\(\int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\) [571]

Optimal result

Integrand size = 33, antiderivative size = 126 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 \left (a^2 A-A b^2-2 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (6 a A b+3 a^2 B+b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b^2 B \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.83 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2 \left (3 \left (a^2 A-A b^2-2 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (6 a A b+3 a^2 B+b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {b (b B+3 (A b+2 a B) \cos (c+d x)) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 129, 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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {(a \cos (c+d x)+b)^2 (A \cos (c+d x)+B)}{\cos ^{\frac {5}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\)

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(649\) vs. \(2(121)=242\).

Time = 4.89 (sec) , antiderivative size = 650, normalized size of antiderivative = 5.16

method	result	size
default	\(-\frac {\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 (\frac {2 B \,a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {2 b \left (A b +2 B a \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \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}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+2 b^{2} B \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{6 \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {1}{2}\right )^{2}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )-\frac {2 A \,a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {2 A \,a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}+\frac {4 A a b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \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 )}{\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\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}\)	\(650\)

Input:

int(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B*a^2*(sin(1 
/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 
/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+ 
2*b*(A*b+2*B*a)/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/ 
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d 
*x+1/2*c)^2-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))+2*b^2*B*(-1/6*cos(1/2*d*x+1/2*c)*(- 
2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1 
/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/( 
-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+ 
1/2*c),2^(1/2)))-2*A*a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2* 
c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti 
cF(cos(1/2*d*x+1/2*c),2^(1/2))+2*A*a^2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co 
s(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) 
^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c) 
,2^(1/2)))+4*A*a*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1 
)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.02 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {\sqrt {2} {\left (-3 i \, B a^{2} - 6 i \, A a b - i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, B a^{2} + 6 i \, A a b + i \, B b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-i \, A a^{2} + 2 i \, B a b + i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (i \, A a^{2} - 2 i \, B a b - i \, A b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (B b^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorith 
m="fricas")
 

Output:

1/3*(sqrt(2)*(-3*I*B*a^2 - 6*I*A*a*b - I*B*b^2)*cos(d*x + c)^2*weierstrass 
PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(3*I*B*a^2 + 6*I* 
A*a*b + I*B*b^2)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c)) - 3*sqrt(2)*(-I*A*a^2 + 2*I*B*a*b + I*A*b^2)*cos(d*x + c)^ 
2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d 
*x + c))) - 3*sqrt(2)*(I*A*a^2 - 2*I*B*a*b - I*A*b^2)*cos(d*x + c)^2*weier 
strassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) 
)) + 2*(B*b^2 + 3*(2*B*a*b + A*b^2)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d 
*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F]

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

integrate(cos(d*x+c)**(1/2)*(a+b*sec(d*x+c))**2*(A+B*sec(d*x+c)),x)
 

Output:

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**2*sqrt(cos(c + d*x)), 
x)
 

Maxima [F]

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

integrate(cos(d*x+c)^(1/2)*(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*sqrt(cos(d*x + c)), 
x)
 

Giac [F]

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

integrate(cos(d*x+c)^(1/2)*(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*sqrt(cos(d*x + c)), 
x)
 

Mupad [B] (verification not implemented)

Time = 13.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.54 \[ \int \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx=\frac {2\,A\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,B\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,A\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,B\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(cos(c + d*x)^(1/2)*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2,x)
 

Output:

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

Reduce [F]

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

int(cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)
 

Output:

int(sqrt(cos(c + d*x)),x)*a**3 + int(sqrt(cos(c + d*x))*sec(c + d*x)**3,x) 
*b**3 + 3*int(sqrt(cos(c + d*x))*sec(c + d*x)**2,x)*a*b**2 + 3*int(sqrt(co 
s(c + d*x))*sec(c + d*x),x)*a**2*b