\(\int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx\) [558]

Optimal result

Integrand size = 35, antiderivative size = 223 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\frac {(19 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sin (c+d x)}{16 a d \sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{3/2}}-\frac {(9 A-B) \sin (c+d x)}{16 a^2 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

1/32*(19*A+5*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)*2^(1/2)/a^(5/2)/d+ 
1/4*(A-B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2)+1/16*(5*A+3 
*B)*sin(d*x+c)/a/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(3/2)-1/16*(9*A-B)*si 
n(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(481\) vs. \(2(223)=446\).

Time = 6.12 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.16 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=-\frac {A \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2}}-\frac {9 A (1+\sec (c+d x)) \sin (c+d x)}{16 d \cos ^{\frac {3}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2}}-\frac {19 A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))^2 \sin (c+d x)}{16 \sqrt {2} d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}}-\frac {B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} (1+\sec (c+d x))^{5/2} \left (\frac {8 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{5/2}}+5 \left (\frac {2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (1+\sec (c+d x))^{3/2}}-\frac {\left (2 \arcsin \left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {2 \left (\arcsin \left (\sqrt {1-\sec (c+d x)}\right )+\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{32 (a (1+\sec (c+d x)))^{5/2}} \] Input:

Integrate[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(5 
/2)),x]
 

Output:

-1/4*(A*Sin[c + d*x])/(d*Cos[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(5/2)) 
- (9*A*(1 + Sec[c + d*x])*Sin[c + d*x])/(16*d*Cos[c + d*x]^(3/2)*(a*(1 + S 
ec[c + d*x]))^(5/2)) - (19*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - 
Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2* 
Sin[c + d*x])/(16*Sqrt[2]*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^ 
(5/2)) - (B*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])^(5/2) 
*((8*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + 5*((2 
*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) - ((2*ArcSi 
n[Sqrt[Sec[c + d*x]]] - Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 
 - Sec[c + d*x]]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c 
+ d*x]]) - (2*(ArcSin[Sqrt[1 - Sec[c + d*x]]] + Sqrt[1 - Sec[c + d*x]]*Sqr 
t[Sec[c + d*x]])*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + 
d*x]]))))/(32*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.32 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3434, 3042, 4507, 27, 3042, 4508, 27, 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.

\(\displaystyle \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3434

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(\sec (c+d x) a+a)^{5/2}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\int \frac {a (A-B)-4 a (A+B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\int \frac {a (A-B)-4 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}\right )\)

\(\Big \downarrow \) 4508

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\int \frac {a^2 (9 A-B)-2 a^2 (5 A+3 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\int \frac {a^2 (9 A-B)-2 a^2 (5 A+3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\int \frac {a^2 (9 A-B)-2 a^2 (5 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

\(\Big \downarrow \) 4501

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\frac {2 a^2 (9 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (19 A+5 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\frac {2 a^2 (9 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a^2 (19 A+5 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\frac {2 a^2 (19 A+5 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (9 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac {\frac {\frac {2 a^2 (9 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} a^{3/2} (19 A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}\right )\)

Input:

Int[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)),x 
]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + 
 d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - (-1/2*(a*(5*A + 3*B)*Sqrt[Sec[c 
+ d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-((Sqrt[2]*a^(3/2) 
*(19*A + 5*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*S 
qrt[a + a*Sec[c + d*x]])])/d) + (2*a^2*(9*A - B)*Sqrt[Sec[c + d*x]]*Sin[c 
+ d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))/(8*a^2))
 

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 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]
 

rule 4507

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

rule 4508

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

Maple [A] (verified)

Time = 1.61 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.14

method	result	size
default	\(\frac {\left (\left (19 \cos \left (d x +c \right )^{2}+38 \cos \left (d x +c \right )+19\right ) A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) B \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-13 \cos \left (d x +c \right )-9\right ) \sin \left (d x +c \right ) A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\left (5 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {2}}{32 d \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) a^{3} \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(255\)

Input:

int((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/32/d*((19*cos(d*x+c)^2+38*cos(d*x+c)+19)*A*arctan(1/2*2^(1/2)*(-csc(d*x+ 
c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+(5*cos(d*x+c)^2+10*cos(d*x+c)+5) 
*B*arctan(1/2*2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+ 
(-13*cos(d*x+c)-9)*sin(d*x+c)*A*(-2/(1+cos(d*x+c)))^(1/2)+(5*cos(d*x+c)+1) 
*sin(d*x+c)*B*(-2/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c) 
))^(1/2)*2^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)/a^3/(-1/(1+c 
os(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.20 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {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 ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (13 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 9 \, A - 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 )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, B\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {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 )}{2 \, {\left (a \cos \left (d x + c\right ) + a\right )}}\right ) + 2 \, {\left ({\left (13 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 9 \, A - 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 )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="fricas")
 

Output:

[1/64*(sqrt(2)*((19*A + 5*B)*cos(d*x + c)^3 + 3*(19*A + 5*B)*cos(d*x + c)^ 
2 + 3*(19*A + 5*B)*cos(d*x + c) + 19*A + 5*B)*sqrt(a)*log(-(a*cos(d*x + c) 
^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d* 
x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x 
 + c) + 1)) - 4*((13*A - 5*B)*cos(d*x + c) + 9*A - B)*sqrt((a*cos(d*x + c) 
 + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 
 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)* 
((19*A + 5*B)*cos(d*x + c)^3 + 3*(19*A + 5*B)*cos(d*x + c)^2 + 3*(19*A + 5 
*B)*cos(d*x + c) + 19*A + 5*B)*sqrt(-a)*arctan(1/2*sqrt(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) + a)) + 2*((13*A - 5*B)*cos(d*x + c) + 9*A - B)*sqrt((a*cos(d*x + 
c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c) 
^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)**(1/2)/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 5924 vs. \(2 (188) = 376\).

Time = 0.70 (sec) , antiderivative size = 5924, normalized size of antiderivative = 26.57 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="maxima")
 

Output:

1/32*((19*(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) - 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))*cos(4*d*x + 4*c)^2 + 304*(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) - 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)) 
*cos(3*d*x + 3*c)^2 + 684*(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) - 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))*cos(2*d*x + 2*c)^2 + 304*(lo 
g(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) - 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))*cos(d*x + c)^2 + 19*(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) - 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))*sin(4*d*x + 4*c)^ 
2 + 304*(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) - 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))*sin(3*d*x + 3*c)^2 + 684*(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) - 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))*s 
in(2*d*x + 2*c)^2 + 304*(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) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2...
 

Giac [A] (verification not implemented)

Time = 167.48 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (11 \, A a^{5} - 3 \, B a^{5}\right )}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {\sqrt {2} {\left (19 \, A + 5 \, B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \] Input:

integrate((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="giac")
 

Output:

1/32*(sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5)*tan(1/ 
2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(11*A*a^5 - 3*B*a^5)/(a 
^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) - sqrt(2)*(19*A + 5*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)))/(a^( 
5/2)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))^(5/2)),x 
)
 

Output:

int((A + B/cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + a/cos(c + d*x))^(5/2)), 
x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\cos (c+d x)} (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\cos \left (d x +c \right )}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right ) \sec \left (d x +c \right )^{3}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \sec \left (d x +c \right )+\cos \left (d x +c \right )}d x \right ) a \right )}{a^{3}} \] Input:

int((A+B*sec(d*x+c))/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x))*sec(c + d*x))/(co 
s(c + d*x)*sec(c + d*x)**3 + 3*cos(c + d*x)*sec(c + d*x)**2 + 3*cos(c + d* 
x)*sec(c + d*x) + cos(c + d*x)),x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(co 
s(c + d*x)))/(cos(c + d*x)*sec(c + d*x)**3 + 3*cos(c + d*x)*sec(c + d*x)** 
2 + 3*cos(c + d*x)*sec(c + d*x) + cos(c + d*x)),x)*a))/a**3