Integrand size = 26, antiderivative size = 173 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {2 \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}+\frac {26 \sin (c+d x)}{15 d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]
-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)+2/5*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a-a*cos(d*x+c) )^(1/2)+2/15*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a-a*cos(d*x+c))^(1/2)+26/15*si n(d*x+c)/d/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.75 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {e^{-\frac {5}{2} i (c+d x)} \left (2 \sqrt {1+e^{2 i (c+d x)}} \left (13+15 e^{i (c+d x)}+40 e^{2 i (c+d x)}+40 e^{3 i (c+d x)}+15 e^{4 i (c+d x)}+13 e^{5 i (c+d x)}\right )-15 \sqrt {2} \left (1+e^{2 i (c+d x)}\right )^3 \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{60 d \sqrt {1+e^{2 i (c+d x)}} \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \]
((2*Sqrt[1 + E^((2*I)*(c + d*x))]*(13 + 15*E^(I*(c + d*x)) + 40*E^((2*I)*( c + d*x)) + 40*E^((3*I)*(c + d*x)) + 15*E^((4*I)*(c + d*x)) + 13*E^((5*I)* (c + d*x))) - 15*Sqrt[2]*(1 + E^((2*I)*(c + d*x)))^3*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])*Sin[(c + d*x)/2])/(60*d *E^(((5*I)/2)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c + d*x]^(5/2)* Sqrt[a - a*Cos[c + d*x]])
Time = 0.88 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3463, 27, 3042, 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.
| \(\displaystyle \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3258 |
| \(\displaystyle \frac {\int \frac {4 \cos (c+d x) a+a}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}dx}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {4 \sin \left (c+d x+\frac {\pi }{2}\right ) a+a}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}-\frac {2 \int -\frac {2 \cos (c+d x) a^2+13 a^2}{2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}dx}{3 a}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {2 \cos (c+d x) a^2+13 a^2}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}dx}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {2 \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+13 a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3463 |
| \(\displaystyle \frac {\frac {\frac {26 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {2 \int -\frac {15 a^3}{2 \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx}{a}}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {15 a^2 \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx+\frac {26 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {26 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {\frac {\frac {26 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {30 a^3 \int \frac {1}{2 a^2-\frac {a^3 \sin (c+d x) \tan (c+d x)}{a-a \cos (c+d x)}}d\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{d}}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {26 a^2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {15 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}}{3 a}+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}}{5 a}+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}}\) |
(2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a - a*Cos[c + d*x]]) + ((2*a *Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a - a*Cos[c + d*x]]) + ((-15*S qrt[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 + (26*a^2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x] ]*Sqrt[a - a*Cos[c + d*x]]))/(3*a))/(5*a)
3.3.81.3.1 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/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-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] - Simp[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]
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_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])
Time = 6.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(-\frac {\left (15 \,\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-13 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-\sqrt {2}\, \cos \left (d x +c \right )-3 \sqrt {2}\right ) \sin \left (d x +c \right ) \sqrt {2}}{15 d \sqrt {-a \left (\cos \left (d x +c \right )-1\right )}\, \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(120\) |
-1/15/d*(15*arctanh(1/2*2^(1/2)/(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)-13*2^(1/2)*cos(d*x+c)^2-2^(1/2)*co s(d*x+c)-3*2^(1/2))*sin(d*x+c)/(-a*(cos(d*x+c)-1))^(1/2)/cos(d*x+c)^(5/2)* 2^(1/2)
Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {15 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{3} \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, {\left (13 \, \cos \left (d x + c\right )^{3} + 14 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 3\right )} \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{30 \, a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )} \]
1/30*(15*sqrt(2)*sqrt(a)*cos(d*x + c)^3*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) + 4*(13*c os(d*x + c)^3 + 14*cos(d*x + c)^2 + 4*cos(d*x + c) + 3)*sqrt(-a*cos(d*x + c) + a)*sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^3*sin(d*x + c))
Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 0.44 (sec) , antiderivative size = 692, normalized size of antiderivative = 4.00 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\text {Too large to display} \]
-1/15*(15*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqr t(2)*cos(2*d*x + 2*c) + sqrt(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)*arctan2(2*sqrt(2)*(cos(2*d*x + 2*c)^2 + si n(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))/(sqrt(a)*abs(e^(I*d*x + I*c) - 1)), 2*(sqr t(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)*sqrt(a)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqr t(-a)*abs(e^(I*d*x + I*c) - 1) + 2*sqrt(a))/(a*abs(e^(I*d*x + I*c) - 1))) - 26*(cos(2*d*x + 2*c)^2*sin(d*x + c) + sin(2*d*x + 2*c)^2*sin(d*x + c) + 2*cos(2*d*x + 2*c)*sin(d*x + c) + sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 24*cos(5/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c) + 1))*sin(d*x + c) - 24*(cos(d*x + c) + 1)*sin(5/2*arctan2(si n(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 2*((13*cos(d*x + c) + 15)*cos(2*d *x + 2*c)^2 + (13*cos(d*x + c) + 15)*sin(2*d*x + 2*c)^2 + 2*(13*cos(d*x + c) + 15)*cos(2*d*x + 2*c) + 13*cos(d*x + c) + 15)*sin(1/2*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c) + 1)) - 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*(7*cos(3/2*arctan2(sin(2*d*x + 2*c), cos (2*d*x + 2*c) + 1))*sin(d*x + c) - (7*cos(d*x + c) + 5)*sin(3/2*arctan2(si n(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)^(5/4)*sqrt(-a)*d)
Time = 0.79 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.53 \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} {\left (\frac {4 \, {\left ({\left ({\left ({\left ({\left (17 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 165\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 650\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 650\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 165\right )} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - 17\right )}}{{\left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1\right )}^{\frac {5}{2}}} - 15 \, \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right ) + 15 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right ) + 15 \, \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )\right )}}{30 \, \sqrt {a} d \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \]
-1/30*sqrt(2)*(4*(((((17*tan(1/4*d*x + 1/4*c)^2 - 165)*tan(1/4*d*x + 1/4*c )^2 + 650)*tan(1/4*d*x + 1/4*c)^2 - 650)*tan(1/4*d*x + 1/4*c)^2 + 165)*tan (1/4*d*x + 1/4*c)^2 - 17)/(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c) ^2 + 1)^(5/2) - 15*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) + 15*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)) + 15 *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)*d*sgn(sin(1/2*d*x + 1/2*c)))
Timed out. \[ \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {a-a\,\cos \left (c+d\,x\right )}} \,d x \]