\(\int \frac {\cos (c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx\) [168]

Optimal result

Integrand size = 32, antiderivative size = 115 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {3 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}-\frac {2 \sqrt {2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {A \left (3 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right )+\cos (c+d x) \sqrt {1+\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \] Input:

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

Output:

(A*(3*ArcTanh[Sqrt[1 + Sec[c + d*x]]] - 2*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + 
 d*x]]/Sqrt[2]] + Cos[c + d*x]*Sqrt[1 + Sec[c + d*x]])*Tan[c + d*x])/(d*Sq 
rt[1 + Sec[c + d*x]]*Sqrt[a - a*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3042, 4510, 27, 3042, 4408, 3042, 4261, 216, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

\(\displaystyle \frac {4 a A \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}}dx+3 A \int \sqrt {a-a \sec (c+d x)}dx}{2 a}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 a A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+3 A \int \sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {6 a A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{a-a \sec (c+d x)}+a}d\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}}{d}+4 a A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {4 a A \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a-a \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}}{2 a}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {6 \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {8 a A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{a-a \sec (c+d x)}+2 a}d\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}}{d}}{2 a}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {6 \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}-\frac {4 \sqrt {2} \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{d}}{2 a}+\frac {A \sin (c+d x)}{d \sqrt {a-a \sec (c+d x)}}\)

Input:

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

Output:

((6*Sqrt[a]*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/d - 
 (4*Sqrt[2]*Sqrt[a]*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a - a*Se 
c[c + d*x]])])/d)/(2*a) + (A*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 216

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

rule 3042

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

rule 4261

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

rule 4282

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

rule 4408

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

rule 4510

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[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 - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 

Maple [F]

\[\int \frac {\cos \left (d x +c \right ) \left (A +A \sec \left (d x +c \right )\right )}{\sqrt {a -a \sec \left (d x +c \right )}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 435, normalized size of antiderivative = 3.78 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\left [\frac {2 \, \sqrt {2} A a \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, A \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 2 \, {\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{2 \, a d \sin \left (d x + c\right )}, \frac {2 \, \sqrt {2} A \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, A \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{a d \sin \left (d x + c\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((A*sec(d*x + c) + A)*cos(d*x + c)/sqrt(-a*sec(d*x + c) + a), x)
 

Giac [A] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99 \[ \int \frac {\cos (c+d x) (A+A \sec (c+d x))}{\sqrt {a-a \sec (c+d x)}} \, dx=\frac {\frac {2 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{\sqrt {a}} - \frac {3 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{\sqrt {a}} - \frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{d} \] Input:

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

Output:

(2*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) - 
3*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/sqrt(a) 
 - sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A/(a*tan(1/2*d*x + 1/2*c)^2 
+ a))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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