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

Optimal. Leaf size=37 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2775, 207} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/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 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]

Rubi steps

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

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Mathematica [C]  time = 0.51, size = 277, normalized size = 7.49 \[ \frac {2 e^{i d x} \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \sqrt {\cos (c)-i \sin (c)} \sqrt {1-\cos (c+d x)} \sqrt {e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )} \left (\tanh ^{-1}\left (\frac {e^{i d x}}{\sqrt {\cos (c)-i \sin (c)} \sqrt {e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}\right )+\tanh ^{-1}\left (\frac {\sqrt {e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}{\sqrt {\cos (c)-i \sin (c)}}\right )\right )}{d \left (i \cos \left (\frac {c}{2}\right ) \left (-1+e^{i d x}\right )-\sin \left (\frac {c}{2}\right ) \left (1+e^{i d x}\right )\right ) \sqrt {2 i \sin (c) \left (-1+e^{2 i d x}\right )+2 \cos (c) \left (1+e^{2 i d x}\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - Cos[c + d*x]]/Sqrt[Cos[c + d*x]],x]

[Out]

(2*E^(I*d*x)*(ArcTanh[E^(I*d*x)/(Sqrt[Cos[c] - I*Sin[c]]*Sqrt[Cos[c] + E^((2*I)*d*x)*(Cos[c] + I*Sin[c]) - I*S
in[c]])] + ArcTanh[Sqrt[Cos[c] + E^((2*I)*d*x)*(Cos[c] + I*Sin[c]) - I*Sin[c]]/Sqrt[Cos[c] - I*Sin[c]]])*Sqrt[
1 - Cos[c + d*x]]*(Cos[c/2] + I*Sin[c/2])*Sqrt[Cos[c] - I*Sin[c]]*Sqrt[((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E
^((2*I)*d*x))*Sin[c])/E^(I*d*x)])/(d*(I*(-1 + E^(I*d*x))*Cos[c/2] - (1 + E^(I*d*x))*Sin[c/2])*Sqrt[2*(1 + E^((
2*I)*d*x))*Cos[c] + (2*I)*(-1 + E^((2*I)*d*x))*Sin[c]])

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fricas [A]  time = 0.87, size = 64, normalized size = 1.73 \[ \frac {\log \left (-\frac {2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 1.40, size = 119, normalized size = 3.22 \[ \frac {2 \, \log \left (\frac {2 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 2 \, \sqrt {2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right )}}{{\left | -2 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 4 \, \sqrt {2} + 2 \, \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} - 2 \right |}}\right ) \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(2*(tan(1/4*d*x + 1/4*c)^2 + 2*sqrt(2) - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1)
/abs(-2*tan(1/4*d*x + 1/4*c)^2 + 4*sqrt(2) + 2*sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) - 2
))*sgn(sin(1/2*d*x + 1/2*c))/d

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maple [B]  time = 0.09, size = 83, normalized size = 2.24 \[ \frac {\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2-2 \cos \left (d x +c \right )}\, \sin \left (d x +c \right ) \arctanh \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )}{d \sqrt {\cos \left (d x +c \right )}\, \left (-1+\cos \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.07, size = 221, normalized size = 5.97 \[ \frac {2 \, \operatorname {arsinh}\relax (1) + \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + \sqrt {\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1} {\left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )^{2}\right )} + 2 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} {\left (\cos \left (d x + c\right ) \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + \sin \left (d x + c\right ) \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*arcsinh(1) + log(cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos
(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c) + 1))^2) + 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)*(c
os(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) + 1)))))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {1-\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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