\(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [1357]

Optimal result

Integrand size = 45, antiderivative size = 291 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \left (8 A b^3-5 a^3 B-10 a b^2 B+a^2 b (7 A+15 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d} \] Output:

-2/15*(8*A*b^3-5*B*a^3-10*B*a*b^2+a^2*b*(7*A+15*C))*((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^3/d/cos( 
d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/15*(8*A*b^2-10*B*a*b+3*a^2*(3*A+5*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^3/d/((b+a*cos(d*x+c))/(a+b))^(1/2)-2/15*(4*A*b-5*B 
*a)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/a^2/d+2/5*A*cos(d*x 
+c)^(3/2)*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/a/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.52 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.30 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 a (b+a \cos (c+d x)) (-4 A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)+\frac {2 \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (i (a+b) \left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right ) E\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-i a \left (8 A b^2+2 a b (A-5 B)+a^2 (9 A+5 (B+3 C))\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right ) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \] Input:

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqr 
t[a + b*Sec[c + d*x]],x]
 

Output:

(2*a*(b + a*Cos[c + d*x])*(-4*A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d* 
x] + (2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(I*(a + b)*(8*A*b^2 - 10*a 
*b*B + 3*a^2*(3*A + 5*C))*EllipticE[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*(8*A*b^2 + 2*a*b*(A - 5*B) + a^2*(9*A + 5*(B + 3*C)))*Elli 
pticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sq 
rt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + (8*A*b^2 - 10*a*b* 
B + 3*a^2*(3*A + 5*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))/(15*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + 
b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 2.52 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.11, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4753, 3042, 4592, 27, 3042, 4592, 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.

\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^{5/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\sqrt {a+b \sec (c+d x)}}dx\)

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 4523

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 4343

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3134

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3132

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 4345

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3142

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3140

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\)

Input:

Int[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + 
b*Sec[c + d*x]],x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sqrt[a + b*Sec[c + d*x]]*Sin[c 
 + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (-1/3*((-2*(8*A*b^3 - 5*a^3*B - 10*a 
*b^2*B + a^2*b*(7*A + 15*C))*Sqrt[(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*(8*A*b^2 - 10*a*b*B + 3*a^2*(3*A + 5*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 + (2*(4*A*b - 5*a*B)*Sqrt[a + b*Sec[c + d*x] 
]*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/(5*a))
 

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 4592

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d 
*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m 
*(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* 
Csc[e + f*x] + A*b*(m + n + 2)*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] && LeQ[n, -1]
 

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. \(1919\) vs. \(2(272)=544\).

Time = 25.09 (sec) , antiderivative size = 1920, normalized size of antiderivative = 6.60

method	result	size
default	\(\text {Expression too large to display}\)	\(1920\)

Input:

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 
),x,method=_RETURNVERBOSE)
 

Output:

2/15/d*((9*cos(d*x+c)^2+18*cos(d*x+c)+9)*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^3*EllipticE(((a-b)/(a+b))^(1/2 
)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-9*cos(d*x+c)^2-18*cos(d* 
x+c)-9)*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*b*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+ 
b)/(a-b))^(1/2))+(8*cos(d*x+c)^2+16*cos(d*x+c)+8)*A*(1/(a+b)*(b+a*cos(d*x+ 
c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(((a-b)/ 
(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-8*cos(d*x+c)^ 
2-16*cos(d*x+c)-8)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(c 
os(d*x+c)+1))^(1/2)*b^3*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+ 
c)),(-(a+b)/(a-b))^(1/2))+(-10*cos(d*x+c)^2-20*cos(d*x+c)-10)*B*(1/(a+b)*( 
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^2*b*Ellip 
ticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(10 
*cos(d*x+c)^2+20*cos(d*x+c)+10)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) 
)^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE(((a-b)/(a+b))^(1/2)*(csc( 
d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(15*cos(d*x+c)^2+30*cos(d*x+c)+15 
)*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^3*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b) 
)^(1/2))+(-15*cos(d*x+c)^2-30*cos(d*x+c)-15)*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*b*EllipticE(((a-b)/(a...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) 
)^(1/2),x, algorithm="fricas")
 

Output:

2/45*(3*(3*A*a^3*cos(d*x + c) + 5*B*a^3 - 4*A*a^2*b)*sqrt((a*cos(d*x + c) 
+ b)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - sqrt(1/2)*(15*I*B*a^3 
 - 6*I*(2*A + 5*C)*a^2*b + 20*I*B*a*b^2 - 16*I*A*b^3)*sqrt(a)*weierstrassP 
Inverse(-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)*(-15*I*B*a^3 + 6*I*(2 
*A + 5*C)*a^2*b - 20*I*B*a*b^2 + 16*I*A*b^3)*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)*(-3*I*(3*A + 5*C)*a^3 + 10*I 
*B*a^2*b - 8*I*A*a*b^2)*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)) - 3*sqrt(1/2)*(3*I*(3*A + 5*C)*a^3 - 10*I*B*a^2*b + 8*I*A*a*b^2)*s 
qrt(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^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+ 
c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*cos(d*x + c)^(5/2)/sqrt( 
b*sec(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*cos(d*x + c)^(5/2)/sqrt( 
b*sec(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \] Input:

int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co 
s(c + d*x))^(1/2),x)
 

Output:

int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co 
s(c + d*x))^(1/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right ) b +a}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )}{\sec \left (d x +c \right ) b +a}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\sec \left (d x +c \right ) b +a}d x \right ) a \] Input:

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 
),x)
 

Output:

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