3.6.70 \(\int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [570]

3.6.70.1 Optimal result

Integrand size = 33, antiderivative size = 121 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 \left (2 a A b+a^2 B-b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (a^2 A+3 A b^2+6 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}+\frac {2 a^2 A \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d} \]

output

2*(2*A*a*b+B*a^2-B*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El 
lipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(A*a^2+3*A*b^2+6*B*a*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))/d+2*b^2*B*sin(d*x+c)/d/cos(d*x+c)^(1/2)+2/3*a^2*A*sin(d*x+c)*cos(d*x+c 
)^(1/2)/d
 

3.6.70.2 Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.84 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 \left (3 \left (2 a A b+a^2 B-b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (a^2 A+3 A b^2+6 a b B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {\left (3 b^2 B+a^2 A \cos (c+d x)\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{3 d} \]

input

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

output

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

3.6.70.3 Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3433, 3042, 3467, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {2}{3} \int \frac {A a^2+6 b B a+3 A b^2+3 \left (B a^2+2 A b a-b^2 B\right ) \cos (c+d x)}{2 \sqrt {\cos (c+d x)}}dx+\frac {2 a^2 A \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}+\frac {2 b^2 B \sin (c+d x)}{d \sqrt {\cos (c+d x)}}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

input

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

output

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

3.6.70.3.1 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 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 3467

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f 
_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ 
(B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 
*(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2))   Int[(c + d* 
Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c 
- 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* 
a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 
1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ 
n, -1]
 

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]
 

3.6.70.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(403\) vs. \(2(165)=330\).

Time = 10.36 (sec) , antiderivative size = 404, normalized size of antiderivative = 3.34

method	result	size
default	\(-\frac {2 \left (4 A \,a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}+A \,a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \,b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -6 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}+6 B a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+3 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}\right )}{3 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(404\)

input

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

output

-2/3*(4*A*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-2*A*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*a^2+A*a^2*(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))+3*A*b^2*(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))-6*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))*a*b-6*B*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*b^2+6*B*a*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))-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+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))*b^2)/sin(1/2*d*x+1/2*c 
)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

3.6.70.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.98 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\sqrt {2} {\left (-i \, A a^{2} - 6 i \, B a b - 3 i \, A b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (i \, A a^{2} + 6 i \, B a b + 3 i \, A b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, B a^{2} - 2 i \, A a b + i \, B b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (i \, B a^{2} + 2 i \, A a b - i \, B b^{2}\right )} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (A a^{2} \cos \left (d x + c\right ) + 3 \, B b^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )} \]

input

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

output

1/3*(sqrt(2)*(-I*A*a^2 - 6*I*B*a*b - 3*I*A*b^2)*cos(d*x + c)*weierstrassPI 
nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(I*A*a^2 + 6*I*B*a* 
b + 3*I*A*b^2)*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*si 
n(d*x + c)) - 3*sqrt(2)*(-I*B*a^2 - 2*I*A*a*b + I*B*b^2)*cos(d*x + c)*weie 
rstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c 
))) - 3*sqrt(2)*(I*B*a^2 + 2*I*A*a*b - I*B*b^2)*cos(d*x + c)*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*( 
A*a^2*cos(d*x + c) + 3*B*b^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x 
+ c))
 

3.6.70.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.6.70.7 Maxima [F]

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

input

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

output

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

3.6.70.8 Giac [F]

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

input

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

output

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

3.6.70.9 Mupad [B] (verification not implemented)

Time = 16.65 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.31 \[ \int \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {A\,a^2\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input

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

output

(A*a^2*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2 
, 2))/3))/d + (2*B*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (2*A*b^2*ellipticF 
(c/2 + (d*x)/2, 2))/d + (4*A*a*b*ellipticE(c/2 + (d*x)/2, 2))/d + (4*B*a*b 
*ellipticF(c/2 + (d*x)/2, 2))/d + (2*B*b^2*sin(c + d*x)*hypergeom([-1/4, 1 
/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))