\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [567]

Optimal result

Integrand size = 33, antiderivative size = 210 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {2 (A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {2 b (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a+b) d}-\frac {2 (A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 13.65 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {\cot (c+d x) \left (-a^2 A \sec ^{\frac {5}{2}}(c+d x)+a^2 A \cos (2 (c+d x)) \sec ^{\frac {5}{2}}(c+d x)-6 a (-A b+a B) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}-2 \left (3 A b^2+a^2 (A-3 B)+3 a b (A-B)\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+6 A b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-6 a b B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{3 a^3 d} \] Input:

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

Output:

-1/3*(Cot[c + d*x]*(-(a^2*A*Sec[c + d*x]^(5/2)) + a^2*A*Cos[2*(c + d*x)]*S 
ec[c + d*x]^(5/2) - 6*a*(-(A*b) + a*B)*EllipticE[ArcSin[Sqrt[Sec[c + d*x]] 
], -1]*Sqrt[-Tan[c + d*x]^2] - 2*(3*A*b^2 + a^2*(A - 3*B) + 3*a*b*(A - B)) 
*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] + 6*A*b^2 
*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] 
- 6*a*b*B*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + 
 d*x]^2]))/(a^3*d)
 

Rubi [A] (verified)

Time = 1.95 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.606, Rules used = {3042, 3439, 3042, 4521, 27, 3042, 4590, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3439

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4590

\(\displaystyle \frac {\frac {2 \int \frac {\left (A a^2-3 b B a+3 A b^2\right ) \sec ^2(c+d x)+a (4 A b-3 a B) \sec (c+d x)+3 b (A b-a B)}{2 \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{a}-\frac {6 (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (A a^2-3 b B a+3 A b^2\right ) \sec ^2(c+d x)+a (4 A b-3 a B) \sec (c+d x)+3 b (A b-a B)}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{a}-\frac {6 (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (A a^2-3 b B a+3 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A b-3 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 b (A b-a B)}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}-\frac {6 (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {\frac {\frac {\int \frac {3 (A b-a B) b^2+a A \sec (c+d x) b^2}{\sqrt {\sec (c+d x)}}dx}{b^2}+3 b (A b-a B) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)}dx}{a}-\frac {6 (A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}}{3 a}+\frac {2 A \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 4336

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

(2*A*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*a*d) + ((((6*b^2*(A*b - a*B)*Sqrt 
[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a*A*b^ 
2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/b^2 
+ (6*b*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/ 
2, 2]*Sqrt[Sec[c + d*x]])/((a + b)*d))/a - (6*(A*b - a*B)*Sqrt[Sec[c + d*x 
]]*Sin[c + d*x])/(a*d))/(3*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 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 4521

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 
2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* 
(m + n))), x] + Simp[d^2/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e 
+ f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + 
 n) - a*B*(n - 1))*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] && GtQ[n, 1] && NeQ[m + 
n, 0] &&  !IGtQ[m, 1]
 

rule 4590

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[(-C)*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1 
)*((d*Csc[e + f*x])^(n - 1)/(b*f*(m + n + 1))), x] + Simp[d/(b*(m + n + 1)) 
   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[a*C*(n - 1) + ( 
A*b*(m + n + 1) + b*C*(m + n))*Csc[e + f*x] + (b*B*(m + n + 1) - a*C*n)*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] && GtQ[n, 0]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(440\) vs. \(2(195)=390\).

Time = 15.18 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.10

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 A \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{a}-\frac {2 \left (A b -B a \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}-\frac {4 b^{2} \left (A b -B a \right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )}{a^{2} \left (-2 a b +2 b^{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(441\)

Input:

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

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a*(-1/6*co 
s(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos 
(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+ 
1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell 
ipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(A*b-B*a)/a^2/sin(1/2*d*x+1/2*c)^2/( 
2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(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))) 
-4*b^2*(A*b-B*a)/a^2/(-2*a*b+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1 
/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1 
/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/sin(1/2*d*x+1/2*c)/ 
(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)