Integrand size = 26, antiderivative size = 185 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{4 \sqrt {a} d}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {\cos (c+d x)} \sin (c+d x)}{4 d \sqrt {a-a \cos (c+d x)}}+\frac {\cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a-a \cos (c+d x)}} \]
7/4*arctanh(sin(d*x+c)*a^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2))/d/ a^(1/2)-arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos(d *x+c))^(1/2))*2^(1/2)/d/a^(1/2)+1/2*cos(d*x+c)^(3/2)*sin(d*x+c)/d/(a-a*cos (d*x+c))^(1/2)+1/4*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a-a*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 1.11 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=-\frac {i e^{-2 i (c+d x)} \left (-1+e^{i (c+d x)}\right ) \left (7 \sqrt {2} e^{2 i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-16 e^{2 i (c+d x)} \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {2} \left (\sqrt {1+e^{2 i (c+d x)}} \left (1+2 e^{i (c+d x)}+2 e^{2 i (c+d x)}+e^{3 i (c+d x)}\right )+7 e^{2 i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )\right ) \sqrt {\cos (c+d x)}}{8 \sqrt {2} d \sqrt {1+e^{2 i (c+d x)}} \sqrt {a-a \cos (c+d x)}} \]
((-1/8*I)*(-1 + E^(I*(c + d*x)))*(7*Sqrt[2]*E^((2*I)*(c + d*x))*ArcSinh[E^ (I*(c + d*x))] - 16*E^((2*I)*(c + d*x))*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqr t[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + Sqrt[2]*(Sqrt[1 + E^((2*I)*(c + d*x ))]*(1 + 2*E^(I*(c + d*x)) + 2*E^((2*I)*(c + d*x)) + E^((3*I)*(c + d*x))) + 7*E^((2*I)*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))*Sqrt[Cos[ c + d*x]])/(Sqrt[2]*d*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sq rt[a - a*Cos[c + d*x]])
Time = 1.00 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3257, 3042, 3462, 27, 3042, 3461, 3042, 3254, 220, 3261, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3257 |
| \(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x)} (\cos (c+d x) a+3 a)}{\sqrt {a-a \cos (c+d x)}}dx}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+3 a\right )}{\sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle \frac {\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}-\frac {\int -\frac {7 \cos (c+d x) a^2+a^2}{2 \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx}{a}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {7 \cos (c+d x) a^2+a^2}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {7 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3461 |
| \(\displaystyle \frac {\frac {8 a^2 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx-7 a \int \frac {\sqrt {a-a \cos (c+d x)}}{\sqrt {\cos (c+d x)}}dx}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {8 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-7 a \int \frac {\sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3254 |
| \(\displaystyle \frac {\frac {8 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {14 a^2 \int \frac {1}{\frac {a^2 \sin (c+d x) \tan (c+d x)}{a-a \cos (c+d x)}-a}d\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{d}}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 220 |
| \(\displaystyle \frac {\frac {8 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {14 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {\frac {14 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {16 a^3 \int \frac {1}{2 a^2-\frac {a^3 \sin (c+d x) \tan (c+d x)}{a-a \cos (c+d x)}}d\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{d}}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {14 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}}{2 a}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}}{4 a}+\frac {\sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a-a \cos (c+d x)}}\) |
(Cos[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[a - a*Cos[c + d*x]]) + (((14*a ^(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/d - (8*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2] *Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])])/d)/(2*a) + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a - a*Cos[c + d*x]]))/(4*a)
3.3.75.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-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]
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] - Simp[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]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-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*S in[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]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim p[(A*b - a*B)/b Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) , x], x] + Simp[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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Time = 12.35 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.10
| method | result | size |
| default | \(\frac {\sin \left (d x +c \right ) \left (2 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+3 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+7 \sqrt {2}\, \operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )+\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-8 \,\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {2}}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-a \left (\cos \left (d x +c \right )-1\right )}}\) | \(203\) |
1/8/d*sin(d*x+c)*(2*cos(d*x+c)^2*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2) +3*cos(d*x+c)*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+7*2^(1/2)*arctanh( (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)-8*arctanh(1/2*2^(1/2)/(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)/(-a*(cos(d*x+c)-1))^( 1/2)*2^(1/2)
Time = 0.27 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\frac {4 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 7 \, \sqrt {a} \log \left (-\frac {2 \, \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )} + {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{8 \, a d \sin \left (d x + c\right )} \]
1/8*(4*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) + 7*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*c os(d*x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) + 2*sqrt(-a*cos( d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(cos(d*x + c)))/ (a*d*sin(d*x + c))
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {-a \cos \left (d x + c\right ) + a}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 779 vs. \(2 (152) = 304\).
Time = 6.91 (sec) , antiderivative size = 779, normalized size of antiderivative = 4.21 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\text {Too large to display} \]
1/4*(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)/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))) - 2 *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))/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))) - 2 *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))/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))) + 7 *log(1/8*abs(8*tan(1/4*d*x + 1/4*c)^2 - 16*sqrt(2) - 8*sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 8)/(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) )/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))) - 4*sqrt(2)*(17*(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))^7*sqrt( a) - 73*(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))^6*sqrt(a) + 157*(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))^5*sqrt(a) - 597*(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))^4*sqrt(a) + 1603*(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*sqrt(a) - 875*(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*sqrt(a) - 1585*(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))*sqrt(a) + 1737*sqrt(a))/(((tan(1/4*d*x + 1/4*c)^2 -...
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a-a \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{\sqrt {a-a\,\cos \left (c+d\,x\right )}} \,d x \]