3.7.92 \(\int (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx\) [692]

3.7.92.1 Optimal result
3.7.92.2 Mathematica [A] (verified)
3.7.92.3 Rubi [A] (verified)
3.7.92.4 Maple [A] (verified)
3.7.92.5 Fricas [C] (verification not implemented)
3.7.92.6 Sympy [F]
3.7.92.7 Maxima [F]
3.7.92.8 Giac [F]
3.7.92.9 Mupad [F(-1)]

3.7.92.1 Optimal result

Integrand size = 21, antiderivative size = 97 \[ \int (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=-\frac {2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

output
2*A*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2*A*(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)*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
 
3.7.92.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.73 \[ \int (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {2 \sqrt {\sec (c+d x)} \left (-A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+A \sin (c+d x)\right )}{d} \]

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

Time = 0.59 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.01, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3717

\(\displaystyle \int \sqrt {\sec (c+d x)} (A \sec (c+d x)+B)dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

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

3.7.92.3.1 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]
 
3.7.92.4 Maple [A] (verified)

Time = 5.72 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55

method result size
default \(\frac {4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(150\)
parts \(-\frac {2 A \left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 B \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(317\)

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

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

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.28 \[ \int (A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \, dx=\frac {-i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {2} A {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + i \, \sqrt {2} A {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, A \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{d} \]

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2),x, algorithm="fricas")
 
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)) - I* 
sqrt(2)*A*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + 
 I*sin(d*x + c))) + I*sqrt(2)*A*weierstrassZeta(-4, 0, weierstrassPInverse 
(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*A*sin(d*x + c)/sqrt(cos(d*x + 
c)))/d
 
3.7.92.6 Sympy [F]

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

input
integrate((A+B*cos(d*x+c))*sec(d*x+c)**(3/2),x)
 
output
Integral((A + B*cos(c + d*x))*sec(c + d*x)**(3/2), x)
 
3.7.92.7 Maxima [F]

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

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

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

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

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

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