3.9.13 \(\int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx\) [813]

3.9.13.1 Optimal result
3.9.13.2 Mathematica [A] (verified)
3.9.13.3 Rubi [A] (verified)
3.9.13.4 Maple [A] (verified)
3.9.13.5 Fricas [C] (verification not implemented)
3.9.13.6 Sympy [F(-1)]
3.9.13.7 Maxima [F]
3.9.13.8 Giac [F]
3.9.13.9 Mupad [B] (verification not implemented)

3.9.13.1 Optimal result

Integrand size = 23, antiderivative size = 126 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {2 b \left (3 a^2-b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 a \left (a^2+9 b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {2 b \left (a^2-3 b^2\right ) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)}}+\frac {2 a^2 \sqrt {\cos (c+d x)} (a+b \sec (c+d x)) \sin (c+d x)}{3 d} \]

output
2*b*(3*a^2-b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE( 
sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*a*(a^2+9*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/ 
2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d-2/3*b*(a^2-3 
*b^2)*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*a^2*(a+b*sec(d*x+c))*sin(d*x+c)*co 
s(d*x+c)^(1/2)/d
 
3.9.13.2 Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.69 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {2 \left (\left (9 a^2 b-3 b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (a^3+9 a b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (3 b^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{3 d} \]

input
Integrate[Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^3,x]
 
output
(2*((9*a^2*b - 3*b^3)*EllipticE[(c + d*x)/2, 2] + (a^3 + 9*a*b^2)*Elliptic 
F[(c + d*x)/2, 2] + ((3*b^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/Sqrt[Cos[c + 
 d*x]]))/(3*d)
 
3.9.13.3 Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.49, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {3042, 4752, 3042, 4328, 27, 3042, 4535, 3042, 4258, 3042, 3120, 4534, 3042, 4258, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4752

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4328

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{3} \int \frac {8 b a^2+\left (a^2+9 b^2\right ) \sec (c+d x) a-b \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \int \frac {8 b a^2+\left (a^2+9 b^2\right ) \sec (c+d x) a-b \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \int \frac {8 b a^2+\left (a^2+9 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a-b \left (a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 4535

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (\int \frac {8 a^2 b-b \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)}}dx+a \left (a^2+9 b^2\right ) \int \sqrt {\sec (c+d x)}dx\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (a \left (a^2+9 b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\int \frac {8 a^2 b-b \left (a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (\int \frac {8 a^2 b-b \left (a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (\int \frac {8 a^2 b-b \left (a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (\int \frac {8 a^2 b-b \left (a^2-3 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 4534

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (3 b \left (3 a^2-b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-\frac {2 b \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (3 b \left (3 a^2-b^2\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 b \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (3 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\frac {2 b \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (3 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{3} \left (-\frac {2 b \left (a^2-3 b^2\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {2 a \left (a^2+9 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 b \left (3 a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a^2 \sin (c+d x) (a+b \sec (c+d x))}{3 d \sqrt {\sec (c+d x)}}\right )\)

input
Int[Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^3,x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a^2*(a + b*Sec[c + d*x])*Sin[c + 
 d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((6*b*(3*a^2 - b^2)*Sqrt[Cos[c + d*x]]*E 
llipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a*(a^2 + 9*b^2)*Sqrt[C 
os[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (2*b*(a^2 - 
 3*b^2)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d)/3)
 

3.9.13.3.1 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 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4328
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[a^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* 
((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m 
 - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2* 
(n + 1))*Csc[e + f*x] - b*(b^2*n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], 
 x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && ((Int 
egerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
 

rule 4534
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

rule 4535
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4752
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 
3.9.13.4 Maple [A] (verified)

Time = 8.75 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.40

method result size
default \(-\frac {2 \left (4 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{3}-6 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{3}+a^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 a \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 a^{2} b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{3}\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(303\)

input
int(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
 
output
-2/3*(4*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*cos(1/2*d*x+1/2*c)*s 
in(1/2*d*x+1/2*c)^2*a^3-6*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^3+a^3* 
(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co 
s(1/2*d*x+1/2*c),2^(1/2))+9*a*b^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* 
d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*a^2*b*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2* 
d*x+1/2*c),2^(1/2))+3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2 
-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^3)/sin(1/2*d*x+1/2*c)/(2 
*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 
3.9.13.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.70 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {\sqrt {2} {\left (-i \, a^{3} - 9 i \, a b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, a^{3} + 9 i \, a b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-3 i \, a^{2} b + i \, b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, a^{2} b - i \, b^{3}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (a^{3} \cos \left (d x + c\right ) + 3 \, b^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \]

input
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3,x, algorithm="fricas")
 
output
1/3*(sqrt(2)*(-I*a^3 - 9*I*a*b^2)*cos(d*x + c)*weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*a^3 + 9*I*a*b^2)*cos(d*x + c)* 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-3* 
I*a^2*b + I*b^3)*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(- 
4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(3*I*a^2*b - I*b^3)*cos( 
d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c))) + 2*(a^3*cos(d*x + c) + 3*b^3)*sqrt(cos(d*x + c))*sin(d*x 
 + c))/(d*cos(d*x + c))
 
3.9.13.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**(3/2)*(a+b*sec(d*x+c))**3,x)
 
output
Timed out
 
3.9.13.7 Maxima [F]

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

input
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3,x, algorithm="maxima")
 
output
integrate((b*sec(d*x + c) + a)^3*cos(d*x + c)^(3/2), x)
 
3.9.13.8 Giac [F]

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

input
integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^3,x, algorithm="giac")
 
output
integrate((b*sec(d*x + c) + a)^3*cos(d*x + c)^(3/2), x)
 
3.9.13.9 Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.98 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {2\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {6\,a^2\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {6\,a\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,a^3\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}+\frac {2\,b^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int(cos(c + d*x)^(3/2)*(a + b/cos(c + d*x))^3,x)
 
output
(2*a^3*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (6*a^2*b*ellipticE(c/2 + (d*x) 
/2, 2))/d + (6*a*b^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*a^3*cos(c + d*x)^ 
(1/2)*sin(c + d*x))/(3*d) + (2*b^3*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4 
, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))