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

Optimal. Leaf size=85 \[ \frac {2 \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2777, 2775, 207, 2782, 206} \[ \frac {2 \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

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]

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

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Mathematica [C]  time = 0.12, size = 160, normalized size = 1.88 \[ -\frac {i \left (-1+e^{i (c+d x)}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\sinh ^{-1}\left (e^{i (c+d x)}\right )-\sqrt {2} \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

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fricas [B]  time = 1.20, size = 170, normalized size = 2.00 \[ \frac {\sqrt {2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) + 2 \, \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.80, size = 105, normalized size = 1.24 \[ -\frac {\sqrt {2} {\left (\sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{\sqrt {2} + \sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}\right ) + \log \left (\sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 1\right ) - \log \left (-\sqrt {-\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 1\right )\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*(sqrt(2)*log((sqrt(2) - sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1))/(sqrt(2) + sqrt(-tan(1/2*d*x + 1/2*c)^
2 + 1))) + log(sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1) + 1) - log(-sqrt(-tan(1/2*d*x + 1/2*c)^2 + 1) + 1))/d

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maple [A]  time = 0.10, size = 117, normalized size = 1.38 \[ -\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right ) \left (-\sqrt {2}\, \arctanh \left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 \arctanh \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \sqrt {2}}{d \sin \left (d x +c \right ) \sqrt {2-2 \cos \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(1/2)/(1-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)))/sin(d*x+c)/(2-2*cos(d*x+c))^(1/2)/(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 \[ \int \frac {\sqrt {\cos \left (d x + c\right )}}{\sqrt {-\cos \left (d x + c\right ) + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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