3.3.51 \(\int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx\) [251]

3.3.51.1 Optimal result

Integrand size = 35, antiderivative size = 99 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}} \]

output

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

3.3.51.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (2 A \sqrt {1-\sec (c+d x)}+\sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \]

input

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

output

((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]])*Tan[c + d*x])/(d*Sqrt 
[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c + d*x])])
 

3.3.51.3 Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 99, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

input

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

output

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

3.3.51.3.1 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 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]
 

3.3.51.4 Maple [A] (verified)

Time = 4.13 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.61

method	result	size
default	\(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}-B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {2}+2 A \cot \left (d x +c \right )-2 A \csc \left (d x +c \right )\right )}{d a \sqrt {\sec \left (d x +c \right )}}\)	\(159\)
parts	\(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}+2 \cot \left (d x +c \right )-2 \csc \left (d x +c \right )\right )}{d a \sqrt {\sec \left (d x +c \right )}}-\frac {B \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{d a \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(200\)

input

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

output

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

3.3.51.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 306, normalized size of antiderivative = 3.09 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) + 2 \, A \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d}\right ] \]

input

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(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(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)) + 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)]
 

3.3.51.6 Sympy [F]

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

input

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

output

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

3.3.51.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (84) = 168\).

Time = 0.52 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.97 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} A}{\sqrt {a}} - \frac {{\left (\sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - \sqrt {2} \log \left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} B}{\sqrt {a}}}{2 \, d} \]

input

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

3.3.51.8 Giac [F]

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

input

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

output

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

3.3.51.9 Mupad [F(-1)]

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

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)