\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\) [1291]

Optimal result

Integrand size = 45, antiderivative size = 294 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {(2 B-5 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {(3 A-43 B+115 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \] Output:

(2*B-5*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1 
/2)*sec(d*x+c)^(1/2)/a^(5/2)/d+1/32*(3*A-43*B+115*C)*arctanh(1/2*a^(1/2)*s 
ec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2 
)*sec(d*x+c)^(1/2)*2^(1/2)/a^(5/2)/d-1/4*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^( 
7/2)/(a+a*sec(d*x+c))^(5/2)+1/16*(A+7*B-15*C)*sin(d*x+c)/a/d/cos(d*x+c)^(5 
/2)/(a+a*sec(d*x+c))^(3/2)+1/16*(3*A-11*B+35*C)*sin(d*x+c)/a^2/d/cos(d*x+c 
)^(3/2)/(a+a*sec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 6.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((6 A-86 B+230 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 \sqrt {2} (2 B-5 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} (3 A-11 B+67 C+2 (7 A-15 B+55 C) \cos (c+d x)+(3 A-11 B+35 C) \cos (2 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d \sqrt {\cos (c+d x)} (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{5/2}} \] Input:

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

Output:

(Cos[(c + d*x)/2]^5*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((6*A - 86*B + 
 230*C)*ArcTanh[Sin[(c + d*x)/2]] + 32*Sqrt[2]*(2*B - 5*C)*ArcTanh[Sqrt[2] 
*Sin[(c + d*x)/2]] + ((3*A - 11*B + 67*C + 2*(7*A - 15*B + 55*C)*Cos[c + d 
*x] + (3*A - 11*B + 35*C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*Sec[c + d*x 
]*Tan[(c + d*x)/2])/2))/(4*d*Sqrt[Cos[c + d*x]]*(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 = 2.00 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.99, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 4753, 3042, 4572, 27, 3042, 4507, 27, 3042, 4509, 3042, 4511, 3042, 4288, 222, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/4*((A - B + C)*Sec[c + d*x]^(7/2 
)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + ((a*(A + 7*B - 15*C)*Sec[ 
c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (((32*a^(5 
/2)*(2*B - 5*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/ 
d + (Sqrt[2]*a^(5/2)*(3*A - 43*B + 115*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d* 
x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a + (2*a^2*(3*A 
- 11*B + 35*C)*Sec[c + d*x]^(3/2)*Sin[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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

rule 4295

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f))   Subst[Int[1/(2*b - d*x^2), x], 
x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; 
 FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
 

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 4509

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

rule 4511

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)/b   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b 
 Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b 
, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 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)]
 

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(795\) vs. \(2(249)=498\).

Time = 4.45 (sec) , antiderivative size = 796, normalized size of antiderivative = 2.71

method	result	size
default	\(\text {Expression too large to display}\)	\(796\)

Input:

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

Output:

1/8*2^(1/2)*(1/4*C/d/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^( 
1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(5/2)/a^2*((140*cos(1/2*d*x+1/2*c)^6-100*c 
os(1/2*d*x+1/2*c)^4+11*cos(1/2*d*x+1/2*c)^2+2)*tan(1/2*d*x+1/2*c)*sec(1/2* 
d*x+1/2*c)^2+ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1)*(460*cos(1/2*d*x 
+1/2*c)^5-460*cos(1/2*d*x+1/2*c)^3+115*cos(1/2*d*x+1/2*c))+2^(1/2)*arctanh 
(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)+1))*(320*cos(1/2*d*x+1 
/2*c)^5-320*cos(1/2*d*x+1/2*c)^3+80*cos(1/2*d*x+1/2*c))+2^(1/2)*arctanh(1/ 
2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)-1))*(320*cos(1/2*d*x+1/2* 
c)^5-320*cos(1/2*d*x+1/2*c)^3+80*cos(1/2*d*x+1/2*c))+ln(-cot(1/2*d*x+1/2*c 
)+csc(1/2*d*x+1/2*c)-1)*(-460*cos(1/2*d*x+1/2*c)^5+460*cos(1/2*d*x+1/2*c)^ 
3-115*cos(1/2*d*x+1/2*c)))-1/4*B/d/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/(a/(2* 
cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(32*2^(1/2)*cos(1/ 
2*d*x+1/2*c)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-csc(1/2*d*x+1/2*c)+1) 
)+32*2^(1/2)*cos(1/2*d*x+1/2*c)*arctanh(1/2*2^(1/2)*(cot(1/2*d*x+1/2*c)-cs 
c(1/2*d*x+1/2*c)-1))-43*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1)*cos(1 
/2*d*x+1/2*c)+43*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1)*cos(1/2*d*x+ 
1/2*c)+11*tan(1/2*d*x+1/2*c)+2*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^2)+1/ 
4*A/d/(a/(2*cos(1/2*d*x+1/2*c)^2-1)*cos(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2 
*d*x+1/2*c)^2-1)^(1/2)/a^2*(3*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)+1) 
*cos(1/2*d*x+1/2*c)-3*ln(-cot(1/2*d*x+1/2*c)+csc(1/2*d*x+1/2*c)-1)*cos(...
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 892, normalized size of antiderivative = 3.03 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/64*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^4 + 3*(3*A - 43*B + 115* 
C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + (3*A - 43*B + 
115*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sq 
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 2* 
a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((3*A - 1 
1*B + 35*C)*cos(d*x + c)^2 + (7*A - 15*B + 55*C)*cos(d*x + c) + 16*C)*sqrt 
((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 16*( 
(2*B - 5*C)*cos(d*x + c)^4 + 3*(2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)* 
cos(d*x + c)^2 + (2*B - 5*C)*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 + 
 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt 
(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + 
cos(d*x + c)^2)))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d 
*cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/32*(sqrt(2)*((3*A - 43*B + 115*C 
)*cos(d*x + c)^4 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 
 115*C)*cos(d*x + c)^2 + (3*A - 43*B + 115*C)*cos(d*x + c))*sqrt(-a)*arcta 
n(1/2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d* 
x + c))*sin(d*x + c)/(a*cos(d*x + c) + a)) - 2*((3*A - 11*B + 35*C)*cos(d* 
x + c)^2 + (7*A - 15*B + 55*C)*cos(d*x + c) + 16*C)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 16*((2*B - 5*C)*cos(d* 
x + c)^4 + 3*(2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*x + c)^2 ...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [A] (verification not implemented)

Time = 165.95 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.59 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(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)*(5*A*a^5 - 13 
*B*a^5 + 21*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) - sqrt(2) 
*(3*A*sqrt(a) - 43*B*sqrt(a) + 115*C*sqrt(a))*log((sqrt(a)*tan(1/2*d*x + 1 
/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(a^3*sgn(cos(d*x + c))) + 3 
2*(2*B - 5*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))^2 - a*(2*sqrt(2) + 3)))/(a^(5/2)*sgn(cos(d*x + c))) - 32*( 
2*B - 5*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))^2 + a*(2*sqrt(2) - 3)))/(a^(5/2)*sgn(cos(d*x + c))) + 128*sqr 
t(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*C*sqrt(a) - C*a^(3/2))/(((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)*a^2*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Reduce [F]

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

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

Output:

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