\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [1166]

Optimal result

Integrand size = 37, antiderivative size = 285 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(8 A+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 a^{3/2} d}-\frac {(5 A+13 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(8*A+19*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^(3/2)/d-1/4*(5*A+13*C)*arctanh(1/2*a^(1/2)*sec 
(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^(3/2)/d-1/2*(A+C)*sin(d*x+c)/d/cos(d*x+c)^(7/2) 
/(a+a*sec(d*x+c))^(3/2)+1/2*(A+2*C)*sin(d*x+c)/a/d/cos(d*x+c)^(5/2)/(a+a*s 
ec(d*x+c))^(1/2)-1/4*(2*A+7*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*sec(d* 
x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 3.90 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.75 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\left (C+A \cos ^2(c+d x)\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left ((5 A+13 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right )^2-\frac {(8 A+19 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right )^2}{\sqrt {2}}+(2 A+3 C+6 C \cos (c+d x)+(2 A+7 C) \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{4 a d \cos ^{\frac {5}{2}}(c+d x) (A+2 C+A \cos (2 (c+d x))) \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^ 
(3/2)),x]
 

Output:

-1/4*((C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]*((5*A + 13*C)*ArcTanh[Sin[(c 
 + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2])^2 - ((8*A + 19*C)*Ar 
cTanh[Sqrt[2]*Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2])^ 
2)/Sqrt[2] + (2*A + 3*C + 6*C*Cos[c + d*x] + (2*A + 7*C)*Cos[2*(c + d*x)]) 
*Sin[(c + d*x)/2]))/(a*d*Cos[c + d*x]^(5/2)*(A + 2*C + A*Cos[2*(c + d*x)]) 
*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.88 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4753, 3042, 4573, 27, 3042, 4509, 27, 3042, 4509, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int -\frac {\sqrt {\sec (c+d x)} \left (a^3 (2 A+7 C)-a^3 (8 A+19 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^3 (2 A+7 C)-a^3 (8 A+19 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (2 A+7 C)-a^3 (8 A+19 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4511

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A+13 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-a^2 (8 A+19 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-a^2 (8 A+19 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {2 a^2 (8 A+19 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^{5/2} (8 A+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {-\frac {4 a^3 (5 A+13 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {2 a^{5/2} (8 A+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {2 \sqrt {2} a^{5/2} (5 A+13 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^{5/2} (8 A+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

Input:

Int[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)) 
,x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((A + C)*Sec[c + d*x]^(7/2)*Si 
n[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) - ((-2*a*(A + 2*C)*Sec[c + d*x] 
^(5/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) - (-1/2*((-2*a^(5/2)*(8* 
A + 19*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2 
*Sqrt[2]*a^(5/2)*(5*A + 13*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + 
d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a - (a^2*(2*A + 7*C)*Sec[c + 
 d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(4*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 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 4573

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) 
*(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*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, 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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(488\) vs. \(2(238)=476\).

Time = 4.21 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.72

method	result	size
default	\(-\frac {\left (\sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (10 \cos \left (d x +c \right )^{3}+10 \cos \left (d x +c \right )^{2}\right )+\sqrt {2}\, C \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (26 \cos \left (d x +c \right )^{3}+26 \cos \left (d x +c \right )^{2}\right )+A \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-8 \cos \left (d x +c \right )^{3}-8 \cos \left (d x +c \right )^{2}\right )+C \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-19 \cos \left (d x +c \right )^{3}-19 \cos \left (d x +c \right )^{2}\right )+A \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-8 \cos \left (d x +c \right )^{3}-8 \cos \left (d x +c \right )^{2}\right )+C \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (-19 \cos \left (d x +c \right )^{3}-19 \cos \left (d x +c \right )^{2}\right )+2 A \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{2}+\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )-2\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{8 d \cos \left (d x +c \right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) a^{2} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(489\)

Input:

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x,method=_R 
ETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-1/8/d*(2^(1/2)*A*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c 
)+cot(d*x+c)))*(10*cos(d*x+c)^3+10*cos(d*x+c)^2)+2^(1/2)*C*arctan(1/2*2^(1 
/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))*(26*cos(d*x+c)^3+2 
6*cos(d*x+c)^2)+A*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1)) 
^(1/2))*(-8*cos(d*x+c)^3-8*cos(d*x+c)^2)+C*arctan(1/2*(cot(d*x+c)-csc(d*x+ 
c)-1)/(-1/(cos(d*x+c)+1))^(1/2))*(-19*cos(d*x+c)^3-19*cos(d*x+c)^2)+A*arct 
an(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))*(-8*cos(d*x+c) 
^3-8*cos(d*x+c)^2)+C*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2)*(cot(d*x+c)-csc( 
d*x+c)+1))*(-19*cos(d*x+c)^3-19*cos(d*x+c)^2)+2*A*2^(1/2)*sin(d*x+c)*(-2/( 
cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2+sin(d*x+c)*(7*cos(d*x+c)^2+3*cos(d*x+c)- 
2)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2))*(a*(1+sec(d*x+c)))^(1/2)/cos(d*x+c 
)^(3/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)/a^2/(-1/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 762, normalized size of antiderivative = 2.67 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, al 
gorithm="fricas")
 

Output:

[1/16*(2*sqrt(2)*((5*A + 13*C)*cos(d*x + c)^4 + 2*(5*A + 13*C)*cos(d*x + c 
)^3 + (5*A + 13*C)*cos(d*x + c)^2)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 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) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) 
 - 4*((2*A + 7*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x 
 + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 19*C)*c 
os(d*x + c)^4 + 2*(8*A + 19*C)*cos(d*x + c)^3 + (8*A + 19*C)*cos(d*x + c)^ 
2)*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^2*d*cos(d*x + c)^4 
+ 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2), 1/8*(2*sqrt(2)*((5*A + 1 
3*C)*cos(d*x + c)^4 + 2*(5*A + 13*C)*cos(d*x + c)^3 + (5*A + 13*C)*cos(d*x 
 + c)^2)*sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/co 
s(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c) + a)) - 2*((2* 
A + 7*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 19*C)*cos(d*x + 
c)^4 + 2*(8*A + 19*C)*cos(d*x + c)^3 + (8*A + 19*C)*cos(d*x + c)^2)*sqrt(- 
a)*arctan(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)^2 - a*cos(d*x + c) - 2*a)))/(a^2*d*cos( 
d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((A+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 10551 vs. \(2 (238) = 476\).

Time = 0.93 (sec) , antiderivative size = 10551, normalized size of antiderivative = 37.02 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, al 
gorithm="maxima")
 

Output:

1/16*(4*(4*(sin(2*d*x + 2*c) + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c))))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqr 
t(2)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c)))^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(2*d*x + 2*c) 
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*sqrt(2)*sin(1/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*(sqrt(2)*cos(2*d*x + 2* 
c) + sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqr 
t(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2* 
sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 2*(sqr 
t(2)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c)))^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(2*d*x + 2*c) 
*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*sqrt(2)*sin(1/2* 
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*(sqrt(2)*cos(2*d*x + 2* 
c) + sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqr 
t(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), c 
os(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) 
))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2* 
sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 2*(...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, al 
gorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionDegree mismatch inside factorisation over extensionUnable 
to divide
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^(3/2)) 
,x)
 

Output:

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^(3/2)) 
, x)
 

Reduce [F]

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

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),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)**2 + 2*cos(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)))/(cos(c + 
 d*x)**3*sec(c + d*x)**2 + 2*cos(c + d*x)**3*sec(c + d*x) + cos(c + d*x)** 
3),x)*a))/a**2