\(\int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx\) [147]

Optimal result

Integrand size = 33, antiderivative size = 85 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=-\frac {(A-3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

-(A-3*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+(A-B)*(cos(
1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a/d+(A-B)*sin(d*x+c)*cos(d*x+
c)^(1/2)/d/(a+a*cos(d*x+c))

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3056, 2827, 2720, 2719} \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {(A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}-\frac {(A-3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)} \]

[In]

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

[Out]

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

Rule 2719

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 2720

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 2827

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

Rule 3056

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

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int \frac {\frac {1}{2} a (A-B)-\frac {1}{2} a (A-3 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{a^2} \\ & = \frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))}-\frac {(A-3 B) \int \sqrt {\cos (c+d x)} \, dx}{2 a}+\frac {(A-B) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{2 a} \\ & = -\frac {(A-3 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(A-B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{d (a+a \cos (c+d x))} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 6.91 (sec) , antiderivative size = 862, normalized size of antiderivative = 10.14 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (-\frac {2 (-A+B+2 B \cos (c)) \csc (c)}{d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )-B \sin \left (\frac {d x}{2}\right )\right )}{d}\right )}{a+a \cos (c+d x)}-\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (a+a \cos (c+d x)) \sqrt {1+\cot ^2(c)}}+\frac {B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (a+a \cos (c+d x)) \sqrt {1+\cot ^2(c)}}+\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (a+a \cos (c+d x))}-\frac {3 B \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (a+a \cos (c+d x))} \]

[In]

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

[Out]

(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((-2*(-A + B + 2*B*Cos[c])*Csc[c])/d + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]
*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2]))/d))/(a + a*Cos[c + d*x]) - (A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hypergeometric
PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcT
an[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])
/(d*(a + a*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2},
{5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqr
t[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c
+ d*x])*Sqrt[1 + Cot[c]^2]) + (A*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/
4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1
+ Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) -
((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]
^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(a + a*Cos[c + d*
x])) - (3*B*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[T
an[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Ta
n[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan
[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c
]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(a + a*Cos[c + d*x]))

Maple [A] (verified)

Time = 4.19 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.87

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+\left (-2 B +2 A \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-A +B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(244\)

[In]

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

[Out]

-((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)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-
B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+(-2*B+2*A)*sin(1/2*d*x+1/2*
c)^4+(-A+B)*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)/si
n(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 higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.79 \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\frac {2 \, {\left (A - B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A + i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + 3 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - 3 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

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

[Out]

1/2*(2*(A - B)*sqrt(cos(d*x + c))*sin(d*x + c) + (sqrt(2)*(-I*A + I*B)*cos(d*x + c) + sqrt(2)*(-I*A + I*B))*we
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - I*B)*cos(d*x + c) + sqrt(2)*(I*A - I
*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (sqrt(2)*(-I*A + 3*I*B)*cos(d*x + c) + sqrt(2
)*(-I*A + 3*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + (sqrt(2)
*(I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(I*A - 3*I*B))*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. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} (A+B \cos (c+d x))}{a+a \cos (c+d x)} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{a+a\,\cos \left (c+d\,x\right )} \,d x \]

[In]

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

[Out]

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