3.4.52 \(\int \frac {(a+a \cos (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [352]

3.4.52.1 Optimal result
3.4.52.2 Mathematica [A] (verified)
3.4.52.3 Rubi [A] (verified)
3.4.52.4 Maple [A] (verified)
3.4.52.5 Fricas [A] (verification not implemented)
3.4.52.6 Sympy [F]
3.4.52.7 Maxima [B] (verification not implemented)
3.4.52.8 Giac [F]
3.4.52.9 Mupad [F(-1)]

3.4.52.1 Optimal result

Integrand size = 25, antiderivative size = 180 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {11 a^{3/2} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{8 d}+\frac {a^2 \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {5}{2}}(c+d x)}+\frac {11 a^2 \sin (c+d x)}{12 d \sqrt {a+a \cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x)}+\frac {11 a^2 \sin (c+d x)}{8 d \sqrt {a+a \cos (c+d x)} \sqrt {\sec (c+d x)}} \]

output
1/3*a^2*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+11/12*a^2*sin 
(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+11/8*a^2*sin(d*x+c)/d/(a 
+a*cos(d*x+c))^(1/2)/sec(d*x+c)^(1/2)+11/8*a^(3/2)*arcsin(sin(d*x+c)*a^(1/ 
2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d
 
3.4.52.2 Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.70 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a \sqrt {\cos (c+d x)} \sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (33 \sqrt {2} \arcsin \left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )+2 \sqrt {\cos (c+d x)} \left (26 \sin \left (\frac {1}{2} (c+d x)\right )+9 \sin \left (\frac {3}{2} (c+d x)\right )+2 \sin \left (\frac {5}{2} (c+d x)\right )\right )\right )}{48 d} \]

input
Integrate[(a + a*Cos[c + d*x])^(3/2)/Sec[c + d*x]^(3/2),x]
 
output
(a*Sqrt[Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*Sqrt[Sec 
[c + d*x]]*(33*Sqrt[2]*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]] + 2*Sqrt[Cos[c + d 
*x]]*(26*Sin[(c + d*x)/2] + 9*Sin[(3*(c + d*x))/2] + 2*Sin[(5*(c + d*x))/2 
])))/(48*d)
 
3.4.52.3 Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4710, 3042, 3242, 27, 2011, 3042, 3249, 3042, 3249, 3042, 3253, 223}

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.

\(\displaystyle \int \frac {(a \cos (c+d x)+a)^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\)

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)^{3/2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx\)

\(\Big \downarrow \) 3242

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \int \frac {11 \cos ^{\frac {3}{2}}(c+d x) \left (\cos (c+d x) a^2+a^2\right )}{2 \sqrt {\cos (c+d x) a+a}}dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\cos (c+d x) a^2+a^2\right )}{\sqrt {\cos (c+d x) a+a}}dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 2011

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \int \cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3249

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \left (\frac {3}{4} \int \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \left (\frac {3}{4} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3249

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\cos (c+d x) a+a}}{\sqrt {\cos (c+d x)}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {11}{6} a \left (\frac {3}{4} \left (\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a^2 \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{3 d \sqrt {a \cos (c+d x)+a}}+\frac {11}{6} a \left (\frac {3}{4} \left (\frac {\sqrt {a} \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a \cos (c+d x)+a}}\right )+\frac {a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{2 d \sqrt {a \cos (c+d x)+a}}\right )\right )\)

input
Int[(a + a*Cos[c + d*x])^(3/2)/Sec[c + d*x]^(3/2),x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((a^2*Cos[c + d*x]^(5/2)*Sin[c + d*x 
])/(3*d*Sqrt[a + a*Cos[c + d*x]]) + (11*a*((a*Cos[c + d*x]^(3/2)*Sin[c + d 
*x])/(2*d*Sqrt[a + a*Cos[c + d*x]]) + (3*((Sqrt[a]*ArcSin[(Sqrt[a]*Sin[c + 
 d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/ 
(d*Sqrt[a + a*Cos[c + d*x]])))/4))/6)
 

3.4.52.3.1 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 223
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 2011
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3242
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 - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m 
+ n))   Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* 
(m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*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] && GtQ[m, 1] &&  !LtQ[ 
n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ 
c, 0]))
 

rule 3249
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 
2*n + 1)))   Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), 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, 0] && IntegerQ[2*n]
 

rule 3253
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

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]
 
3.4.52.4 Maple [A] (verified)

Time = 14.72 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.99

method result size
default \(\frac {\sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, \left (8 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+22 \sin \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+33 \tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+33 \sec \left (d x +c \right ) \arctan \left (\tan \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) a}{24 d \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) \(179\)

input
int((a+cos(d*x+c)*a)^(3/2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 
output
1/24/d*(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(3/2)/(cos(d*x+c 
)/(1+cos(d*x+c)))^(1/2)*(8*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c) 
))^(1/2)+22*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+33*tan(d*x+c)*(co 
s(d*x+c)/(1+cos(d*x+c)))^(1/2)+33*sec(d*x+c)*arctan(tan(d*x+c)*(cos(d*x+c) 
/(1+cos(d*x+c)))^(1/2)))*a
 
3.4.52.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {33 \, {\left (a \cos \left (d x + c\right ) + a\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \frac {{\left (8 \, a \cos \left (d x + c\right )^{3} + 22 \, a \cos \left (d x + c\right )^{2} + 33 \, a \cos \left (d x + c\right )\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]

input
integrate((a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="fricas")
 
output
-1/24*(33*(a*cos(d*x + c) + a)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqr 
t(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - (8*a*cos(d*x + c)^3 + 22*a*cos(d 
*x + c)^2 + 33*a*cos(d*x + c))*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt( 
cos(d*x + c)))/(d*cos(d*x + c) + d)
 
3.4.52.6 Sympy [F]

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

input
integrate((a+a*cos(d*x+c))**(3/2)/sec(d*x+c)**(3/2),x)
 
output
Integral((a*(cos(c + d*x) + 1))**(3/2)/sec(c + d*x)**(3/2), x)
 
3.4.52.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1942 vs. \(2 (150) = 300\).

Time = 0.53 (sec) , antiderivative size = 1942, normalized size of antiderivative = 10.79 \[ \int \frac {(a+a \cos (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

input
integrate((a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="maxima")
 
output
1/96*(4*(a*cos(3/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3 
*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1))*sin(3*d* 
x + 3*c) - (a*cos(3*d*x + 3*c) - a)*sin(3/2*arctan2(sin(2/3*arctan2(sin(3* 
d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d* 
x + 3*c))) + 1)))*(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 
+ sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arcta 
n2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(3/4)*sqrt(a) + 6*(cos(2/3*ar 
ctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 
 3*c), cos(3*d*x + 3*c)))^2 + 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d* 
x + 3*c))) + 1)^(1/4)*((3*a*sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 
3*c))) + 11*a*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))*cos(1/ 
2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3*ar 
ctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)) - (3*a*cos(2/3*arctan2(si 
n(3*d*x + 3*c), cos(3*d*x + 3*c))) + 5*a*cos(1/3*arctan2(sin(3*d*x + 3*c), 
 cos(3*d*x + 3*c))) - 8*a)*sin(1/2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c 
), cos(3*d*x + 3*c))), cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)) 
) + 1)))*sqrt(a) + 33*(a*arctan2(-(cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3 
*d*x + 3*c)))^2 + sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c)))^2 + 
 2*cos(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))) + 1)^(1/4)*(cos(1/ 
2*arctan2(sin(2/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))), cos(2/3...
 
3.4.52.8 Giac [F]

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

input
integrate((a+a*cos(d*x+c))^(3/2)/sec(d*x+c)^(3/2),x, algorithm="giac")
 
output
integrate((a*cos(d*x + c) + a)^(3/2)/sec(d*x + c)^(3/2), x)
 
3.4.52.9 Mupad [F(-1)]

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

input
int((a + a*cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(3/2),x)
 
output
int((a + a*cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(3/2), x)