\(\int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\) [144]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right )}{a^{5/2} d}-\frac {43 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {\sin (c+d x)}{4 d (a+a \cos (c+d x))^{5/2}}-\frac {11 \sin (c+d x)}{16 a d (a+a \cos (c+d x))^{3/2}} \] Output:

2*arctanh(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))/a^(5/2)/d-43/32*arcta 
nh(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/a^(5/2)/ 
d-1/4*sin(d*x+c)/d/(a+a*cos(d*x+c))^(5/2)-11/16*sin(d*x+c)/a/d/(a+a*cos(d* 
x+c))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.58 (sec) , antiderivative size = 603, normalized size of antiderivative = 4.19 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {4 \sqrt {2} \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {i e^{i c} x}{-1+e^{i c}}-\frac {\log \left (i-\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )}{d}\right )}{(a (1+\cos (c+d x)))^{5/2}}+\frac {4 \sqrt {2} \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {i e^{i c} x}{-1+e^{i c}}+\frac {\log \left (i+\sqrt {2} e^{\frac {1}{2} i (c+d x)}-i e^{i (c+d x)}\right )}{d}\right )}{(a (1+\cos (c+d x)))^{5/2}}+\frac {43 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )}{4 d (a (1+\cos (c+d x)))^{5/2}}-\frac {43 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )}{4 d (a (1+\cos (c+d x)))^{5/2}}-\frac {\cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d (a (1+\cos (c+d x)))^{5/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^4}-\frac {11 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d (a (1+\cos (c+d x)))^{5/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )-\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^2}+\frac {\cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d (a (1+\cos (c+d x)))^{5/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^4}+\frac {11 \cos ^5\left (\frac {c}{2}+\frac {d x}{2}\right )}{8 d (a (1+\cos (c+d x)))^{5/2} \left (\cos \left (\frac {c}{4}+\frac {d x}{4}\right )+\sin \left (\frac {c}{4}+\frac {d x}{4}\right )\right )^2} \] Input:

Integrate[Sec[c + d*x]/(a + a*Cos[c + d*x])^(5/2),x]
 

Output:

(4*Sqrt[2]*Cos[c/2 + (d*x)/2]^5*((I*E^(I*c)*x)/(-1 + E^(I*c)) - Log[I - Sq 
rt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x))]/d))/(a*(1 + Cos[c + d*x]))^ 
(5/2) + (4*Sqrt[2]*Cos[c/2 + (d*x)/2]^5*(((-I)*E^(I*c)*x)/(-1 + E^(I*c)) + 
 Log[I + Sqrt[2]*E^((I/2)*(c + d*x)) - I*E^(I*(c + d*x))]/d))/(a*(1 + Cos[ 
c + d*x]))^(5/2) + (43*Cos[c/2 + (d*x)/2]^5*Log[Cos[c/4 + (d*x)/4] - Sin[c 
/4 + (d*x)/4]])/(4*d*(a*(1 + Cos[c + d*x]))^(5/2)) - (43*Cos[c/2 + (d*x)/2 
]^5*Log[Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4]])/(4*d*(a*(1 + Cos[c + d*x 
]))^(5/2)) - Cos[c/2 + (d*x)/2]^5/(8*d*(a*(1 + Cos[c + d*x]))^(5/2)*(Cos[c 
/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])^4) - (11*Cos[c/2 + (d*x)/2]^5)/(8*d*(a 
*(1 + Cos[c + d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] - Sin[c/4 + (d*x)/4])^2) + 
Cos[c/2 + (d*x)/2]^5/(8*d*(a*(1 + Cos[c + d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] 
 + Sin[c/4 + (d*x)/4])^4) + (11*Cos[c/2 + (d*x)/2]^5)/(8*d*(a*(1 + Cos[c + 
 d*x]))^(5/2)*(Cos[c/4 + (d*x)/4] + Sin[c/4 + (d*x)/4])^2)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3245, 27, 3042, 3457, 27, 3042, 3464, 3042, 3128, 219, 3252, 219}

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.

\(\displaystyle \int \frac {\sec (c+d x)}{(a \cos (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 3245

\(\displaystyle \frac {\int \frac {(8 a-3 a \cos (c+d x)) \sec (c+d x)}{2 (\cos (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(8 a-3 a \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {8 a-3 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {\left (32 a^2-11 a^2 \cos (c+d x)\right ) \sec (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (32 a^2-11 a^2 \cos (c+d x)\right ) \sec (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {32 a^2-11 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {32 a \int \sqrt {\cos (c+d x) a+a} \sec (c+d x)dx-43 a^2 \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {32 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-43 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {86 a^2 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+32 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {32 a \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {-\frac {64 a^2 \int \frac {1}{a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {43 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {11 a \sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}}{8 a^2}-\frac {\sin (c+d x)}{4 d (a \cos (c+d x)+a)^{5/2}}\)

Input:

Int[Sec[c + d*x]/(a + a*Cos[c + d*x])^(5/2),x]
 

Output:

-1/4*Sin[c + d*x]/(d*(a + a*Cos[c + d*x])^(5/2)) + (((64*a^(3/2)*ArcTanh[( 
Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d - (43*Sqrt[2]*a^(3/2)*A 
rcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d)/(4*a 
^2) - (11*a*Sin[c + d*x])/(2*d*(a + a*Cos[c + d*x])^(3/2)))/(8*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 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

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 3252

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b*c + a*d - d*x^2), 
x], x, b*(Cos[e + f*x]/Sqrt[a + b*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 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 3464

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(324\) vs. \(2(119)=238\).

Time = 0.95 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.26

method	result	size
default	\(-\frac {\sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (43 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-32 \ln \left (-\frac {4 \left (a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}-2 a \right )}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {2}}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -32 \ln \left (\frac {4 a \sqrt {2}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+4 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}+8 a}{2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {2}}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +11 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \sqrt {2}\, \sqrt {a}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+2 \sqrt {2}\, \sqrt {a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {a}\right )}{32 a^{\frac {7}{2}} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, d}\)	\(325\)

Input:

int(sec(d*x+c)/(a+a*cos(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/32/a^(7/2)/cos(1/2*d*x+1/2*c)^3*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(43*2^(1 
/2)*ln(2*(2*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c) 
)*a*cos(1/2*d*x+1/2*c)^4-32*ln(-4/(2*cos(1/2*d*x+1/2*c)-2^(1/2))*(a*2^(1/2 
)*cos(1/2*d*x+1/2*c)-2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-2*a))* 
cos(1/2*d*x+1/2*c)^4*a-32*ln(4/(2*cos(1/2*d*x+1/2*c)+2^(1/2))*(a*2^(1/2)*c 
os(1/2*d*x+1/2*c)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)+2*a))*cos 
(1/2*d*x+1/2*c)^4*a+11*cos(1/2*d*x+1/2*c)^2*2^(1/2)*a^(1/2)*(a*sin(1/2*d*x 
+1/2*c)^2)^(1/2)+2*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/sin(1/2 
*d*x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (119) = 238\).

Time = 0.10 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.07 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {43 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 32 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (11 \, \cos \left (d x + c\right ) + 15\right )} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:

integrate(sec(d*x+c)/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

1/64*(43*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)* 
sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a 
)*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) 
+ 1)) + 32*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(a 
)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*sqrt(a*cos(d*x + c) + a)* 
sqrt(a)*(cos(d*x + c) - 2)*sin(d*x + c) + 8*a)/(cos(d*x + c)^3 + cos(d*x + 
 c)^2)) - 4*sqrt(a*cos(d*x + c) + a)*(11*cos(d*x + c) + 15)*sin(d*x + c))/ 
(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^ 
3*d)
 

Sympy [F]

\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(sec(d*x+c)/(a+a*cos(d*x+c))**(5/2),x)
 

Output:

Integral(sec(c + d*x)/(a*(cos(c + d*x) + 1))**(5/2), x)
 

Maxima [F]

\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (a \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(sec(d*x+c)/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate(sec(d*x + c)/(a*cos(d*x + c) + a)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(sec(d*x+c)/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \] Input:

int(1/(cos(c + d*x)*(a + a*cos(c + d*x))^(5/2)),x)
 

Output:

int(1/(cos(c + d*x)*(a + a*cos(c + d*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

int(sec(d*x+c)/(a+a*cos(d*x+c))^(5/2),x)
 

Output:

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