\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\) [932]

Optimal result

Integrand size = 44, antiderivative size = 393 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (5 a^2 b B-8 b^3 B-3 a^3 C+6 a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^4 \sqrt {a+b} d}+\frac {2 (a+2 b) (4 b B+a (B-3 C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 \sqrt {a+b} d}+\frac {2 b (b B-a C) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (a^2 B-4 b^2 B+3 a b C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) d \cos ^{\frac {3}{2}}(c+d x)} \] Output:

-2/3*(5*B*a^2*b-8*B*b^3-3*C*a^3+6*C*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^4/(a+b)^(1/2)/d+2/3* 
(a+2*b)*(4*B*b+a*(B-3*C))*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^3/(a+b)^(1/2)/d+2*b*(B*b-C*a)*sin(d*x 
+c)/a/(a^2-b^2)/d/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(1/2)+2/3*(B*a^2-4*B*b 
^2+3*C*a*b)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/a^2/(a^2-b^2)/d/cos(d*x+c)^( 
3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.12 (sec) , antiderivative size = 1357, normalized size of antiderivative = 3.45 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*C 
os[c + d*x])^(3/2)),x]
 

Output:

((-4*a*(a^4*B + 7*a^2*b^2*B - 8*b^4*B - 6*a^3*b*C + 6*a*b^3*C)*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)*Sqrt[Cos[c + d*x]]*Sqrt[ 
a + b*Cos[c + d*x]]) - 4*a*(5*a^3*b*B - 8*a*b^3*B - 3*a^4*C + 6*a^2*b^2*C) 
*((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)*Sqrt[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), ArcSi 
n[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*(5*a^2*b^2*B - 8*b^4*B - 3*a^3*b*C + 6*a*b^3*C)*((I*Cos[(c + d*x)/2]*S 
qrt[a + b*Cos[c + d*x]]*EllipticE[I*ArcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + 
d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x])/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + 
d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[c + d*x])/(a + b)]) + (2*a*((a*Sqrt[( 
(a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[...
 

Rubi [A] (verified)

Time = 1.85 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.295, Rules used = {3042, 3508, 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.

\(\displaystyle \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\frac {2 \int -\frac {-3 C a^3+5 b B a^2+6 b^2 C a-\left (B a^2-3 b C a+2 b^2 B\right ) \cos (c+d x) a-8 b^3 B}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}+\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}}{a \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 C a^3+5 b B a^2+6 b^2 C a-\left (B a^2-3 b C a+2 b^2 B\right ) \cos (c+d x) a-8 b^3 B}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{3 a}}{a \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 C a^3+5 b B a^2+6 b^2 C a-\left (B a^2-3 b C a+2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-8 b^3 B}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{a \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3477

\(\displaystyle \frac {\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^3 C+5 a^2 b B+6 a b^2 C-8 b^3 B\right ) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-(a-b) (a+2 b) (a B-3 a C+4 b B) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{3 a}}{a \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^3 C+5 a^2 b B+6 a b^2 C-8 b^3 B\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) (a+2 b) (a B-3 a C+4 b B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{a \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\left (-3 a^3 C+5 a^2 b B+6 a b^2 C-8 b^3 B\right ) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} (a+2 b) (a B-3 a C+4 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}}{a \left (a^2-b^2\right )}+\frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {2 b (b B-a C) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}+\frac {\frac {2 \left (a^2 B+3 a b C-4 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 a d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (a-b) \sqrt {a+b} \left (-3 a^3 C+5 a^2 b B+6 a b^2 C-8 b^3 B\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d}-\frac {2 (a-b) \sqrt {a+b} (a+2 b) (a B-3 a C+4 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{3 a}}{a \left (a^2-b^2\right )}\)

Input:

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + 
 d*x])^(3/2)),x]
 

Output:

(2*b*(b*B - a*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)*Sqrt[a 
+ b*Cos[c + d*x]]) + (-1/3*((2*(a - b)*Sqrt[a + b]*(5*a^2*b*B - 8*b^3*B - 
3*a^3*C + 6*a*b^2*C)*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*(a 
- b)*Sqrt[a + b]*(a + 2*b)*(a*B + 4*b*B - 3*a*C)*Cot[c + d*x]*EllipticF[Ar 
cSin[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*(a^2*B - 4*b^2*B + 3*a*b*C)*Sqrt[a + b*Cos[c + d 
*x]]*Sin[c + d*x])/(3*a*d*Cos[c + d*x]^(3/2)))/(a*(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 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 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 3508

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[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 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])))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1923\) vs. \(2(357)=714\).

Time = 18.88 (sec) , antiderivative size = 1924, normalized size of antiderivative = 4.90

method	result	size
parts	\(\text {Expression too large to display}\)	\(1924\)
default	\(\text {Expression too large to display}\)	\(1970\)

Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(3/2), 
x,method=_RETURNVERBOSE)
 

Output:

-2/3*B/d/(a+b)/(a-b)/a^3*((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)*a^3*b*EllipticE(-csc(d*x+c)+cot(d*x+c),( 
-(a-b)/(a+b))^(1/2))*(5*cos(d*x+c)^3+10*cos(d*x+c)^2+5*cos(d*x+c))+(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) 
*a^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(5*cos(d*x 
+c)^3+10*cos(d*x+c)^2+5*cos(d*x+c))+(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)*a*b^3*EllipticE(-csc(d*x+c)+co 
t(d*x+c),(-(a-b)/(a+b))^(1/2))*(-8*cos(d*x+c)^3-16*cos(d*x+c)^2-8*cos(d*x+ 
c))+(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)*b^4*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-8 
*cos(d*x+c)^3-16*cos(d*x+c)^2-8*cos(d*x+c))+(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)*a^4*EllipticF(-csc(d*x 
+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(cos(d*x+c)^3+2*cos(d*x+c)^2+cos(d*x+ 
c))+(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)*a^3*b*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*( 
-5*cos(d*x+c)^3-10*cos(d*x+c)^2-5*cos(d*x+c))+(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)*a^2*b^2*EllipticF(-c 
sc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(2*cos(d*x+c)^3+4*cos(d*x+c)^2+ 
2*cos(d*x+c))+(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)*a*b^3*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a...
 

Fricas [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^ 
(3/2),x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c) + B)*sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c))/ 
(b^2*cos(d*x + c)^5 + 2*a*b*cos(d*x + c)^4 + a^2*cos(d*x + c)^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(7/2)/(a+b*cos(d*x+c) 
)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^ 
(3/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^(3/2)* 
cos(d*x + c)^(7/2)), x)
 

Giac [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^ 
(3/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/((b*cos(d*x + c) + a)^(3/2)* 
cos(d*x + c)^(7/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(7/2)*(a + b*cos(c + 
 d*x))^(3/2)),x)
 

Output:

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^(7/2)*(a + b*cos(c + 
 d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))^{3/2}} \, dx=\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{5} b^{2}+2 \cos \left (d x +c \right )^{4} a b +\cos \left (d x +c \right )^{3} a^{2}}d x \right ) b +\left (\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4} b^{2}+2 \cos \left (d x +c \right )^{3} a b +\cos \left (d x +c \right )^{2} a^{2}}d x \right ) c \] Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2)/(a+b*cos(d*x+c))^(3/2), 
x)
 

Output:

int((sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)))/(cos(c + d*x)**5*b**2 + 
2*cos(c + d*x)**4*a*b + cos(c + d*x)**3*a**2),x)*b + int((sqrt(cos(c + d*x 
)*b + a)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4*b**2 + 2*cos(c + d*x)**3*a*b 
 + cos(c + d*x)**2*a**2),x)*c