\(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx\) [590]

Optimal result

Integrand size = 33, antiderivative size = 338 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=-\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{4 a b \left (a^2-b^2\right )^2 d}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (3 a^4 A b+10 a^2 A b^3-A b^5+a^5 B-10 a^3 b^2 B-3 a b^4 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{4 a^2 (a-b)^2 b (a+b)^3 d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (b+a \cos (c+d x))^2}+\frac {\left (5 a^2 A b+A b^3-a^3 B-5 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{4 b \left (a^2-b^2\right )^2 d (b+a \cos (c+d x))} \] Output:

-1/4*(5*A*a^2*b+A*b^3-B*a^3-5*B*a*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2 
))/a/b/(a^2-b^2)^2/d-1/4*(7*A*a^2*b-A*b^3-3*B*a^3-3*B*a*b^2)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))/a^2/(a^2-b^2)^2/d+1/4*(3*A*a^4*b+10*A*a^2*b^3-A* 
b^5+B*a^5-10*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^2/(a-b)^2/b/(a+b)^3/d-1/2*(A*b-B*a)*cos(d*x+c)^(1/2)*sin(d*x+c) 
/(a^2-b^2)/d/(b+a*cos(d*x+c))^2+1/4*(5*A*a^2*b+A*b^3-B*a^3-5*B*a*b^2)*cos( 
d*x+c)^(1/2)*sin(d*x+c)/b/(a^2-b^2)^2/d/(b+a*cos(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 3.24 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3} \, dx=\frac {\frac {2 \sqrt {\cos (c+d x)} \left (b \left (3 a^2 A b+3 A b^3+a^3 B-7 a b^2 B\right )-a \left (-5 a^2 A b-A b^3+a^3 B+5 a b^2 B\right ) \cos (c+d x)\right ) \sin (c+d x)}{\left (a^2-b^2\right )^2 (b+a \cos (c+d x))^2}+\frac {\frac {\left (a^2 A b+5 A b^3+3 a^3 B-9 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 b \left (-3 a A b+a^2 B+2 b^2 B\right ) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a (a+b)}+\frac {\left (-5 a^2 A b-A b^3+a^3 B+5 a b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b)^2 (a+b)^2}}{8 b d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 2.17 (sec) , antiderivative size = 336, normalized size of antiderivative = 0.99, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3433, 3042, 3478, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3478

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

-1/2*((A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(b + a*C 
os[c + d*x])^2) - (-1/2*((-2*(5*a^2*A*b + A*b^3 - a^3*B - 5*a*b^2*B)*Ellip 
ticE[(c + d*x)/2, 2])/(a*d) + ((-2*b*(7*a^2*A*b - A*b^3 - 3*a^3*B - 3*a*b^ 
2*B)*EllipticF[(c + d*x)/2, 2])/(a*d) + (2*(3*a^4*A*b + 10*a^2*A*b^3 - A*b 
^5 + a^5*B - 10*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)/(b*(a^2 - b^2)) - ((5*a^2*A*b + A*b^3 - a^3*B - 
5*a*b^2*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(b + a*Cos[c 
+ d*x])))/(4*(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 3478

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1872\) vs. \(2(329)=658\).

Time = 6.56 (sec) , antiderivative size = 1873, normalized size of antiderivative = 5.54

method	result	size
default	\(\text {Expression too large to display}\)	\(1873\)

Input:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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