\(\int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [158]

Optimal result

Integrand size = 33, antiderivative size = 221 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(19 A-12 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 A-9 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 A-6 B) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(19*A-12*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/ 
d-1/4*(13*A-9*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1 
/2))*2^(1/2)/a^(3/2)/d-1/2*(A-B)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^ 
(3/2)-1/4*(7*A-6*B)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/2*(2*A-B)*cos( 
d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.60 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sec (c+d x) \left (-52 \sqrt {2} A \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \sin (c+d x)+36 \sqrt {2} B \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \sin (c+d x)-13 A \sqrt {1-\sec (c+d x)} \sin (c+d x)+24 B \sqrt {1-\sec (c+d x)} \sin (c+d x)+18 A \cos ^2(c+d x) \sqrt {1-\sec (c+d x)} \sin (c+d x)+\frac {13}{2} A \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))+8 B \sqrt {1-\sec (c+d x)} \sin (2 (c+d x))-52 \sqrt {2} A \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \tan (c+d x)+36 \sqrt {2} B \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \tan (c+d x)+(91 A-48 B) \text {arctanh}\left (\sqrt {1-\sec (c+d x)}\right ) (\sin (c+d x)+\tan (c+d x))-40 A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},1-\sec (c+d x)\right ) \sqrt {1-\sec (c+d x)} (\sin (c+d x)+\tan (c+d x))\right )}{16 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(Sec[c + d*x]*(-52*Sqrt[2]*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Sin[c 
 + d*x] + 36*Sqrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Sin[c + d*x 
] - 13*A*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 24*B*Sqrt[1 - Sec[c + d*x]] 
*Sin[c + d*x] + 18*A*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 
(13*A*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)])/2 + 8*B*Sqrt[1 - Sec[c + d* 
x]]*Sin[2*(c + d*x)] - 52*Sqrt[2]*A*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2] 
]*Tan[c + d*x] + 36*Sqrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Tan[ 
c + d*x] + (91*A - 48*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]]*(Sin[c + d*x] + T 
an[c + d*x]) - 40*A*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]*Sqrt[ 
1 - Sec[c + d*x]]*(Sin[c + d*x] + Tan[c + d*x])))/(16*d*Sqrt[1 - Sec[c + d 
*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4508, 27, 3042, 4510, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\int -\frac {2 \cos (c+d x) \left (a^2 (7 A-6 B)-3 a^2 (2 A-B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {\cos (c+d x) \left (a^2 (7 A-6 B)-3 a^2 (2 A-B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (7 A-6 B)-3 a^2 (2 A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {\int -\frac {a^3 (19 A-12 B)-a^3 (7 A-6 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (19 A-12 B)-a^3 (7 A-6 B) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (19 A-12 B)-a^3 (7 A-6 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (19 A-12 B) \int \sqrt {\sec (c+d x) a+a}dx-2 a^3 (13 A-9 B) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (19 A-12 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-2 a^3 (13 A-9 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-2 a^3 (13 A-9 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^3 (19 A-12 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (19 A-12 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-2 a^3 (13 A-9 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {4 a^3 (13 A-9 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{5/2} (19 A-12 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {2 a (2 A-B) \sin (c+d x) \cos (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {a^2 (7 A-6 B) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^{5/2} (19 A-12 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 \sqrt {2} a^{5/2} (13 A-9 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}}{a}}{4 a^2}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((A - B)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + 
((2*a*(2*A - B)*Cos[c + d*x]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) - 
(-1/2*((2*a^(5/2)*(19*A - 12*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*S 
ec[c + d*x]]])/d - (2*Sqrt[2]*a^(5/2)*(13*A - 9*B)*ArcTan[(Sqrt[a]*Tan[c + 
 d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (a^2*(7*A - 6*B)*Sin[c 
+ d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(4*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 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 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]
 

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 [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(453\) vs. \(2(190)=380\).

Time = 2.14 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.05

method	result	size
default	\(-\frac {\left (\left (19 \cos \left (d x +c \right )^{2}+38 \cos \left (d x +c \right )+19\right ) A \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (-12 \cos \left (d x +c \right )^{2}-24 \cos \left (d x +c \right )-12\right ) B \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right )+\left (13 \cos \left (d x +c \right )^{2}+26 \cos \left (d x +c \right )+13\right ) \sqrt {2}\, A \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-9 \cos \left (d x +c \right )^{2}-18 \cos \left (d x +c \right )-9\right ) \sqrt {2}\, B \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-2 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+7\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-4 \cos \left (d x +c \right )-6\right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(454\)

Input:

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

Output:

-1/4/d/a^2*((19*cos(d*x+c)^2+38*cos(d*x+c)+19)*A*(-cos(d*x+c)/(1+cos(d*x+c 
)))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+c))/(cot(d*x+c)^2-2*csc(d*x 
+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2))+(-12*cos(d*x+c)^2-24*cos(d*x+c)-12)* 
B*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(2^(1/2)*(-csc(d*x+c)+cot(d*x+ 
c))/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2))+(13*cos(d 
*x+c)^2+26*cos(d*x+c)+13)*2^(1/2)*A*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln( 
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-cot(d*x+c)+csc(d*x+c))+(-9*cos(d*x+c) 
^2-18*cos(d*x+c)-9)*2^(1/2)*B*(-cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln((-2*co 
s(d*x+c)/(1+cos(d*x+c)))^(1/2)-cot(d*x+c)+csc(d*x+c))+sin(d*x+c)*cos(d*x+c 
)*(-2*cos(d*x+c)^2+3*cos(d*x+c)+7)*A+sin(d*x+c)*cos(d*x+c)*(-4*cos(d*x+c)- 
6)*B)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 3.04 (sec) , antiderivative size = 644, normalized size of antiderivative = 2.91 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/8*(sqrt(2)*((13*A - 9*B)*cos(d*x + c)^2 + 2*(13*A - 9*B)*cos(d*x + c) + 
 13*A - 9*B)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/co 
s(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + 
 c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + ((19*A - 12*B)*cos(d*x + 
 c)^2 + 2*(19*A - 12*B)*cos(d*x + c) + 19*A - 12*B)*sqrt(-a)*log((2*a*cos( 
d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + 
c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) + 2*(2*A*cos(d*x 
 + c)^3 - (3*A - 4*B)*cos(d*x + c)^2 - (7*A - 6*B)*cos(d*x + c))*sqrt((a*c 
os(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2 
*d*cos(d*x + c) + a^2*d), 1/4*(sqrt(2)*((13*A - 9*B)*cos(d*x + c)^2 + 2*(1 
3*A - 9*B)*cos(d*x + c) + 13*A - 9*B)*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))) - ((19*A - 
 12*B)*cos(d*x + c)^2 + 2*(19*A - 12*B)*cos(d*x + c) + 19*A - 12*B)*sqrt(a 
)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin 
(d*x + c))) + (2*A*cos(d*x + c)^3 - (3*A - 4*B)*cos(d*x + c)^2 - (7*A - 6* 
B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^ 
2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 638 vs. \(2 (190) = 380\).

Time = 1.01 (sec) , antiderivative size = 638, normalized size of antiderivative = 2.89 \[ \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(sqrt(2)*(13*A - 9*B)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan 
(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a)*a*sgn(cos(d*x + c))) + (19*A - 12*B 
)*log(abs(147573952589676412928*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t 
an(1/2*d*x + 1/2*c)^2 + a))^2 - 295147905179352825856*sqrt(2)*abs(a) - 442 
721857769029238784*a)/abs(147573952589676412928*(sqrt(-a)*tan(1/2*d*x + 1/ 
2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 295147905179352825856*sqrt 
(2)*abs(a) - 442721857769029238784*a))/(sqrt(-a)*abs(a)*sgn(cos(d*x + c))) 
 - 2*(sqrt(2)*A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c)))*sqrt( 
-a*tan(1/2*d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3 - 4*sqrt(2)*(29*(s 
qrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A - 
12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6 
*B - 133*(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 + 76*(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 + 55*(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 - 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*t 
an(1/2*d*x + 1/2*c)^2 + a))^2*B*a^2 - 7*A*a^3 + 4*B*a^3)/(((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*sqrt 
(-a)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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