3.6.71 \(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [571]

3.6.71.1 Optimal result

Integrand size = 33, antiderivative size = 197 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 d}-\frac {2 \left (3 a A b-3 a^2 B-b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^3 d}+\frac {2 a^2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}+\frac {2 B \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}} \]

output

2/3*B*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)+2*(A*b-B*a)*(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))*cos(d*x+c)^( 
1/2)*sec(d*x+c)^(1/2)/b^2/d-2/3*(3*A*a*b-3*B*a^2-B*b^2)*(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))*cos(d* 
x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/d+2*a^2*(A*b-B*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*b/(a+b),2^(1/2))*c 
os(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^3/(a+b)/d
 

3.6.71.2 Mathematica [A] (verified)

Time = 5.01 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.41 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \csc (c+d x) \left (-3 A b^2+3 a b B+3 A b^2 \sec ^2(c+d x)-3 a b B \sec ^2(c+d x)+b^2 B \sin (c+d x) \tan (c+d x)-3 b (A b-a B) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+b (3 A b-3 a B+b B) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-3 a A b \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+3 a^2 B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{3 b^3 d \sec ^{\frac {3}{2}}(c+d x)} \]

input

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

output

(2*Csc[c + d*x]*(-3*A*b^2 + 3*a*b*B + 3*A*b^2*Sec[c + d*x]^2 - 3*a*b*B*Sec 
[c + d*x]^2 + b^2*B*Sin[c + d*x]*Tan[c + d*x] - 3*b*(A*b - a*B)*EllipticE[ 
ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 
 b*(3*A*b - 3*a*B + b*B)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Se 
c[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 3*a*A*b*EllipticPi[-(a/b), ArcSin[Sqrt 
[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 3*a^2*B*El 
lipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[- 
Tan[c + d*x]^2]))/(3*b^3*d*Sec[c + d*x]^(3/2))
 

3.6.71.3 Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 3439, 3042, 4522, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3439

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {3 a^2 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}+\frac {\frac {6 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\left (-3 a^2 B+3 a A b-b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2}}{3 b}+\frac {2 B \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 4336

\(\displaystyle \frac {\frac {3 a^2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b^2}+\frac {\frac {6 b (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (-3 a^2 B+3 a A b-b^2 B\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{b^2}}{3 b}+\frac {2 B \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

input

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

output

(((6*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c 
 + d*x]])/d - (2*(3*a*A*b - 3*a^2*B - b^2*B)*Sqrt[Cos[c + d*x]]*EllipticF[ 
(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/b^2 + (6*a^2*(A*b - a*B)*Sqrt[Cos[c 
 + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(b^ 
2*(a + b)*d))/(3*b) + (2*B*Sin[c + d*x])/(3*b*d*Sqrt[Sec[c + d*x]])
 

3.6.71.3.1 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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

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 3439

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^(m + n)   Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + 
c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4336

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

rule 4522

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), 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] + Sim 
p[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*(n + 1)*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, m}, x] && NeQ[A*b - a*B, 0] 
 && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

rule 4594

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

3.6.71.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(821\) vs. \(2(257)=514\).

Time = 6.48 (sec) , antiderivative size = 822, normalized size of antiderivative = 4.17

method	result	size
default	\(\text {Expression too large to display}\)	\(822\)

input

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

output

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*B*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2-4*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1 
/2*c)^4*b^3+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2*b-3*A*a^2*b*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/ 
2*d*x+1/2*c),2^(1/2))+3*A*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d* 
x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2))*a*b^2+3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) 
^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3-2*B*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*a*b^2+2*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2* 
b^3-3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli 
pticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^3+3*B*a^3*(sin(1/2*d*x+1/2 
*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c) 
,2^(1/2))-3*B*a^2*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1 
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+B*a*b^2*(sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^( 
1/2))-B*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)* 
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2...
 

3.6.71.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.6.71.6 Sympy [F]

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

input

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

output

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

3.6.71.7 Maxima [F]

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

input

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

output

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

3.6.71.8 Giac [F]

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

input

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

output

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

3.6.71.9 Mupad [F(-1)]

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

input

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

output

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