3.269 \(\int \sqrt{1-\cos (c+d x)} \cos ^{\frac{3}{2}}(c+d x) \, dx\)

Optimal. Leaf size=114 \[ -\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{3 \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d \sqrt{1-\cos (c+d x)}}-\frac{3 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 d} \]

[Out]

(-3*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/(4*d) + (3*Sqrt[Cos[c + d*x]]*Sin[c + d
*x])/(4*d*Sqrt[1 - Cos[c + d*x]]) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[1 - Cos[c + d*x]])

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Rubi [A]  time = 0.153723, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2770, 2775, 207} \[ -\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{3 \sin (c+d x) \sqrt{\cos (c+d x)}}{4 d \sqrt{1-\cos (c+d x)}}-\frac{3 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2),x]

[Out]

(-3*ArcTanh[Sin[c + d*x]/(Sqrt[1 - Cos[c + d*x]]*Sqrt[Cos[c + d*x]])])/(4*d) + (3*Sqrt[Cos[c + d*x]]*Sin[c + d
*x])/(4*d*Sqrt[1 - Cos[c + d*x]]) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[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)} \cos ^{\frac{3}{2}}(c+d x) \, dx &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}-\frac{3}{4} \int \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)} \, dx\\ &=\frac{3 \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{3}{8} \int \frac{\sqrt{1-\cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx\\ &=\frac{3 \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}+\frac{3 \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 )}{4 d}\\ &=-\frac{3 \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{4 d}+\frac{3 \sqrt{\cos (c+d x)} \sin (c+d x)}{4 d \sqrt{1-\cos (c+d x)}}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt{1-\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.477127, size = 284, normalized size = 2.49 \[ -\frac{\sqrt{-(\cos (c+d x)-1) \cos (c+d x)} \left (2 \sqrt{2} \left (\cos \left (\frac{3}{2} (c+d x)\right )-2 \cos \left (\frac{1}{2} (c+d x)\right )\right ) \csc \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x) (\cos (d x)+i \sin (d x))}+3 \sqrt{\cos (c)-i \sin (c)} \left (\cot \left (\frac{1}{2} (c+d x)\right )+i\right ) \tanh ^{-1}\left (\frac{e^{i d x}}{\sqrt{\cos (c)-i \sin (c)} \sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}\right )+3 \sqrt{\cos (c)-i \sin (c)} \left (\cot \left (\frac{1}{2} (c+d x)\right )+i\right ) \tanh ^{-1}\left (\frac{\sqrt{e^{2 i d x} (\cos (c)+i \sin (c))-i \sin (c)+\cos (c)}}{\sqrt{\cos (c)-i \sin (c)}}\right )\right )}{8 d \sqrt{i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[1 - Cos[c + d*x]]*Cos[c + d*x]^(3/2),x]

[Out]

-(Sqrt[-((-1 + Cos[c + d*x])*Cos[c + d*x])]*(3*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]] + 3*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]*(-2*Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2])*Csc[(c + d*x)/2]*Sqrt[Cos[c + d*x]*(Cos[
d*x] + I*Sin[d*x])]))/(8*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 = 0.338, size = 164, normalized size = 1.4 \begin{align*}{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) }{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( 2\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) -3\,\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+3\,{\it Artanh} \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}} \right ) \right ) \sqrt{2-2\,\cos \left ( dx+c \right ) } \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

1/8/d*2^(1/2)*(-1+cos(d*x+c))*(2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2-(cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*cos(d*x+c)-3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+3*arctanh((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))*(2-2*cos(d*x+
c))^(1/2)*cos(d*x+c)^(3/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)/sin(d*x+c)^3

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Maxima [B]  time = 2.00639, size = 1762, normalized size = 15.46 \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)^(3/2),x, algorithm="maxima")

[Out]

1/32*(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(2*d*x + 2*c) - 2)*cos(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(2*d*x + 2*c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) + cos(2*d*x + 2*c) - 2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - (cos(1/2*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c) - 2)*sin(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c))) - sin(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 3*log(s
qrt(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(si
n(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) - 3*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) + 3*log(((cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2
+ (cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*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(
1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + si
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) +
1) - 3*log(((cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c)))^2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + (cos(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*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*(co
s(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))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos
(2*d*x + 2*c) + 1))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))) + 1))/d

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Fricas [A]  time = 2.1249, size = 332, normalized size = 2.91 \begin{align*} -\frac{2 \,{\left (2 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) - 3\right )} \sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )} - 3 \, \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 )}{8 \, 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)^(3/2),x, algorithm="fricas")

[Out]

-1/8*(2*(2*cos(d*x + c)^2 - cos(d*x + c) - 3)*sqrt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)) - 3*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(3/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)^(3/2),x, algorithm="giac")

[Out]

Timed out