\(\int \frac {a+a \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx\) [149]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 35 \[ \int \frac {a+a \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 a E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d} \] Output:

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

Mathematica [C] (warning: unable to verify)

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

Time = 2.13 (sec) , antiderivative size = 155, normalized size of antiderivative = 4.43 \[ \int \frac {a+a \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {a \sqrt {\cos (c+d x)} (1+\cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-2 \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec (d x-\arctan (\cot (c))) \sin (c)+\tan (d x+\arctan (\tan (c)))-\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \tan (d x+\arctan (\tan (c)))}{\sqrt {\sin ^2(d x+\arctan (\tan (c)))}}\right )}{2 d} \] Input:

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

Output:

(a*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[(c + d*x)/2]^2*(-2*Sqrt[Cos[d 
*x - ArcTan[Cot[c]]]^2]*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4} 
, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + Tan[d*x 
+ ArcTan[Tan[c]]] - (HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcT 
an[Tan[c]]]^2]*Tan[d*x + ArcTan[Tan[c]]])/Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2 
]))/(2*d)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle a \int \frac {1}{\sqrt {\cos (c+d x)}}dx+a \int \sqrt {\cos (c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(2*a*EllipticE[(c + d*x)/2, 2])/d + (2*a*EllipticF[(c + d*x)/2, 2])/d
 

Defintions of rubi rules used

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(149\) vs. \(2(38)=76\).

Time = 1.95 (sec) , antiderivative size = 150, normalized size of antiderivative = 4.29

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(150\)
parts \(\frac {2 a \,\operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d}+\frac {2 a \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(153\)
risch \(-\frac {i \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{-i \left (d x +c \right )}}}-\frac {i \left (\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )}{\sqrt {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}}-\frac {2 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{\sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{i \left (d x +c \right )}}}+\frac {i \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {2}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, \left (-2 i \operatorname {EllipticE}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )+i \operatorname {EllipticF}\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}}}\right ) a \sqrt {2}\, \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{i \left (d x +c \right )}}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \sqrt {\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) {\mathrm e}^{-i \left (d x +c \right )}}}\) \(407\)

Input:

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

Output:

-2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4 
+sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-Ellipt 
icE(cos(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 = 107, normalized size of antiderivative = 3.06 \[ \int \frac {a+a \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {-i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} a {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {2} a {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 41.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {a+a \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,a\,\left (\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{d} \] Input:

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

Output:

(2*a*(ellipticE(c/2 + (d*x)/2, 2) + ellipticF(c/2 + (d*x)/2, 2)))/d
 

Reduce [F]

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

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

Output:

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