\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [582]

Optimal result

Integrand size = 23, antiderivative size = 112 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}+\frac {2 \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}-\frac {2 a^3 \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d} \] Output:

-2*a*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/d+2/3*(3*a^2+b^2)*InverseJa 
cobiAM(1/2*d*x+1/2*c,2^(1/2))/b^3/d-2*a^3*EllipticPi(sin(1/2*d*x+1/2*c),2* 
b/(a+b),2^(1/2))/b^3/(a+b)/d+2/3*cos(d*x+c)^(1/2)*sin(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 12.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {6 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+4 \sqrt {\cos (c+d x)} \sin (c+d x)-\frac {6 \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{b^2 \sqrt {\sin ^2(c+d x)}}}{6 b d} \] Input:

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

Output:

(4*EllipticF[(c + d*x)/2, 2] - (6*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 
 2])/(a + b) + 4*Sqrt[Cos[c + d*x]]*Sin[c + d*x] - (6*(-2*a*b*EllipticE[Ar 
cSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + 
d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]] 
, -1])*Sin[c + d*x])/(b^2*Sqrt[Sin[c + d*x]^2]))/(6*b*d)
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 3272, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3272

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {2 \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {3 a^3 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}-\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {\frac {2 \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}-\frac {6 a^3 \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}-\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{3 b}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 b d}\)

Input:

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

Output:

((-6*a*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(3*a^2 + b^2)*EllipticF[(c + 
 d*x)/2, 2])/(b*d) - (6*a^3*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b* 
(a + b)*d))/b)/(3*b) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*b*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3272

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
 + n))   Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d 
*(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 
 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m 
] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] 
&& NeQ[c, 0])))
 

rule 3284

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

rule 3481

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

rule 3538

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(551\) vs. \(2(111)=222\).

Time = 6.25 (sec) , antiderivative size = 552, normalized size of antiderivative = 4.93

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a \,b^{2}-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b^{3}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a \,b^{2}+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}+3 a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 a^{2} b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+b^{2} a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-b^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2} b -3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \,b^{2}-3 a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )\right )}{3 b^{3} \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(552\)

Input:

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

Output:

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4*b^3-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a*b^2+2*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*b^3+3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* 
d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*a^2*b*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2* 
d*x+1/2*c),2^(1/2))+b^2*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* 
c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-b^3*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2 
^(1/2))+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El 
lipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*( 
2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^ 
2-3*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli 
pticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/b^3/(a-b)/(-2*sin(1/2*d*x+1 
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2* 
c)^2-1)^(1/2)/d
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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