\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) [1323]

Optimal result

Integrand size = 43, antiderivative size = 236 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=-\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (b B-a C) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.47 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.41 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=-\frac {\frac {2 \left (-45 a^2 b B-10 b^3 B+45 a^3 C+a b^2 (45 A+19 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {2 b \left (15 A b^2-20 a b B+20 a^2 C+9 b^2 C\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {6 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}-\frac {2 \left (10 b (b B-a C) \sin (c+d x)+3 \left (5 A b^2-5 a b B+5 a^2 C+3 b^2 C\right ) \sin (2 (c+d x))+6 b^2 C \tan (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)}}{30 b^3 d} \] Input:

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

Output:

-1/30*((2*(-45*a^2*b*B - 10*b^3*B + 45*a^3*C + a*b^2*(45*A + 19*C))*Ellipt 
icPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (2*b*(15*A*b^2 - 20*a*b*B + 
 20*a^2*C + 9*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 + (6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 
3*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*E 
llipticF[ArcSin[Sqrt[Cos[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] 
) - (2*(10*b*(b*B - a*C)*Sin[c + d*x] + 3*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3 
*b^2*C)*Sin[2*(c + d*x)] + 6*b^2*C*Tan[c + d*x]))/Cos[c + d*x]^(3/2))/(b^3 
*d)
 

Rubi [A] (verified)

Time = 2.27 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.465, Rules used = {3042, 4600, 3042, 3534, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

\(\displaystyle \int \frac {A \cos ^2(c+d x)+B \cos (c+d x)+C}{\cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}dx\)

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {2 \int \frac {5 a (b B-a C) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {5 a (b B-a C) \cos ^2(c+d x)+b (5 b B+4 a C) \cos (c+d x)+3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right )}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {5 a (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 b B+4 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\frac {2 \int \frac {-3 a \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)-b \left (20 C a^2-20 b B a+15 A b^2+9 b^2 C\right ) \cos (c+d x)+5 \left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right )}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {-3 a \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \cos ^2(c+d x)-b \left (20 C a^2-20 b B a+15 A b^2+9 b^2 C\right ) \cos (c+d x)+5 \left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {-3 a \left (5 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b \left (20 C a^2-20 b B a+15 A b^2+9 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {\frac {-3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {5 \left (b (b B-a C) \cos (c+d x) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\frac {5 \int \frac {b (b B-a C) \cos (c+d x) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {5 \int \frac {b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-3 \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\frac {5 \int \frac {b (b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (-3 C a^3+3 b B a^2-b^2 (3 A+C) a+b^3 B\right ) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\frac {5 \left (a b (b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {5 \left (a b (b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\frac {5 \left (\frac {2 a b (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a^2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}+\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {\frac {6 \sin (c+d x) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{b d \sqrt {\cos (c+d x)}}+\frac {\frac {5 \left (\frac {2 a b (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}\right )}{a}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 C-5 a b B+5 A b^2+3 b^2 C\right )}{d}}{b}}{3 b}+\frac {10 (b B-a C) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 C \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\)

Input:

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

Output:

(2*C*Sin[c + d*x])/(5*b*d*Cos[c + d*x]^(5/2)) + ((10*(b*B - a*C)*Sin[c + d 
*x])/(3*b*d*Cos[c + d*x]^(3/2)) + (((-6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3*b 
^2*C)*EllipticE[(c + d*x)/2, 2])/d + (5*((2*a*b*(b*B - a*C)*EllipticF[(c + 
 d*x)/2, 2])/d - (6*a^2*(A*b^2 - a*(b*B - a*C))*EllipticPi[(2*a)/(a + b), 
(c + d*x)/2, 2])/((a + b)*d)))/a)/b + (6*(5*A*b^2 - 5*a*b*B + 5*a^2*C + 3* 
b^2*C)*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/(3*b))/(5*b)
 

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 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. \(772\) vs. \(2(225)=450\).

Time = 7.22 (sec) , antiderivative size = 773, normalized size of antiderivative = 3.28

method	result	size
default	\(\text {Expression too large to display}\)	\(773\)

Input:

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

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*C/b/(8*sin 
(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 
2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(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)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4+12*(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)*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c) 
^2*cos(1/2*d*x+1/2*c)-3*(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*sin(1/2*d*x+1/2*c)^ 
4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(B*b-C*a)/b^2*(-1/6*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)/(cos(1/2*d*x+1/2*c)^2-1/ 
2)^2+1/3*(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)*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2)))+2*(A*b^2-B*a*b+C*a^2)/b^3/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*si 
n(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)*Ell 
ipticE(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- 
B*a*b+C*a^2)*a^2/b^3/(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)*E 
llipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/(2*...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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