3.7.14 \(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [614]

3.7.14.1 Optimal result
3.7.14.2 Mathematica [A] (warning: unable to verify)
3.7.14.3 Rubi [A] (verified)
3.7.14.4 Maple [B] (verified)
3.7.14.5 Fricas [F]
3.7.14.6 Sympy [F(-1)]
3.7.14.7 Maxima [F]
3.7.14.8 Giac [F]
3.7.14.9 Mupad [F(-1)]

3.7.14.1 Optimal result

Integrand size = 35, antiderivative size = 403 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (9 a^2 A+8 A b^2-10 a b B\right ) \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^4 d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {a+b} \left (8 A b^2+a^2 (9 A-5 B)-2 a b (A+5 B)\right ) \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{15 a^3 d \sqrt {\sec (c+d x)}}-\frac {2 (4 A b-5 a B) \sqrt {a+b \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}+\frac {2 A \sqrt {a+b \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d} \]

output
-2/15*(4*A*b-5*B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a^2 
/d+2/5*A*sec(d*x+c)^(5/2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/a/d+2/15*(a-b) 
*(9*A*a^2+8*A*b^2-10*B*a*b)*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+b)^(1/2)*cos(d*x+c)^(1 
/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a^4/d/se 
c(d*x+c)^(1/2)-2/15*(8*A*b^2+a^2*(9*A-5*B)-2*a*b*(A+5*B))*csc(d*x+c)*Ellip 
ticF((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+b)^(1/2)*cos(d*x+c)^(1/2)*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec 
(d*x+c))/(a-b))^(1/2)/a^3/d/sec(d*x+c)^(1/2)
 
3.7.14.2 Mathematica [A] (warning: unable to verify)

Time = 13.68 (sec) , antiderivative size = 385, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \left (-\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (9 a^2 A+8 A b^2-10 a b B\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-2 a \left (8 A b^2+2 a b (A-5 B)+a^2 (9 A+5 B)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {b+a \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+\left (9 a^2 A+8 A b^2-10 a b B\right ) \cos (c+d x) (a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+(a+b \cos (c+d x)) \sqrt {\sec (c+d x)} \left (\left (9 a^2 A+8 A b^2-10 a b B\right ) \sin (c+d x)+a (-4 A b+5 a B+3 a A \sec (c+d x)) \tan (c+d x)\right )\right )}{15 a^3 d \sqrt {a+b \cos (c+d x)}} \]

input
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2))/Sqrt[a + b*Cos[c + d*x 
]],x]
 
output
(2*(-((Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(9*a^2*A + 8*A*b^2 
 - 10*a*b*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 
 + Sec[c + d*x])^(-1)]*Sqrt[(b + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x 
]))] - 2*a*(8*A*b^2 + 2*a*b*(A - 5*B) + a^2*(9*A + 5*B))*EllipticF[ArcSin[ 
Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(b 
 + a*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + (9*a^2*A + 8*A*b^2 - 10 
*a*b*B)*Cos[c + d*x]*(a + b*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x) 
/2]))/Sqrt[Sec[(c + d*x)/2]^2]) + (a + b*Cos[c + d*x])*Sqrt[Sec[c + d*x]]* 
((9*a^2*A + 8*A*b^2 - 10*a*b*B)*Sin[c + d*x] + a*(-4*A*b + 5*a*B + 3*a*A*S 
ec[c + d*x])*Tan[c + d*x])))/(15*a^3*d*Sqrt[a + b*Cos[c + d*x]])
 
3.7.14.3 Rubi [A] (verified)

Time = 1.67 (sec) , antiderivative size = 396, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3440, 3042, 3479, 27, 3042, 3534, 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.

\(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3440

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3479

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 \int -\frac {-2 A b \cos ^2(c+d x)-3 a A \cos (c+d x)+4 A b-5 a B}{2 \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{5 a}+\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-2 A b \cos ^2(c+d x)-3 a A \cos (c+d x)+4 A b-5 a B}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\int \frac {-2 A b \sin \left (c+d x+\frac {\pi }{2}\right )^2-3 a A \sin \left (c+d x+\frac {\pi }{2}\right )+4 A b-5 a B}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}\right )\)

\(\Big \downarrow \) 3534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {9 A a^2-10 b B a+(2 A b+5 a B) \cos (c+d x) a+8 A b^2}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}}{5 a}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {9 A a^2-10 b B a+(2 A b+5 a B) \cos (c+d x) a+8 A b^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {9 A a^2-10 b B a+(2 A b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\left (a^2 (9 A-5 B)-2 a b (A+5 B)+8 A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (a^2 (9 A-5 B)-2 a b (A+5 B)+8 A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2 (9 A-5 B)-2 a b (A+5 B)+8 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}}{5 a}\right )\)

\(\Big \downarrow \) 3473

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (9 a^2 A-10 a b B+8 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 \sqrt {a+b} \left (a^2 (9 A-5 B)-2 a b (A+5 B)+8 A b^2\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}}{5 a}\right )\)

input
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(7/2))/Sqrt[a + b*Cos[c + d*x]],x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*Sqrt[a + b*Cos[c + d*x]]*Sin[c 
 + d*x])/(5*a*d*Cos[c + d*x]^(5/2)) - (-1/3*((2*(a - b)*Sqrt[a + b]*(9*a^2 
*A + 8*A*b^2 - 10*a*b*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 - 
Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a^2*d) - (2 
*Sqrt[a + b]*(8*A*b^2 + a^2*(9*A - 5*B) - 2*a*b*(A + 5*B))*Cot[c + d*x]*El 
lipticF[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*(4*A*b - 5*a*B)*Sqrt[a + b*Cos[c + d*x 
]]*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)))/(5*a))
 

3.7.14.3.1 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 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]) 
))
 

rule 3534
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(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[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 
NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 
3.7.14.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3319\) vs. \(2(363)=726\).

Time = 25.69 (sec) , antiderivative size = 3320, normalized size of antiderivative = 8.24

method result size
parts \(\text {Expression too large to display}\) \(3320\)
default \(\text {Expression too large to display}\) \(3350\)

input
int((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+cos(d*x+c)*b)^(1/2),x,method=_RET 
URNVERBOSE)
 
output
-2/15*A/d*sec(d*x+c)^(7/2)/(1+cos(d*x+c))/(a+cos(d*x+c)*b)^(1/2)*(-16*Elli 
pticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c 
)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*b^2*cos(d*x+c)^4 
+4*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+co 
s(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a^2*b*cos(d 
*x+c)^4+16*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+ 
c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a*b 
^2*cos(d*x+c)^4-9*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*(c 
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a+b))^(1 
/2)*a^2*b*cos(d*x+c)^3-8*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1 
/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x+c))/(a 
+b))^(1/2)*a*b^2*cos(d*x+c)^3+2*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a 
+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+cos(d*x 
+c))/(a+b))^(1/2)*a^2*b*cos(d*x+c)^3+8*EllipticF(cot(d*x+c)-csc(d*x+c),(-( 
a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+ 
cos(d*x+c))/(a+b))^(1/2)*a*b^2*cos(d*x+c)^3+9*EllipticF(cot(d*x+c)-csc(d*x 
+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c) 
*b)/(1+cos(d*x+c))/(a+b))^(1/2)*a^3*cos(d*x+c)^3-3*sin(d*x+c)*cos(d*x+c)*a 
^3+sin(d*x+c)*cos(d*x+c)^2*a^2*b-18*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b 
)/(a+b))^(1/2))*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((a+cos(d*x+c)*b)/(1+...
 
3.7.14.5 Fricas [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(1/2),x, algo 
rithm="fricas")
 
output
integral((B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/sqrt(b*cos(d*x + c) + a), 
 x)
 
3.7.14.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\text {Timed out} \]

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)**(7/2)/(a+b*cos(d*x+c))**(1/2),x)
 
output
Timed out
 
3.7.14.7 Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(1/2),x, algo 
rithm="maxima")
 
output
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/sqrt(b*cos(d*x + c) + a) 
, x)
 
3.7.14.8 Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(1/2),x, algo 
rithm="giac")
 
output
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/sqrt(b*cos(d*x + c) + a) 
, x)
 
3.7.14.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}}{\sqrt {a+b\,\cos \left (c+d\,x\right )}} \,d x \]

input
int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + b*cos(c + d*x))^(1/ 
2),x)
 
output
int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(7/2))/(a + b*cos(c + d*x))^(1/ 
2), x)