3.2.40 \(\int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [140]

3.2.40.1 Optimal result
3.2.40.2 Mathematica [A] (verified)
3.2.40.3 Rubi [A] (verified)
3.2.40.4 Maple [A] (verified)
3.2.40.5 Fricas [A] (verification not implemented)
3.2.40.6 Sympy [F(-1)]
3.2.40.7 Maxima [F(-1)]
3.2.40.8 Giac [A] (verification not implemented)
3.2.40.9 Mupad [F(-1)]

3.2.40.1 Optimal result

Integrand size = 23, antiderivative size = 145 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {75 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}+\frac {13 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}}+\frac {9 \sin (c+d x)}{4 a^2 d \sqrt {a+a \cos (c+d x)}} \]

output
-1/4*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)+13/16*sin(d*x+c)/a/d 
/(a+a*cos(d*x+c))^(3/2)-75/32*arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a* 
cos(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+9/4*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c)) 
^(1/2)
 
3.2.40.2 Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\left (-150 \sqrt {2} \text {arctanh}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+\sqrt {1-\cos (c+d x)} \left (49+85 \cos (c+d x)+32 \cos ^2(c+d x)\right )\right ) \sin (c+d x)}{16 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{5/2}} \]

input
Integrate[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^(5/2),x]
 
output
((-150*Sqrt[2]*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2]]*Cos[(c + d*x)/2]^4 + Sqrt 
[1 - Cos[c + d*x]]*(49 + 85*Cos[c + d*x] + 32*Cos[c + d*x]^2))*Sin[c + d*x 
])/(16*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(5/2))
 
3.2.40.3 Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3244, 27, 3042, 3447, 3042, 3498, 27, 3042, 3230, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int \frac {\cos (c+d x) (4 a-9 a \cos (c+d x))}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {\cos (c+d x) (4 a-9 a \cos (c+d x))}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

\(\displaystyle -\frac {\int \frac {4 a \cos (c+d x)-9 a \cos ^2(c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3498

\(\displaystyle -\frac {-\frac {\int -\frac {3 \left (13 a^2-12 a^2 \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {3 \int \frac {13 a^2-12 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3 \int \frac {13 a^2-12 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3230

\(\displaystyle -\frac {\frac {3 \left (25 a^2 \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {24 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {3 \left (25 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {24 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle -\frac {\frac {3 \left (-\frac {50 a^2 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {24 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {3 \left (\frac {25 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {24 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\right )}{4 a^2}-\frac {13 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

input
Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^(5/2),x]
 
output
-1/4*(Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(5/2)) - ((-13* 
a*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)) + (3*((25*Sqrt[2]*a^(3/2) 
*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - ( 
24*a^2*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])))/(4*a^2))/(8*a^2)
 

3.2.40.3.1 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 219
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3230
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 3244
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e 
+ f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(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 - 2)* 
Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) 
 + b*c*(m + n))*Sin[e + f*x], x], 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] && LtQ[m, -1] 
&& GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3447
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

rule 3498
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* 
B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 
/(a^2*(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b 
*B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
 B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
 
3.2.40.4 Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.43

method result size
default \(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-75 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+64 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+21 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}-2 \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{32 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a^{\frac {7}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(208\)

input
int(cos(d*x+c)^3/(a+cos(d*x+c)*a)^(5/2),x,method=_RETURNVERBOSE)
 
output
1/32/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-75*2^(1/2)*ln(2 
*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*a*cos( 
1/2*d*x+1/2*c)^4+64*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*cos(1/2 
*d*x+1/2*c)^4+21*cos(1/2*d*x+1/2*c)^2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*2^(1/ 
2)*a^(1/2)-2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(7/2)/sin(1 
/2*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 
3.2.40.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {75 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (32 \, \cos \left (d x + c\right )^{2} + 85 \, \cos \left (d x + c\right ) + 49\right )} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

input
integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")
 
output
1/64*(75*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)* 
sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a 
)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) 
+ 1)) + 4*sqrt(a*cos(d*x + c) + a)*(32*cos(d*x + c)^2 + 85*cos(d*x + c) + 
49)*sin(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)
 
3.2.40.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**3/(a+a*cos(d*x+c))**(5/2),x)
 
output
Timed out
 
3.2.40.7 Maxima [F(-1)]

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

input
integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")
 
output
Timed out
 
3.2.40.8 Giac [A] (verification not implemented)

Time = 2.12 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.11 \[ \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=-\frac {\frac {75 \, \sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {75 \, \sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {128 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} {\left (21 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 19 \, \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{64 \, d} \]

input
integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")
 
output
-1/64*(75*sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(1/2*d*x + 
 1/2*c))) - 75*sqrt(2)*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^(5/2)*sgn(cos(1/2 
*d*x + 1/2*c))) - 128*sqrt(2)*sin(1/2*d*x + 1/2*c)/(a^(5/2)*sgn(cos(1/2*d* 
x + 1/2*c))) + 2*sqrt(2)*(21*sin(1/2*d*x + 1/2*c)^3 - 19*sin(1/2*d*x + 1/2 
*c))/((sin(1/2*d*x + 1/2*c)^2 - 1)^2*a^(5/2)*sgn(cos(1/2*d*x + 1/2*c))))/d
 
3.2.40.9 Mupad [F(-1)]

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

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