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

Optimal. Leaf size=83 \[ \frac{2 \sin (c+d x)}{d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{2} \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0925313, antiderivative size = 83, 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 = {2779, 2782, 206} \[ \frac{2 \sin (c+d x)}{d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sin (c+d x)}{\sqrt{2} \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2779

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/
(2*b*(n + 1)*(c^2 - d^2)), Int[((c + d*Sin[e + f*x])^(n + 1)*Simp[a*d - 2*b*c*(n + 1) + b*d*(2*n + 3)*Sin[e +
f*x], x])/Sqrt[a + b*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] && LtQ[n, -1] && IntegerQ[2*n]

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]

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])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{1-\cos (c+d x)} \cos ^{\frac{3}{2}}(c+d x)} \, dx &=\frac{2 \sin (c+d x)}{d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}+\int \frac{1}{\sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}} \, dx\\ &=\frac{2 \sin (c+d x)}{d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}-\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{\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}+\frac{2 \sin (c+d x)}{d \sqrt{1-\cos (c+d x)} \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.138378, size = 152, normalized size = 1.83 \[ \frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) \left (2 \sqrt{1+e^{2 i (c+d x)}} \cos \left (\frac{1}{2} (c+d x)\right )-\frac{e^{-\frac{1}{2} i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )}{\sqrt{2}}\right )}{d \sqrt{1+e^{2 i (c+d x)}} \sqrt{-(\cos (c+d x)-1) \cos (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.305, size = 159, normalized size = 1.9 \begin{align*}{\frac{\sqrt{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}-1 \right ) } \left ( \cos \left ( dx+c \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}}{2}{\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \right ) \sqrt{2}+ \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}}{2}{\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}}}} \right ) \sqrt{2}-2\,\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{2-2\,\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [C]  time = 1.9527, size = 540, normalized size = 6.51 \begin{align*} -\frac{\sqrt{2}{\left (2 \, \sqrt{2} \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 ) + 2 \,{\left (\sqrt{2} \cos \left (d x + c\right ) + \sqrt{2}\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 ) -{\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}} \log \left (\frac{4 \,{\left ({\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |}^{2} + 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 (\sqrt{2}{\left | i \, e^{\left (i \, d x + i \, c\right )} - i \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 ) - 2 \, \sqrt{2} \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 )\right )}{\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}} + 4\right )}}{{\left | i \, e^{\left (i \, d x + i \, c\right )} - i \right |}^{2}}\right )\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}} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*(2*sqrt(2)*sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*(sqrt(2)*cos
(d*x + c) + sqrt(2))*cos(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)*log(4*(abs(I*e^(I*d*x + I*c) - I)^2 + 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)*(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*(sqrt(2)*abs(I*e^(I*d*x + I*c) - I)*sin(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 2*sqrt(2)*cos(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) + 4)/abs(I*e^(I*d*x + I*c) - I)^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)*d)

________________________________________________________________________________________

Fricas [A]  time = 2.2223, size = 396, normalized size = 4.77 \begin{align*} \frac{\sqrt{2} \cos \left (d x + c\right ) \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 ) \sin \left (d x + c\right ) + 4 \,{\left (\cos \left (d x + c\right ) + 1\right )} \sqrt{-\cos \left (d x + c\right ) + 1} \sqrt{\cos \left (d x + c\right )}}{2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*cos(d*x + c)*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)))*sin(d*x + c) + 4*(cos(d*x + c) + 1)*sq
rt(-cos(d*x + c) + 1)*sqrt(cos(d*x + c)))/(d*cos(d*x + c)*sin(d*x + c))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{1 - \cos{\left (c + d x \right )}} \cos ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))**(1/2)/cos(d*x+c)**(3/2),x)

[Out]

Integral(1/(sqrt(1 - cos(c + d*x))*cos(c + d*x)**(3/2)), x)

________________________________________________________________________________________

Giac [A]  time = 1.94227, size = 97, normalized size = 1.17 \begin{align*} \frac{\sqrt{2}{\left (\frac{4}{\sqrt{-\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}} - \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*(4/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