\(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx\) [175]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {5 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {5 \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

-3*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+5/3*InverseJacobiAM(1/2*d*x+1 
/2*c,2^(1/2))/a/d+5/3*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d-cos(d*x+c)^(3/2)*sin 
(d*x+c)/d/(a+a*cos(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.23 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.89 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {2 i \sqrt {2} e^{-i (c+d x)} \left (9 \left (1+e^{2 i (c+d x)}\right )+9 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+5 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{d \left (-1+e^{2 i c}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}+\frac {2 \sqrt {\cos (c+d x)} \csc (c) \left (3+6 \cos (c)+2 \cos (d x) \sin ^2(c)+6 \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\sin (2 c) \sin (d x)\right )}{d}\right )}{3 a (1+\cos (c+d x))} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*(((-2*I)*Sqrt[2]*(9*(1 + E^((2*I)*(c + d*x))) + 9*(-1 
+ E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 
3/4, -E^((2*I)*(c + d*x))] + 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + 
 E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x) 
)]))/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/ 
E^(I*(c + d*x))]) + (2*Sqrt[Cos[c + d*x]]*Csc[c]*(3 + 6*Cos[c] + 2*Cos[d*x 
]*Sin[c]^2 + 6*Sec[(c + d*x)/2]*Sin[c/2]*Sin[(d*x)/2] + Sin[2*c]*Sin[d*x]) 
)/d))/(3*a*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 3246, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3246

\(\displaystyle -\frac {\int \frac {1}{2} \sqrt {\cos (c+d x)} (3 a-5 a \cos (c+d x))dx}{a^2}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {6 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\)

Input:

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

Output:

-((Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) - ((6*a*Elli 
pticE[(c + d*x)/2, 2])/d - 5*a*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*S 
qrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/(2*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 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3246

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(214\) vs. \(2(95)=190\).

Time = 4.32 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.15

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (5 \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+18 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(215\)

Input:

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

Output:

-1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+ 
1/2*c)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(5*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2) 
))-8*sin(1/2*d*x+1/2*c)^6+18*sin(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^2)/ 
a/cos(1/2*d*x+1/2*c)/(-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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.98 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2 \, {\left (2 \, \cos \left (d x + c\right ) + 5\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/6*(2*(2*cos(d*x + c) + 5)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(I*sqrt(2) 
*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin 
(d*x + c)) - 5*(-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(- 
4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*(I*sqrt(2)*cos(d*x + c) + I*sqrt( 
2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin 
(d*x + c))) - 9*(-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 
0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a*d*cos(d* 
x + c) + a*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x) + 1),x)/a