3.2.74 \(\int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \, dx\) [174]

3.2.74.1 Optimal result
3.2.74.2 Mathematica [C] (verified)
3.2.74.3 Rubi [A] (verified)
3.2.74.4 Maple [B] (verified)
3.2.74.5 Fricas [C] (verification not implemented)
3.2.74.6 Sympy [F]
3.2.74.7 Maxima [F]
3.2.74.8 Giac [F]
3.2.74.9 Mupad [F(-1)]

3.2.74.1 Optimal result

Integrand size = 23, antiderivative size = 131 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \, dx=-\frac {4 a^2 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {8 a^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^2 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {2 a^2 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d} \]

output
2/3*a^2*sec(d*x+c)^(3/2)*sin(d*x+c)/d+4*a^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d- 
4*a^2*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+8/3*a^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))* 
cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d
 
3.2.74.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.71 (sec) , antiderivative size = 264, normalized size of antiderivative = 2.02 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \, dx=\frac {1}{3} a^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^2 \left (-\frac {i \sqrt {2} e^{-i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \cos ^2(c+d x) \left (3 \left (1+e^{2 i (c+d x)}\right )+3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+2 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right )}{d \left (-1+e^{2 i c}\right )}+\frac {6 \cos (d x) \csc (c)+\tan (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x)}\right ) \]

input
Integrate[Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2,x]
 
output
(a^2*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(((-I)*Sqrt[2]*Sqrt[E^(I*(c + 
 d*x))/(1 + E^((2*I)*(c + d*x)))]*Cos[c + d*x]^2*(3*(1 + E^((2*I)*(c + d*x 
))) + 3*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1 
[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] + 2*E^(I*(c + d*x))*(-1 + E^((2*I)* 
c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2* 
I)*(c + d*x))]))/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))) + (6*Cos[d*x]*Csc[ 
c] + Tan[c + d*x])/(2*d*Sec[c + d*x]^(3/2))))/3
 
3.2.74.3 Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4275

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 4534

\(\displaystyle \frac {4}{3} a^2 \int \sqrt {\sec (c+d x)}dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+2 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{3} a^2 \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+2 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {4}{3} a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+2 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4}{3} a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+2 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 a^2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {8 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+2 a^2 \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\)

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

3.2.74.3.1 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]
 
3.2.74.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(370\) vs. \(2(167)=334\).

Time = 18.74 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.83

method result size
default \(-\frac {4 a^{2} \sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (12 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) \(371\)
parts \(-\frac {2 a^{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}}\, \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}}\, \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}-\frac {2 a^{2} \left (-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}-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{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 )\right ) \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}}}{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}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) d}-\frac {4 a^{2} \left (-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{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}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\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 )\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}}\, \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}\) \(536\)

input
int(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
-4/3*a^2*(-(-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*(12*sin(1/ 
2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-4*(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)^2-6*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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)^2-7*cos(1/2*d*x+1/ 
2*c)*sin(1/2*d*x+1/2*c)^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))+3*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*EllipticE(cos(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*cos(1/2*d 
*x+1/2*c)^2-1)^(1/2)/d
 
3.2.74.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.37 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \, dx=-\frac {2 \, {\left (2 i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 2 i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (6 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d \cos \left (d x + c\right )} \]

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

\[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \, dx=a^{2} \left (\int \sqrt {\sec {\left (c + d x \right )}}\, dx + \int 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int \sec ^{\frac {5}{2}}{\left (c + d x \right )}\, dx\right ) \]

input
integrate(sec(d*x+c)**(1/2)*(a+a*sec(d*x+c))**2,x)
 
output
a**2*(Integral(sqrt(sec(c + d*x)), x) + Integral(2*sec(c + d*x)**(3/2), x) 
 + Integral(sec(c + d*x)**(5/2), x))
 
3.2.74.7 Maxima [F]

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

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

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

input
integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^2,x, algorithm="giac")
 
output
integrate((a*sec(d*x + c) + a)^2*sqrt(sec(d*x + c)), x)
 
3.2.74.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}} \,d x \]

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