\(\int \frac {A+B \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [696]

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

Optimal result

Integrand size = 21, antiderivative size = 151 \[ \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {6 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {10 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 B \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 A \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {10 B \sin (c+d x)}{21 d \sqrt {\sec (c+d x)}} \] Output:

6/5*A*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1 
/2)/d+10/21*B*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec( 
d*x+c)^(1/2)/d+2/7*B*sin(d*x+c)/d/sec(d*x+c)^(5/2)+2/5*A*sin(d*x+c)/d/sec( 
d*x+c)^(3/2)+10/21*B*sin(d*x+c)/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.66 \[ \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\sqrt {\sec (c+d x)} \left (252 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+100 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(65 B+42 A \cos (c+d x)+15 B \cos (2 (c+d x))) \sin (2 (c+d x))\right )}{210 d} \] Input:

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

Output:

(Sqrt[Sec[c + d*x]]*(252*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 
100*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (65*B + 42*A*Cos[c + 
d*x] + 15*B*Cos[2*(c + d*x)])*Sin[2*(c + d*x)]))/(210*d)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 3717, 3042, 4274, 3042, 4256, 3042, 4256, 3042, 4258, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

\(\displaystyle \int \frac {A \sec (c+d x)+B}{\sec ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle A \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx+B \int \frac {1}{\sec ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle B \left (\frac {5}{7} \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+A \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle A \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )+B \left (\frac {2 \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {5}{7} \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )\)

Input:

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

Output:

A*((6*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5* 
d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))) + B*((2*Sin[c + d*x])/(7*d 
*Sec[c + d*x]^(5/2)) + (5*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] 
*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])))/7 
)
 

Defintions of rubi rules used

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 3717
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Csc[e + f*x])^(m - n*p 
)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

rule 4256
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

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 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(289\) vs. \(2(130)=260\).

Time = 18.73 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.92

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (240 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-168 A -360 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (168 A +280 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-42 A -80 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-63 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(290\)
parts \(-\frac {2 A \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}-\frac {2 B \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (48 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-72 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(403\)

Input:

int((A+B*cos(d*x+c))/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*B*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-168*A-360*B)*sin(1/2*d*x+1/2*c)^6*co 
s(1/2*d*x+1/2*c)+(168*A+280*B)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-4 
2*A-80*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-63*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))+25*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+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)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03 \[ \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-25 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 25 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {2} A {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {2} A {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (15 \, B \cos \left (d x + c\right )^{3} + 21 \, A \cos \left (d x + c\right )^{2} + 25 \, B \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d} \] Input:

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

Output:

1/105*(-25*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x 
 + c)) + 25*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c)) + 63*I*sqrt(2)*A*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
 cos(d*x + c) + I*sin(d*x + c))) - 63*I*sqrt(2)*A*weierstrassZeta(-4, 0, w 
eierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(15*B*cos(d* 
x + c)^3 + 21*A*cos(d*x + c)^2 + 25*B*cos(d*x + c))*sin(d*x + c)/sqrt(cos( 
d*x + c)))/d
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x))/sec(c + d*x)**(5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B*cos(c + d*x))/(1/cos(c + d*x))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int((A + B*cos(c + d*x))/(1/cos(c + d*x))^(5/2), x)
 

Reduce [F]

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

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

Output:

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