\(\int \frac {\cos (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [532]

Optimal result

Integrand size = 41, antiderivative size = 217 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(5 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(115 A-43 B+3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(15 A-7 B-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

-(5*A-2*B)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d+1/3 
2*(115*A-43*B+3*C)*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^(5/2)/d-1/4*(A-B+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1 
/16*(15*A-7*B-C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/16*(35*A-11*B+3*C 
)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 5.00 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.83 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {32 (5 A-2 B) \arctan \left (\sqrt {-1+\sec (c+d x)}\right ) \sqrt {-1+\sec (c+d x)} \sin (c+d x)-\sqrt {2} (115 A-43 B+3 C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \sqrt {-1+\sec (c+d x)} \sin (c+d x)-2 (43 A-11 B+3 C+(55 A-15 B+7 C) \cos (c+d x)+8 A \cos (2 (c+d x))) \tan ^3\left (\frac {1}{2} (c+d x)\right )}{32 a^2 d (-1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(32*(5*A - 2*B)*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Sqrt[-1 + Sec[c + d*x]]*Si 
n[c + d*x] - Sqrt[2]*(115*A - 43*B + 3*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/S 
qrt[2]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x] - 2*(43*A - 11*B + 3*C + (55* 
A - 15*B + 7*C)*Cos[c + d*x] + 8*A*Cos[2*(c + d*x)])*Tan[(c + d*x)/2]^3)/( 
32*a^2*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.46 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.390, Rules used = {3042, 4572, 27, 3042, 4508, 27, 3042, 4510, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (35 A-11 B+3 C)-3 a^2 (15 A-7 B-C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (35 A-11 B+3 C)-3 a^2 (15 A-7 B-C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 a^2 (35 A-11 B+3 C)-3 a^2 (15 A-7 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {\frac {\int -\frac {16 a^3 (5 A-2 B)-a^3 (35 A-11 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {16 a^3 (5 A-2 B)-a^3 (35 A-11 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {16 a^3 (5 A-2 B)-a^3 (35 A-11 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {16 a^2 (5 A-2 B) \int \sqrt {\sec (c+d x) a+a}dx-a^3 (115 A-43 B+3 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {16 a^2 (5 A-2 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^3 (115 A-43 B+3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {-\left (a^3 (115 A-43 B+3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {32 a^3 (5 A-2 B) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {32 a^{5/2} (5 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^3 (115 A-43 B+3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (115 A-43 B+3 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {32 a^{5/2} (5 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {2 a^2 (35 A-11 B+3 C) \sin (c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {\frac {32 a^{5/2} (5 A-2 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{5/2} (115 A-43 B+3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}}{4 a^2}-\frac {a (15 A-7 B-C) \sin (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A - B + C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (-1/2*(a* 
(15*A - 7*B - C)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + (-(((32*a^ 
(5/2)*(5*A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]) 
/d - (Sqrt[2]*a^(5/2)*(115*A - 43*B + 3*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/( 
Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a) + (2*a^2*(35*A - 11*B + 3*C)*Sin 
[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a^2))/(8*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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]
 

rule 4572

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(615\) vs. \(2(188)=376\).

Time = 6.76 (sec) , antiderivative size = 616, normalized size of antiderivative = 2.84

method	result	size
default	\(-\frac {\left (\left (80 \cos \left (d x +c \right )^{3}+240 \cos \left (d x +c \right )^{2}+240 \cos \left (d x +c \right )+80\right ) \sqrt {2}\, A \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-32 \cos \left (d x +c \right )^{3}-96 \cos \left (d x +c \right )^{2}-96 \cos \left (d x +c \right )-32\right ) \sqrt {2}\, B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\csc \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\cot \left (d x +c \right )^{2}-1}}\right )+\left (-115 \cos \left (d x +c \right )^{3}-345 \cos \left (d x +c \right )^{2}-345 \cos \left (d x +c \right )-115\right ) A \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (43 \cos \left (d x +c \right )^{3}+129 \cos \left (d x +c \right )^{2}+129 \cos \left (d x +c \right )+43\right ) B \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+\left (-3 \cos \left (d x +c \right )^{3}-9 \cos \left (d x +c \right )^{2}-9 \cos \left (d x +c \right )-3\right ) C \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\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 (-32 \cos \left (d x +c \right )^{2}-110 \cos \left (d x +c \right )-70\right ) A +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (30 \cos \left (d x +c \right )+22\right ) B +\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-14 \cos \left (d x +c \right )-6\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \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 )}\)	\(616\)

Input:

int(cos(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-1/32/d/a^3*((80*cos(d*x+c)^3+240*cos(d*x+c)^2+240*cos(d*x+c)+80)*2^(1/2)* 
A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d*x+c)^2-2*csc 
(d*x+c)*cot(d*x+c)+cot(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(-32*cos 
(d*x+c)^3-96*cos(d*x+c)^2-96*cos(d*x+c)-32)*2^(1/2)*B*(-2*cos(d*x+c)/(cos( 
d*x+c)+1))^(1/2)*arctanh(2^(1/2)/(csc(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+cot 
(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(-115*cos(d*x+c)^3-345*cos(d*x 
+c)^2-345*cos(d*x+c)-115)*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((-2*co 
s(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(43*cos(d*x+c)^3+129 
*cos(d*x+c)^2+129*cos(d*x+c)+43)*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln 
((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+(-3*cos(d*x+c 
)^3-9*cos(d*x+c)^2-9*cos(d*x+c)-3)*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+sin(d*x+c)* 
cos(d*x+c)*(-32*cos(d*x+c)^2-110*cos(d*x+c)-70)*A+sin(d*x+c)*cos(d*x+c)*(3 
0*cos(d*x+c)+22)*B+sin(d*x+c)*cos(d*x+c)*(-14*cos(d*x+c)-6)*C)*(a*(1+sec(d 
*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)
 

Fricas [A] (verification not implemented)

Time = 24.74 (sec) , antiderivative size = 774, normalized size of antiderivative = 3.57 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/64*(sqrt(2)*((115*A - 43*B + 3*C)*cos(d*x + c)^3 + 3*(115*A - 43*B + 3 
*C)*cos(d*x + c)^2 + 3*(115*A - 43*B + 3*C)*cos(d*x + c) + 115*A - 43*B + 
3*C)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(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)) - 32*((5*A - 2*B)*cos(d*x + c)^3 + 
 3*(5*A - 2*B)*cos(d*x + c)^2 + 3*(5*A - 2*B)*cos(d*x + c) + 5*A - 2*B)*sq 
rt(-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)) - 4*(16*A*cos(d*x + c)^3 + (55*A - 15*B + 7*C)*cos(d*x + c)^2 + (35*A 
- 11*B + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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)*((115*A - 43*B + 3*C)*cos(d*x + c)^3 + 3*(115 
*A - 43*B + 3*C)*cos(d*x + c)^2 + 3*(115*A - 43*B + 3*C)*cos(d*x + c) + 11 
5*A - 43*B + 3*C)*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))) - 32*((5*A - 2*B)*cos(d*x + c) 
^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + 3*(5*A - 2*B)*cos(d*x + c) + 5*A - 2*B 
)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqr 
t(a)*sin(d*x + c))) - 2*(16*A*cos(d*x + c)^3 + (55*A - 15*B + 7*C)*cos(d*x 
 + c)^2 + (35*A - 11*B + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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 ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (188) = 376\).

Time = 1.16 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.20 \[ \int \frac {\cos (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/64*(2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(A*a^5 - B*a^5 + C 
*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x + c))) - sqrt(2)*(21*A*a^5 - 
 13*B*a^5 + 5*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt( 
2)*(115*A - 43*B + 3*C)*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^2*sgn(cos(d*x + c))) - 32*(5*A - 2* 
B)*log(abs(-562949953421312*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1 
/2*d*x + 1/2*c)^2 + a))^2 - 1125899906842624*sqrt(2)*abs(a) + 168884986026 
3936*a)/abs(-562949953421312*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan( 
1/2*d*x + 1/2*c)^2 + a))^2 + 1125899906842624*sqrt(2)*abs(a) + 16888498602 
63936*a))/(sqrt(-a)*a*abs(a)*sgn(cos(d*x + c))) - 128*sqrt(2)*(3*(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*a)/(( 
(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)*sqrt(-a)*a*sgn(cos(d*x + c))))/d
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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