\(\int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\) [819]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 31, antiderivative size = 176 \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=-\frac {6 A b^2 \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 b^3 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^5 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/2}}+\frac {2 b^4 B \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}}+\frac {6 A b^3 \sin (c+d x)}{5 d \sqrt {b \cos (c+d x)}} \] Output:

-6/5*A*b^2*(b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/co 
s(d*x+c)^(1/2)+2/3*b^3*B*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^ 
(1/2))/d/(b*cos(d*x+c))^(1/2)+2/5*A*b^5*sin(d*x+c)/d/(b*cos(d*x+c))^(5/2)+ 
2/3*b^4*B*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)+6/5*A*b^3*sin(d*x+c)/d/(b*cos( 
d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.58 \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=-\frac {2 b^4 \left (9 A \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-5 B \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-5 B \sin (c+d x)-\frac {9}{2} A \sin (2 (c+d x))-3 A \tan (c+d x)\right )}{15 d (b \cos (c+d x))^{3/2}} \] Input:

Integrate[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]
 

Output:

(-2*b^4*(9*A*Cos[c + d*x]^(3/2)*EllipticE[(c + d*x)/2, 2] - 5*B*Cos[c + d* 
x]^(3/2)*EllipticF[(c + d*x)/2, 2] - 5*B*Sin[c + d*x] - (9*A*Sin[2*(c + d* 
x)])/2 - 3*A*Tan[c + d*x]))/(15*d*(b*Cos[c + d*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.07, 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.

\(\displaystyle \int \sec ^6(c+d x) (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^6 \int \frac {A+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{7/2}}dx\)

\(\Big \downarrow \) 3227

\(\displaystyle b^6 \left (A \int \frac {1}{(b \cos (c+d x))^{7/2}}dx+\frac {B \int \frac {1}{(b \cos (c+d x))^{5/2}}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (A \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx+\frac {B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx}{b}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^6 \left (A \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \cos (c+d x)}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (A \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^6 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b^6 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^6 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b^6 \left (\frac {B \left (\frac {\sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}+A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b^6 \left (A \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )+\frac {B \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )}{b}\right )\)

Input:

Int[(b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]
 

Output:

b^6*((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)))
 

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(578\) vs. \(2(155)=310\).

Time = 2354.18 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.29

method result size
default \(-\frac {2 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{2} \left (72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-36 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-72 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+36 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-20 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-9 A \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 )+10 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-5 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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\) \(579\)
parts \(-\frac {2 A \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{2} \left (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \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 \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 )\right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b}}{5 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}-\frac {2 B \left (-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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\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 )\right ) b^{3} \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) b}\, d}\) \(610\)

Input:

int((cos(d*x+c)*b)^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,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^2/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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d* 
x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-20*B*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*si 
n(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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d 
*x+1/2*c)^2-1)^(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)*EllipticF(cos(1/2*d*x+1/2*c 
),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(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*si 
n(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*s 
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*b* 
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2*b)^(1/2)/((2*cos(1/2*d*x+1/2*c)^ 
2-1)*b)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.20 \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=-\frac {2 \, {\left (5 i \, \sqrt {\frac {1}{2}} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {\frac {1}{2}} B b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {\frac {1}{2}} A b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {\frac {1}{2}} A b^{\frac {5}{2}} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (9 \, A b^{2} \cos \left (d x + c\right )^{2} + 5 \, B b^{2} \cos \left (d x + c\right ) + 3 \, A b^{2}\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm= 
"fricas")
 

Output:

-2/15*(5*I*sqrt(1/2)*B*b^(5/2)*cos(d*x + c)^3*weierstrassPInverse(-4, 0, c 
os(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(1/2)*B*b^(5/2)*cos(d*x + c)^3*wei 
erstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*I*sqrt(1/2)*A*b 
^(5/2)*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co 
s(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(1/2)*A*b^(5/2)*cos(d*x + c)^3*wei 
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
c))) - (9*A*b^2*cos(d*x + c)^2 + 5*B*b^2*cos(d*x + c) + 3*A*b^2)*sqrt(b*co 
s(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)**6,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{6} \,d x } \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm= 
"maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^6, x)
 

Giac [F]

\[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sec \left (d x + c\right )^{6} \,d x } \] Input:

integrate((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm= 
"giac")
 

Output:

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(5/2)*sec(d*x + c)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^6} \,d x \] Input:

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^6,x)
                                                                                    
                                                                                    
 

Output:

int(((b*cos(c + d*x))^(5/2)*(A + B*cos(c + d*x)))/cos(c + d*x)^6, x)
 

Reduce [F]

\[ \int (b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\sqrt {b}\, b^{2} \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{6}d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{6}d x \right ) a \right ) \] Input:

int((b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c))*sec(d*x+c)^6,x)
 

Output:

sqrt(b)*b**2*(int(sqrt(cos(c + d*x))*cos(c + d*x)**3*sec(c + d*x)**6,x)*b 
+ int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**6,x)*a)