\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [621]

Optimal result

Integrand size = 35, antiderivative size = 345 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (a^2 A-2 A b^2+a b B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^3 \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 (2 A b+a (A-B)) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

2*(A*a^2-2*A*b^2+B*a*b)*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^3/(a+b)^(1/2)/d/sec(d* 
x+c)^(1/2)-2*(2*A*b+a*(A-B))*cos(d*x+c)^(1/2)*csc(d*x+c)*EllipticF((a+b*co 
s(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-s 
ec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/(a+b)^(1/2)/d/s 
ec(d*x+c)^(1/2)+2*b*(A*b-B*a)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/(a^2-b^2)/d/(a 
+b*cos(d*x+c))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2988\) vs. \(2(345)=690\).

Time = 22.96 (sec) , antiderivative size = 2988, normalized size of antiderivative = 8.66 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(a^2*A - 2*A*b^2 + a*b*B) 
*Sin[c + d*x])/(a^2*(a^2 - b^2)) - (2*(-(A*b^2*Sin[c + d*x]) + a*b*B*Sin[c 
 + d*x]))/(a*(a^2 - b^2)*(a + b*Cos[c + d*x]))))/d + (2*(-((a*A)/((a^2 - b 
^2)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]])) + (2*A*b^2)/(a*(a^2 - b^ 
2)*Sqrt[a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (b*B)/((a^2 - b^2)*Sqrt[ 
a + b*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*A*b*Sqrt[Sec[c + d*x]])/((a^2 
 - b^2)*Sqrt[a + b*Cos[c + d*x]]) + (2*A*b^3*Sqrt[Sec[c + d*x]])/(a^2*(a^2 
 - b^2)*Sqrt[a + b*Cos[c + d*x]]) + (a*B*Sqrt[Sec[c + d*x]])/((a^2 - b^2)* 
Sqrt[a + b*Cos[c + d*x]]) - (b^2*B*Sqrt[Sec[c + d*x]])/(a*(a^2 - b^2)*Sqrt 
[a + b*Cos[c + d*x]]) - (A*b*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/((a^2 - 
b^2)*Sqrt[a + b*Cos[c + d*x]]) + (2*A*b^3*Cos[2*(c + d*x)]*Sqrt[Sec[c + d* 
x]])/(a^2*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]) - (b^2*B*Cos[2*(c + d*x)]* 
Sqrt[Sec[c + d*x]])/(a*(a^2 - b^2)*Sqrt[a + b*Cos[c + d*x]]))*Sqrt[Cos[(c 
+ d*x)/2]^2*Sec[c + d*x]]*(-2*(a + b)*(a^2*A - 2*A*b^2 + a*b*B)*Sqrt[Cos[c 
 + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c 
+ d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] + 2*a*(a + 
 b)*(-2*A*b + a*(A + B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(a + b 
*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x 
)/2]], (-a + b)/(a + b)] - (a^2*A - 2*A*b^2 + a*b*B)*Cos[c + d*x]*(a + b*C 
os[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(a^2*(a^2 - b^2)*d*S...
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3440, 3042, 3479, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3440

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 A+a b B-2 A b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) (a (A-B)+2 A b) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*(a - b)*Sqrt[a + b]*(a^2*A - 2* 
A*b^2 + a*b*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqr 
t[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^2*d) - (2*(a - b)*S 
qrt[a + b]*(2*A*b + a*(A - B))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Co 
s[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*(a^2 - b^2)) + (2*b*(A*b - a*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[Co 
s[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 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 3440

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p   Int[(a + b*Sin[e + f*x])^m*((c + 
d*Sin[e + f*x])^n/(g*Sin[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g 
, m, n, p}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && I 
ntegerQ[n])
 

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 3479

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[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin 
[e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 
1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e 
 + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 
2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) 
*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n 
}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat 
ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(I 
ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0]) 
))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1299\) vs. \(2(311)=622\).

Time = 12.94 (sec) , antiderivative size = 1300, normalized size of antiderivative = 3.77

method	result	size
default	\(\text {Expression too large to display}\)	\(1300\)
parts	\(\text {Expression too large to display}\)	\(1379\)

Input:

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

Output:

-2/d/(a+b)/(a-b)/a^2*sec(d*x+c)^(3/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(( 
(1-cos(d*x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*a^3*A+(2*(1-cos(d*x+c 
))^3*csc(d*x+c)^3-2*csc(d*x+c)+2*cot(d*x+c))*b^3*A+(-(1-cos(d*x+c))^3*csc( 
d*x+c)^3+csc(d*x+c)-cot(d*x+c))*b*a^2*A-2*A*a*b^2*(1-cos(d*x+c))^3*csc(d*x 
+c)^3+((1-cos(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*b*a^2*B+(-(1-c 
os(d*x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*b^2*a*B-2*A*(cos(d*x+c)/( 
1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Ellip 
ticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3+2*A*(cos(d*x+c)/(1+c 
os(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elliptic 
F(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b+4*A*(cos(d*x+c)/(1+co 
s(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF 
(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2+2*A*(cos(d*x+c)/(1+cos 
(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE( 
-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^3+2*A*(cos(d*x+c)/(1+cos(d* 
x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-cs 
c(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a^2*b-4*A*(cos(d*x+c)/(1+cos(d*x 
+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc 
(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b^2-4*A*(cos(d*x+c)/(1+cos(d*x+ 
c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc( 
d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*b^3-2*B*(cos(d*x+c)/(1+cos(d*x+...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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