\(\int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \, dx\) [148]

Optimal result

Integrand size = 21, antiderivative size = 61 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \, dx=\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 5.10 (sec) , antiderivative size = 222, normalized size of antiderivative = 3.64 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \, dx=\frac {a (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {3 (3 \cos (c-d x-\arctan (\tan (c)))+\cos (c+d x+\arctan (\tan (c)))) \csc (c) \sec (c)}{\sqrt {\sec ^2(c)}}-4 \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)-4 \cos (c+d x) (3 \cot (c)-\sin (c+d x))-6 \cos (c) \csc (d x+\arctan (\tan (c))) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{12 d \sqrt {\cos (c+d x)}} \] Input:

Integrate[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]),x]
 

Output:

(a*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*((3*(3*Cos[c - d*x - ArcTan[Tan[c 
]]] + Cos[c + d*x + ArcTan[Tan[c]]])*Csc[c]*Sec[c])/Sqrt[Sec[c]^2] - 4*Cos 
[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricP 
FQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c] 
]]*Sin[c] - 4*Cos[c + d*x]*(3*Cot[c] - Sin[c + d*x]) - 6*Cos[c]*Csc[d*x + 
ArcTan[Tan[c]]]*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Ta 
n[c]]]^2]*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(12*d*Sqrt[Co 
s[c + d*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {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.

\(\displaystyle \int \sqrt {\cos (c+d x)} (a \cos (c+d x)+a) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x]),x]
 

Output:

(2*a*EllipticE[(c + d*x)/2, 2])/d + a*((2*EllipticF[(c + d*x)/2, 2])/(3*d) 
 + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d))
 

Defintions of rubi rules used

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(224\) vs. \(2(58)=116\).

Time = 4.55 (sec) , antiderivative size = 225, normalized size of antiderivative = 3.69

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(225\)
parts	\(-\frac {2 a \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}+\frac {2 a \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(314\)

Input:

int(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c)),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)*a*(4*sin(1/2* 
d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^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(co 
s(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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 = 125, normalized size of antiderivative = 2.05 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \, dx=\frac {2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c)),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c)),x, algorithm="maxima")
 

Output:

integrate((a*cos(d*x + c) + a)*sqrt(cos(d*x + c)), x)
 

Giac [F]

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

integrate(cos(d*x+c)^(1/2)*(a+a*cos(d*x+c)),x, algorithm="giac")
 

Output:

integrate((a*cos(d*x + c) + a)*sqrt(cos(d*x + c)), x)
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \sqrt {\cos (c+d x)} (a+a \cos (c+d x)) \, dx=\frac {2\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,a\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \] Input:

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

Output:

(2*a*ellipticE(c/2 + (d*x)/2, 2))/d + (2*a*ellipticF(c/2 + (d*x)/2, 2))/(3 
*d) + (2*a*cos(c + d*x)^(1/2)*sin(c + d*x))/(3*d)
 

Reduce [F]

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

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

Output:

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