\(\int \frac {\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [406]

Optimal result

Integrand size = 40, antiderivative size = 164 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 B-3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 B-3 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \] Output:

2*B*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/32*(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*(B-C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(11*B-3*C)* 
tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)
 

Mathematica [A] (warning: unable to verify)

Time = 7.30 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.82 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left ((-43 B+3 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+32 \sqrt {2} B \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec ^{\frac {5}{2}}(c+d x) \sqrt {1+\sec (c+d x)}}{4 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{5/2}}+\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\frac {1}{2} (-15 B+7 C) \sin \left (\frac {1}{2} (c+d x)\right )+\frac {1}{4} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (19 B \sin \left (\frac {1}{2} (c+d x)\right )-11 C \sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-B \sin \left (\frac {1}{2} (c+d x)\right )+C \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(((-43*B + 3*C)*ArcSin[Tan[(c + d*x)/2]] + 32*Sqrt[2]*B*ArcTan[Tan[(c + d* 
x)/2]/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]])*Cos[(c + d*x)/2]^4*Sqrt[Cos[ 
c + d*x]/(1 + Cos[c + d*x])]*Sec[c + d*x]^(5/2)*Sqrt[1 + Sec[c + d*x]])/(4 
*d*Sqrt[Sec[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/2)) + (Cos[(c + d*x) 
/2]^5*Sec[c + d*x]^3*(((-15*B + 7*C)*Sin[(c + d*x)/2])/2 + (Sec[(c + d*x)/ 
2]^2*(19*B*Sin[(c + d*x)/2] - 11*C*Sin[(c + d*x)/2]))/4 + (Sec[(c + d*x)/2 
]^4*(-(B*Sin[(c + d*x)/2]) + C*Sin[(c + d*x)/2]))/2))/(d*(a*(1 + Sec[c + d 
*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4560, 3042, 4410, 27, 3042, 4410, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

\(\displaystyle \frac {-\frac {\int -\frac {32 a^2 B-a^2 (11 B-3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {32 a^2 B-a^2 (11 B-3 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

\(\displaystyle \frac {\frac {32 a B \int \sqrt {\sec (c+d x) a+a}dx-a^2 (43 B-3 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

\(\displaystyle \frac {\frac {-\left (a^2 (43 B-3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {64 a^2 B \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^2 (43 B-3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {\frac {2 a^2 (43 B-3 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {64 a^{3/2} B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} B \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (43 B-3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 B-3 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((B - C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (((64*a^(3/2) 
*B*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a 
^(3/2)*(43*B - 3*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[ 
c + d*x]])])/d)/(4*a^2) - (a*(11*B - 3*C)*Tan[c + d*x])/(2*d*(a + a*Sec[c 
+ d*x])^(3/2)))/(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 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 4410

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-(b*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + 
f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e + 
f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x] 
, x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && 
EqQ[a^2 - b^2, 0] && IntegerQ[2*m]
 

rule 4560

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(139)=278\).

Time = 0.98 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.29

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-2 B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+2 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+32 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+11 B \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-3 C \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-43 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+3 C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\)	\(376\)

Input:

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

Output:

1/32/d/a^3*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^ 
2*csc(d*x+c)^2-1)^(1/2)*(-2*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(csc 
(d*x+c)-cot(d*x+c))+2*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(csc(d*x+c 
)-cot(d*x+c))+32*B*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2- 
1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+11*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^( 
1/2)*(csc(d*x+c)-cot(d*x+c))-3*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*( 
csc(d*x+c)-cot(d*x+c))-43*B*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc 
(d*x+c)^2-1)^(1/2))+3*C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x 
+c)^2-1)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (139) = 278\).

Time = 3.21 (sec) , antiderivative size = 670, normalized size of antiderivative = 4.09 \[ \int \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)*(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)*((43*B - 3*C)*cos(d*x + c)^3 + 3*(43*B - 3*C)*cos(d*x + c)^ 
2 + 3*(43*B - 3*C)*cos(d*x + c) + 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)) - 64*(B*cos(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B) 
*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) 
 + 1)) - 4*((15*B - 7*C)*cos(d*x + c)^2 + (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)*((43* 
B - 3*C)*cos(d*x + c)^3 + 3*(43*B - 3*C)*cos(d*x + c)^2 + 3*(43*B - 3*C)*c 
os(d*x + c) + 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))) - 64*(B*cos(d*x + c)^3 
 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + 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*((15*B 
 - 7*C)*cos(d*x + c)^2 + (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)]
 

Sympy [F(-1)]

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

integrate(cos(d*x+c)*(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]

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

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Reduce [F]

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

int(cos(d*x+c)*(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))/a**3