\(\int \cos ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1182]

Optimal result

Integrand size = 31, antiderivative size = 65 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 (A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

2*B*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(A+3*C)*InverseJacobiAM(1/ 
2*d*x+1/2*c,2^(1/2))/d+2/3*A*cos(d*x+c)^(1/2)*sin(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 4.58 (sec) , antiderivative size = 295, normalized size of antiderivative = 4.54 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-6 B \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sec (c) \sin (d x+\arctan (\tan (c)))+\left (9 B \cos (c-d x-\arctan (\tan (c))) \csc (c) \sec (c)+3 B \cos (c+d x+\arctan (\tan (c))) \csc (c) \sec (c)-12 B \cos (c+d x) \cot (c) \sqrt {\sec ^2(c)}-4 (A+3 C) \cos (c+d x) \sqrt {\cos ^2(d x-\arctan (\cot (c)))} \sqrt {\csc ^2(c)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sqrt {\sec ^2(c)} \sec (d x-\arctan (\cot (c))) \sin (c)+4 A \cos (c+d x) \sqrt {\sec ^2(c)} \sin (c+d x)\right ) \sqrt {\sin ^2(d x+\arctan (\tan (c)))}\right )}{3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sqrt {\sec ^2(c)} \sqrt {\sin ^2(d x+\arctan (\tan (c)))}} \] Input:

Integrate[Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

(Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-6*B*Hypergeo 
metricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sec[c]*Sin[d*x 
 + ArcTan[Tan[c]]] + (9*B*Cos[c - d*x - ArcTan[Tan[c]]]*Csc[c]*Sec[c] + 3* 
B*Cos[c + d*x + ArcTan[Tan[c]]]*Csc[c]*Sec[c] - 12*B*Cos[c + d*x]*Cot[c]*S 
qrt[Sec[c]^2] - 4*(A + 3*C)*Cos[c + d*x]*Sqrt[Cos[d*x - ArcTan[Cot[c]]]^2] 
*Sqrt[Csc[c]^2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[ 
c]]]^2]*Sqrt[Sec[c]^2]*Sec[d*x - ArcTan[Cot[c]]]*Sin[c] + 4*A*Cos[c + d*x] 
*Sqrt[Sec[c]^2]*Sin[c + d*x])*Sqrt[Sin[d*x + ArcTan[Tan[c]]]^2]))/(3*d*(A 
+ 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sqrt[Sec[c]^2]*Sqrt[Sin[d*x 
 + ArcTan[Tan[c]]]^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4552, 3042, 3502, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4552

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \int \frac {A+3 C+3 B \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

Int[Cos[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

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

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 3502

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

rule 4552

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*( 
x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b^2   Int[(b*Cos 
[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ 
[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(273\) vs. \(2(62)=124\).

Time = 3.62 (sec) , antiderivative size = 274, normalized size of antiderivative = 4.22

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A +A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(274\)

Input:

int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBO 
SE)
 

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*A*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*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)*Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))-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))+3*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(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 = 0.08 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.08 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, A \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \] Input:

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F(-1)]

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

integrate(cos(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="m 
axima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2), x)
 

Giac [F]

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

integrate(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="g 
iac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(3/2), x)
 

Mupad [B] (verification not implemented)

Time = 13.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int \cos ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,B\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d} \] Input:

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

Output:

(2*A*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*B*ellipticE(c/2 + (d*x)/2, 2) 
)/d + (2*C*ellipticF(c/2 + (d*x)/2, 2))/d + (2*A*cos(c + d*x)^(1/2)*sin(c 
+ d*x))/(3*d)
 

Reduce [F]

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

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

Output:

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