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

Optimal. Leaf size=141 \[ \frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a-a \cos (c+d x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.295414, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2778, 2982, 2782, 208, 2775, 207} \[ \frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a-a \cos (c+d x)}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2778

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-2*d*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]]), x] - Dist[1/(b*(2*n
- 1)), Int[((c + d*Sin[e + f*x])^(n - 2)*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*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] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e
, f, A, B}, 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]

Rule 208

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

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 \frac{\cos ^{\frac{3}{2}}(c+d x)}{\sqrt{a-a \cos (c+d x)}} \, dx &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a-a \cos (c+d x)}}+\frac{\int \frac{a+a \cos (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}} \, dx}{2 a}\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a-a \cos (c+d x)}}-\frac{\int \frac{\sqrt{a-a \cos (c+d x)}}{\sqrt{\cos (c+d x)}} \, dx}{2 a}+\int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}} \, dx\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a-a \cos (c+d x)}}-\frac{\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{(2 a) \operatorname{Subst}\left (\int \frac{1}{2 a^2-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{\tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{\sqrt{a} d}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a-a \cos (c+d x)}}\right )}{\sqrt{a} d}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{d \sqrt{a-a \cos (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.869351, size = 228, normalized size = 1.62 \[ -\frac{i e^{-i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \sqrt{\cos (c+d x)} \left (\sqrt{2} e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-4 e^{i (c+d x)} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+\sqrt{2} \left (\sqrt{1+e^{2 i (c+d x)}} \left (1+e^{i (c+d x)}\right )+e^{i (c+d x)} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )\right )}{2 \sqrt{2} d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a-a \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*(-1 + E^(I*(c + d*x)))*(Sqrt[2]*E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))] - 4*E^(I*(c + d*x))*ArcTanh[(
1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + Sqrt[2]*((1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*
I)*(c + d*x))] + E^(I*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))*Sqrt[Cos[c + d*x]])/(Sqrt[2]*d*E^(I*
(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a - a*Cos[c + d*x]])

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Maple [A]  time = 0.364, size = 162, normalized size = 1.2 \begin{align*}{\frac{\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{d \left ( \sin \left ( dx+c \right ) \right ) ^{3}} \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}\cos \left ( dx+c \right ) -{\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}+{\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 ) }}} \right ) \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-2\,a \left ( -1+\cos \left ( dx+c \right ) \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(d*x + c)^(3/2)/sqrt(-a*cos(d*x + c) + a), x)

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Fricas [A]  time = 2.00281, size = 593, normalized size = 4.21 \begin{align*} \frac{\sqrt{2} \sqrt{a} \log \left (-\frac{\frac{2 \, \sqrt{2} \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 )}}{\sqrt{a}} -{\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 ) + \sqrt{a} \log \left (-\frac{2 \, \sqrt{-a \cos \left (d x + c\right ) + a} \sqrt{a}{\left (\cos \left (d x + c\right ) + 1\right )} \sqrt{\cos \left (d x + c\right )} +{\left (2 \, 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 ) + 2 \, \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 )}}{2 \, a 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)^(3/2)/(a-a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 2.19494, size = 171, normalized size = 1.21 \begin{align*} -\frac{\sqrt{2}{\left (\frac{\sqrt{2} \arctan \left (\frac{\sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{2 \, \arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} - \frac{2 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )} a}\right )}{\left | a \right |}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a) - 2*arctan
(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a) - 2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/((a*tan(1/
2*d*x + 1/2*c)^2 + a)*a))*abs(a)/d