3.13.58 \(\int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx\) [1258]

3.13.58.1 Optimal result
3.13.58.2 Mathematica [A] (verified)
3.13.58.3 Rubi [A] (verified)
3.13.58.4 Maple [B] (verified)
3.13.58.5 Fricas [C] (verification not implemented)
3.13.58.6 Sympy [F]
3.13.58.7 Maxima [F]
3.13.58.8 Giac [F]
3.13.58.9 Mupad [F(-1)]

3.13.58.1 Optimal result

Integrand size = 30, antiderivative size = 127 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\frac {6 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 C \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 B \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

output
2/5*C*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2/3*B*sin(d*x+c)/d/sec(d*x+c)^(1/2)+6/ 
5*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+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)*sec(d*x+c)^(1/2)/d
 
3.13.58.2 Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.69 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\frac {\sqrt {\sec (c+d x)} \left (18 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(5 B+3 C \cos (c+d x)) \sin (2 (c+d x))\right )}{15 d} \]

input
Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Sec[c + d*x]],x]
 
output
(Sqrt[Sec[c + d*x]]*(18*C*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 1 
0*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (5*B + 3*C*Cos[c + d*x] 
)*Sin[2*(c + d*x)]))/(15*d)
 
3.13.58.3 Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 4709, 3042, 3489, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {B \cos (c+d x)+C \cos (c+d x)^2}{\sqrt {\sec (c+d x)}}dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)} \left (C \cos ^2(c+d x)+B \cos (c+d x)\right )dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3489

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

input
Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)/Sqrt[Sec[c + d*x]],x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(B*((2*EllipticF[(c + d*x)/2, 2])/(3 
*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) + C*((6*EllipticE[(c + d* 
x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))
 

3.13.58.3.1 Defintions of rubi rules used

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 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 3489
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + 
(C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b   Int[(b*Sin[e + f* 
x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
 

rule 4709
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 
3.13.58.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(382\) vs. \(2(159)=318\).

Time = 2.31 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.02

method result size
default \(-\frac {2 B \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}-\frac {2 C \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(383\)
parts \(-\frac {2 B \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}-\frac {2 C \sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(383\)

input
int((B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE 
)
 
output
-2/3*B*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*sin(1/2 
*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*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)*EllipticF(c 
os(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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d-2/5*C*((-1+2 
*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-8*cos(1/2*d*x+1/2*c)* 
sin(1/2*d*x+1/2*c)^6+8*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d 
*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(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)/sin(1/2*d*x+1/2*c)/(-1+2*cos(1/2*d* 
x+1/2*c)^2)^(1/2)/d
 
3.13.58.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.14 \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\frac {-5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} C {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} C {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + 5 \, B \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d} \]

input
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="fri 
cas")
 
output
1/15*(-5*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
 c)) + 5*I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + 
 c)) + 9*I*sqrt(2)*C*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(2)*C*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*C*cos(d*x + c) 
^2 + 5*B*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/d
 
3.13.58.6 Sympy [F]

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

input
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)/sec(d*x+c)**(1/2),x)
 
output
Integral((B + C*cos(c + d*x))*cos(c + d*x)/sqrt(sec(c + d*x)), x)
 
3.13.58.7 Maxima [F]

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

input
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="max 
ima")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/sqrt(sec(d*x + c)), x)
 
3.13.58.8 Giac [F]

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

input
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)/sec(d*x+c)^(1/2),x, algorithm="gia 
c")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))/sqrt(sec(d*x + c)), x)
 
3.13.58.9 Mupad [F(-1)]

Timed out. \[ \int \frac {B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

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