3.14.60 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\) [1360]

3.14.60.1 Optimal result

Integrand size = 45, antiderivative size = 260 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\frac {(2 A+C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {(2 b B-a C) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{b d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {C \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{b d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {C \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{b d \sqrt {\cos (c+d x)}} \]

output

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

3.14.60.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 39.13 (sec) , antiderivative size = 60622, normalized size of antiderivative = 233.16 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \]

input

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

output

Result too large to show
 

3.14.60.3 Rubi [A] (verified)

Time = 2.75 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.10, number of steps used = 25, number of rules used = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4753, 3042, 4590, 27, 3042, 4596, 3042, 4346, 3042, 3286, 3042, 3284, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4590

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4596

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4346

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

\(\Big \downarrow \) 4523

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4343

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 4345

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

input

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

output

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((-2*b*(2*A + C)*Sqrt[(b + a*C 
os[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d 
*x]])/(d*Sqrt[a + b*Sec[c + d*x]]) - (2*(2*b*B - a*C)*Sqrt[(b + a*Cos[c + 
d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x] 
])/(d*Sqrt[a + b*Sec[c + d*x]]) + (2*C*EllipticE[(c + d*x)/2, (2*a)/(a + b 
)]*Sqrt[a + b*Sec[c + d*x]])/(d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Se 
c[c + d*x]]))/b + (C*Sqrt[Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d 
*x])/(b*d))
 

3.14.60.3.1 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 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 3286

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + 
 d))*Sin[e + f*x]]), x], 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 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 4346

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_)], x_Symbol] :> Simp[d*Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x 
]]/Sqrt[a + b*Csc[e + f*x]])   Int[1/(Sin[e + f*x]*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 4590

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[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 
)*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) 
   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( 
A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*Csc 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 
 - b^2, 0] && GtQ[n, 0]
 

rule 4596

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)]), x_Symbol] :> Simp[C/d^2   Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a + b*C 
sc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[ 
a + b*Csc[e + f*x]]), x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - 
 b^2, 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 
]
 

3.14.60.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 7.04 (sec) , antiderivative size = 1362, normalized size of antiderivative = 5.24

method	result	size
default	\(\text {Expression too large to display}\)	\(1362\)

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))^(1/2 
),x,method=_RETURNVERBOSE)
 

output

1/d*(2*A*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elli 
pticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b 
-2*B*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elliptic 
F(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b+4*B 
*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(( 
(a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(a+b)/(a-b),I/((a-b)/(a+b))^(1 
/2))*b+2*C*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*El 
lipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2)) 
*a-2*C*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ellipt 
icPi(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(a+b)/(a-b),I/((a-b)/(a+ 
b))^(1/2))*a-C*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2 
)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1 
/2))*a+C*cos(d*x+c)^2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elli 
pticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b 
+C*cos(d*x+c)*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a+2* 
A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a 
-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b-2*B*cos( 
d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a 
+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b+4*B*cos(d*x+c) 
*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(((a-b)/(a+b...
 

3.14.60.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.14.60.6 Sympy [F]

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

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))**(1/2),x)
 

output

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

3.14.60.7 Maxima [F]

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

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) 
)^(1/2),x, algorithm="maxima")
 

output

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

3.14.60.8 Giac [F]

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

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) 
)^(1/2),x, algorithm="giac")
 

output

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

3.14.60.9 Mupad [F(-1)]

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

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))^(1/2)),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))^(1/2)), x)