\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx\) [1331]

Optimal result

Integrand size = 43, antiderivative size = 423 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=-\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^2 b \left (a^2-b^2\right )^2 d}+\frac {\left (3 A b^4-7 a^3 b B+a b^3 B-a^2 b^2 (5 A-3 C)+a^4 (8 A+3 C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (3 A b^6-3 a^5 b B-10 a^3 b^3 B+a b^5 B-3 a^2 b^4 (2 A-C)-a^6 C+5 a^4 b^2 (3 A+2 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^3 (a-b)^2 b (a+b)^3 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (3 A b^4+5 a^3 b B+a b^3 B-a^4 C-a^2 b^2 (9 A+5 C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \] Output:

-1/4*(3*A*b^4+5*B*a^3*b+B*a*b^3-a^4*C-a^2*b^2*(9*A+5*C))*EllipticE(sin(1/2 
*d*x+1/2*c),2^(1/2))/a^2/b/(a^2-b^2)^2/d+1/4*(3*A*b^4-7*B*a^3*b+B*a*b^3-a^ 
2*b^2*(5*A-3*C)+a^4*(8*A+3*C))*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/ 
(a^2-b^2)^2/d-1/4*(3*A*b^6-3*a^5*b*B-10*a^3*b^3*B+a*b^5*B-3*a^2*b^4*(2*A-C 
)-a^6*C+5*a^4*b^2*(3*A+2*C))*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/ 
2))/a^3/(a-b)^2/b/(a+b)^3/d+1/2*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(1/2)*sin(d 
*x+c)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+1/4*(3*A*b^4+5*B*a^3*b+B*a*b^3-a^4* 
C-a^2*b^2*(9*A+5*C))*cos(d*x+c)^(1/2)*sin(d*x+c)/a/b/(a^2-b^2)^2/d/(b+a*co 
s(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 4.45 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.00 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 \sqrt {\cos (c+d x)} \left (-b \left (A b^4+3 a^3 b B+3 a b^3 B+a^4 C-7 a^2 b^2 (A+C)\right )+a \left (-3 A b^4-5 a^3 b B-a b^3 B+a^4 C+a^2 b^2 (9 A+5 C)\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (-A b^4+a^3 b B+5 a b^3 B+3 a^4 C-a^2 b^2 (5 A+9 C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 b \left (-3 a b B+a^2 (2 A+C)+b^2 (A+2 C)\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (-3 A b^4-5 a^3 b B-a b^3 B+a^4 C+a^2 b^2 (9 A+5 C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a b d} \] Input:

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

Output:

((-2*Sqrt[Cos[c + d*x]]*(-(b*(A*b^4 + 3*a^3*b*B + 3*a*b^3*B + a^4*C - 7*a^ 
2*b^2*(A + C))) + a*(-3*A*b^4 - 5*a^3*b*B - a*b^3*B + a^4*C + a^2*b^2*(9*A 
 + 5*C))*Cos[c + d*x])*Sin[c + d*x])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2 
) + (((-(A*b^4) + a^3*b*B + 5*a*b^3*B + 3*a^4*C - a^2*b^2*(5*A + 9*C))*Ell 
ipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) + (8*b*(-3*a*b*B + a^2*(2* 
A + C) + b^2*(A + 2*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[ 
(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + ((-3*A*b^4 - 5*a^3*b*B - a*b^3* 
B + a^4*C + a^2*b^2*(9*A + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x 
]]], -1] + 2*b*(a + b)*EllipticF[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^2*b*Sqrt[Sin[c + d*x]^2]))/((a - b)^2*(a + b)^2))/(8*a*b*d)
 

Rubi [A] (verified)

Time = 2.75 (sec) , antiderivative size = 420, normalized size of antiderivative = 0.99, 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, 3526, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}-\frac {\int -\frac {C a^4+3 b B a^3-7 b^2 (A+C) a^2+3 b^3 B a-4 b \left (-\left ((2 A+C) a^2\right )+3 b B a-b^2 (A+2 C)\right ) \cos (c+d x) a+A b^4-\left (-C a^4+5 b B a^3-b^2 (9 A+5 C) a^2+b^3 B a+3 A b^4\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {C a^4+3 b B a^3-7 b^2 (A+C) a^2+3 b^3 B a-4 b \left (-\left ((2 A+C) a^2\right )+3 b B a-b^2 (A+2 C)\right ) \cos (c+d x) a+A b^4-\left (-C a^4+5 b B a^3-b^2 (9 A+5 C) a^2+b^3 B a+3 A b^4\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {C a^4+3 b B a^3-7 b^2 (A+C) a^2+3 b^3 B a-4 b \left (-\left ((2 A+C) a^2\right )+3 b B a-b^2 (A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a+A b^4+\left (C a^4-5 b B a^3+b^2 (9 A+5 C) a^2-b^3 B a-3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (C a^4+3 b B a^3-7 b^2 (A+C) a^2+3 b^3 B a+A b^4\right )+b \left ((8 A+3 C) a^4-7 b B a^3-b^2 (5 A-3 C) a^2+b^3 B a+3 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{a d}}{2 b \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\frac {b \left (a^4 (8 A+3 C)-7 a^3 b B-a^2 b^2 (5 A-3 C)+a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {\left (a^6 (-C)-3 a^5 b B+5 a^4 b^2 (3 A+2 C)-10 a^3 b^3 B-3 a^2 b^4 (2 A-C)+a b^5 B+3 A b^6\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{a d}}{2 b \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {b \left (a^4 (8 A+3 C)-7 a^3 b B-a^2 b^2 (5 A-3 C)+a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (a^6 (-C)-3 a^5 b B+5 a^4 b^2 (3 A+2 C)-10 a^3 b^3 B-3 a^2 b^4 (2 A-C)+a b^5 B+3 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{a d}}{2 b \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 (8 A+3 C)-7 a^3 b B-a^2 b^2 (5 A-3 C)+a b^3 B+3 A b^4\right )}{a d}-\frac {\left (a^6 (-C)-3 a^5 b B+5 a^4 b^2 (3 A+2 C)-10 a^3 b^3 B-3 a^2 b^4 (2 A-C)+a b^5 B+3 A b^6\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{a d}}{2 b \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{b d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\frac {\frac {2 b \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^4 (8 A+3 C)-7 a^3 b B-a^2 b^2 (5 A-3 C)+a b^3 B+3 A b^4\right )}{a d}-\frac {2 \left (a^6 (-C)-3 a^5 b B+5 a^4 b^2 (3 A+2 C)-10 a^3 b^3 B-3 a^2 b^4 (2 A-C)+a b^5 B+3 A b^6\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^4 (-C)+5 a^3 b B-a^2 b^2 (9 A+5 C)+a b^3 B+3 A b^4\right )}{a d}}{2 b \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

Input:

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

Output:

((A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*a*(a^2 - b^2) 
*d*(b + a*Cos[c + d*x])^2) + (((-2*(3*A*b^4 + 5*a^3*b*B + a*b^3*B - a^4*C 
- a^2*b^2*(9*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(a*d) + ((2*b*(3*A*b^4 - 
 7*a^3*b*B + a*b^3*B - a^2*b^2*(5*A - 3*C) + a^4*(8*A + 3*C))*EllipticF[(c 
 + d*x)/2, 2])/(a*d) - (2*(3*A*b^6 - 3*a^5*b*B - 10*a^3*b^3*B + a*b^5*B - 
3*a^2*b^4*(2*A - C) - a^6*C + 5*a^4*b^2*(3*A + 2*C))*EllipticPi[(2*a)/(a + 
 b), (c + d*x)/2, 2])/(a*(a + b)*d))/a)/(2*b*(a^2 - b^2)) + ((3*A*b^4 + 5* 
a^3*b*B + a*b^3*B - a^4*C - a^2*b^2*(9*A + 5*C))*Sqrt[Cos[c + d*x]]*Sin[c 
+ d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c + d*x])))/(4*a*(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 3526

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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{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] && GtQ[m, 0] && LtQ[n, -1]
 

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. \(1972\) vs. \(2(414)=828\).

Time = 7.17 (sec) , antiderivative size = 1973, normalized size of antiderivative = 4.66

method	result	size
default	\(\text {Expression too large to display}\)	\(1973\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,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*A/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 
2/a^2*(3*A*b-B*a)/(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)*Elli 
pticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2/a^3*(3*A*b^2-2*B*a*b+C*a^2) 
*(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+s 
in(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)*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) 
*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*c 
os(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*b*(A*b^2-B*...
 

Fricas [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^3,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+b*sec(d*x+ 
c))**3,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-1)]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^3,x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^3,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b/cos 
(c + d*x))^3),x)
 

Output:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^(1/2)*(a + b/cos 
(c + d*x))^3), x)
 

Reduce [F]

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

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^3,x)
 

Output:

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