\(\int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [241]

Optimal result

Integrand size = 35, antiderivative size = 201 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {(7 A+C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 (5 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \] Output:

-(7*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)/a^2/d+2/3*(5*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)/a^2/d+2/3*(5*A+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2 
)-1/3*(7*A+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)/(1+sec(d*x+c))-1/3*(A+C)*s 
in(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 7.82 (sec) , antiderivative size = 538, normalized size of antiderivative = 2.68 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-14 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-2 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-\frac {\left (2 (36 A+5 C) \cos \left (\frac {1}{2} (c-d x)\right )+(54 A+8 C) \cos \left (\frac {1}{2} (3 c+d x)\right )+33 A \cos \left (\frac {1}{2} (c+3 d x)\right )+6 C \cos \left (\frac {1}{2} (c+3 d x)\right )+9 A \cos \left (\frac {1}{2} (5 c+3 d x)\right )+A \cos \left (\frac {1}{2} (3 c+5 d x)\right )-A \cos \left (\frac {1}{2} (7 c+5 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right )}{4 \sqrt {\sec (c+d x)}}-40 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-8 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}\right ) \left (A+C \sec ^2(c+d x)\right )}{3 a^2 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \] Input:

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

Output:

(-2*Cos[(c + d*x)/2]^4*((-14*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)* 
(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c 
 + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7 
/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - (2*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/ 
(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 
 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2 
F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) - ((2*(36*A + 5*C)*Cos 
[(c - d*x)/2] + (54*A + 8*C)*Cos[(3*c + d*x)/2] + 33*A*Cos[(c + 3*d*x)/2] 
+ 6*C*Cos[(c + 3*d*x)/2] + 9*A*Cos[(5*c + 3*d*x)/2] + A*Cos[(3*c + 5*d*x)/ 
2] - A*Cos[(7*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^3)/(4*Sqrt 
[Sec[c + d*x]]) - 40*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[S 
ec[c + d*x]] - 8*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c 
 + d*x]])*(A + C*Sec[c + d*x]^2))/(3*a^2*d*(A + 2*C + A*Cos[2*(c + d*x)])* 
(1 + Sec[c + d*x])^2)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4573, 27, 3042, 4508, 27, 3042, 4274, 3042, 4256, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {3 a (3 A+C)-a (5 A-C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {2 a^2 (5 A+C)-a^2 (7 A+C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \int \frac {2 a^2 (5 A+C)-a^2 (7 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-a^2 (7 A+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4256

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/3*((A + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) 
+ ((-2*(7*A + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) + 
 (3*((-2*a^2*(7*A + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[S 
ec[c + d*x]])/d + 2*a^2*(5*A + C)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d* 
x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d* 
x]]))))/a^2)/(6*a^2)
 

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 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 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 4508

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

rule 4573

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) 
*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m 
+ 1))), x] + Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - 
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(436\) vs. \(2(182)=364\).

Time = 3.61 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.17

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+20 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+42 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-48 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+21 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+9 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(437\)

Input:

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

Output:

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*A*cos(1/2 
*d*x+1/2*c)^8+12*A*cos(1/2*d*x+1/2*c)^6+20*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*co 
s(1/2*d*x+1/2*c)^3+42*A*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+12 
*C*cos(1/2*d*x+1/2*c)^6+4*C*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) 
)+6*C*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-48*A*cos(1/2*d*x+1/2 
*c)^4-20*C*cos(1/2*d*x+1/2*c)^4+21*A*cos(1/2*d*x+1/2*c)^2+9*C*cos(1/2*d*x+ 
1/2*c)^2-A-C)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(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 = 375, normalized size of antiderivative = 1.87 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (\sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (7 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (7 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-7 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-7 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (2 \, A \cos \left (d x + c\right )^{3} + {\left (13 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

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

Output:

-1/6*(2*(sqrt(2)*(5*I*A + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(5*I*A + I*C)*co 
s(d*x + c) + sqrt(2)*(5*I*A + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c 
) + I*sin(d*x + c)) + 2*(sqrt(2)*(-5*I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2) 
*(-5*I*A - I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A - I*C))*weierstrassPInverse 
(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(7*I*A + I*C)*cos(d*x 
+ c)^2 + 2*sqrt(2)*(7*I*A + I*C)*cos(d*x + c) + sqrt(2)*(7*I*A + I*C))*wei 
erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
c))) + 3*(sqrt(2)*(-7*I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-7*I*A - I*C) 
*cos(d*x + c) + sqrt(2)*(-7*I*A - I*C))*weierstrassZeta(-4, 0, weierstrass 
PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(2*A*cos(d*x + c)^3 + 
(13*A + 3*C)*cos(d*x + c)^2 + 2*(5*A + C)*cos(d*x + c))*sin(d*x + c)/sqrt( 
cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \] Input:

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

Output:

(Integral(A/(sec(c + d*x)**(7/2) + 2*sec(c + d*x)**(5/2) + sec(c + d*x)**( 
3/2)), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**(7/2) + 2*sec(c + d* 
x)**(5/2) + sec(c + d*x)**(3/2)), x))/a**2
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

\[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}+2 \sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) c}{a^{2}} \] Input:

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

Output:

(int(sqrt(sec(c + d*x))/(sec(c + d*x)**4 + 2*sec(c + d*x)**3 + sec(c + d*x 
)**2),x)*a + int(sqrt(sec(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1) 
,x)*c)/a**2