3.13.62 \(\int (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\) [1262]

3.13.62.1 Optimal result
3.13.62.2 Mathematica [A] (verified)
3.13.62.3 Rubi [A] (verified)
3.13.62.4 Maple [A] (verified)
3.13.62.5 Fricas [C] (verification not implemented)
3.13.62.6 Sympy [F(-1)]
3.13.62.7 Maxima [F]
3.13.62.8 Giac [F]
3.13.62.9 Mupad [F(-1)]

3.13.62.1 Optimal result

Integrand size = 31, antiderivative size = 101 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

output
2*A*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*(A-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*B*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt 
icF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d
 
3.13.62.2 Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 \sqrt {\sec (c+d x)} \left (-\left ((A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )+B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+A \sin (c+d x)\right )}{d} \]

input
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(3/2),x]
 
output
(2*Sqrt[Sec[c + d*x]]*(-((A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 
 2]) + B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + A*Sin[c + d*x]))/d
 
3.13.62.3 Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4709, 3042, 3500, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

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

3.13.62.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 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 3500
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 
 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* 
(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A 
*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 
B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
 

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]
 
3.13.62.4 Maple [A] (verified)

Time = 3.47 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.93

method result size
default \(\frac {4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A -2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) \(195\)
parts \(-\frac {2 A \left (-2 \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 )}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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}\, \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 )}\, 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 )+\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 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 )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}{\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 )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{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 )}{\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}\) \(450\)

input
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x,method=_RETURNVERBO 
SE)
 
output
2*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-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)) 
-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)*EllipticF 
(cos(1/2*d*x+1/2*c),2^(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)*EllipticE(cos(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
 
3.13.62.5 Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.33 \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {-i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (i \, A - i \, C\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, A \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{d} \]

input
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="f 
ricas")
 
output
(-I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 
I*sqrt(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + sq 
rt(2)*(-I*A + I*C)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c))) + sqrt(2)*(I*A - I*C)*weierstrassZeta(-4, 0, we 
ierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*A*sin(d*x + c 
)/sqrt(cos(d*x + c)))/d
 
3.13.62.6 Sympy [F(-1)]

Timed out. \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\text {Timed out} \]

input
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(3/2),x)
 
output
Timed out
 
3.13.62.7 Maxima [F]

\[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

input
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="m 
axima")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2), x)
 
3.13.62.8 Giac [F]

\[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

input
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(3/2),x, algorithm="g 
iac")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(3/2), x)
 
3.13.62.9 Mupad [F(-1)]

Timed out. \[ \int \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]

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