\(\int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [573]

Optimal result

Integrand size = 33, antiderivative size = 214 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 b (A b+2 a B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \] Output:

-2/5*(5*A*a^2+3*A*b^2+6*B*a*b)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/2 
1*(14*A*a*b+7*B*a^2+5*B*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/7* 
b^2*B*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*b*(A*b+2*B*a)*sin(d*x+c)/d/cos(d*x 
+c)^(5/2)+2/21*(14*A*a*b+7*B*a^2+5*B*b^2)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/ 
5*(5*A*a^2+3*A*b^2+6*B*a*b)*sin(d*x+c)/d/cos(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 3.69 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \left (-21 \left (5 a^2 A+3 A b^2+6 a b B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {15 b^2 B \sin (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}+\frac {21 b (A b+2 a B) \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {5 \left (14 a A b+7 a^2 B+5 b^2 B\right ) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {21 \left (5 a^2 A+3 A b^2+6 a b B\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}\right )}{105 d} \] Input:

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

Output:

(2*(-21*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*EllipticE[(c + d*x)/2, 2] + 5*(14*a* 
A*b + 7*a^2*B + 5*b^2*B)*EllipticF[(c + d*x)/2, 2] + (15*b^2*B*Sin[c + d*x 
])/Cos[c + d*x]^(7/2) + (21*b*(A*b + 2*a*B)*Sin[c + d*x])/Cos[c + d*x]^(5/ 
2) + (5*(14*a*A*b + 7*a^2*B + 5*b^2*B)*Sin[c + d*x])/Cos[c + d*x]^(3/2) + 
(21*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x])/Sqrt[Cos[c + d*x]]))/(105* 
d)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.89, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3433, 3042, 3467, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3433

\(\displaystyle \int \frac {(a \cos (c+d x)+b)^2 (A \cos (c+d x)+B)}{\cos ^{\frac {9}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )^2 \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\)

\(\Big \downarrow \) 3467

\(\displaystyle \frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2}{7} \int -\frac {7 a^2 A \cos ^2(c+d x)+\left (7 B a^2+14 A b a+5 b^2 B\right ) \cos (c+d x)+7 b (A b+2 a B)}{2 \cos ^{\frac {7}{2}}(c+d x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \frac {7 a^2 A \cos ^2(c+d x)+\left (7 B a^2+14 A b a+5 b^2 B\right ) \cos (c+d x)+7 b (A b+2 a B)}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \frac {7 a^2 A \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (7 B a^2+14 A b a+5 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+7 b (A b+2 a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3500

\(\displaystyle \frac {1}{7} \left (\frac {2}{5} \int \frac {5 \left (7 B a^2+14 A b a+5 b^2 B\right )+7 \left (5 A a^2+6 b B a+3 A b^2\right ) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 \left (7 B a^2+14 A b a+5 b^2 B\right )+7 \left (5 A a^2+6 b B a+3 A b^2\right ) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \int \frac {5 \left (7 B a^2+14 A b a+5 b^2 B\right )+7 \left (5 A a^2+6 b B a+3 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx+7 \left (5 a^2 A+6 a b B+3 A b^2\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+7 \left (5 a^2 A+6 a b B+3 A b^2\right ) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (5 a^2 A+6 a b B+3 A b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (5 a^2 A+6 a b B+3 A b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (5 a^2 A+6 a b B+3 A b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left (\frac {1}{5} \left (5 \left (7 a^2 B+14 a A b+5 b^2 B\right ) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 \left (5 a^2 A+6 a b B+3 A b^2\right ) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {14 b (2 a B+A b) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 b^2 B \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\)

Input:

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

Output:

(2*b^2*B*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ((14*b*(A*b + 2*a*B)*Sin 
[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (5*(14*a*A*b + 7*a^2*B + 5*b^2*B)*(( 
2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3 
/2))) + 7*(5*a^2*A + 3*A*b^2 + 6*a*b*B)*((-2*EllipticE[(c + d*x)/2, 2])/d 
+ (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/5)/7
 

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 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

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. \(831\) vs. \(2(197)=394\).

Time = 8.05 (sec) , antiderivative size = 832, normalized size of antiderivative = 3.89

method	result	size
default	\(\text {Expression too large to display}\)	\(832\)

Input:

int((a+b*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(3/2),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*A*a^2/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*a*(2*A*b+B*a)*(-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) 
))+2/5*b*(A*b+2*B*a)/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1 
/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2 
*d*x+1/2*c)-12*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)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d 
*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/ 
2*c)^2+8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-3*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*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*b^2*B*(-1/56*co 
s(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)^4-5/42*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+5/21*(sin(1/2*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.47 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, \sqrt {2} {\left (7 i \, B a^{2} + 14 i \, A a b + 5 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-7 i \, B a^{2} - 14 i \, A a b - 5 i \, B b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 \, \sqrt {2} {\left (5 i \, A a^{2} + 6 i \, B a b + 3 i \, A b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 \, \sqrt {2} {\left (-5 i \, A a^{2} - 6 i \, B a b - 3 i \, A b^{2}\right )} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, {\left (21 \, {\left (5 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 15 \, B b^{2} + 5 \, {\left (7 \, B a^{2} + 14 \, A a b + 5 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 21 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

-1/105*(5*sqrt(2)*(7*I*B*a^2 + 14*I*A*a*b + 5*I*B*b^2)*cos(d*x + c)^4*weie 
rstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-7*I*B* 
a^2 - 14*I*A*a*b - 5*I*B*b^2)*cos(d*x + c)^4*weierstrassPInverse(-4, 0, co 
s(d*x + c) - I*sin(d*x + c)) + 21*sqrt(2)*(5*I*A*a^2 + 6*I*B*a*b + 3*I*A*b 
^2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c))) + 21*sqrt(2)*(-5*I*A*a^2 - 6*I*B*a*b - 3*I*A*b^ 
2)*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(d*x + c))) - 2*(21*(5*A*a^2 + 6*B*a*b + 3*A*b^2)*cos(d*x + 
c)^3 + 15*B*b^2 + 5*(7*B*a^2 + 14*A*a*b + 5*B*b^2)*cos(d*x + c)^2 + 21*(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)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Mupad [B] (verification not implemented)

Time = 13.86 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b \sec (c+d x))^2 (A+B \sec (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx=\frac {6\,A\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {30\,B\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )+70\,B\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+84\,B\,a\,b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{105\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}} \] Input:

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

Output:

(6*A*b^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 30*A* 
a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2 
) + 20*A*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + 
 d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (30*B*b^2 
*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2) + 70*B*a^2*cos( 
c + d*x)^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 84*B 
*a*b*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2 
))/(105*d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2))
 

Reduce [F]

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

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

Output:

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