\(\int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [226]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 78 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {\sqrt {a} (2 A+B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a B \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}} \] Output:

a^(1/2)*(2*A+B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+a*B*s 
ec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {a \left (B \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-2 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a*(B*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 2*A*ArcSin[Sqrt[Sec[c + d*x]]] + B* 
Sqrt[-((-1 + Sec[c + d*x])*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 = 0.43 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4504, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{2} (2 A+B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} (2 A+B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {(2 A+B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\sqrt {a} (2 A+B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\)

Input:

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

Output:

(Sqrt[a]*(2*A + B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]] 
])/d + (a*B*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])
 

Defintions of rubi rules used

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 4504
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C 
ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] 
 + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1))   Int[Sqrt[a + b*Csc[e + f* 
x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ 
[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && 
!LtQ[n, 0]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(243\) vs. \(2(68)=136\).

Time = 1.77 (sec) , antiderivative size = 244, normalized size of antiderivative = 3.13

method result size
default \(\frac {\left (B \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-2 A \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-2 A \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-B \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-B \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{2 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\) \(244\)
parts \(-\frac {A \left (\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \cos \left (d x +c \right )}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-\cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-\cos \left (d x +c \right )^{2} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right )}{2 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\) \(292\)

Input:

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

Output:

1/2/d*(B*sin(d*x+c)*2^(1/2)*(-2/(1+cos(d*x+c)))^(1/2)-2*A*cos(d*x+c)*arcta 
n(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))-2*A*cos(d*x+c) 
*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))-B*cos(d* 
x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))-B*co 
s(d*x+c)*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)*sec(d*x+c)^(1/2)*(a*(1+sec(d*x+c 
)))^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (68) = 136\).

Time = 0.14 (sec) , antiderivative size = 319, normalized size of antiderivative = 4.09 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\left [\frac {{\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right ) + 2 \, A + B\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, B \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {{\left ({\left (2 \, A + B\right )} \cos \left (d x + c\right ) + 2 \, A + B\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, B \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

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

Output:

[1/4*(((2*A + B)*cos(d*x + c) + 2*A + 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*co 
s(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)) + 4*B*sqrt((a*cos(d*x + c) + a)/cos(d*x + c) 
)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/2*(((2*A + B)*c 
os(d*x + c) + 2*A + 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))) + 2*B*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/ 
sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 905 vs. \(2 (68) = 136\).

Time = 0.30 (sec) , antiderivative size = 905, normalized size of antiderivative = 11.60 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(2*A*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 
+ 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - l 
og(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2 
*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + log(2*cos(1/2*d*x + 
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*s 
qrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/ 
2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x 
+ 1/2*c) + 2)) - (4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*s 
in(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))*s 
in(2*d*x + 2*c) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 
 2*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) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c 
) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*ar 
ctan2(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) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 
1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan 
2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 305 vs. \(2 (68) = 136\).

Time = 1.73 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.91 \[ \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, A \sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B \sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right ) - {\left (2 \, A \sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + B \sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right ) + \frac {4 \, {\left (3 \, \sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} B a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}}}{2 \, d} \] Input:

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

Output:

1/2*((2*A*sqrt(a)*sgn(cos(d*x + c)) + B*sqrt(a)*sgn(cos(d*x + c)))*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))) - (2*A*sqrt(a)*sgn(cos(d*x + c)) + B*sqrt(a)*sgn(cos(d* 
x + c)))*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))) + 4*(3*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1 
/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(3/2)*sgn(cos(d*x + c)) 
- sqrt(2)*B*a^(5/2)*sgn(cos(d*x + c)))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sq 
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sq 
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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