\(\int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx\) [388]

Optimal result

Integrand size = 25, antiderivative size = 197 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {63 \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{64 \sqrt {2} a^{7/2} d}-\frac {\sin (c+d x)}{6 d (a+a \cos (c+d x))^{7/2} \sqrt {\sec (c+d x)}}-\frac {5 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{5/2} \sqrt {\sec (c+d x)}}-\frac {103 \sin (c+d x)}{192 a^2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \] Output:

63/128*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x 
+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/a^(7/2)/d-1/6*sin(d* 
x+c)/d/(a+a*cos(d*x+c))^(7/2)/sec(d*x+c)^(1/2)-5/16*sin(d*x+c)/a/d/(a+a*co 
s(d*x+c))^(5/2)/sec(d*x+c)^(1/2)-103/192*sin(d*x+c)/a^2/d/(a+a*cos(d*x+c)) 
^(3/2)/sec(d*x+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 4.10 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {\sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-2 (493+532 \cos (c+d x)+103 \cos (2 (c+d x))) \sqrt {2-2 \sec (c+d x)}+6048 \text {arctanh}\left (\sqrt {-\sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{3072 \sqrt {2} a^3 d \sqrt {a (1+\cos (c+d x))} \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}} \] Input:

Integrate[Sqrt[Sec[c + d*x]]/(a + a*Cos[c + d*x])^(7/2),x]
 

Output:

(Sec[(c + d*x)/2]^4*(-2*(493 + 532*Cos[c + d*x] + 103*Cos[2*(c + d*x)])*Sq 
rt[2 - 2*Sec[c + d*x]] + 6048*ArcTanh[Sqrt[-(Sec[c + d*x]*Sin[(c + d*x)/2] 
^2)]]*Cos[(c + d*x)/2]^6*Sec[c + d*x])*Tan[(c + d*x)/2])/(3072*Sqrt[2]*a^3 
*d*Sqrt[a*(1 + Cos[c + d*x])]*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3042, 4710, 3042, 3245, 27, 3042, 3457, 27, 3042, 3457, 27, 3042, 3261, 218}

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.

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)}}{(a \cos (c+d x)+a)^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{7/2}}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{7/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{7/2}}dx\)

\(\Big \downarrow \) 3245

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {11 a-4 a \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{6 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {11 a-4 a \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{5/2}}dx}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {11 a-4 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}dx}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 3457

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {73 a^2-30 a^2 \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {73 a^2-30 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {73 a^2-30 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 3457

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {\int \frac {189 a^3}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {103 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {189}{4} a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx-\frac {103 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {189}{4} a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {103 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 3261

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {-\frac {189 a^2 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{2 d}-\frac {103 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {189 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{2 \sqrt {2} d}-\frac {103 a^2 \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {15 a \sin (c+d x) \sqrt {\cos (c+d x)}}{4 d (a \cos (c+d x)+a)^{5/2}}}{12 a^2}-\frac {\sin (c+d x) \sqrt {\cos (c+d x)}}{6 d (a \cos (c+d x)+a)^{7/2}}\right )\)

Input:

Int[Sqrt[Sec[c + d*x]]/(a + a*Cos[c + d*x])^(7/2),x]
 

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/6*(Sqrt[Cos[c + d*x]]*Sin[c + d* 
x])/(d*(a + a*Cos[c + d*x])^(7/2)) + ((-15*a*Sqrt[Cos[c + d*x]]*Sin[c + d* 
x])/(4*d*(a + a*Cos[c + d*x])^(5/2)) + ((189*Sqrt[a]*ArcTan[(Sqrt[a]*Sin[c 
 + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(2*Sqrt[2 
]*d) - (103*a^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x]) 
^(3/2)))/(8*a^2))/(12*a^2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3245

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( 
a*(2*m + 1)*(b*c - a*d))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + 
f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (Intege 
rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3261

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]
 

rule 3457

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 7.56 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.98

method	result	size
default	\(-\frac {\left (\sin \left (d x +c \right ) \left (103 \cos \left (d x +c \right )^{2}+266 \cos \left (d x +c \right )+195\right ) \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+\left (189 \cos \left (d x +c \right )^{3}+567 \cos \left (d x +c \right )^{2}+567 \cos \left (d x +c \right )+189\right ) \arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )\right ) \sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \cos \left (d x +c \right ) \sqrt {\sec \left (d x +c \right )}}{384 d \left (\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, a^{4}}\)	\(194\)

Input:

int(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-1/384/d*(sin(d*x+c)*(103*cos(d*x+c)^2+266*cos(d*x+c)+195)*2^(1/2)*(cos(d* 
x+c)/(cos(d*x+c)+1))^(1/2)+(189*cos(d*x+c)^3+567*cos(d*x+c)^2+567*cos(d*x+ 
c)+189)*arcsin(cot(d*x+c)-csc(d*x+c)))*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*co 
s(d*x+c)*sec(d*x+c)^(1/2)/(cos(d*x+c)^4+4*cos(d*x+c)^3+6*cos(d*x+c)^2+4*co 
s(d*x+c)+1)/(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/a^4
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=-\frac {189 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (103 \, \cos \left (d x + c\right )^{3} + 266 \, \cos \left (d x + c\right )^{2} + 195 \, \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{384 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

integrate(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

-1/384*(189*sqrt(2)*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 
+ 4*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt 
(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*(103*cos(d*x + c)^3 + 266*cos(d 
*x + c)^2 + 195*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(c 
os(d*x + c)))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos 
(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\int { \frac {\sqrt {\sec \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(sec(d*x + c))/(a*cos(d*x + c) + a)^(7/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{7/2}} \,d x \] Input:

int((1/cos(c + d*x))^(1/2)/(a + a*cos(c + d*x))^(7/2),x)
 

Output:

int((1/cos(c + d*x))^(1/2)/(a + a*cos(c + d*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{7/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\cos \left (d x +c \right )+1}}{\cos \left (d x +c \right )^{4}+4 \cos \left (d x +c \right )^{3}+6 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+1}d x \right )}{a^{4}} \] Input:

int(sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(7/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x)**4 
+ 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1),x))/a**4