\(\int \frac {\cos (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [515]

Optimal result

Integrand size = 41, antiderivative size = 120 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {(A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {A \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}} \] Output:

-(A-2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/d+2^(1/ 
2)*(A-B+C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a 
^(1/2)/d+A*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 2.30 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (A (-1+\cos (c+d x))+(A-2 B) \arctan \left (\sqrt {-1+\sec (c+d x)}\right ) \sqrt {-1+\sec (c+d x)}-\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \sqrt {-1+\sec (c+d x)}\right ) \sin (c+d x)}{d (-1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

((A*(-1 + Cos[c + d*x]) + (A - 2*B)*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Sqrt[- 
1 + Sec[c + d*x]] - Sqrt[2]*(A - B + C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqr 
t[2]]*Sqrt[-1 + Sec[c + d*x]])*Sin[c + d*x])/(d*(-1 + Cos[c + d*x])*Sqrt[a 
*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4574, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

\(\displaystyle \frac {\int -\frac {a (A-2 B)-a (A+2 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-2 B)-a (A+2 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

\(\displaystyle \frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {(A-2 B) \int \sqrt {\sec (c+d x) a+a}dx-2 a (A-B+C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

\(\displaystyle \frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\frac {2 a (A-2 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-2 a (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \sqrt {a} (A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-2 a (A-B+C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a (A-B+C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 \sqrt {a} (A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {A \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \sqrt {a} (A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 \sqrt {2} \sqrt {a} (A-B+C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}\)

Input:

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

Output:

-1/2*((2*Sqrt[a]*(A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c 
+ d*x]]])/d - (2*Sqrt[2]*Sqrt[a]*(A - B + C)*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]])
 

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 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 4574

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), 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 - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
&&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(426\) vs. \(2(103)=206\).

Time = 6.71 (sec) , antiderivative size = 427, normalized size of antiderivative = 3.56

method	result	size
default	\(\frac {\left (\left (\cos \left (d x +c \right )+1\right ) \sqrt {2}\, A \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (2 \cos \left (d x +c \right )+2\right ) A \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+2 A \sin \left (d x +c \right ) \cos \left (d x +c \right )+\left (-2 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-2 \cos \left (d x +c \right )-2\right ) B \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (2 \cos \left (d x +c \right )+2\right ) C \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d a \left (\cos \left (d x +c \right )+1\right )}\)	\(427\)

Input:

int(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x,me 
thod=_RETURNVERBOSE)
 

Output:

1/2/d/a*((cos(d*x+c)+1)*2^(1/2)*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arc 
tanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c 
)+cot(d*x+c)^2-1)^(1/2))+(2*cos(d*x+c)+2)*A*ln((-2*cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)-cot(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+2*A* 
sin(d*x+c)*cos(d*x+c)+(-2*cos(d*x+c)-2)*2^(1/2)*B*(-2*cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(csc(d*x+c)^2-2*csc( 
d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2))+(-2*cos(d*x+c)-2)*B*ln((-2*cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)+(2*cos(d*x+c)+2)*C*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-co 
t(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(a*(1+sec(d*x+c 
)))^(1/2)/(cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 3.17 (sec) , antiderivative size = 462, normalized size of antiderivative = 3.85 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + \sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right ) + A - 2 \, B\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right ) + A - 2 \, B\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {\sqrt {2} {\left ({\left (A - B + C\right )} a \cos \left (d x + c\right ) + {\left (A - B + C\right )} a\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )}{\sqrt {a}}}{a d \cos \left (d x + c\right ) + a d}\right ] \] Input:

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

Output:

[1/2*(2*A*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c 
) + sqrt(2)*((A - B + C)*a*cos(d*x + c) + (A - B + 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)) + ((A - 2*B)*cos(d*x + c) + A - 2*B)*sqrt(-a)*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)))/(a*d*cos(d* 
x + c) + a*d), (A*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin 
(d*x + c) + ((A - 2*B)*cos(d*x + c) + A - 2*B)*sqrt(a)*arctan(sqrt((a*cos( 
d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - sqrt(2) 
*((A - B + C)*a*cos(d*x + c) + (A - B + 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)]
 

Sympy [F]

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

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

Output:

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)/sqrt(a*(sec 
(c + d*x) + 1)), x)
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (103) = 206\).

Time = 0.92 (sec) , antiderivative size = 366, normalized size of antiderivative = 3.05 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {\sqrt {2} {\left (A - B + C\right )} \log \left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {{\left (A - 2 \, B\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left (A - 2 \, B\right )} \log \left ({\left | {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {4 \, \sqrt {2} {\left (3 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} A \sqrt {-a} - A \sqrt {-a} a\right )}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{2 \, d} \] Input:

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

Output:

-1/2*(sqrt(2)*(A - B + C)*log((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)*sgn(cos(d*x + c))) + (A - 2*B)*log( 
abs((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^ 
2 - a*(2*sqrt(2) + 3)))/(sqrt(-a)*sgn(cos(d*x + c))) - (A - 2*B)*log(abs(( 
sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a 
*(2*sqrt(2) - 3)))/(sqrt(-a)*sgn(cos(d*x + c))) + 4*sqrt(2)*(3*(sqrt(-a)*t 
an(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a) - 
A*sqrt(-a)*a)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2 
*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1 
/2*c)^2 + a))^2*a + a^2)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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