\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx\) [1327]

Optimal result

Integrand size = 43, antiderivative size = 303 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=-\frac {\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a (b B-a C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a b \left (a^2-b^2\right ) d}+\frac {\left (A b^4+a^3 b B-3 a b^3 B-3 a^4 C+a^2 b^2 (A+5 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a-b) b^2 (a+b)^2 d}+\frac {\left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)}}-\frac {\left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (b+a \cos (c+d x))} \] Output:

-(A*b^2-B*a*b+3*C*a^2-2*C*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/( 
a^2-b^2)/d-(A*b^2-a*(B*b-C*a))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/b/ 
(a^2-b^2)/d+(A*b^4+B*a^3*b-3*B*a*b^3-3*a^4*C+a^2*b^2*(A+5*C))*EllipticPi(s 
in(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a/(a-b)/b^2/(a+b)^2/d+(A*b^2-B*a*b+3* 
C*a^2-2*C*b^2)*sin(d*x+c)/b^2/(a^2-b^2)/d/cos(d*x+c)^(1/2)-(A*b^2-a*(B*b-C 
*a))*sin(d*x+c)/b/(a^2-b^2)/d/cos(d*x+c)^(1/2)/(b+a*cos(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 3.95 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2} \, dx=\frac {-\frac {\frac {2 \left (-3 a^2 b B+4 b^3 B+9 a^3 C-a b^2 (A+10 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {4 b \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {2 \left (A b^2-a b B+3 a^2 C-2 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}+4 \sqrt {\cos (c+d x)} \left (\frac {a \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+2 C \tan (c+d x)\right )}{4 b^2 d} \] Input:

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

Output:

(-(((2*(-3*a^2*b*B + 4*b^3*B + 9*a^3*C - a*b^2*(A + 10*C))*EllipticPi[(2*a 
)/(a + b), (c + d*x)/2, 2])/(a + b) + (4*b*(A*b^2 - a*b*B + 2*a^2*C - b^2* 
C)*(2*EllipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x) 
/2, 2])/(a + b)))/a + (2*(A*b^2 - a*b*B + 3*a^2*C - 2*b^2*C)*(-2*a*b*Ellip 
ticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[C 
os[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + 
d*x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b))) 
+ 4*Sqrt[Cos[c + d*x]]*((a*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x])/((a^2 
- b^2)*(b + a*Cos[c + d*x])) + 2*C*Tan[c + d*x]))/(4*b^2*d)
 

Rubi [A] (verified)

Time = 2.25 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.92, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4600, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (-3 C a^3+b B a^2+b^2 (A+4 C) a-2 b^3 B\right )-a b \left (A b^2-a (b B-a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{d}}{b}+\frac {2 \sin (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\left (-3 a^4 C+a^3 b B+a^2 b^2 (A+5 C)-3 a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx-b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{d}}{b}+\frac {2 \sin (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-3 a^4 C+a^3 b B+a^2 b^2 (A+5 C)-3 a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-b \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{d}}{b}+\frac {2 \sin (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\left (-3 a^4 C+a^3 b B+a^2 b^2 (A+5 C)-3 a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx-\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (A b^2-a (b B-a C)\right )}{d}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{d}}{b}+\frac {2 \sin (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {2 \sin (c+d x) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}+\frac {\frac {\frac {2 \left (-3 a^4 C+a^3 b B+a^2 b^2 (A+5 C)-3 a b^3 B+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}-\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (A b^2-a (b B-a C)\right )}{d}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (3 a^2 C-a b B+A b^2-2 b^2 C\right )}{d}}{b}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a \cos (c+d x)+b)}\)

Input:

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

Output:

-(((A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[Cos[c + d*x 
]]*(b + a*Cos[c + d*x]))) + (((-2*(A*b^2 - a*b*B + 3*a^2*C - 2*b^2*C)*Elli 
pticE[(c + d*x)/2, 2])/d + ((-2*b*(A*b^2 - a*(b*B - a*C))*EllipticF[(c + d 
*x)/2, 2])/d + (2*(A*b^4 + a^3*b*B - 3*a*b^3*B - 3*a^4*C + a^2*b^2*(A + 5* 
C))*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/((a + b)*d))/a)/b + (2*(A*b 
^2 - a*b*B + 3*a^2*C - 2*b^2*C)*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(2 
*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* 
c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int 
[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

rule 4600

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(868\) vs. \(2(304)=608\).

Time = 6.38 (sec) , antiderivative size = 869, normalized size of antiderivative = 2.87

method	result	size
default	\(\text {Expression too large to display}\)	\(869\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*sec(d*x+c))^2,x, 
method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C/b^2/sin(1/ 
2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2 
*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(2*sin(1/2 
*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2))-2*(A*b^2-C*a^2)/b^2/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2 
)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 
/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*(A*b^2-B 
*a*b+C*a^2)/a/b*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c) 
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/b* 
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2 
*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^( 
1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c 
)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2 
-a*b)*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/( 
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x 
+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+si...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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