\(\int \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\) [519]

Optimal result

Integrand size = 35, antiderivative size = 96 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \sqrt {a} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{d}+\frac {2 a A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

2*a^(1/2)*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)/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 = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.98 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {2 \sqrt {\cos (c+d x)} \left (A \sqrt {1-\sec (c+d x)}-B \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}\right ) \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {1-\sec (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3434, 3042, 4503, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4503

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*Sqrt[a]*B*ArcSinh[(Sqrt[a]*Tan[c 
 + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (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 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 3434

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Csc[e + f*x])^m*((c + 
d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

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 4503

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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 [A] (verified)

Time = 2.01 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.51

method	result	size
default	\(\frac {\left (2 A \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(145\)

Input:

int(cos(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/d*(2*A*(-1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-B*arctan(1/2*(-cot(d*x+c)+cs 
c(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)))*(a*(1+sec(d*x+c)))^(1/2)*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 = 0.10 (sec) , antiderivative size = 298, normalized size of antiderivative = 3.10 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\left [\frac {4 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + B\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{2 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (B \cos \left (d x + c\right ) + B\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{d \cos \left (d x + c\right ) + d}\right ] \] Input:

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

Output:

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

Sympy [F]

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

integrate(cos(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(cos(c + d*x) 
), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (82) = 164\).

Time = 0.22 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.73 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {4 \, \sqrt {2} A \sqrt {a} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B \sqrt {a} {\left (\log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) + \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right ) - \log \left (2 \, \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sqrt {2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2\right )\right )}}{2 \, d} \] Input:

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

Output:

1/2*(4*sqrt(2)*A*sqrt(a)*sin(1/2*d*x + 1/2*c) + B*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) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*s 
in(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*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*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)))/d
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.70 \[ \int \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx=\frac {\frac {2 \, \sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {B a^{\frac {3}{2}} \log \left (\frac {{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}}}{d} \] Input:

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

Output:

(2*sqrt(2)*A*a*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 
 1/2*c)^2 + a) + B*a^(3/2)*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)*a 
bs(a) - 6*a))*sgn(cos(d*x + c))/abs(a))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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