\(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [1365]

Optimal result

Integrand size = 45, antiderivative size = 249 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 (2 A b-a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (2 A b^2-a b B-a^2 (A-C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \] Output:

-2*(2*A*b-B*a)*((b+a*cos(d*x+c))/(a+b))^(1/2)*InverseJacobiAM(1/2*d*x+1/2* 
c,2^(1/2)*(a/(a+b))^(1/2))/a^2/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-2 
*(2*A*b^2-B*a*b-a^2*(A-C))*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2 
^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a^2/(a^2-b^2)/d/((b+a*cos(d 
*x+c))/(a+b))^(1/2)+2*(A*b^2-a*(B*b-C*a))*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x 
+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 18.94 (sec) , antiderivative size = 408, normalized size of antiderivative = 1.64 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {4 \sqrt {\cos (c+d x)} (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (a \left (A b^2+a (-b B+a C)\right ) \sin (c+d x)-\frac {\left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i (a+b) \left (-2 A b^2+a b B+a^2 (A-C)\right ) E\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+i a (a+b) (-2 A b+a (A+B-C)) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (2 A b^2-a b B+a^2 (-A+C)\right ) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}\right )}{a^2 \left (a^2-b^2\right ) d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^{3/2}} \] Input:

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

Output:

(4*Sqrt[Cos[c + d*x]]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + 
 d*x]^2)*(a*(A*b^2 + a*(-(b*B) + a*C))*Sin[c + d*x] - ((Cos[(c + d*x)/2]^2 
*Sec[c + d*x])^(3/2)*((-I)*(a + b)*(-2*A*b^2 + a*b*B + a^2*(A - C))*Ellipt 
icE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt 
[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(a + b)*(-2*A*b 
+ a*(A + B - C))*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]* 
Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] 
 + (2*A*b^2 - a*b*B + a^2*(-A + C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2] 
^2)^(3/2)*Tan[(c + d*x)/2]))/Sec[c + d*x]^(3/2)))/(a^2*(a^2 - b^2)*d*(A + 
2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 2.11 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.14, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4753, 3042, 4588, 27, 3042, 4523, 3042, 4343, 3042, 3134, 3042, 3132, 4345, 3042, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4588

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4523

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4343

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 4345

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((2*(a^2 - b^2)*(2*A*b - a*B)*Sqr 
t[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt 
[Sec[c + d*x]])/(a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(2*A*b^2 - a*b*B - a^2 
*(A - C))*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/ 
(a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]))/(a*(a^2 - b^2 
))) + (2*(A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*(a^2 
- b^2)*d*Sqrt[a + b*Sec[c + d*x]]))
 

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 3132

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

rule 3134

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

rule 3140

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

rule 3142

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

rule 4343

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] 
*(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S 
qrt[b + a*Sin[e + f*x]])   Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a 
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4345

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/S 
qrt[a + b*Csc[e + f*x]])   Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[ 
{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4523

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d 
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Simp[A/a   I 
nt[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Simp[(A*b - a*B) 
/(a*d)   Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ 
[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
 

rule 4588

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc 
[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Sim 
p[1/(a*(m + 1)*(a^2 - b^2))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f 
*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*(m + n + 
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m 
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x 
] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])
 

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1080\) vs. \(2(242)=484\).

Time = 11.48 (sec) , antiderivative size = 1081, normalized size of antiderivative = 4.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1081\)

Input:

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

Output:

-2/d*((-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*( 
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*EllipticE(((a-b)/(a+b))^(1/2)*(c 
sc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2 
)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/ 
2)*b^2*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b) 
)^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b 
)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b*EllipticE(((a-b)/(a+b))^(1/2) 
*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+ 
1)*C*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1 
/2)*a^2*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b 
))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b 
)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*EllipticF(((a-b)/(a+b))^(1/2) 
*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c 
)+2)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^ 
(1/2)*a*b*EllipticF(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a 
-b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(cos(d*x+c)+1))^(1/2)*(1/( 
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*EllipticF(((a-b)/(a+b))^(1 
/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x 
+c)-1)*C*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) 
)^(1/2)*a^2*EllipticF(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 716, normalized size of antiderivative = 2.88 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

2/3*(3*(C*a^4 - B*a^3*b + A*a^2*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c 
))*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(1/2)*(-3*I*B*a^3*b + I*(5*A + C) 
*a^2*b^2 + 2*I*B*a*b^3 - 4*I*A*b^4 + (-3*I*B*a^4 + I*(5*A + C)*a^3*b + 2*I 
*B*a^2*b^2 - 4*I*A*a*b^3)*cos(d*x + c))*sqrt(a)*weierstrassPInverse(-4/3*( 
3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3* 
I*a*sin(d*x + c) + 2*b)/a) + sqrt(1/2)*(3*I*B*a^3*b - I*(5*A + C)*a^2*b^2 
- 2*I*B*a*b^3 + 4*I*A*b^4 + (3*I*B*a^4 - I*(5*A + C)*a^3*b - 2*I*B*a^2*b^2 
 + 4*I*A*a*b^3)*cos(d*x + c))*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4* 
b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d* 
x + c) + 2*b)/a) + 3*sqrt(1/2)*(I*(A - C)*a^3*b + I*B*a^2*b^2 - 2*I*A*a*b^ 
3 + (I*(A - C)*a^4 + I*B*a^3*b - 2*I*A*a^2*b^2)*cos(d*x + c))*sqrt(a)*weie 
rstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstr 
assPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a 
*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) + 3*sqrt(1/2)*(-I*(A - C)*a^ 
3*b - I*B*a^2*b^2 + 2*I*A*a*b^3 + (-I*(A - C)*a^4 - I*B*a^3*b + 2*I*A*a^2* 
b^2)*cos(d*x + c))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27* 
(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27* 
(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a 
)))/((a^6 - a^4*b^2)*d*cos(d*x + c) + (a^5*b - a^3*b^3)*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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