\(\int \frac {\sec ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [403]

Optimal result

Integrand size = 42, antiderivative size = 169 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(19 B-75 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(B-C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(5 B-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(B-9 C) \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

1/32*(19*B-75*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)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c) 
)^(5/2)-1/16*(5*B-13*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)-1/4*(B-9*C)* 
tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

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

Time = 10.43 (sec) , antiderivative size = 628, normalized size of antiderivative = 3.72 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/4*(B*Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*Sin[(c + d*x)/2]*Sqrt[(1 - 2*S 
in[(c + d*x)/2]^2)^(-1)]*(11 - 31*Sin[(c + d*x)/2]^2 + 18*Sin[(c + d*x)/2] 
^4 - (19*ArcTanh[Sqrt[-(Sin[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]]*C 
os[(c + d*x)/2]^4)/Sqrt[-(Sin[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]) 
)/(d*(a*(1 + Sec[c + d*x]))^(5/2)) + (2*C*Cos[(c + d*x)/2]*Sec[c + d*x]^(5 
/2)*Sin[(c + d*x)/2]*((1 - 2*Sin[(c + d*x)/2]^2)^(-1))^(3/2)*((8*Cos[(c + 
d*x)/2]^6*HypergeometricPFQ[{2, 2, 2, 5/2}, {1, 1, 11/2}, Sin[(c + d*x)/2] 
^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*Sin[(c + d*x)/2]^2)/(315*(-1 + 2*Sin[(c + 
d*x)/2]^2)) + (Csc[(c + d*x)/2]^8*(1 - 2*Sin[(c + d*x)/2]^2)^2*Sqrt[Sin[(c 
 + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*(-15*ArcTanh[Sqrt[Sin[(c + d*x)/ 
2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]]*Cos[(c + d*x)/2]^4*(-343 + 1465*Sin[(c 
+ d*x)/2]^2 - 2021*Sin[(c + d*x)/2]^4 + 824*Sin[(c + d*x)/2]^6) + Sqrt[Sin 
[(c + d*x)/2]^2/(-1 + 2*Sin[(c + d*x)/2]^2)]*(-5145 + 33980*Sin[(c + d*x)/ 
2]^2 - 87764*Sin[(c + d*x)/2]^4 + 109737*Sin[(c + d*x)/2]^6 - 66122*Sin[(c 
 + d*x)/2]^8 + 15344*Sin[(c + d*x)/2]^10)))/120))/(d*(a*(1 + Sec[c + d*x]) 
)^(5/2))
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4560, 3042, 4507, 27, 3042, 4496, 27, 3042, 4489, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (B-C)-a (B-9 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (4 a (B-C)-a (B-9 C) \sec (c+d x))}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4496

\(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (3 a^2 (5 B-13 C)-4 a^2 (B-9 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (3 a^2 (5 B-13 C)-4 a^2 (B-9 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a^2 (5 B-13 C)-4 a^2 (B-9 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {a^2 (19 B-75 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx-\frac {8 a^2 (B-9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (19 B-75 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {8 a^2 (B-9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {-\frac {2 a^2 (19 B-75 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {8 a^2 (B-9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\frac {\sqrt {2} a^{3/2} (19 B-75 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 a^2 (B-9 C) \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {a (5 B-13 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

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

Output:

((B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ( 
-1/2*(a*(5*B - 13*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) + ((Sqrt 
[2]*a^(3/2)*(19*B - 75*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + 
a*Sec[c + d*x]])])/d - (8*a^2*(B - 9*C)*Tan[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 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 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 4489

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

rule 4496

Int[csc[(e_.) + (f_.)*(x_)]^2*(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/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b 
*B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne 
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-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 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 [A] (warning: unable to verify)

Time = 1.68 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {\left (\frac {\left (-2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+83 \csc \left (d x +c \right )-83 \cot \left (d x +c \right )\right ) C}{32}+\frac {\left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+9 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-11 \csc \left (d x +c \right )+11 \cot \left (d x +c \right )\right ) B}{32}+\frac {19 B \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}-\frac {75 C \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\)	\(267\)
parts	\(\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2 \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \csc \left (d x +c \right )^{3}+11 \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+19 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3}}+\frac {C \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}-\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}-\frac {75 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\)	\(300\)

Input:

int(sec(d*x+c)^2*(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/d*(1/32*(-2*(1-cos(d*x+c))^5*csc(d*x+c)^5-17*(1-cos(d*x+c))^3*csc(d*x+c) 
^3+83*csc(d*x+c)-83*cot(d*x+c))*C+1/32*(2*(1-cos(d*x+c))^5*csc(d*x+c)^5+9* 
(1-cos(d*x+c))^3*csc(d*x+c)^3-11*csc(d*x+c)+11*cot(d*x+c))*B+19/32*B*ln((- 
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/( 
cos(d*x+c)+1))^(1/2)-75/32*C*ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d 
*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/a^3*(a*(1+sec(d*x+ 
c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 484, normalized size of antiderivative = 2.86 \[ \int \frac {\sec ^2(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=\left [\frac {\sqrt {2} {\left ({\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right ) + 19 \, B - 75 \, C\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (13 \, B - 85 \, C\right )} \cos \left (d x + c\right ) - 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\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 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right ) + 19 \, B - 75 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (13 \, B - 85 \, C\right )} \cos \left (d x + c\right ) - 32 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\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(sec(d*x+c)^2*(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)*((19*B - 75*C)*cos(d*x + c)^3 + 3*(19*B - 75*C)*cos(d*x + c 
)^2 + 3*(19*B - 75*C)*cos(d*x + c) + 19*B - 75*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)) - 4*((9*B - 49*C)*cos(d*x + c)^2 + (13*B - 85*C)*cos(d*x + c) 
- 32*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)*((19*B - 75*C)*cos(d*x + c)^3 + 3*(19*B - 75*C)*cos(d*x + c)^2 + 
3*(19*B - 75*C)*cos(d*x + c) + 19*B - 75*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))) + 2* 
((9*B - 49*C)*cos(d*x + c)^2 + (13*B - 85*C)*cos(d*x + c) - 32*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]

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

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

Output:

Integral((B + C*sec(c + d*x))*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(5/2 
), x)
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 1.47 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.53 \[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left ({\left (\frac {2 \, {\left (\sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} + \frac {9 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 17 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {11 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 83 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} + \frac {{\left (19 \, \sqrt {2} B - 75 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \] Input:

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

Output:

-1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*(sqrt(2)*B*a^6*sgn(cos(d*x 
+ c)) - sqrt(2)*C*a^6*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a^8 + (9*s 
qrt(2)*B*a^6*sgn(cos(d*x + c)) - 17*sqrt(2)*C*a^6*sgn(cos(d*x + c)))/a^8)* 
tan(1/2*d*x + 1/2*c)^2 - (11*sqrt(2)*B*a^6*sgn(cos(d*x + c)) - 83*sqrt(2)* 
C*a^6*sgn(cos(d*x + c)))/a^8)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c) 
^2 - a) + (19*sqrt(2)*B - 75*sqrt(2)*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)))/(sqrt(-a)*a^2*sgn(cos(d*x + c 
))))/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b \right )}{a^{3}} \] Input:

int(sec(d*x+c)^2*(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)*sec(c + d*x)**4)/(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) 
*sec(c + d*x)**3)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 
1),x)*b))/a**3