\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [588]

Optimal result

Integrand size = 33, antiderivative size = 367 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (24 a^4 A b-33 a^2 A b^3+15 A b^5-8 a^5 B+5 a^3 b^2 B-3 a b^4 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^4 \left (a^2-b^2\right )^2 d}+\frac {b \left (35 a^4 A b-38 a^2 A b^3+15 A b^5-15 a^5 B+6 a^3 b^2 B-3 a b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^4 (a-b)^2 (a+b)^3 d}+\frac {b (A b-a B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {b \left (11 a^2 A b-5 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 a^2 \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \] Output:

1/4*(8*A*a^4-29*A*a^2*b^2+15*A*b^4+9*B*a^3*b-3*B*a*b^3)*EllipticE(sin(1/2* 
d*x+1/2*c),2^(1/2))/a^3/(a^2-b^2)^2/d-1/4*(24*A*a^4*b-33*A*a^2*b^3+15*A*b^ 
5-8*B*a^5+5*B*a^3*b^2-3*B*a*b^4)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^ 
4/(a^2-b^2)^2/d+1/4*b*(35*A*a^4*b-38*A*a^2*b^3+15*A*b^5-15*B*a^5+6*B*a^3*b 
^2-3*B*a*b^4)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^4/(a-b)^2 
/(a+b)^3/d+1/2*b*(A*b-B*a)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/(a^2-b^2)/d/(b+a* 
cos(d*x+c))^2+1/4*b*(11*A*a^2*b-5*A*b^3-7*B*a^3+B*a*b^2)*cos(d*x+c)^(1/2)* 
sin(d*x+c)/a^2/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 3.61 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {-\frac {2 b \sqrt {\cos (c+d x)} \left (b \left (-11 a^2 A b+5 A b^3+7 a^3 B-a b^2 B\right )+a \left (-13 a^2 A b+7 A b^3+9 a^3 B-3 a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (8 a^4 A-7 a^2 A b^2+5 A b^4-5 a^3 b B-a b^3 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 \left (-4 a^2 A b+A b^3+2 a^3 B+a b^2 B\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {\left (8 a^4 A-29 a^2 A b^2+15 A b^4+9 a^3 b B-3 a b^3 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 a^2 d} \] Input:

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

Output:

((-2*b*Sqrt[Cos[c + d*x]]*(b*(-11*a^2*A*b + 5*A*b^3 + 7*a^3*B - a*b^2*B) + 
 a*(-13*a^2*A*b + 7*A*b^3 + 9*a^3*B - 3*a*b^2*B)*Cos[c + d*x])*Sin[c + d*x 
])/((a^2 - b^2)^2*(b + a*Cos[c + d*x])^2) + (((8*a^4*A - 7*a^2*A*b^2 + 5*A 
*b^4 - 5*a^3*b*B - a*b^3*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a 
+ b) + (8*(-4*a^2*A*b + A*b^3 + 2*a^3*B + a*b^2*B)*((a + b)*EllipticF[(c + 
 d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]))/(a + b) + ((8* 
a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3*a*b^3*B)*(-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^2*d)
 

Rubi [A] (verified)

Time = 2.33 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.01, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3433, 3042, 3468, 27, 3042, 3526, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

\(\displaystyle \frac {\frac {\int \frac {\left (8 A a^4+9 b B a^3-29 A b^2 a^2-3 b^3 B a+15 A b^4\right ) \cos ^2(c+d x)-4 a \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \cos (c+d x)+b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (8 A a^4+9 b B a^3-29 A b^2 a^2-3 b^3 B a+15 A b^4\right ) \cos ^2(c+d x)-4 a \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \cos (c+d x)+b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (8 A a^4+9 b B a^3-29 A b^2 a^2-3 b^3 B a+15 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-4 a \left (-2 B a^3+4 A b a^2-b^2 B a-A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3538

\(\displaystyle \frac {\frac {\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) \int \sqrt {\cos (c+d x)}dx}{a}-\frac {\int -\frac {a b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )-\left (-8 B a^5+24 A b a^4+5 b^2 B a^3-33 A b^3 a^2-3 b^4 B a+15 A b^5\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) \int \sqrt {\cos (c+d x)}dx}{a}+\frac {\int \frac {a b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )-\left (-8 B a^5+24 A b a^4+5 b^2 B a^3-33 A b^3 a^2-3 b^4 B a+15 A b^5\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {\int \frac {a b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )+\left (8 B a^5-24 A b a^4-5 b^2 B a^3+33 A b^3 a^2+3 b^4 B a-15 A b^5\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {\int \frac {a b \left (-7 B a^3+11 A b a^2+b^2 B a-5 A b^3\right )+\left (8 B a^5-24 A b a^4-5 b^2 B a^3+33 A b^3 a^2+3 b^4 B a-15 A b^5\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {2 \left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\frac {b \left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}-\frac {\left (-8 a^5 B+24 a^4 A b+5 a^3 b^2 B-33 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}}{a}+\frac {2 \left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {b \left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {\left (-8 a^5 B+24 a^4 A b+5 a^3 b^2 B-33 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{a}+\frac {2 \left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {\frac {b \left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {2 \left (-8 a^5 B+24 a^4 A b+5 a^3 b^2 B-33 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}}{a}+\frac {2 \left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}}{4 a \left (a^2-b^2\right )}+\frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {b (A b-a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac {\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B-5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\frac {2 \left (8 a^4 A+9 a^3 b B-29 a^2 A b^2-3 a b^3 B+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {\frac {2 b \left (-15 a^5 B+35 a^4 A b+6 a^3 b^2 B-38 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a d (a+b)}-\frac {2 \left (-8 a^5 B+24 a^4 A b+5 a^3 b^2 B-33 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}}{a}}{2 a \left (a^2-b^2\right )}}{4 a \left (a^2-b^2\right )}\)

Input:

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

Output:

(b*(A*b - a*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*a*(a^2 - b^2)*d*(b + a* 
Cos[c + d*x])^2) + (((2*(8*a^4*A - 29*a^2*A*b^2 + 15*A*b^4 + 9*a^3*b*B - 3 
*a*b^3*B)*EllipticE[(c + d*x)/2, 2])/(a*d) + ((-2*(24*a^4*A*b - 33*a^2*A*b 
^3 + 15*A*b^5 - 8*a^5*B + 5*a^3*b^2*B - 3*a*b^4*B)*EllipticF[(c + d*x)/2, 
2])/(a*d) + (2*b*(35*a^4*A*b - 38*a^2*A*b^3 + 15*A*b^5 - 15*a^5*B + 6*a^3* 
b^2*B - 3*a*b^4*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a*(a + b)*d 
))/a)/(2*a*(a^2 - b^2)) + (b*(11*a^2*A*b - 5*A*b^3 - 7*a^3*B + a*b^2*B)*Sq 
rt[Cos[c + d*x]]*Sin[c + d*x])/(a*(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 3433

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 3468

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

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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1999\) vs. \(2(358)=716\).

Time = 8.36 (sec) , antiderivative size = 2000, normalized size of antiderivative = 5.45

method	result	size
default	\(\text {Expression too large to display}\)	\(2000\)

Input:

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

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^4/(-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)^2-1)^( 
1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(3*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2 
))*b+A*a*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*a*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2)))+2*b^3*(A*b-B*a)/a^4*(1/2*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)^2+3/4*a^2*(a^2-3*b^2)/b^2/(a^2-b^2)^2*cos(1/2*d*x+1/2*c)*(-2*s 
in(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)-3/8/(a+b)/(a^2-b^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)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-1/4/(a+b)/(a^2-b^2)/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))*a+7/8/( 
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))+3/8*a^3/b^2/(a^2-b^2)^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))-9/8*a/(a^2-b^2)^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/...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral((A + B*sec(c + d*x))*sqrt(cos(c + d*x))/(a + b*sec(c + d*x))**3, 
x)
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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