\(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [250]

Optimal result

Integrand size = 35, antiderivative size = 100 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {2 B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d} \] Output:

2*B*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/d+2^(1/2)*( 
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))/a^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (-2 B \arcsin \left (\sqrt {\sec (c+d x)}\right )+\sqrt {2} (-A+B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*B*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) 
 + (Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqr 
t[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)
 

Defintions of rubi rules used

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(210\) vs. \(2(83)=166\).

Time = 1.82 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.11

method	result	size
default	\(-\frac {\cos \left (d x +c \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (A \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-B \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right )}{d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(211\)
parts	\(-\frac {A \sqrt {2}\, \sqrt {\sec \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \cos \left (d x +c \right )}{d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right )}{d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(265\)

Input:

int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 363, normalized size of antiderivative = 3.63 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {\sqrt {2} {\left (A - B\right )} \sqrt {a} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - B \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \, a d}, -\frac {\sqrt {2} {\left (A - B\right )} a \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - B \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right )}{a d}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((A + B*sec(c + d*x))*sqrt(sec(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. 567 vs. \(2 (83) = 166\).

Time = 0.27 (sec) , antiderivative size = 567, normalized size of antiderivative = 5.67 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="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))*A/sqrt(a) - (sqrt(2)*log(cos(1/2* 
arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), cos 
(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - sqrt 
(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(s 
in(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + 
c))) + 1) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1 
/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin( 
d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x 
 + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*si 
n(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(s 
in(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos( 
d*x + c))) + 2) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2 
*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan 
2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), c 
os(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 
+ 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arc 
tan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c) 
, cos(d*x + c))) + 2))*B/sqrt(a))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (83) = 166\).

Time = 1.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.77 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {\sqrt {2} {\left (A \sqrt {a} - B \sqrt {a}\right )} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a} - \frac {2 \, B \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {a}} + \frac {2 \, B \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {a}}}{2 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

-1/2*(sqrt(2)*(A*sqrt(a) - B*sqrt(a))*log((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*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)))/s 
qrt(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))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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