\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx\) [176]

Optimal result

Integrand size = 23, antiderivative size = 72 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {3 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {\operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \] Output:

3*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d-InverseJacobiAM(1/2*d*x+1/2*c, 
2^(1/2))/a/d-cos(d*x+c)^(1/2)*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 = 3.59 (sec) , antiderivative size = 264, normalized size of antiderivative = 3.67 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {2 i \sqrt {2} e^{-i (c+d x)} \left (3 \left (1+e^{2 i (c+d x)}\right )+3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{d \left (-1+e^{2 i c}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}-\frac {2 \sqrt {\cos (c+d x)} \left (2 \cot (c)+\csc (c)+\sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )\right )}{d}\right )}{a (1+\cos (c+d x))} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*(((2*I)*Sqrt[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-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))] + 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]]*(2*Cot[c] + Csc[c] + Sec[c/2]*Sec[( 
c + d*x)/2]*Sin[(d*x)/2]))/d))/(a*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3246, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3246

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/2*((-6*a*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2]) 
/d)/a^2 - (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))
 

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 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. \(198\) vs. \(2(73)=146\).

Time = 2.64 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.76

method	result	size
default	\(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(199\)

Input:

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

Output:

((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)*(EllipticF 
(cos(1/2*d*x+1/2*c),2^(1/2))+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*si 
n(1/2*d*x+1/2*c)^4-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 = 0.08 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.58 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/2*((I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c)) + (-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*(-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))) - 3*(I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierst 
rassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) 
 - 2*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

Integral(cos(c + d*x)**(3/2)/(cos(c + d*x) + 1), x)/a
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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