\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [529]

Optimal result

Integrand size = 43, antiderivative size = 179 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(5 A+19 B-75 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(3 A+5 B-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+9 C) \tan (c+d x)}{4 a^2 d \sqrt {a+a \sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 10.75 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sqrt {2} (5 A+19 B-75 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)}+\frac {1}{4} (A-9 B+113 C+2 (5 A-13 B+85 C) \cos (c+d x)+(A-9 B+49 C) \cos (2 (c+d x))) \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(Sqrt[2]*(5*A 
+ 19*B - 75*C)*ArcSin[Tan[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Sqrt[Cos[c + d* 
x]/(1 + Cos[c + d*x])^2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]] + ((A - 
 9*B + 113*C + 2*(5*A - 13*B + 85*C)*Cos[c + d*x] + (A - 9*B + 49*C)*Cos[2 
*(c + d*x)])*Sec[c + d*x]*Tan[(c + d*x)/2])/4))/(2*d*(A + 2*C + 2*B*Cos[c 
+ d*x] + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4572, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4496

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4489

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4282

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

\(\Big \downarrow \) 216

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

Input:

Int[(Sec[c + d*x]^2*(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)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 
2)) + (-1/2*(a*(3*A + 5*B - 13*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(3 
/2)) + ((Sqrt[2]*a^(3/2)*(5*A + 19*B - 75*C)*ArcTan[(Sqrt[a]*Tan[c + d*x]) 
/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (8*a^2*(A - 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 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. \(389\) vs. \(2(156)=312\).

Time = 3.46 (sec) , antiderivative size = 390, normalized size of antiderivative = 2.18

method	result	size
default	\(\frac {\left (\frac {\left (-2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+83 \csc \left (d x +c \right )-83 \cot \left (d x +c \right )\right ) C}{32}+\frac {\left (-2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+3 \csc \left (d x +c \right )-3 \cot \left (d x +c \right )\right ) A}{32}+\frac {\left (2 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+9 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-11 \csc \left (d x +c \right )+11 \cot \left (d x +c \right )\right ) B}{32}+\frac {5 A \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}+\frac {19 B \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}-\frac {75 C \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\)	\(390\)
parts	\(\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-\frac {4 \sqrt {2}\, \left (\cos \left (d x +c \right )-1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cot \left (d x +c \right )}{\cos \left (d x +c \right )+1}-5 \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+5 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3}}+\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (2 \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \csc \left (d x +c \right )^{3}+11 \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+19 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3}}+\frac {C \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}}{16}-\frac {17 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{32}-\frac {75 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{32}+\frac {83 \csc \left (d x +c \right )}{32}-\frac {83 \cot \left (d x +c \right )}{32}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\)	\(466\)

Input:

int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, 
method=_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-( 
1-cos(d*x+c))^3*csc(d*x+c)^3+3*csc(d*x+c)-3*cot(d*x+c))*A+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+5/32*A*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)+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 = 516, normalized size of antiderivative = 2.88 \[ \int \frac {\sec ^2(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=\left [\frac {\sqrt {2} {\left ({\left (5 \, A + 19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 19 \, B - 75 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 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 (A - 9 \, B + 49 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 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 (5 \, A + 19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 19 \, B - 75 \, C\right )} \cos \left (d x + c\right ) + 5 \, A + 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 (A - 9 \, B + 49 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, A - 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*(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)*((5*A + 19*B - 75*C)*cos(d*x + c)^3 + 3*(5*A + 19*B - 75*C) 
*cos(d*x + c)^2 + 3*(5*A + 19*B - 75*C)*cos(d*x + c) + 5*A + 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)/(co 
s(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((A - 9*B + 49*C)*cos(d*x + c)^2 + 
 (5*A - 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)*((5*A + 19*B - 75*C)*cos(d*x + 
c)^3 + 3*(5*A + 19*B - 75*C)*cos(d*x + c)^2 + 3*(5*A + 19*B - 75*C)*cos(d* 
x + c) + 5*A + 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*((A - 9*B + 49*C 
)*cos(d*x + c)^2 + (5*A - 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*co 
s(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

integrate(sec(d*x+c)^2*(A+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 = 0.87 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left ({\left (\frac {2 \, {\left (\sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8}} + \frac {\sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 9 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 17 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {3 \, \sqrt {2} A a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 11 \, \sqrt {2} B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 83 \, \sqrt {2} C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{8}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} - \frac {{\left (5 \, \sqrt {2} A + 19 \, \sqrt {2} B - 75 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \] Input:

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

Output:

1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*((2*(sqrt(2)*A*a^6*sgn(cos(d*x + 
 c)) - 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 + (sqrt(2)*A*a^6*sgn(cos(d*x + c)) - 9*sqrt(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 - (3*sqrt(2)*A*a^6*sgn(cos(d*x + c)) - 11*sqrt(2)*B*a^6*sg 
n(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) - (5*sqrt(2)*A + 19*sqrt(2)*B - 75*sq 
rt(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. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + 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((A + 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]

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

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