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

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

[Out]

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

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Rubi [A]  time = 0.121755, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2770, 2775, 207} \[ \frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{d}-\frac{a \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a-a \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2770

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(2*n*(b*c + a*d)
)/(b*(2*n + 1)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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] && GtQ[n, 0] && IntegerQ[2*n]

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 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])

Rubi steps

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

Mathematica [C]  time = 0.720696, size = 264, normalized size = 3.11 \[ \frac{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)} \left (-2 \sqrt{2} \cot \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x) (\cos (d x)+i \sin (d x))}+\sqrt{\cos (c)-i \sin (c)} \left (\cot \left (\frac{1}{2} (c+d x)\right )+i\right ) \tanh ^{-1}\left (\frac{e^{i d x}}{\sqrt{\cos (c)-i \sin (c)} \sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}\right )+\sqrt{\cos (c)-i \sin (c)} \left (\cot \left (\frac{1}{2} (c+d x)\right )+i\right ) \tanh ^{-1}\left (\frac{\sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}{\sqrt{\cos (c)-i \sin (c)}}\right )\right )}{2 d \sqrt{i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.354, size = 95, normalized size = 1.1 \begin{align*}{\frac{\sqrt{2} \left ( 1+\cos \left ( dx+c \right ) \right ) }{2\,d\sin \left ( dx+c \right ) } \left ({\it Artanh} \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}-\cos \left ( dx+c \right ) \right ) \sqrt{-2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) }{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.01623, size = 1073, normalized size = 12.62 \begin{align*} \text{result too large to display} \end{align*}

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]

-1/4*(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)*(cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (cos(d*x + c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2
*c) + 1)))*sqrt(-a) + sqrt(-a)*(arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1
/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4
)*(cos(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))) + 1) - arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2
*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c
) + 1)^(1/4)*(cos(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*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) + arctan2((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*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2
*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) -
arctan2((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*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1)))/d

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Fricas [A]  time = 2.25367, size = 389, normalized size = 4.58 \begin{align*} \frac{\sqrt{a} \log \left (-\frac{4 \, \sqrt{-a \cos \left (d x + c\right ) + a}{\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt{a} \sqrt{\cos \left (d x + c\right )} +{\left (8 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, \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 )}}{4 \, d \sin \left (d x + c\right )} \end{align*}

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]

1/4*(sqrt(a)*log(-(4*sqrt(-a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a)*sqrt(cos(d*x +
c)) + (8*a*cos(d*x + c)^2 + 8*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*sqrt(-a*cos(d*x
+ c) + a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c)))/(d*sin(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- a \left (\cos{\left (c + d x \right )} - 1\right )} \sqrt{\cos{\left (c + d x \right )}}\, dx \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{-a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}\,{d x} \end{align*}

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]

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