\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [544]

Optimal result

Integrand size = 35, antiderivative size = 119 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

-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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(1/2)/d+2*A*sin(d 
*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \left (2 A \sqrt {1-\sec (c+d x)}+\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) (A+B \sec (c+d x)) \sin (c+d x)}{d (B+A \cos (c+d x)) \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 120, 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, 4501, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a] 
*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sq 
rt[a]*d)) + (2*A*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d* 
x]]))
 

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 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 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 4501

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

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.59

method	result	size
default	\(\frac {\sqrt {-a \left (-1-\sec \left (d x +c \right )\right )}\, \left (-\frac {B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}}{2 \sqrt {\cos \left (d x +c \right )}\, \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}-A \sqrt {\cos \left (d x +c \right )}\, \left (-\arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-2 \csc \left (d x +c \right )+2 \cot \left (d x +c \right )\right )\right )}{d a}\)	\(189\)

Input:

int(cos(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*(-a*(-1-sec(d*x+c)))^(1/2)*(-1/2*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1) 
*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))/co 
s(d*x+c)^(1/2)*2^(1/2)/(-1/(1+cos(d*x+c)))^(1/2)-A*cos(d*x+c)^(1/2)*(-arct 
an(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))*(-2/(1+ 
cos(d*x+c)))^(1/2)-2*csc(d*x+c)+2*cot(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 315, normalized size of antiderivative = 2.65 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right ) + 2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d}\right ] \] Input:

integrate(cos(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*(4*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*cos(d*x + c) + (A - B)*a)*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))*s 
in(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), (sqrt(2)*((A - B)*a*cos(d*x + c) 
 + (A - B)*a)*sqrt(-1/a)*arctan(1/2*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)/(cos(d*x + c) + 1)) + 
 2*A*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + 
c))/(a*d*cos(d*x + c) + a*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A}{\sqrt {a}} - \frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} B}{\sqrt {a}}}{2 \, d} \] Input:

integrate(cos(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) - 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))/d
 

Giac [A] (verification not implemented)

Time = 167.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\frac {2 \, \sqrt {2} A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} + \frac {{\left (\sqrt {2} A - \sqrt {2} B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{d} \] Input:

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

Output:

(2*sqrt(2)*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))) + (sqrt(2)*A - sqrt(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)))/(sqrt(a)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

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