + d*x)]*(2*a*(a - b)*EllipticE[I*ArcSinh[Sqrt[ 
-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - b)] + b*(2*a*Ellipt 
icF[I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos[c + d*x]]], (a + b)/(a - 
b)] - b*EllipticPi[(a + b)/a, I*ArcSinh[Sqrt[-(a + b)^(-1)]*Sqrt[a + b*Cos 
[c + d*x]]], (a + b)/(a - b)]))*Sin[c + d*x])/(a*Sqrt[-(a + b)^(-1)]*Sqrt[ 
1 - Cos[c + d*x]^2]*Sqrt[-((a^2 - b^2 - 2*a*(a + b*Cos[c + d*x]) + (a + b* 
Cos[c + d*x])^2)/b^2)]*(2*a^2 - b^2 - 4*a*(a + b*Cos[c + d*x]) + 2*(a + b* 
Cos[c + d*x])^2))))/(2*a*d*(2*A + C + C*Cos[2*c + 2*d*x]))
 

Rubi [A] (verified)

Time = 1.87 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.06, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 3535, 27, 3042, 3538, 25, 3042, 3134, 3042, 3132, 3481, 3042, 3142, 3042, 3140, 3286, 3042, 3284}

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.

\(\displaystyle \int \frac {\sec ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3535

\(\displaystyle \frac {\int -\frac {\left (A b \cos ^2(c+d x)-2 a C \cos (c+d x)+A b\right ) \sec (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{a}+\frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {\left (A b \cos ^2(c+d x)-2 a C \cos (c+d x)+A b\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\int \frac {A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a C \sin \left (c+d x+\frac {\pi }{2}\right )+A b}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {A \int \sqrt {a+b \cos (c+d x)}dx-\frac {\int -\frac {\left (A b^2-a b (A+2 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {\left (A b^2-a b (A+2 C) \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{b}+A \int \sqrt {a+b \cos (c+d x)}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {A b^2-a b (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+A \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {A b^2-a b (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {A \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {A b^2-a b (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {A \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\int \frac {A b^2-a b (A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {A b^2 \int \frac {\sec (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx-a b (A+2 C) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {A b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a b (A+2 C) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {A b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b (A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {A b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a b (A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {A b^2 \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a b (A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {\sec (c+d x)}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b (A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}-\frac {2 a b (A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {A \tan (c+d x) \sqrt {a+b \cos (c+d x)}}{a d}-\frac {\frac {\frac {2 A b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}-\frac {2 a b (A+2 C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{b}+\frac {2 A \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{2 a}\)

Input:

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2)/Sqrt[a + b*Cos[c + d*x]],x]
 

Output:

-1/2*((2*A*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)]) 
/(d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + ((-2*a*b*(A + 2*C)*Sqrt[(a + b*C 
os[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b 
*Cos[c + d*x]]) + (2*A*b^2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticPi[2 
, (c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[c + d*x]]))/b)/a + (A*Sqr 
t[a + b*Cos[c + d*x]]*Tan[c + d*x])/(a*d)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3284
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3286
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*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] && NeQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]
 

rule 3481
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3535
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m 
+ 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin 
[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 0])))
 

rule 3538
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(637\) vs. \(2(214)=428\).

Time = 2.66 (sec) , antiderivative size = 638, normalized size of antiderivative = 2.98

method result size
default \(-\frac {\sqrt {-\left (-2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+2 A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{a \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{2 \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{2 \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{2 a \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right )}{2 a \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a +b}\, d}\) \(638\)
parts \(-\frac {A \sqrt {-\left (-2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{a \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{a \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \sqrt {-\frac {2 b}{a -b}}\right )}{a \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a +b}\, d}-\frac {2 C \sqrt {\left (2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{\sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a +b}\, d}\) \(699\)

Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b* 
sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d 
*x+1/2*c),(-2*b/(a-b))^(1/2))+2*A*(-cos(1/2*d*x+1/2*c)/a*(-2*b*sin(1/2*d*x 
+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)+1/2 
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2) 
/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(co 
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2* 
b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)* 
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2 
))+1/2/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b 
))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*El 
lipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+1/2/a*b*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1 
/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2, 
(-2*b/(a-b))^(1/2))))/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^( 
1/2)/d
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^(1/2),x, algori 
thm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \] Input:

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**(1/2),x)
 

Output:

Integral((A + C*cos(c + d*x)**2)*sec(c + d*x)**2/sqrt(a + b*cos(c + d*x)), 
 x)
 

Maxima [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^(1/2),x, algori 
thm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^2/sqrt(b*cos(d*x + c) + a), 
x)
 

Giac [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \] Input:

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^(1/2),x, algori 
thm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)^2/sqrt(b*cos(d*x + c) + a), 
x)
 

Mupad [F(-1)]

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

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^(1/2)),x)
 

Output:

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^2*(a + b*cos(c + d*x))^(1/2)), x)
 

Reduce [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) b +a}d x \right ) a \] Input:

int((A+C*cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^(1/2),x)
 

Output:

int((sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2*sec(c + d*x)**2)/(cos(c + d* 
x)*b + a),x)*c + int((sqrt(cos(c + d*x)*b + a)*sec(c + d*x)**2)/(cos(c + d 
*x)*b + a),x)*a