\(\int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [762]

Optimal result

Integrand size = 25, antiderivative size = 525 \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {\left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a b^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {(3 a+b) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {3 a \sqrt {a+b} \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d \sqrt {\sec (c+d x)}}-\frac {2 a^2 \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}}+\frac {\left (3 a^2-b^2\right ) \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d} \] Output:

-(3*a^2-b^2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticE((a+b*cos(d*x+c))^(1/2)/ 
(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b) 
)^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b^2/(a+b)^(1/2)/d/sec(d*x+c)^(1/2 
)+(3*a+b)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+ 
b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^( 
1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^2/(a+b)^(1/2)/d/sec(d*x+c)^(1/2)+3*a 
*(a+b)^(1/2)*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticPi((a+b*cos(d*x+c))^(1/2) 
/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+ 
c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/b^3/d/sec(d*x+c)^(1/2)-2*a 
^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+(3*a^2 
-b^2)*(a+b*cos(d*x+c))^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)/b^2/(a^2-b^2)/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 13.44 (sec) , antiderivative size = 1025, normalized size of antiderivative = 1.95 \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[1/((a + b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(5/2)),x]
 

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*a^2*Sin[c + d*x])/(b^2*( 
a^2 - b^2)) - (2*a^3*Sin[c + d*x])/(b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x])) 
))/d - (Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b + a*Tan[(c + d*x)/ 
2]^2 - b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-3*a^3*Tan[(c + d* 
x)/2] - 3*a^2*b*Tan[(c + d*x)/2] + a*b^2*Tan[(c + d*x)/2] + b^3*Tan[(c + d 
*x)/2] + 6*a^2*b*Tan[(c + d*x)/2]^3 - 2*b^3*Tan[(c + d*x)/2]^3 + 3*a^3*Tan 
[(c + d*x)/2]^5 - 3*a^2*b*Tan[(c + d*x)/2]^5 - a*b^2*Tan[(c + d*x)/2]^5 + 
b^3*Tan[(c + d*x)/2]^5 + 6*a^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (- 
a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x) 
/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - 6*a*b^2*EllipticPi[-1, ArcSin[Tan 
[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b 
 + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 6*a^3*EllipticP 
i[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[ 
1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d* 
x)/2]^2)/(a + b)] - 6*a*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + 
 b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b + 
 a*Tan[(c + d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] - (3*a^3 + 3*a^2*b 
- a*b^2 - b^3)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[ 
1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b + a*Tan[(c + 
d*x)/2]^2 - b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a*b*(a + b)*EllipticF[Ar...
 

Rubi [A] (verified)

Time = 2.24 (sec) , antiderivative size = 515, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4710, 3042, 3271, 27, 3042, 3540, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}

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.

\(\displaystyle \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3271

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3540

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3532

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {2 b \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (3 a^2-b^2\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {a \left (3 a^2-b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a (a-b) (3 a+b) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx-\frac {6 a \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {a \left (3 a^2-b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-a (a-b) (3 a+b) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {a \left (3 a^2-b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} (3 a+b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (3 a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {6 a \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} (3 a+b) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}}{b \left (a^2-b^2\right )}-\frac {2 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\right )\)

Input:

Int[1/((a + b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(5/2)),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((-2*a^2*Sqrt[Cos[c + d*x]]*Sin[c + 
d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (((2*(a - b)*Sqrt[a + b 
]*(3*a^2 - b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sq 
rt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d 
*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d) - (2*(a - b)*Sq 
rt[a + b]*(3*a + b)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]] 
/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c 
 + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/d - (6*a*Sqrt[a + 
 b]*(a^2 - b^2)*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c 
 + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 
 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(b*d))/(2 
*b) - ((3*a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[Cos[ 
c + d*x]]))/(b*(a^2 - b^2)))
 

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 3271

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co 
s[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f* 
(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin 
[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^ 
2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*d*(a^2 
+ b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - 
 d^2) - m*(b*c - a*d)^2 + d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, 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[m, 2] && LtQ[n, -1] && (IntegerQ[m] || 
IntegersQ[2*m, 2*n])
 

rule 3288

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c 
*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti 
cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + 
 d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - 
 d^2, 0] && PosQ[(c + d)/b]
 

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

rule 3532

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2   Int[Sqrt[a + b*Sin[e + f*x]] 
/Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2   Int[(A*b^2 - a^2*C + b*(b* 
B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x 
]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] & 
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3540

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f 
*x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d)   Int[(1/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 
 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + 
 f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 4710

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(960\) vs. \(2(472)=944\).

Time = 17.94 (sec) , antiderivative size = 961, normalized size of antiderivative = 1.83

method	result	size
default	\(\text {Expression too large to display}\)	\(961\)

Input:

int(1/(a+cos(d*x+c)*b)^(3/2)/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

1/d*sec(d*x+c)^(1/2)*((6*cos(d*x+c)^2+12*cos(d*x+c)+6)*((a+cos(d*x+c)*b)/( 
cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*EllipticP 
i(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(-6*cos(d*x+c)^2-12*cos(d 
*x+c)-6)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2)*a*b^2*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1 
/2))+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+ 
b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*EllipticE(cot(d*x+c)-csc(d 
*x+c),(-(a-b)/(a+b))^(1/2))+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*((a+cos(d*x+c 
)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*E 
llipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d 
*x+c)+1)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d* 
x+c)+1))^(1/2)*a*b^2*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2)) 
+(cos(d*x+c)^2+2*cos(d*x+c)+1)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/ 
2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3*EllipticE(cot(d*x+c)-csc(d*x+c),( 
-(a-b)/(a+b))^(1/2))+(2+2*cos(d*x+c)^2+4*cos(d*x+c))*(cos(d*x+c)/(cos(d*x+ 
c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*a^2*b*EllipticF 
(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+(2+2*cos(d*x+c)^2+4*cos(d*x+c 
))*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1 
))^(1/2)*a*b^2*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))+3*a^3 
*cos(d*x+c)*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)-2)*a^2*b-a*b^2...
 

Fricas [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

integral(sqrt(b*cos(d*x + c) + a)/((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c 
) + a^2)*sec(d*x + c)^(5/2)), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cos(d*x+c))**(3/2)/sec(d*x+c)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^(5/2)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

integrate(1/((b*cos(d*x + c) + a)^(3/2)*sec(d*x + c)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {1}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int(1/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^(3/2)),x)
 

Output:

int(1/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b \cos (c+d x))^{3/2} \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right ) b +a}}{\cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{3} b^{2}+2 \cos \left (d x +c \right ) \sec \left (d x +c \right )^{3} a b +\sec \left (d x +c \right )^{3} a^{2}}d x \] Input:

int(1/(a+b*cos(d*x+c))^(3/2)/sec(d*x+c)^(5/2),x)
 

Output:

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