\(\int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [564]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 31, antiderivative size = 75 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 (A b+a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 (a A+3 b B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \] Output:

2*(A*b+B*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(A*a+3*B*b)*Invers 
eJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*a*A*cos(d*x+c)^(1/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.89 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 \left (3 (A b+a B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+(a A+3 b B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+a A \sqrt {\cos (c+d x)} \sin (c+d x)\right )}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 3433, 3042, 3447, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {(a \cos (c+d x)+b) (A \cos (c+d x)+B)}{\sqrt {\cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

((6*(A*b + a*B)*EllipticE[(c + d*x)/2, 2])/d + (2*(a*A + 3*b*B)*EllipticF[ 
(c + d*x)/2, 2])/d)/3 + (2*a*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 3433
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* 
(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + 
c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 3447
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
 

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(72)=144\).

Time = 4.08 (sec) , antiderivative size = 326, normalized size of antiderivative = 4.35

method result size
default \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 A a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+A a \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b +3 B b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a \right )}{3 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(326\)

Input:

int(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
 

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*a*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*A*a*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)*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b+3*B*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(co 
s(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))*a)/(-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 = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.08 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 \, A a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + \sqrt {2} {\left (-i \, A a - 3 i \, B b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a + 3 i \, B b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a - i \, A b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a + i \, A b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{3 \, d} \] Input:

integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= 
"fricas")
 

Output:

1/3*(2*A*a*sqrt(cos(d*x + c))*sin(d*x + c) + sqrt(2)*(-I*A*a - 3*I*B*b)*we 
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*A*a + 
 3*I*B*b)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*sq 
rt(2)*(-I*B*a - I*A*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c 
os(d*x + c) + I*sin(d*x + c))) - 3*sqrt(2)*(I*B*a + I*A*b)*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= 
"maxima")
 

Output:

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

Giac [F]

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

integrate(cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= 
"giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.62 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2\,A\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,A\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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