\(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx\) [936]

Optimal result

Integrand size = 44, antiderivative size = 429 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^3 (a-b) (a+b)^{3/2} d}-\frac {2 \left (2 b^2 B-3 a^2 (B+C)+a b (3 B+C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{3 a^2 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 b (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (6 a^2 b B-2 b^3 B-3 a^3 C-a b^2 C\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.00 (sec) , antiderivative size = 1384, normalized size of antiderivative = 3.23 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*C 
os[c + d*x])^(5/2)),x]
 

Output:

(Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*((-2*(-(b^2*B*Sin[c + d*x]) + 
 a*b*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) - (2*(-6*a^ 
2*b^2*B*Sin[c + d*x] + 2*b^4*B*Sin[c + d*x] + 3*a^3*b*C*Sin[c + d*x] + a*b 
^3*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(a + b*Cos[c + d*x]))))/d + ((-4* 
a*(3*a^4*B - 5*a^2*b^2*B + 2*b^4*B - a^3*b*C + 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]*Ellipt 
icF[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*C 
os[c + d*x]]) - 4*a*(-6*a^3*b*B + 2*a*b^3*B + 3*a^4*C + 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), 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)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2*(-6* 
a^2*b^2*B + 2*b^4*B + 3*a^3*b*C + a*b^3*C)*((I*Cos[(c + d*x)/2]*Sqrt[a ...
 

Rubi [A] (verified)

Time = 1.86 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3508, 3042, 3479, 27, 3042, 3472, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3479

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3472

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

\(\Big \downarrow \) 3473

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

Input:

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(3/2)*(a + b*Cos[c + 
 d*x])^(5/2)),x]
 

Output:

(2*b*(b*B - a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*(a + 
b*Cos[c + d*x])^(3/2)) + (((2*(a - b)*Sqrt[a + b]*(6*a^2*b*B - 2*b^3*B - 3 
*a^3*C - 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]*(2*b^2*B - 3*a^2*(B + C) + a*b*(3*B + C))*Cot[c + d*x]*Ellip 
ticF[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 - b^2) - (2*(6*a^2*b*B - 2*b^3*B - 3*a^3*C 
- a*b^2*C)*Sin[c + d*x])/((a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[ 
c + d*x]]))/(3*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 3472

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2438\) vs. \(2(391)=782\).

Time = 12.72 (sec) , antiderivative size = 2439, normalized size of antiderivative = 5.69

method	result	size
default	\(\text {Expression too large to display}\)	\(2439\)
parts	\(\text {Expression too large to display}\)	\(2500\)

Input:

int((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^(5/2), 
x,method=_RETURNVERBOSE)
 

Output:

-2/3/d*(6*B*a^4*b*cos(d*x+c)*sin(d*x+c)-3*C*a^5*cos(d*x+c)*sin(d*x+c)+sin( 
d*x+c)*cos(d*x+c)*(-6*cos(d*x+c)-2)*B*a^2*b^3+sin(d*x+c)*cos(d*x+c)*(-cos( 
d*x+c)+3)*a*b^4*B+sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)+4)*C*a^4*b+sin(d*x+ 
c)*cos(d*x+c)*(3*cos(d*x+c)-1)*C*a^3*b^2+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*B 
*(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^5*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-3*co 
s(d*x+c)^2-6*cos(d*x+c)-3)*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^5*EllipticF(-csc(d*x+c)+cot(d*x+c), 
(-(a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*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^5*Ellipt 
icE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+B*(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^5*EllipticE(- 
csc(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))+sin(d*x+c)*cos(d*x+c)*(5*cos(d*x+c)-7)*a^3*b^2*B+2*sin(d*x+ 
c)*cos(d*x+c)^2*B*b^5-2*C*sin(d*x+c)*cos(d*x+c)^2*a^2*b^3+C*cos(d*x+c)^2*s 
in(d*x+c)*a*b^4+B*(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^4*EllipticF(-csc(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))+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 
^3*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))*(-cos(d*x+c)^...
 

Fricas [F]

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

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^ 
(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(3/2)/(a+b*cos(d*x+c) 
)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^ 
(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+b*cos(d*x+c))^ 
(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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