3.2.25 \(\int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\) [125]

3.2.25.1 Optimal result
3.2.25.2 Mathematica [A] (verified)
3.2.25.3 Rubi [A] (verified)
3.2.25.4 Maple [B] (verified)
3.2.25.5 Fricas [A] (verification not implemented)
3.2.25.6 Sympy [F]
3.2.25.7 Maxima [F]
3.2.25.8 Giac [B] (verification not implemented)
3.2.25.9 Mupad [F(-1)]

3.2.25.1 Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {7 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {\cos (c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \]

output
7/4*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)-arctan(1/2 
*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)-1/4* 
sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/2*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d 
*x+c))^(1/2)
 
3.2.25.2 Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (7 \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right )-4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) (-1+2 \cos (c+d x)) \sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[Cos[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]
 
output
((7*ArcTanh[Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*ArcTanh[Sqrt[1 - Sec[c + d 
*x]]/Sqrt[2]] + Cos[c + d*x]*(-1 + 2*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]]) 
*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.2.25.3 Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4310, 3042, 4510, 27, 3042, 4408, 3042, 4261, 216, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^2 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4310

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos (c+d x) (a-3 a \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {7 a^2-a^2 \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {7 a^2-a^2 \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {7 a^2-a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {7 a \int \sqrt {\sec (c+d x) a+a}dx-8 a^2 \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {7 a \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-8 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-8 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {14 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-8 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{4 a}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {16 a^2 \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {14 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\sin (c+d x) \cos (c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {14 a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{4 a}\)

input
Int[Cos[c + d*x]^2/Sqrt[a + a*Sec[c + d*x]],x]
 
output
(Cos[c + d*x]*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) - (-1/2*((14*a^ 
(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (8*Sqrt 
[2]*a^(3/2)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x] 
])])/d)/a + (a*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a)
 

3.2.25.3.1 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 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4261
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 4282
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2/f   Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ 
a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4310
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + 
b*Csc[e + f*x]])), x] + Simp[1/(2*b*d*n)   Int[(d*Csc[e + f*x])^(n + 1)*((a 
 + b*(2*n + 1)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, 
 b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0] && IntegerQ[2*n]
 

rule 4408
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c/a   Int[Sqrt[a + b*Csc[e + f*x]], x], x] - 
Simp[(b*c - a*d)/a   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F 
reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
 

rule 4510
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 
3.2.25.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(342\) vs. \(2(122)=244\).

Time = 25.80 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.33

method result size
default \(-\frac {\left (4 \sqrt {2}\, \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+4 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\cot \left (d x +c \right )^{2}-2 \cot \left (d x +c \right ) \csc \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-7 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-2 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )-7 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+\cos \left (d x +c \right ) \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d a \left (\cos \left (d x +c \right )+1\right )}\) \(343\)

input
int(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/4/d/a*(4*2^(1/2)*ln(csc(d*x+c)-cot(d*x+c)+(cot(d*x+c)^2-2*cot(d*x+c)*cs 
c(d*x+c)+csc(d*x+c)^2-1)^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x 
+c)+4*ln(csc(d*x+c)-cot(d*x+c)+(cot(d*x+c)^2-2*cot(d*x+c)*csc(d*x+c)+csc(d 
*x+c)^2-1)^(1/2))*2^(1/2)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-7*(-cos(d*x+c 
)/(cos(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2))*cos(d*x+c)-2*cos(d*x+c)^2*sin(d*x+c)-7*(-cos(d*x+c)/(c 
os(d*x+c)+1))^(1/2)*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2))+cos(d*x+c)*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c 
)+1)
 
3.2.25.5 Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 446, normalized size of antiderivative = 3.03 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {-\frac {1}{a}} \log \left (\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 7 \, \sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {7 \, \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - {\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) - \frac {4 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

input
integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")
 
output
[1/8*(4*sqrt(2)*(a*cos(d*x + c) + a)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos 
(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos( 
d*x + c)^2 + 2*cos(d*x + c) - 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 
7*sqrt(-a)*(cos(d*x + c) + 1)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a 
*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c 
) - a)/(cos(d*x + c) + 1)) + 2*(2*cos(d*x + c)^2 - cos(d*x + c))*sqrt((a*c 
os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), -1/ 
4*(7*sqrt(a)*(cos(d*x + c) + 1)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + 
 c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (2*cos(d*x + c)^2 - cos(d*x + 
c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 4*sqrt(2)*(a*co 
s(d*x + c) + a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos 
(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)]
 
3.2.25.6 Sympy [F]

\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

input
integrate(cos(d*x+c)**2/(a+a*sec(d*x+c))**(1/2),x)
 
output
Integral(cos(c + d*x)**2/sqrt(a*(sec(c + d*x) + 1)), x)
 
3.2.25.7 Maxima [F]

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

input
integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")
 
output
integrate(cos(d*x + c)^2/sqrt(a*sec(d*x + c) + a), x)
 
3.2.25.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 423 vs. \(2 (122) = 244\).

Time = 1.01 (sec) , antiderivative size = 423, normalized size of antiderivative = 2.88 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {7 \, \sqrt {2} \sqrt {-a} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |}} - \frac {8 \, \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt {-a}} - \frac {8 \, {\left (17 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{6} \sqrt {-a} - 57 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} \sqrt {-a} a + 19 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{2} - 3 \, \sqrt {-a} a^{3}\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )}^{2}}\right )}}{16 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]

input
integrate(cos(d*x+c)^2/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
 
output
-1/16*sqrt(2)*(7*sqrt(2)*sqrt(-a)*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) 
 - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2* 
(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 
4*sqrt(2)*abs(a) - 6*a))/abs(a) - 8*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - s 
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/sqrt(-a) - 8*(17*(sqrt(-a)*tan(1/2* 
d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*sqrt(-a) - 57*(sqrt( 
-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*sqrt(-a) 
*a + 19*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + 
a))^2*sqrt(-a)*a^2 - 3*sqrt(-a)*a^3)/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr 
t(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s 
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2)/(d*sgn(cos(d*x + c)))
 
3.2.25.9 Mupad [F(-1)]

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

input
int(cos(c + d*x)^2/(a + a/cos(c + d*x))^(1/2),x)
 
output
int(cos(c + d*x)^2/(a + a/cos(c + d*x))^(1/2), x)