\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\) [432]

Optimal result

Integrand size = 35, antiderivative size = 545 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (4 A b^3+3 a^3 B-7 a b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^2 (a+b)^{3/2} d}+\frac {2 \left (a A b^2-3 A b^3-3 a^3 B-a^2 b B+6 a b^2 B\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a (a-b) b^2 (a+b)^{3/2} d}-\frac {2 \sqrt {a+b} B \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b^3 d}+\frac {2 a (A b-a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 a \left (4 A b^3+3 a^3 B-7 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \] Output:

2/3*(4*A*b^3+3*B*a^3-7*B*a*b^2)*cot(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/(a-b)/b^2/(a+b)^(3/2)/d+2/3*(A* 
a*b^2-3*A*b^3-3*B*a^3-B*a^2*b+6*B*a*b^2)*cot(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)/a/(a-b)/b^2/(a+b)^(3/2)/ 
d-2*(a+b)^(1/2)*B*cot(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+2/3*a*(A*b-B*a)*cos(d*x+c)^(1/2) 
*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-2/3*a*(4*A*b^3+3*B*a^3-7* 
B*a*b^2)*sin(d*x+c)/b^2/(a^2-b^2)^2/d/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(1 
/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.48 (sec) , antiderivative size = 1342, normalized size of antiderivative = 2.46 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((2*(-(a*A*b*Sin[c + d*x]) + 
a^2*B*Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(a + b*Cos[c + d*x])^2) + (2*(4*A*b 
^3*Sin[c + d*x] + 3*a^3*B*Sin[c + d*x] - 7*a*b^2*B*Sin[c + d*x]))/(3*b*(-a 
^2 + b^2)^2*(a + b*Cos[c + d*x]))))/d - ((-4*a*(-(a^2*A*b) + A*b^3 + a^3*B 
 - a*b^2*B)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Co 
s[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x 
)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c 
 + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*S 
qrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(4*a*A*b^2 - a^2*b*B - 3 
*b^3*B)*((Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[ 
c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/ 
2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + 
 d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqr 
t[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2 
]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[( 
(a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b) 
, ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a 
)/(-a + b)]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d 
*x]])) + 2*(4*A*b^3 + 3*a^3*B - 7*a*b^2*B)*((I*Cos[(c + d*x)/2]*Sqrt[a + b 
*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]]...
 

Rubi [A] (verified)

Time = 2.48 (sec) , antiderivative size = 569, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3468, 27, 3042, 3530, 3042, 3288, 3472, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3530

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

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

\(\Big \downarrow \) 3472

\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (3 B a^3-7 b^2 B a+4 A b^3\right )+b \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right ) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {6 B \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (3 B a^3-7 b^2 B a+4 A b^3\right )+b \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {6 B \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3477

\(\displaystyle \frac {\frac {\frac {a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) \left (3 a^3 B+a^2 b B-a b^2 (A+6 B)+3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {6 B \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (3 a^3 B+a^2 b B-a b^2 (A+6 B)+3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {6 B \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {\frac {\frac {a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (3 a^3 B+a^2 b B-a b^2 (A+6 B)+3 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {6 B \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}}{3 b \left (a^2-b^2\right )}+\frac {2 a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {2 a (A b-a B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {\frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} \left (3 a^3 B+a^2 b B-a b^2 (A+6 B)+3 A b^3\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {2 a \left (3 a^3 B-7 a b^2 B+4 A b^3\right ) \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}}{b}-\frac {6 B \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b^2 d}}{3 b \left (a^2-b^2\right )}\)

Input:

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

Output:

(2*a*(A*b - a*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + 
b*Cos[c + d*x])^(3/2)) + ((-6*Sqrt[a + b]*(a^2 - b^2)*B*Cot[c + d*x]*Ellip 
ticPi[(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^2*d) + (((2*(a - b)*Sqrt[a + b]*(4*A*b^3 
 + 3*a^3*B - 7*a*b^2*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d 
*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - S 
ec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d) - (2*(a 
 - b)*Sqrt[a + b]*(3*A*b^3 + 3*a^3*B + a^2*b*B - a*b^2*(A + 6*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)])/(a*d))/(a^2 - b^2) - (2*a*(4*A*b^3 + 3*a^3*B 
- 7*a*b^2*B)*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Co 
s[c + d*x]]))/b)/(3*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 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 3468

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((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 - 2)*(c + d*Sin[e + f*x])^(n 
+ 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a 
*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) 
 - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - 
a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, 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] && GtQ[m, 1] && LtQ[n, -1]
 

rule 3472

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

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 3530

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2847\) vs. \(2(496)=992\).

Time = 27.51 (sec) , antiderivative size = 2848, normalized size of antiderivative = 5.23

method	result	size
default	\(\text {Expression too large to display}\)	\(2848\)
parts	\(\text {Expression too large to display}\)	\(2901\)

Input:

int(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

-2/3/d*((6*cos(d*x+c)^2+12*cos(d*x+c)+6)*B*((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^5*EllipticPi(cot(d*x+c) 
-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+B*((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^4*b*EllipticPi(cot(d*x+c)-c 
sc(d*x+c),-1,(-(a-b)/(a+b))^(1/2))*(6*cos(d*x+c)^3+12*cos(d*x+c)^2+6*cos(d 
*x+c))-4*A*b^5*cos(d*x+c)^2*sin(d*x+c)+3*B*a^5*cos(d*x+c)*sin(d*x+c)+(-3*c 
os(d*x+c)^2-6*cos(d*x+c)-3)*B*((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^5*EllipticE(cot(d*x+c)-csc(d*x+c),(- 
(a-b)/(a+b))^(1/2))+sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)-1)*A*a^3*b^2+sin(d* 
x+c)*cos(d*x+c)*(-3+5*cos(d*x+c))*A*a*b^4+sin(d*x+c)*cos(d*x+c)*(4*cos(d*x 
+c)-2)*B*a^4*b+sin(d*x+c)*cos(d*x+c)*(-3*cos(d*x+c)-7)*B*a^3*b^2+sin(d*x+c 
)*cos(d*x+c)*(-8*cos(d*x+c)+6)*B*a^2*b^3+B*((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^5*EllipticF(cot(d*x+c)- 
csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-3*cos(d*x+c)^3-6*cos(d*x+c)^2-3*cos(d*x 
+c))+B*((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^5*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,(-(a-b)/(a+b))^(1/2)) 
*(6*cos(d*x+c)^3+12*cos(d*x+c)^2+6*cos(d*x+c))+A*((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^5*EllipticE(cot(d 
*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(-4*cos(d*x+c)^3-8*cos(d*x+c)^2-4*c 
os(d*x+c))+A*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*(cos(d*x+c)/...
 

Fricas [F(-1)]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algo 
rithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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