\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [1285]

Optimal result

Integrand size = 45, antiderivative size = 300 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 a^{3/2} d}-\frac {(5 A-9 B+13 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A-B+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(2 A-6 B+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(8*A-12*B+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-9*B+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-B+C)*sin(d*x+c)/d/cos(d 
*x+c)^(7/2)/(a+a*sec(d*x+c))^(3/2)+1/2*(A-B+2*C)*sin(d*x+c)/a/d/cos(d*x+c) 
^(5/2)/(a+a*sec(d*x+c))^(1/2)-1/4*(2*A-6*B+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 = 2.97 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.80 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (5 A-9 B+13 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {\sqrt {2} (8 A-12 B+19 C) \text {arctanh}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+\frac {1}{2} (-2 A+6 B-3 C+(8 B-6 C) \cos (c+d x)+(-2 A+6 B-7 C) \cos (2 (c+d x))) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )}{-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{3/2}} \] 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])^(3/2)),x]
 

Output:

-((Cos[(c + d*x)/2]^3*Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d 
*x]^2)*(2*(5*A - 9*B + 13*C)*ArcTanh[Sin[(c + d*x)/2]] + (Sqrt[2]*(8*A - 1 
2*B + 19*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2 + ((-2*A 
+ 6*B - 3*C + (8*B - 6*C)*Cos[c + d*x] + (-2*A + 6*B - 7*C)*Cos[2*(c + d*x 
)])*Sec[c + d*x]^2*Sin[(c + d*x)/2])/2)/(-1 + Sin[(c + d*x)/2]^2)))/(d*(A 
+ 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(3/2 
)))
 

Rubi [A] (verified)

Time = 1.98 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.98, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4753, 3042, 4572, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int -\frac {2 \sec ^{\frac {3}{2}}(c+d x) \left (3 a^2 (A-B+2 C)-a^2 (2 A-6 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 a^2 (A-B+2 C)-a^2 (2 A-6 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 a^2 (A-B+2 C)-a^2 (2 A-6 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4509

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\frac {\int -\frac {\sqrt {\sec (c+d x)} \left (a^3 (2 A-6 B+7 C)-a^3 (8 A-12 B+19 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^3 (2 A-6 B+7 C)-a^3 (8 A-12 B+19 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (2 A-6 B+7 C)-a^3 (8 A-12 B+19 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4511

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-a^2 (8 A-12 B+19 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-a^2 (8 A-12 B+19 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {2 a^2 (8 A-12 B+19 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^{5/2} (8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {-\frac {4 a^3 (5 A-9 B+13 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {2 a^{5/2} (8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {-\frac {\frac {2 \sqrt {2} a^{5/2} (5 A-9 B+13 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^{5/2} (8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )\)

Input:

Int[(A + B*Sec[c + d*x] + 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 - B + C)*Sec[c + d*x]^(7/2 
)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(3/2)) - ((-2*a*(A - B + 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 - 12*B + 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 - 9*B + 13*C)*ArcTanh[(Sqrt[a]*Sqrt[S 
ec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a - (a^ 
2*(2*A - 6*B + 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 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. \(1012\) vs. \(2(253)=506\).

Time = 4.49 (sec) , antiderivative size = 1013, normalized size of antiderivative = 3.38

method	result	size
default	\(\text {Expression too large to display}\)	\(1013\)

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))^(3/2 
),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/4*2^(1/2)*(-1/2*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)^(7/2)/a*((56*cos(1/2*d*x+1/2*c)^6-72*cos( 
1/2*d*x+1/2*c)^4+26*cos(1/2*d*x+1/2*c)^2-2)*tan(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)*(208*cos(1/2*d*x+1/2*c)^7-312*cos(1/2*d* 
x+1/2*c)^5+156*cos(1/2*d*x+1/2*c)^3-26*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))*(152*cos(1/2*d*x+1 
/2*c)^7-228*cos(1/2*d*x+1/2*c)^5+114*cos(1/2*d*x+1/2*c)^3-19*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))*(152*cos(1/2*d*x+1/2*c)^7-228*cos(1/2*d*x+1/2*c)^5+114*cos(1/2*d*x+1/2 
*c)^3-19*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)* 
(-208*cos(1/2*d*x+1/2*c)^7+312*cos(1/2*d*x+1/2*c)^5-156*cos(1/2*d*x+1/2*c) 
^3+26*cos(1/2*d*x+1/2*c)))+B/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*((12*cos(1/2*d*x+1/2*c)^ 
4-8*cos(1/2*d*x+1/2*c)^2+1)*tan(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)*(36*cos(1/2*d*x+1/2*c)^5-36*cos(1/2*d*x+1/2*c)^3+9*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)*(-36*cos(1/2*d 
*x+1/2*c)^5+36*cos(1/2*d*x+1/2*c)^3-9*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))*(24*cos(1/2*d*x+1/2 
*c)^5-24*cos(1/2*d*x+1/2*c)^3+6*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))*(24*cos(1/2*d*x+1/2*c)...
 

Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 816, normalized size of antiderivative = 2.72 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] 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) 
)^(3/2),x, algorithm="fricas")
 

Output:

[1/16*(2*sqrt(2)*((5*A - 9*B + 13*C)*cos(d*x + c)^4 + 2*(5*A - 9*B + 13*C) 
*cos(d*x + c)^3 + (5*A - 9*B + 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 - 6*B + 7*C)*cos(d*x + c)^2 - (4*B - 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 - 12*B + 19*C)*cos(d*x + c)^4 + 2*(8*A - 12*B + 19* 
C)*cos(d*x + c)^3 + (8*A - 12*B + 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 - 9*B + 13*C)*cos(d*x + 
c)^4 + 2*(5*A - 9*B + 13*C)*cos(d*x + c)^3 + (5*A - 9*B + 13*C)*cos(d*x + 
c)^2)*sqrt(-a)*arctan(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*((2*A - 
 6*B + 7*C)*cos(d*x + c)^2 - (4*B - 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 - 12*B 
+ 19*C)*cos(d*x + c)^4 + 2*(8*A - 12*B + 19*C)*cos(d*x + c)^3 + (8*A - 12* 
B + 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 - ...
 

Sympy [F(-1)]

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15483 vs. \(2 (253) = 506\).

Time = 1.25 (sec) , antiderivative size = 15483, normalized size of antiderivative = 51.61 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] 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) 
)^(3/2),x, algorithm="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. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c) 
)^(3/2),x, algorithm="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 extensionDegree 
mismatch
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

Reduce [F]

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

int((A+B*sec(d*x+c)+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))*sec(c + d 
*x))/(cos(c + d*x)**3*sec(c + d*x)**2 + 2*cos(c + d*x)**3*sec(c + d*x) + c 
os(c + d*x)**3),x)*b + int((sqrt(sec(c + d*x) + 1)*sqrt(cos(c + d*x)))/(co 
s(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