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

Optimal. Leaf size=72 \[ \frac{\tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{1-\cos (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0859657, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2770, 2775, 207} \[ \frac{\tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{1-\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])]/d - (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqr
t[1 - 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{1-\cos (c+d x)} \sqrt{\cos (c+d x)} \, dx &=-\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{1-\cos (c+d x)}}-\frac{1}{2} \int \frac{\sqrt{1-\cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=-\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{1-\cos (c+d x)}}-\frac{\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{\tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d}-\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{1-\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.589676, size = 252, normalized size = 3.5 \[ \frac{\sqrt{\sin ^2\left (\frac{1}{2} (c+d x)\right ) \cos (c+d x)} \left (-2 \sqrt{2} \cot \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x) (\cos (d x)+i \sin (d x))}+\sqrt{\cos (c)-i \sin (c)} \left (\cot \left (\frac{1}{2} (c+d x)\right )+i\right ) \tanh ^{-1}\left (\frac{e^{i d x}}{\sqrt{\cos (c)-i \sin (c)} \sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}\right )+\sqrt{\cos (c)-i \sin (c)} \left (\cot \left (\frac{1}{2} (c+d x)\right )+i\right ) \tanh ^{-1}\left (\frac{\sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}{\sqrt{\cos (c)-i \sin (c)}}\right )\right )}{2 d \sqrt{\cos (c+d x) (\cos (d x)+i \sin (d x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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])])*Sqrt[Cos[c + d*x]*Sin[(c + d*x)/2]^2])/(2*d*Sqrt[Cos[c + d*x]*(Cos[d*x]
+ I*Sin[d*x])])

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Maple [A]  time = 0.312, size = 94, normalized size = 1.3 \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{2-2\,\cos \left ( dx+c \right ) }{\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}} \end{align*}

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]

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

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Maxima [B]  time = 1.87976, size = 1304, normalized size = 18.11 \begin{align*} \text{result too large to display} \end{align*}

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]

1/8*(4*(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) + 1)*cos(1/2*ar
ctan2(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))) - log(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 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*s
in(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)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + log(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 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*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)*cos(1/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - log(((cos(d*x + c)^2 + sin(d*x + c)^2)*cos(1/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (cos(d*x + c)^2 + sin(d*x + c)^2)*sin(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c) + 1))^2)*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + 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(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) + l
og(((cos(d*x + c)^2 + sin(d*x + c)^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (cos(d*x +
c)^2 + sin(d*x + c)^2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2)*sqrt(cos(2*d*x + 2*c)^2 + s
in(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) - 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(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*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1))/d

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Fricas [A]  time = 2.17102, size = 304, normalized size = 4.22 \begin{align*} -\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 )} - \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 ) \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \end{align*}

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]

-1/2*(2*(cos(d*x + c) + 1)*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - 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))*sin(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{1 - \cos{\left (c + d x \right )}} \sqrt{\cos{\left (c + d x \right )}}\, dx \end{align*}

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

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]

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