\(\int (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx\) [691]

Optimal result

Integrand size = 21, antiderivative size = 123 \[ \int (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=-\frac {2 B \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 B \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 A \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \] Output:

-2*B*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+2/3*A*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*B*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2/3*A*sec(d*x+c)^(3/2)*sin(d 
*x+c)/d
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.69 \[ \int (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {3}{2}}(c+d x) \left (-6 B \cos ^{\frac {3}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 A \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (A+3 B \cos (c+d x)) \sin (c+d x)\right )}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 3717, 3042, 4274, 3042, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

\(\displaystyle \int \sec ^{\frac {3}{2}}(c+d x) (A \sec (c+d x)+B)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

\(\displaystyle A \int \sec ^{\frac {5}{2}}(c+d x)dx+B \int \sec ^{\frac {3}{2}}(c+d x)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

B*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d 
+ (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) + A*((2*Sqrt[Cos[c + d*x]]*Ellipt 
icF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sec[c + d*x]^(3/2)*Sin[ 
c + d*x])/(3*d))
 

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 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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. \(396\) vs. \(2(110)=220\).

Time = 32.92 (sec) , antiderivative size = 397, normalized size of antiderivative = 3.23

method	result	size
default	\(\frac {2 \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(397\)
parts	\(-\frac {2 A \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}-\frac {2 B \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(397\)

Input:

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

Output:

2/3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(4*sin(1/2*d 
*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(2*A*(2*sin(1/2 
*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+ 
1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-12*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1 
/2*c)^4+6*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+2*A*cos(1/2*d*x 
+1/2*c)*sin(1/2*d*x+1/2*c)^2-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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*B*cos(1/2*d*x+1 
/2*c)*sin(1/2*d*x+1/2*c)^2-3*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)))*(-2*sin(1/2*d*x+ 
1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(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 = 0.09 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.36 \[ \int (A+B \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x) \, dx=\frac {-i \, \sqrt {2} A \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} A \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} B \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, B \cos \left (d x + c\right ) + A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, d \cos \left (d x + c\right )} \] Input:

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

Output:

1/3*(-I*sqrt(2)*A*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c)) + I*sqrt(2)*A*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d 
*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*B*cos(d*x + c)*weierstrassZeta(-4, 
 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt( 
2)*B*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(d*x + c))) + 2*(3*B*cos(d*x + c) + A)*sin(d*x + c)/sqrt(cos 
(d*x + c)))/(d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

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

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

Output:

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