\(\int \frac {\cos ^{\frac {3}{2}}(c+d x) (A+C \cos ^2(c+d x))}{(a+a \cos (c+d x))^{5/2}} \, dx\) [211]

Optimal result

Integrand size = 37, antiderivative size = 237 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {5 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A+115 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {(A-15 C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {(3 A+35 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \cos (c+d x)}} \] Output:

-5*C*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d+1/32*(3*A 
+115*C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d* 
x+c))^(1/2))*2^(1/2)/a^(5/2)/d-1/4*(A+C)*cos(d*x+c)^(5/2)*sin(d*x+c)/d/(a+ 
a*cos(d*x+c))^(5/2)+1/16*(A-15*C)*cos(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*cos 
(d*x+c))^(3/2)+1/16*(3*A+35*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(a+a*cos( 
d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 1.45 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.18 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\left (140 C \arcsin \left (\sqrt {1-\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+460 C \arcsin \left (\sqrt {\cos (c+d x)}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-2 \sqrt {2} (3 A+115 C) \arctan \left (\frac {\sqrt {\cos (c+d x)}}{\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+7 A \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+55 C \sqrt {1-\cos (c+d x)} \cos ^{\frac {3}{2}}(c+d x)+16 C \sqrt {1-\cos (c+d x)} \cos ^{\frac {5}{2}}(c+d x)+3 A \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}+35 C \sqrt {-((-1+\cos (c+d x)) \cos (c+d x))}\right ) \sin (c+d x)}{16 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \] Input:

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

Output:

((140*C*ArcSin[Sqrt[1 - Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 + 460*C*ArcSin[S 
qrt[Cos[c + d*x]]]*Cos[(c + d*x)/2]^4 - 2*Sqrt[2]*(3*A + 115*C)*ArcTan[Sqr 
t[Cos[c + d*x]]/Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 + 7*A*Sqrt[1 
- Cos[c + d*x]]*Cos[c + d*x]^(3/2) + 55*C*Sqrt[1 - Cos[c + d*x]]*Cos[c + d 
*x]^(3/2) + 16*C*Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(5/2) + 3*A*Sqrt[-((- 
1 + Cos[c + d*x])*Cos[c + d*x])] + 35*C*Sqrt[-((-1 + Cos[c + d*x])*Cos[c + 
 d*x])])*Sin[c + d*x])/(16*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x])) 
^(5/2))
 

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.405, Rules used = {3042, 3521, 27, 3042, 3456, 27, 3042, 3462, 3042, 3461, 3042, 3253, 223, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (a (3 A-5 C)+2 a (A+5 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\cos ^{\frac {3}{2}}(c+d x) (a (3 A-5 C)+2 a (A+5 C) \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3456

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 (A-15 C) a^2+2 (3 A+35 C) \cos (c+d x) a^2\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\cos (c+d x)} \left (3 (A-15 C) a^2+2 (3 A+35 C) \cos (c+d x) a^2\right )}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 (A-15 C) a^2+2 (3 A+35 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3462

\(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (3 A+35 C)-80 a^3 C \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (3 A+35 C)-80 a^3 C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-80 a^2 C \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-80 a^2 C \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3253

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {160 a^2 C \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\frac {\frac {a^3 (3 A+115 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {160 a^{5/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {\frac {\frac {-\frac {2 a^4 (3 A+115 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {160 a^{5/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\frac {\sqrt {2} a^{5/2} (3 A+115 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {160 a^{5/2} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{a}+\frac {2 a^2 (3 A+35 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}+\frac {a (A-15 C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/4*((A + C)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/ 
2)) + ((a*(A - 15*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Cos[c + 
d*x])^(3/2)) + (((-160*a^(5/2)*C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a* 
Cos[c + d*x]]])/d + (Sqrt[2]*a^(5/2)*(3*A + 115*C)*ArcTan[(Sqrt[a]*Sin[c + 
 d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/d)/a + (2*a 
^2*(3*A + 35*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x 
]]))/(4*a^2))/(8*a^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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 
 EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3456

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( 
a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + 
 b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & 
& NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In 
tegerQ[2*n] || EqQ[c, 0])
 

rule 3461

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[(A*b - a*B)/b   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) 
, x], x] + Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]]/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] && EqQ[ 
a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3462

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Sin[e + f*x])^m*(c + d*S 
in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m 
 + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
 

rule 3521

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

Maple [A] (verified)

Time = 6.06 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.27

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \left (43+8 \cos \left (2 d x +2 c \right )+55 \cos \left (d x +c \right )\right ) \sqrt {2}\, C \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sin \left (d x +c \right ) \left (7 \cos \left (d x +c \right )+3\right ) \sqrt {2}\, A \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-40 \sqrt {2}\, \left (\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )+3\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right ) C +\left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+\left (-115 \cos \left (d x +c \right )^{2}-230 \cos \left (d x +c \right )-115\right ) C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(301\)
parts	\(-\frac {A \left (\sin \left (d x +c \right ) \left (-7 \cos \left (d x +c \right )-3\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}-\frac {C \left (\sin \left (d x +c \right ) \left (-43-8 \cos \left (2 d x +2 c \right )-55 \cos \left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+40 \sqrt {2}\, \left (\cos \left (2 d x +2 c \right )+4 \cos \left (d x +c \right )+3\right ) \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\left (115 \cos \left (d x +c \right )^{2}+230 \cos \left (d x +c \right )+115\right ) \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {2}\, \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(380\)

Input:

int(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/32/d/a^3*(sin(d*x+c)*(43+8*cos(2*d*x+2*c)+55*cos(d*x+c))*2^(1/2)*C*(cos( 
d*x+c)/(1+cos(d*x+c)))^(1/2)+sin(d*x+c)*(7*cos(d*x+c)+3)*2^(1/2)*A*(cos(d* 
x+c)/(1+cos(d*x+c)))^(1/2)-40*2^(1/2)*(cos(2*d*x+2*c)+4*cos(d*x+c)+3)*arct 
an((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c))*C+(-3*cos(d*x+c)^2-6*cos( 
d*x+c)-3)*A*arcsin(-csc(d*x+c)+cot(d*x+c))+(-115*cos(d*x+c)^2-230*cos(d*x+ 
c)-115)*C*arcsin(-csc(d*x+c)+cot(d*x+c)))*2^(1/2)*(a*(1+cos(d*x+c)))^(1/2) 
*cos(d*x+c)^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)/(cos(d*x+c) 
/(1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 5.53 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} {\left ({\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 115 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 115 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )\right )}}\right ) + 2 \, {\left (16 \, C \cos \left (d x + c\right )^{2} + {\left (7 \, A + 55 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + 35 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 160 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} + a \cos \left (d x + c\right )}\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

1/32*(sqrt(2)*((3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c) 
^2 + 3*(3*A + 115*C)*cos(d*x + c) + 3*A + 115*C)*sqrt(a)*arctan(1/2*sqrt(2 
)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos( 
d*x + c)^2 + a*cos(d*x + c))) + 2*(16*C*cos(d*x + c)^2 + (7*A + 55*C)*cos( 
d*x + c) + 3*A + 35*C)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x 
 + c) - 160*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C) 
*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(a)*sqrt(cos(d*x + c))*sin(d* 
x + c)/(a*cos(d*x + c)^2 + a*cos(d*x + c))))/(a^3*d*cos(d*x + c)^3 + 3*a^3 
*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

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

Giac [F(-1)]

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

integrate(cos(d*x+c)^(3/2)*(A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^(5/2),x, al 
gorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right ) c \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*(int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x))/(co 
s(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*a + int((sqrt(c 
os(c + d*x) + 1)*sqrt(cos(c + d*x))*cos(c + d*x)**3)/(cos(c + d*x)**3 + 3* 
cos(c + d*x)**2 + 3*cos(c + d*x) + 1),x)*c))/a**3