\(\int \frac {(a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [217]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 198 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {4 a^2 (A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {8 a^2 (A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (A-5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {2 (A-C) \sqrt {\sec (c+d x)} \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \] Output:

4*a^2*(A-C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x 
+c)^(1/2)/d+8/3*a^2*(A+C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2 
^(1/2))*sec(d*x+c)^(1/2)/d-2/3*a^2*(A-5*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/d+2 
/3*A*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/3*(A-C)*sec(d*x+c) 
^(1/2)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 2.77 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.96 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 e^{-i d x} \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (12 i A-12 i C+12 i A \cos (2 (c+d x))-12 i C \cos (2 (c+d x))+16 (A+C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-4 i (A-C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+A \sin (c+d x)+4 C \sin (c+d x)+12 C \sin (2 (c+d x))+A \sin (3 (c+d x))\right )}{6 d} \] Input:

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

Output:

(a^2*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((12*I)*A - (12*I)*C + (12 
*I)*A*Cos[2*(c + d*x)] - (12*I)*C*Cos[2*(c + d*x)] + 16*(A + C)*Cos[c + d* 
x]^(3/2)*EllipticF[(c + d*x)/2, 2] - (4*I)*(A - C)*(1 + E^((2*I)*(c + d*x) 
))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))] + A*Sin[c 
+ d*x] + 4*C*Sin[c + d*x] + 12*C*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/( 
6*d*E^(I*d*x))
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4575, 27, 3042, 4506, 27, 3042, 4485, 3042, 4274, 3042, 4258, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

\(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^2 (4 a A-3 a (A-C) \sec (c+d x))}{2 \sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^2 (4 a A-3 a (A-C) \sec (c+d x))}{\sqrt {\sec (c+d x)}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (4 a A-3 a (A-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {2}{3} \int \frac {3 (\sec (c+d x) a+a) \left (a^2 (5 A-C)-a^2 (A-5 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a) \left (a^2 (5 A-C)-a^2 (A-5 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (5 A-C)-a^2 (A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {2 \int \frac {3 (A-C) a^3+2 (A+C) \sec (c+d x) a^3}{\sqrt {\sec (c+d x)}}dx-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {3 (A-C) a^3+2 (A+C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {2 \left (3 a^3 (A-C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+2 a^3 (A+C) \int \sqrt {\sec (c+d x)}dx\right )-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (3 a^3 (A-C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a^3 (A+C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 \left (2 a^3 (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (2 a^3 (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {-\frac {2 a^3 (A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{d}+2 \left (\frac {4 a^3 (A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{3 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{3 d \sqrt {\sec (c+d x)}}\)

Input:

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

Output:

(2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*(( 
6*a^3*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d* 
x]])/d + (4*a^3*(A + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[ 
Sec[c + d*x]])/d) - (2*a^3*(A - 5*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d - 
(2*(A - C)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/d)/(3 
*a)
 

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 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 4274
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

rule 4506
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(d*(m + n))   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] 
)^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - 
 a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]
 

rule 4575
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co 
t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( 
b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b 
*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 
 C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || 
 EqQ[m + n + 1, 0])
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(650\) vs. \(2(179)=358\).

Time = 2.62 (sec) , antiderivative size = 651, normalized size of antiderivative = 3.29

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 (4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \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 A \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}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 C \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \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 C \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \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 ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A -2 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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 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 )+7 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C -2 C \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 C \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 )\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}\) \(651\)
parts \(\text {Expression too large to display}\) \(864\)

Input:

int((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x,method=_RETUR 
NVERBOSE)
 

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*(4*A*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-4*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c 
)^4+4*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli 
pticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2-6*A*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)^(1/2)*sin(1/2*d*x+1/2*c)^2-12*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) 
^4+4*C*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellip 
ticF(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+6*C*(2*sin(1/2*d*x+1 
/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c) 
,2^(1/2))*sin(1/2*d*x+1/2*c)^2+sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-2 
*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)*EllipticF 
(cos(1/2*d*x+1/2*c),2^(1/2))+3*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))+7*sin(1/2*d*x+1 
/2*c)^2*cos(1/2*d*x+1/2*c)*C-2*C*(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*C*(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)))*(-2*sin(1/2*d*x+1/2*c)^4+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)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.07 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (2 i \, \sqrt {2} {\left (A + C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 2 i \, \sqrt {2} {\left (A + C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (A - C\right )} a^{2} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (A a^{2} \cos \left (d x + c\right )^{2} + 6 \, C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{3 \, d \cos \left (d x + c\right )} \] Input:

integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algori 
thm="fricas")
 

Output:

-2/3*(2*I*sqrt(2)*(A + C)*a^2*cos(d*x + c)*weierstrassPInverse(-4, 0, cos( 
d*x + c) + I*sin(d*x + c)) - 2*I*sqrt(2)*(A + C)*a^2*cos(d*x + c)*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*(A - C)*a^ 
2*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + 
 c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(A - C)*a^2*cos(d*x + c)*weierstrassZ 
eta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (A 
*a^2*cos(d*x + c)^2 + 6*C*a^2*cos(d*x + c) + C*a^2)*sin(d*x + c)/sqrt(cos( 
d*x + c)))/(d*cos(d*x + c))
 

Sympy [F]

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

integrate((a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2),x)
 

Output:

a**2*(Integral(A/sec(c + d*x)**(3/2), x) + Integral(2*A/sqrt(sec(c + d*x)) 
, x) + Integral(A*sqrt(sec(c + d*x)), x) + Integral(C*sqrt(sec(c + d*x)), 
x) + Integral(2*C*sec(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**(5/2) 
, x))
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algori 
thm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x, algori 
thm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(3/2) 
,x)
 

Output:

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2)/(1/cos(c + d*x))^(3/2) 
, x)
 

Reduce [F]

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

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

Output:

a**2*(int(sqrt(sec(c + d*x))/sec(c + d*x)**2,x)*a + 2*int(sqrt(sec(c + d*x 
))/sec(c + d*x),x)*a + int(sqrt(sec(c + d*x)),x)*a + int(sqrt(sec(c + d*x) 
),x)*c + int(sqrt(sec(c + d*x))*sec(c + d*x)**2,x)*c + 2*int(sqrt(sec(c + 
d*x))*sec(c + d*x),x)*c)