\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [521]

Optimal result

Integrand size = 35, antiderivative size = 151 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {a} (4 A+3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 d}+\frac {a B \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {a (4 A+3 B) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*a^(1/2)*(4*A+3*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*c 
os(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+1/2*a*B*sin(d*x+c)/d/cos(d*x+c)^(5/2)/( 
a+a*sec(d*x+c))^(1/2)+1/4*a*(4*A+3*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*s 
ec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left ((4 A+3 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+(4 A+3 B) \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/ 
2),x]
 

Output:

(a*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*((4*A + 3*B)*ArcSin[Sqrt[1 - Sec[ 
c + d*x]]] + 2*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + (4*A + 3*B)*S 
qrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Sin[c + d*x])/(4*d*Sqrt[1 - Sec[ 
c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3434, 3042, 4504, 3042, 4290, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4504

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} (4 A+3 B) \int \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} (4 A+3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 4290

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} (4 A+3 B) \left (\frac {1}{2} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} (4 A+3 B) \left (\frac {1}{2} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} (4 A+3 B) \left (\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{4} (4 A+3 B) \left (\frac {\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

Input:

Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a*B*Sec[c + d*x]^(5/2)*Sin[c + d*x 
])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + ((4*A + 3*B)*((Sqrt[a]*ArcSinh[(Sqrt[a 
]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a*Sec[c + d*x]^(3/2)*Sin[c 
 + d*x])/(d*Sqrt[a + a*Sec[c + d*x]])))/4)
 

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 4290

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( 
f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 
1)))   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre 
eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
 

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. \(293\) vs. \(2(127)=254\).

Time = 2.20 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.95

method	result	size
default	\(\frac {\left (4 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+3 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+4 A \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+3 B \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+4 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (3 \cos \left (d x +c \right )+2\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{8 d \cos \left (d x +c \right )^{\frac {3}{2}} \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(294\)

Input:

int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/8/d*(4*A*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(1+cos(d* 
x+c)))^(1/2))+3*B*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(1 
+cos(d*x+c)))^(1/2))+4*A*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d*x+c)+1) 
/(-1/(1+cos(d*x+c)))^(1/2))+3*B*cos(d*x+c)^2*arctan(1/2*(cot(d*x+c)-csc(d* 
x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+4*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2/( 
1+cos(d*x+c)))^(1/2)+sin(d*x+c)*(3*cos(d*x+c)+2)*2^(1/2)*B*(-2/(1+cos(d*x+ 
c)))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)/cos(d*x+c)^(3/2)/(1+cos(d*x+c))/(-1/( 
1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.66 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {4 \, {\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, {\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algo 
rithm="fricas")
 

Output:

[1/16*(4*((4*A + 3*B)*cos(d*x + c) + 2*B)*sqrt((a*cos(d*x + c) + a)/cos(d* 
x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((4*A + 3*B)*cos(d*x + c)^3 + (4 
*A + 3*B)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt(( 
a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*si 
n(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)^3 + d*cos(d*x + c)^2), 1/8*(2*((4*A + 3*B)*cos(d*x + c) + 
 2*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + 
 c) + ((4*A + 3*B)*cos(d*x + c)^3 + (4*A + 3*B)*cos(d*x + c)^2)*sqrt(-a)*a 
rctan(2*sqrt(-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) 
^3 + d*cos(d*x + c)^2)]
 

Sympy [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c))/cos(d*x+c)**(3/2),x)
 

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x))/cos(c + d*x)**(3/ 
2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1927 vs. \(2 (127) = 254\).

Time = 0.29 (sec) , antiderivative size = 1927, normalized size of antiderivative = 12.76 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algo 
rithm="maxima")
                                                                                    
                                                                                    
 

Output:

-1/16*(4*(4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x 
 + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(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*arctan 
2(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)*l 
og(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(si 
n(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*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) + (c 
os(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) - 4*(sqr 
t(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c 
))) + 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x + ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (127) = 254\).

Time = 0.48 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.69 \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x, algo 
rithm="giac")
 

Output:

1/8*((4*A*sqrt(a)*sgn(cos(d*x + c)) + 3*B*sqrt(a)*sgn(cos(d*x + c)))*log(a 
bs((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*A*sqrt(a)*sgn(cos(d*x + c)) + 3*B*sqrt(a)*sgn(co 
s(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*sqrt(2)*(12*(sqrt(a)*tan(1/2*d* 
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(3/2)*sgn(cos(d*x + 
 c)) - 5*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a 
))^6*B*a^(3/2)*sgn(cos(d*x + c)) - 76*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt 
(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*a^(5/2)*sgn(cos(d*x + c)) - 19*(sqrt(a 
)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(5/2)*s 
gn(cos(d*x + c)) + 36*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 
 1/2*c)^2 + a))^2*A*a^(7/2)*sgn(cos(d*x + c)) + 17*(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^(7/2)*sgn(cos(d*x + c)) 
 - 4*A*a^(9/2)*sgn(cos(d*x + c)) - B*a^(9/2)*sgn(cos(d*x + c)))/((sqrt(a)* 
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)* 
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x)^(3/2),x 
)
 

Output:

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(1/2))/cos(c + d*x)^(3/2), 
x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{2}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{2}}d x \right ) a \right ) \] Input:

int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),x)
 

Output:

sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x))/cos( 
c + d*x)**2,x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x)))/cos(c + 
 d*x)**2,x)*a)