\(\int \frac {\sqrt {\cos (c+d x)} (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [1279]

Optimal result

Integrand size = 54, antiderivative size = 184 \[ \int \frac {\sqrt {\cos (c+d x)} \left (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 b B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}-\frac {\sqrt {2} (a-b) (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

2*b*B*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)* 
sec(d*x+c)^(1/2)/a^(1/2)/d-2^(1/2)*(a-b)*(A-B)*arctanh(1/2*a^(1/2)*sec(d*x 
+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec( 
d*x+c)^(1/2)/a^(1/2)/d+2*a*A*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^(1/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {\cos (c+d x)} \left (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (2 a A \sqrt {1-\sec (c+d x)}-2 b B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}+\sqrt {2} (a-b) (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) \sin (c+d x)}{d \sqrt {-1+\cos (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

((2*a*A*Sqrt[1 - Sec[c + d*x]] - 2*b*B*ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Sec 
[c + d*x]] + Sqrt[2]*(a - b)*(A - B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/S 
qrt[1 - Sec[c + d*x]]]*Sqrt[Sec[c + d*x]])*Sin[c + d*x])/(d*Sqrt[-1 + Cos[ 
c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.92, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4753, 3042, 4574, 27, 3042, 4511, 3042, 4288, 222, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 a (a-b) (A-B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 \sqrt {a} b B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \sqrt {a} b B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} \sqrt {a} (a-b) (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}+\frac {2 a A \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*Sqrt[a]*b*B*ArcSinh[(Sqrt[a]*Ta 
n[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*Sqrt[a]*(a - b)*(A - B 
)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Se 
c[c + d*x]])])/d)/a + (2*a*A*Sqrt[Sec[c + d*x]]*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 222

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

rule 3042

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

rule 4288

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

rule 4295

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

rule 4511

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

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(347\) vs. \(2(155)=310\).

Time = 3.39 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.89

method	result	size
default	\(-\frac {\left (\sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) A a \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-2 A \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, a \sin \left (d x +c \right )+A \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, b +B \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, a -B \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, b +B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) b +B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) b \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{d \left (\cos \left (d x +c \right )+1\right ) a \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(348\)

Input:

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

Output:

-1/d*((-2/(cos(d*x+c)+1))^(1/2)*(-1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1)*A 
*a*arctan(1/2*2^(1/2)*(csc(d*x+c)-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))-2 
*A*(-1/(cos(d*x+c)+1))^(1/2)*a*sin(d*x+c)+A*arctan(1/2*2^(1/2)*(csc(d*x+c) 
-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*b+B*arctan(1/2*2^(1/2)*(cs 
c(d*x+c)-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*a-B*arctan(1/2*2^( 
1/2)*(csc(d*x+c)-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*b+B*arctan 
(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*b+B*arctan(1/2* 
(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*b)*(a*(1+sec(d*x+c)) 
)^(1/2)*cos(d*x+c)^(1/2)/(cos(d*x+c)+1)/a/(-1/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 562, normalized size of antiderivative = 3.05 \[ \int \frac {\sqrt {\cos (c+d x)} \left (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/2*(4*A*a*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin 
(d*x + c) + (B*b*cos(d*x + c) + B*b)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqr 
t(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d 
*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d* 
x + c)^2)) + sqrt(2)*((A - B)*a^2 - (A - B)*a*b + ((A - B)*a^2 - (A - B)*a 
*b)*cos(d*x + c))*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + 
a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) 
- 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a 
*d), (2*A*a*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin 
(d*x + c) + sqrt(2)*((A - B)*a^2 - (A - B)*a*b + ((A - B)*a^2 - (A - B)*a* 
b)*cos(d*x + c))*sqrt(-1/a)*arctan(1/2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))*sin(d*x + c)/(cos(d*x + c) + 1) 
) + (B*b*cos(d*x + c) + B*b)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + 
c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - 
a*cos(d*x + c) - 2*a)))/(a*d*cos(d*x + c) + a*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 890 vs. \(2 (155) = 310\).

Time = 0.41 (sec) , antiderivative size = 890, normalized size of antiderivative = 4.84 \[ \int \frac {\sqrt {\cos (c+d x)} \left (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin 
(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x 
+ 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 4*sqrt(2)*sin(1/2*d*x + 1/2*c)) 
*A*sqrt(a) - (sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 
+ 2*sin(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1 
/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*B*sqrt(a) - (sqrt(2)*log( 
cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 
 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin( 
1/2*d*x + 1/2*c) + 1))*A*b/sqrt(a) + (sqrt(2)*log(cos(1/3*arctan2(sin(3/2* 
d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2 
*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co 
s(3/2*d*x + 3/2*c))) + 1) - sqrt(2)*log(cos(1/3*arctan2(sin(3/2*d*x + 3/2* 
c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3 
/2*d*x + 3/2*c)))^2 - 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x 
+ 3/2*c))) + 1) - log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x 
+ 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c 
)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* 
c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c 
))) + 2) + log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c 
)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^...
 

Giac [A] (verification not implemented)

Time = 168.16 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {\cos (c+d x)} \left (a A+(A b+a B) \sec (c+d x)+b B \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {4 \, \sqrt {2} A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {2 \, B b \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {2 \, B b \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {\sqrt {2} {\left (A a^{\frac {3}{2}} - B a^{\frac {3}{2}} - A \sqrt {a} b + B \sqrt {a} b\right )} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{2 \, d} \] Input:

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

Output:

1/2*(4*sqrt(2)*A*a*tan(1/2*d*x + 1/2*c)/(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a 
)*sgn(cos(d*x + c))) + 2*B*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))) - 2*B*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))) + s 
qrt(2)*(A*a^(3/2) - B*a^(3/2) - A*sqrt(a)*b + B*sqrt(a)*b)*log((sqrt(a)*ta 
n(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(a*sgn(cos(d*x 
 + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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