3.3.49 \(\int (b \cos (c+d x))^{3/2} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [249]

3.3.49.1 Optimal result
3.3.49.2 Mathematica [A] (verified)
3.3.49.3 Rubi [A] (verified)
3.3.49.4 Maple [A] (verified)
3.3.49.5 Fricas [C] (verification not implemented)
3.3.49.6 Sympy [F(-1)]
3.3.49.7 Maxima [F]
3.3.49.8 Giac [F]
3.3.49.9 Mupad [F(-1)]

3.3.49.1 Optimal result

Integrand size = 39, antiderivative size = 146 \[ \int (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 b (5 A+3 C) \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b B \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 C (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 d} \]

output
2/5*C*(b*cos(d*x+c))^(3/2)*sin(d*x+c)/d+2/3*b^2*B*(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))*cos(d*x+c)^( 
1/2)/d/(b*cos(d*x+c))^(1/2)+2/3*b*B*sin(d*x+c)*(b*cos(d*x+c))^(1/2)/d+2/5* 
b*(5*A+3*C)*(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))*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
 
3.3.49.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.65 \[ \int (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {2 b \sqrt {b \cos (c+d x)} \left (3 (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 B+3 C \cos (c+d x)) \sin (c+d x)\right )}{15 d \sqrt {\cos (c+d x)}} \]

input
Integrate[(b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S 
ec[c + d*x],x]
 
output
(2*b*Sqrt[b*Cos[c + d*x]]*(3*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2] + 5*B*E 
llipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*B + 3*C*Cos[c + d*x])*Sin 
[c + d*x]))/(15*d*Sqrt[Cos[c + d*x]])
 
3.3.49.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 2030, 3502, 27, 3042, 3227, 3042, 3115, 3042, 3121, 3042, 3119, 3120}

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.

\(\displaystyle \int \sec (c+d x) (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b \int \sqrt {b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \left (C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A\right )dx\)

\(\Big \downarrow \) 3502

\(\displaystyle b \left (\frac {2 \int \frac {1}{2} \sqrt {b \cos (c+d x)} (b (5 A+3 C)+5 b B \cos (c+d x))dx}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle b \left (\frac {\int \sqrt {b \cos (c+d x)} (b (5 A+3 C)+5 b B \cos (c+d x))dx}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b (5 A+3 C)+5 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3227

\(\displaystyle b \left (\frac {b (5 A+3 C) \int \sqrt {b \cos (c+d x)}dx+5 B \int (b \cos (c+d x))^{3/2}dx}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {b (5 A+3 C) \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle b \left (\frac {b (5 A+3 C) \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 B \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {b (5 A+3 C) \int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 B \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b \left (\frac {\frac {b (5 A+3 C) \sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)}}+5 B \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b \left (\frac {\frac {b (5 A+3 C) \sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)}}+5 B \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b \left (\frac {5 B \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 b (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b \left (\frac {\frac {2 b (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{d \sqrt {\cos (c+d x)}}+5 B \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )}{5 b}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{3/2}}{5 b d}\right )\)

input
Int[(b*Cos[c + d*x])^(3/2)*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + 
 d*x],x]
 
output
b*((2*C*(b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b*d) + ((2*b*(5*A + 3*C)*S 
qrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(d*Sqrt[Cos[c + d*x]]) + 5* 
B*((2*b^2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c 
+ d*x]]) + (2*b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/(5*b))
 

3.3.49.3.1 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 2030
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

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

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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 3121
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 
3.3.49.4 Maple [A] (verified)

Time = 13.84 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.18

method result size
default \(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-20 B -24 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (10 B +6 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+15 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(319\)
parts \(\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 B \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(549\)

input
int((cos(d*x+c)*b)^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x,meth 
od=_RETURNVERBOSE)
 
output
2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(24*cos 
(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^6+(-20*B-24*C)*sin(1/2*d*x+1/2*c)^4*c 
os(1/2*d*x+1/2*c)+(10*B+6*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+15*A* 
(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(co 
s(1/2*d*x+1/2*c),2^(1/2))-5*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*C*(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*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/sin( 
1/2*d*x+1/2*c)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d
 
3.3.49.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.13 \[ \int (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {-5 i \, \sqrt {2} B b^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B b^{\frac {3}{2}} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, C b \cos \left (d x + c\right ) + 5 \, B b\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d} \]

input
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c), 
x, algorithm="fricas")
 
output
1/15*(-5*I*sqrt(2)*B*b^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s 
in(d*x + c)) + 5*I*sqrt(2)*B*b^(3/2)*weierstrassPInverse(-4, 0, cos(d*x + 
c) - I*sin(d*x + c)) + 3*I*sqrt(2)*(5*A + 3*C)*b^(3/2)*weierstrassZeta(-4, 
 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*I*sqrt( 
2)*(5*A + 3*C)*b^(3/2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c 
os(d*x + c) - I*sin(d*x + c))) + 2*(3*C*b*cos(d*x + c) + 5*B*b)*sqrt(b*cos 
(d*x + c))*sin(d*x + c))/d
 
3.3.49.6 Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{3/2} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Timed out} \]

input
integrate((b*cos(d*x+c))**(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c 
),x)
 
output
Timed out
 
3.3.49.7 Maxima [F]

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

input
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c), 
x, algorithm="maxima")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(3/2)*s 
ec(d*x + c), x)
 
3.3.49.8 Giac [F]

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

input
integrate((b*cos(d*x+c))^(3/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c), 
x, algorithm="giac")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^(3/2)*s 
ec(d*x + c), x)
 
3.3.49.9 Mupad [F(-1)]

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

input
int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x),x)
 
output
int(((b*cos(c + d*x))^(3/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c 
 + d*x), x)