3.9.7 \(\int \frac {(a+b \sec (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx\) [807]

3.9.7.1 Optimal result
3.9.7.2 Mathematica [A] (verified)
3.9.7.3 Rubi [A] (verified)
3.9.7.4 Maple [B] (verified)
3.9.7.5 Fricas [C] (verification not implemented)
3.9.7.6 Sympy [F]
3.9.7.7 Maxima [F]
3.9.7.8 Giac [F]
3.9.7.9 Mupad [B] (verification not implemented)

3.9.7.1 Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \frac {(a+b \sec (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=-\frac {4 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 b^2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a b \sin (c+d x)}{d \sqrt {\cos (c+d x)}} \]

output
-4*a*b*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d 
*x+1/2*c),2^(1/2))/d+2/3*(3*a^2+b^2)*(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+2/3*b^2*sin(d*x+c)/d/co 
s(d*x+c)^(3/2)+4*a*b*sin(d*x+c)/d/cos(d*x+c)^(1/2)
 
3.9.7.2 Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.77 \[ \int \frac {(a+b \sec (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {2 \left (-6 a b E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\left (3 a^2+b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {b (b+6 a \cos (c+d x)) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}\right )}{3 d} \]

input
Integrate[(a + b*Sec[c + d*x])^2/Sqrt[Cos[c + d*x]],x]
 
output
(2*(-6*a*b*EllipticE[(c + d*x)/2, 2] + (3*a^2 + b^2)*EllipticF[(c + d*x)/2 
, 2] + (b*(b + 6*a*Cos[c + d*x])*Sin[c + d*x])/Cos[c + d*x]^(3/2)))/(3*d)
 
3.9.7.3 Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.65, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4752, 3042, 4275, 3042, 4255, 3042, 4258, 3042, 3119, 4534, 3042, 4258, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4752

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4275

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \sqrt {\sec (c+d x)} \left (a^2+b^2 \sec ^2(c+d x)\right )dx+2 a b \int \sec ^{\frac {3}{2}}(c+d x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^2+b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+2 a b \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )\)

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

input
Int[(a + b*Sec[c + d*x])^2/Sqrt[Cos[c + d*x]],x]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*(3*a^2 + b^2)*Sqrt[Cos[c + d*x]] 
*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*b^2*Sec[c + d*x] 
^(3/2)*Sin[c + d*x])/(3*d) + 2*a*b*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + 
d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))
 

3.9.7.3.1 Defintions of rubi rules used

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 4255
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4275
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^2, x_Symbol] :> Simp[2*a*(b/d)   Int[(d*Csc[e + f*x])^(n + 1), x], x] 
 + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, 
 e, f, n}, x]
 

rule 4534
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

rule 4752
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 
3.9.7.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(512\) vs. \(2(139)=278\).

Time = 12.44 (sec) , antiderivative size = 513, normalized size of antiderivative = 5.40

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 (24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a b -6 \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a^{2}-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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}-12 \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b^{2}+3 \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 ) 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 ) b^{2}+6 \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 \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}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(513\)

input
int((a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 
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*sin(1/2* 
d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)^3*(24*cos(1/2*d* 
x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b-6*(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))*sin(1/2*d*x+ 
1/2*c)^2*a^2-2*(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))*sin(1/2*d*x+1/2*c)^2*b^2-12*(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))*sin(1/2*d*x+1/2*c)^2*a*b-12*cos(1/2*d*x+1/2*c)*sin(1/ 
2*d*x+1/2*c)^2*a*b-2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^2+3*(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))*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))*b^2+6*(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)*(-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)^2-1)^(1/2)/d
 
3.9.7.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b \sec (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {-6 i \, \sqrt {2} a b \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 6 i \, \sqrt {2} a b \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \sqrt {2} {\left (-3 i \, a^{2} - i \, b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (3 i \, a^{2} + i \, b^{2}\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (6 \, a b \cos \left (d x + c\right ) + b^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{2}} \]

input
integrate((a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="fricas")
 
output
1/3*(-6*I*sqrt(2)*a*b*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPIn 
verse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 6*I*sqrt(2)*a*b*cos(d*x + c 
)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin 
(d*x + c))) + sqrt(2)*(-3*I*a^2 - I*b^2)*cos(d*x + c)^2*weierstrassPInvers 
e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + sqrt(2)*(3*I*a^2 + I*b^2)*cos(d* 
x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(6* 
a*b*cos(d*x + c) + b^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 
)
 
3.9.7.6 Sympy [F]

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

input
integrate((a+b*sec(d*x+c))**2/cos(d*x+c)**(1/2),x)
 
output
Integral((a + b*sec(c + d*x))**2/sqrt(cos(c + d*x)), x)
 
3.9.7.7 Maxima [F]

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

input
integrate((a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="maxima")
 
output
integrate((b*sec(d*x + c) + a)^2/sqrt(cos(d*x + c)), x)
 
3.9.7.8 Giac [F]

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

input
integrate((a+b*sec(d*x+c))^2/cos(d*x+c)^(1/2),x, algorithm="giac")
 
output
integrate((b*sec(d*x + c) + a)^2/sqrt(cos(d*x + c)), x)
 
3.9.7.9 Mupad [B] (verification not implemented)

Time = 14.59 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14 \[ \int \frac {(a+b \sec (c+d x))^2}{\sqrt {\cos (c+d x)}} \, dx=\frac {2\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,b^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {4\,a\,b\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int((a + b/cos(c + d*x))^2/cos(c + d*x)^(1/2),x)
 
output
(2*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (2*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*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))