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

Optimal. Leaf size=58 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

-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 = 0.07, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2782, 208} \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]]),x]

[Out]

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

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 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 {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \, dx &=-\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 {\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 = 0.33, size = 118, normalized size = 2.03 \[ \frac {i \left (-1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a-a \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 1.33, size = 144, normalized size = 2.48 \[ \left [\frac {\sqrt {2} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right )}{2 \, \sqrt {a} d}, \frac {\sqrt {2} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right )}{d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(2)*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)))/(sqrt(a)*d), sqrt(2)*sqrt(-1/a)*arctan(sqrt(2)*
sqrt(-a*cos(d*x + c) + a)*sqrt(-1/a)*sqrt(cos(d*x + c))/sin(d*x + c))/d]

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giac [B]  time = 0.76, size = 137, normalized size = 2.36 \[ -\frac {\sqrt {2} {\left (\frac {a^{2} {\left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {\arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a}\right )}}{{\left | a \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {{\left (a \arctan \left (\frac {\sqrt {2} \sqrt {a}}{\sqrt {-a}}\right ) - a \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right )\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\sqrt {-a} {\left | a \right |}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 77, normalized size = 1.33 \[ -\frac {2 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )}{d \sqrt {-2 a \left (-1+\cos \left (d x +c \right )\right )}\, \sqrt {\cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.36, size = 209, normalized size = 3.60 \[ -\frac {\sqrt {2} \arctan \left (\frac {2 \, \sqrt {2} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )}{\sqrt {a} {\left | e^{\left (i \, d x + i \, c\right )} - 1 \right |}}, \frac {2 \, {\left (\sqrt {2} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {a} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - \sqrt {-a} {\left | e^{\left (i \, d x + i \, c\right )} - 1 \right |} + 2 \, \sqrt {a}\right )}}{a {\left | e^{\left (i \, d x + i \, c\right )} - 1 \right |}}\right )}{\sqrt {-a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*arctan2(2*sqrt(2)*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))/(sqrt(a)*abs(e^(I*d*x + I*c) - 1)), 2*(sqrt(2)*(cos(2*d*x + 2*c
)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c) + 1)) - sqrt(-a)*abs(e^(I*d*x + I*c) - 1) + 2*sqrt(a))/(a*abs(e^(I*d*x + I*c) - 1)))/(sqrt(-a)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {a-a\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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