\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [582]

Optimal result

Integrand size = 33, antiderivative size = 223 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (2 a^2 A-3 A b^2+a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (4 a^2 A b-3 A b^3-2 a^3 B+a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^3 \left (a^2-b^2\right ) d}+\frac {b \left (5 a^2 A b-3 A b^3-3 a^3 B+a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \] Output:

(2*A*a^2-3*A*b^2+B*a*b)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/(a^2-b^2 
)/d-(4*A*a^2*b-3*A*b^3-2*B*a^3+B*a*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1 
/2))/a^3/(a^2-b^2)/d+b*(5*A*a^2*b-3*A*b^3-3*B*a^3+B*a*b^2)*EllipticPi(sin( 
1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^3/(a-b)/(a+b)^2/d+b*(A*b-B*a)*cos(d*x+ 
c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 1.96 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.26 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 b (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {\frac {2 \left (2 a^2 A-A b^2-a b B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 (-A b+a B) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {2 \left (2 a^2 A-3 A b^2+a b B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 a d} \] Input:

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

Output:

((4*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(b + a*Cos 
[c + d*x])) + ((2*(2*a^2*A - A*b^2 - a*b*B)*EllipticPi[(2*a)/(a + b), (c + 
 d*x)/2, 2])/(a + b) + (8*(-(A*b) + a*B)*((a + b)*EllipticF[(c + d*x)/2, 2 
] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + (2*(2*a^2*A - 
3*A*b^2 + a*b*B)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*( 
a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticP 
i[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c 
 + d*x]^2]))/((a - b)*(a + b)))/(4*a*d)
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3433, 3042, 3468, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B)}{(a \cos (c+d x)+b)^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2}dx\)

\(\Big \downarrow \) 3468

\(\displaystyle \frac {\int \frac {\left (2 A a^2+b B a-3 A b^2\right ) \cos ^2(c+d x)-2 a (A b-a B) \cos (c+d x)+b (A b-a B)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (2 A a^2+b B a-3 A b^2\right ) \cos ^2(c+d x)-2 a (A b-a B) \cos (c+d x)+b (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (2 A a^2+b B a-3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )+b (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {\left (2 a^2 A+a b B-3 A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{a}-\frac {\int -\frac {a b (A b-a B)-\left (-2 B a^3+4 A b a^2+b^2 B a-3 A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\left (2 a^2 A+a b B-3 A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{a}+\frac {\int \frac {a b (A b-a B)-\left (-2 B a^3+4 A b a^2+b^2 B a-3 A b^3\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (2 a^2 A+a b B-3 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {\int \frac {a b (A b-a B)+\left (2 B a^3-4 A b a^2-b^2 B a+3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\int \frac {a b (A b-a B)+\left (2 B a^3-4 A b a^2-b^2 B a+3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {\left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}}{a}+\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {\left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a}+\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 \left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}}{a}+\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {b (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\frac {2 \left (2 a^2 A+a b B-3 A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {\frac {2 b \left (-3 a^3 B+5 a^2 A b+a b^2 B-3 A b^3\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}-\frac {2 \left (-2 a^3 B+4 a^2 A b+a b^2 B-3 A b^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}}{a}}{2 a \left (a^2-b^2\right )}\)

Input:

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

Output:

((2*(2*a^2*A - 3*A*b^2 + a*b*B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((-2*(4 
*a^2*A*b - 3*A*b^3 - 2*a^3*B + a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(a*d) + 
 (2*b*(5*a^2*A*b - 3*A*b^3 - 3*a^3*B + a*b^2*B)*EllipticPi[(2*a)/(a + b), 
(c + d*x)/2, 2])/(a*(a + b)*d))/a)/(2*a*(a^2 - b^2)) + (b*(A*b - a*B)*Sqrt 
[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(b + a*Cos[c + d*x]))
 

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 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 3120

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

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 3433

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 3468

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

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 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. \(842\) vs. \(2(226)=452\).

Time = 5.99 (sec) , antiderivative size = 843, normalized size of antiderivative = 3.78

method	result	size
default	\(\text {Expression too large to display}\)	\(843\)

Input:

int(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x,method=_RETURNV 
ERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/a^3/(-2*sin 
(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2 
))*b+A*a*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*a*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2)))-2/a^2*b*(3*A*b-2*B*a)/(a^2-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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*b^2*(A* 
b-B*a)/a^3*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si 
n(1/2*d*x+1/2*c)^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/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)/(-2*sin(1/2*d*x+ 
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) 
+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1 
)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos 
(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(- 
2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c 
)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b) 
*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*si 
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2* 
c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2...
 

Fricas [F]

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

integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorith 
m="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorith 
m="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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