3.9.26 \(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [826]

3.9.26.1 Optimal result
3.9.26.2 Mathematica [A] (verified)
3.9.26.3 Rubi [A] (verified)
3.9.26.4 Maple [B] (verified)
3.9.26.5 Fricas [C] (verification not implemented)
3.9.26.6 Sympy [F]
3.9.26.7 Maxima [F]
3.9.26.8 Giac [F]
3.9.26.9 Mupad [F(-1)]

3.9.26.1 Optimal result

Integrand size = 31, antiderivative size = 168 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {6 A \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)}}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^2 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {6 A \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \]

output
2/5*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+2/3*b*B*sin(d*x+c)/d/(b*cos(d* 
x+c))^(3/2)+6/5*A*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)+2/3*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)-6/5*A*(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) 
/b/d/cos(d*x+c)^(1/2)
 
3.9.26.2 Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 \left (-9 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+9 A \sin (c+d x)+5 B \tan (c+d x)+3 A \sec (c+d x) \tan (c+d x)\right )}{15 d \sqrt {b \cos (c+d x)}} \]

input
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/Sqrt[b*Cos[c + d*x]],x]
 
output
(2*(-9*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 5*B*Sqrt[Cos[c + d 
*x]]*EllipticF[(c + d*x)/2, 2] + 9*A*Sin[c + d*x] + 5*B*Tan[c + d*x] + 3*A 
*Sec[c + d*x]*Tan[c + d*x]))/(15*d*Sqrt[b*Cos[c + d*x]])
 
3.9.26.3 Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 2030, 3227, 3042, 3116, 3042, 3116, 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.

\(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \cos (c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 2030

\(\displaystyle b^3 \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{7/2}}dx\)

\(\Big \downarrow \) 3227

\(\displaystyle b^3 \left (A \int \frac {1}{(b \cos (c+d x))^{7/2}}dx+\frac {B \int \frac {1}{(b \cos (c+d x))^{5/2}}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^3 \left (A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx+\frac {B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{b}\right )\)

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

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

3.9.26.3.1 Defintions of rubi rules used

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 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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]
 
3.9.26.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(578\) vs. \(2(196)=392\).

Time = 8.49 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.45

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 )}\, \left (72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-36 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 ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 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 ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+36 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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 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 )+10 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{15 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(579\)
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 )}\, \left (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \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 ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{5 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 B \left (-2 \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 ) \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 )}}{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 )}\, \left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\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}\) \(607\)

input
int((A+B*cos(d*x+c))*sec(d*x+c)^3/(cos(d*x+c)*b)^(1/2),x,method=_RETURNVER 
BOSE)
 
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/sin(1/ 
2*d*x+1/2*c)^3/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d 
*x+1/2*c)^2-1)*(72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-36*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(cos(1/2*d* 
x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-20*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))*sin( 
1/2*d*x+1/2*c)^4-72*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+36*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(cos(1/2*d 
*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-20*B*cos(1/2*d*x+1/2*c)*sin(1/2*d* 
x+1/2*c)^4+20*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))*sin(1/2*d*x+1/2*c)^2+24*A*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-9*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(cos(1/2*d*x+1/2*c),2^(1/2))+10*B*cos(1 
/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)))*(-2*sin( 
1/2*d*x+1/2*c)^4*b+b*sin(1/2*d*x+1/2*c)^2)^(1/2)/((2*cos(1/2*d*x+1/2*c)^2- 
1)*b)^(1/2)/d
 
3.9.26.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.22 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {-5 i \, \sqrt {2} B \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 i \, \sqrt {2} A \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 i \, \sqrt {2} A \sqrt {b} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (9 \, A \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right ) + 3 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, b d \cos \left (d x + c\right )^{3}} \]

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(b*cos(d*x+c))^(1/2),x, algorithm= 
"fricas")
 
output
1/15*(-5*I*sqrt(2)*B*sqrt(b)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*B*sqrt(b)*cos(d*x + c)^3*weierst 
rassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*I*sqrt(2)*A*sqrt(b) 
*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c))) + 9*I*sqrt(2)*A*sqrt(b)*cos(d*x + c)^3*weierstrass 
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2 
*(9*A*cos(d*x + c)^2 + 5*B*cos(d*x + c) + 3*A)*sqrt(b*cos(d*x + c))*sin(d* 
x + c))/(b*d*cos(d*x + c)^3)
 
3.9.26.6 Sympy [F]

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

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(b*cos(d*x+c))**(1/2),x)
 
output
Integral((A + B*cos(c + d*x))*sec(c + d*x)**3/sqrt(b*cos(c + d*x)), x)
 
3.9.26.7 Maxima [F]

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

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(b*cos(d*x+c))^(1/2),x, algorithm= 
"maxima")
 
output
integrate((B*cos(d*x + c) + A)*sec(d*x + c)^3/sqrt(b*cos(d*x + c)), x)
 
3.9.26.8 Giac [F]

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

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

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

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