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

Optimal. Leaf size=107 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

2*arctanh(sin(d*x+c)*a^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2))/d/a^(1/2)-arctanh(1/2*sin(d*x+c)*a^(1/2)
*2^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)

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Rubi [A]  time = 0.18, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2777, 2775, 207, 2782, 208} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/(Sqrt[a]*d) - (Sqrt[2]*ArcTa
nh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/(Sqrt[a]*d)

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2775

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]

Rule 2777

Int[Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
d/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[(b*c - a*d)/b, Int[1/(Sqrt[a + b*Sin
[e + f*x]]*Sqrt[c + d*Sin[e + f*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]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*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]

Rubi steps

\begin {align*} \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a-a \cos (c+d x)}} \, dx &=-\frac {\int \frac {\sqrt {a-a \cos (c+d x)}}{\sqrt {\cos (c+d x)}} \, dx}{a}+\int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \, dx\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-a+x^2} \, dx,x,\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{2 a^2-a x^2} \, dx,x,\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}\\ \end {align*}

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Mathematica [C]  time = 0.38, size = 161, normalized size = 1.50 \[ -\frac {i \left (-1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\sinh ^{-1}\left (e^{i (c+d x)}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a-a \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(-1 + E^(I*(c + d*x)))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*(ArcSinh[E^(I*(c + d*x))] - Sqrt[
2]*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x
))]]))/(Sqrt[2]*d*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a - a*Cos[c + d*x]])

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fricas [A]  time = 0.92, size = 162, normalized size = 1.51 \[ \frac {\sqrt {2} \sqrt {a} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {a} \log \left (-\frac {2 \, \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )} + {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right )}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*sqrt(a)*log(-(2*sqrt(2)*sqrt(-a*cos(d*x + c) + a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sqrt(a) -
(3*cos(d*x + c) + 1)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c))) + 2*sqrt(a)*log(-(2*sqrt(-a*cos(d*x + c
) + a)*sqrt(a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c)) + (2*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c)))/(a*
d)

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giac [A]  time = 1.93, size = 86, normalized size = 0.80 \[ -\frac {\sqrt {2} {\left (\frac {\sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {a \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}}\right )} {\left | a \right |}}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*(sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) - a*arctan(sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a))*abs(a)/(a^2*d)

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maple [A]  time = 0.11, size = 118, normalized size = 1.10 \[ -\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right ) \left (-\sqrt {2}\, \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \arctanh \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {2}}{d \sqrt {-2 a \left (-1+\cos \left (d x +c \right )\right )}\, \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*cos(d*x+c)^(1/2)*(-1+cos(d*x+c))*(-2^(1/2)*arctanh(1/2*2^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2*arcta
nh((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))/(-2*a*(-1+cos(d*x+c)))^(1/2)/sin(d*x+c)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {-a \cos \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\cos \left (c+d\,x\right )}}{\sqrt {a-a\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\cos {\left (c + d x \right )}}}{\sqrt {- a \left (\cos {\left (c + d x \right )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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