\(\int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx\) [283]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {1-\cos (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2857, 3061, 2861, 212, 2854, 213} \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {1-\cos (c+d x)}} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2854

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 2857

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)/Sqrt[a + b*Sin[e + f*x]])*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], 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 2861

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 3061

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {1-\cos (c+d x)}}+\frac {1}{2} \int \frac {1+\cos (c+d x)}{\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+\int \frac {1}{\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 {\text {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 {2 \text {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 {\text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {1-\cos (c+d x)} \sqrt {\cos (c+d x)}}\right )}{d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sin (c+d x)}{\sqrt {2} \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] (verified)

Result contains complex when optimal does not.

Time = 0.37 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=-\frac {i e^{-i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \left (\sqrt {2} e^{i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-4 e^{i (c+d x)} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {2} \left (\left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}+e^{i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )\right ) \sqrt {\cos (c+d x)}}{2 \sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {1-\cos (c+d x)}} \]

[In]

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

[Out]

((-1/2*I)*(-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[1 - Cos[c + d*x]])

Maple [A] (verified)

Time = 13.24 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {\sin \left (d x +c \right ) \left (\cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-2 \cos \left (d x +c \right )+2}}\)	\(159\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (103) = 206\).

Time = 0.26 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.91 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\sqrt {2} \log \left (-\frac {2 \, {\left (\sqrt {2} \cos \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - \log \left (\frac {2 \, {\left (\sqrt {-\cos \left (d x + c\right ) + 1} \sqrt {\cos \left (d x + c\right )} - \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right )}{2 \, d \sin \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\int \frac {\cos ^{\frac {3}{2}}{\left (c + d x \right )}}{\sqrt {1 - \cos {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {3}{2}}}{\sqrt {-\cos \left (d x + c\right ) + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 560 vs. \(2 (103) = 206\).

Time = 0.94 (sec) , antiderivative size = 560, normalized size of antiderivative = 4.75 \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\frac {\frac {\sqrt {2} \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right )}{\mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right )}{\mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )}{\mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {2 \, \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 2 \, \sqrt {2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right )}{\mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 2 \, \sqrt {2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} - 1 \right |}\right )}{\mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {8 \, \sqrt {2} {\left (3 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1}\right )}^{3} - 7 \, {\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1}\right )}^{2} + \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 11\right )}}{{\left ({\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1}\right )}^{2} + 2 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 2 \, \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} - 7\right )}^{2} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{2 \, d} \]

[In]

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

[Out]

1/2*(sqrt(2)*log(tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1)/sgn
(sin(1/2*d*x + 1/2*c)) - sqrt(2)*log(abs(-tan(1/4*d*x + 1/4*c)^2 + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x
+ 1/4*c)^2 + 1) + 3))/sgn(sin(1/2*d*x + 1/2*c)) - sqrt(2)*log(abs(-tan(1/4*d*x + 1/4*c)^2 + sqrt(tan(1/4*d*x
+ 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1))/sgn(sin(1/2*d*x + 1/2*c)) - 2*log(tan(1/4*d*x + 1/4*c)^2 + 2*
sqrt(2) - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1)/sgn(sin(1/2*d*x + 1/2*c)) + 2*log(a
bs(-tan(1/4*d*x + 1/4*c)^2 + 2*sqrt(2) + sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) - 1))/sgn
(sin(1/2*d*x + 1/2*c)) - 8*sqrt(2)*(3*(tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x +
1/4*c)^2 + 1))^3 - 7*(tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1))^2
+ tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 11)/(((tan(1/4*d*x +
1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1))^2 + 2*tan(1/4*d*x + 1/4*c)^2 - 2*sqrt(
tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) - 7)^2*sgn(sin(1/2*d*x + 1/2*c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^{\frac {3}{2}}(c+d x)}{\sqrt {1-\cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{3/2}}{\sqrt {1-\cos \left (c+d\,x\right )}} \,d x \]

[In]

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

[Out]

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