\(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [220]

Optimal result

Integrand size = 33, antiderivative size = 222 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(9 A+B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(3 A+B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(3 A+2 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(3 A+B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

-1/10*(9*A+B)*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^3/d+1/6*(3*A+B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2 
*c,2^(1/2))*sec(d*x+c)^(1/2)/a^3/d+1/5*(A-B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d 
/(a+a*sec(d*x+c))^3+1/15*(3*A+2*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a*se 
c(d*x+c))^2+1/6*(3*A+B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))
 

Mathematica [C] (verified)

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

Time = 8.49 (sec) , antiderivative size = 919, normalized size of antiderivative = 4.14 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^(( 
2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*(-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))])*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(5*d 
*E^(I*d*x)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (Sqrt[2]*B*Sqrt[ 
E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*C 
os[c/2 + (d*x)/2]^6*Csc[c/2]*(-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))])*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x]))/(15*d*E^(I*d*x)*(B + A 
*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (2*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Co 
s[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(5/2) 
*(A + B*Sec[c + d*x])*Sin[c])/(d*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x]) 
^3) + (2*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + 
 d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*Sin[c])/(3*d* 
(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*Sec[c 
 + d*x]^(5/2)*(A + B*Sec[c + d*x])*((2*(9*A + B)*Cos[d*x]*Csc[c/2]*Sec[c/2 
])/(5*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(-9*A*Sin[(d*x)/2] + B*Sin[(d*x) 
/2]))/(3*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[ 
(d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-12*A*Sin[(d*x)/2] + 
7*B*Sin[(d*x)/2]))/(15*d) + (4*(-9*A + B)*Tan[c/2])/(3*d) - (4*(-12*A +...
 

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4507, 27, 3042, 4508, 3042, 4508, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\int \frac {a (A-B)-5 a (A+B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^2}dx}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\int \frac {a (A-B)-5 a (A+B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\int \frac {a^2 (6 A-B)-3 a^2 (3 A+2 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\int \frac {a^2 (6 A-B)-3 a^2 (3 A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {\int \frac {3 a^3 (9 A+B)-5 a^3 (3 A+B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx}{a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {\int \frac {3 a^3 (9 A+B)-5 a^3 (3 A+B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {\int \frac {3 a^3 (9 A+B)-5 a^3 (3 A+B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {3 a^3 (9 A+B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 a^3 (3 A+B) \int \sqrt {\sec (c+d x)}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {3 a^3 (9 A+B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a^3 (3 A+B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {3 a^3 (9 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-5 a^3 (3 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {3 a^3 (9 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 a^3 (3 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {\frac {6 a^3 (9 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a^3 (3 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}-\frac {\frac {\frac {\frac {6 a^3 (9 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 a^3 (3 A+B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{2 a^2}-\frac {5 a^2 (3 A+B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (3 A+2 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}\)

Input:

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

Output:

((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ( 
(-2*a*(3*A + 2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x 
])^2) + (((6*a^3*(9*A + B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq 
rt[Sec[c + d*x]])/d - (10*a^3*(3*A + B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + 
d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/(2*a^2) - (5*a^2*(3*A + B)*Sqrt[Sec[c + 
d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(10*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 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 4507

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(201)=402\).

Time = 5.02 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.03

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 (108 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+30 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )+54 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )-198 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+114 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-27 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+17 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A -3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\)	\(451\)

Input:

int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNV 
ERBOSE)
 

Output:

-1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(108*A*c 
os(1/2*d*x+1/2*c)^8+30*A*cos(1/2*d*x+1/2*c)^5*(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))+5 
4*A*cos(1/2*d*x+1/2*c)^5*(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*B*cos(1/2*d*x+1/2*c 
)^8+10*B*cos(1/2*d*x+1/2*c)^5*(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*B*cos(1/2*d*x+1 
/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Ell 
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))-198*A*cos(1/2*d*x+1/2*c)^6-2*B*cos(1/2* 
d*x+1/2*c)^6+114*A*cos(1/2*d*x+1/2*c)^4-24*B*cos(1/2*d*x+1/2*c)^4-27*A*cos 
(1/2*d*x+1/2*c)^2+17*B*cos(1/2*d*x+1/2*c)^2+3*A-3*B)/cos(1/2*d*x+1/2*c)^5/ 
(-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 = 474, normalized size of antiderivative = 2.14 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

-1/60*(5*(sqrt(2)*(3*I*A + I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(3*I*A + I*B)*c 
os(d*x + c)^2 + 3*sqrt(2)*(3*I*A + I*B)*cos(d*x + c) + sqrt(2)*(3*I*A + I* 
B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2) 
*(-3*I*A - I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-3*I*A - I*B)*cos(d*x + c)^2 + 
 3*sqrt(2)*(-3*I*A - I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A - I*B))*weierstra 
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(9*I*A + I*B 
)*cos(d*x + c)^3 + 3*sqrt(2)*(9*I*A + I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(9*I 
*A + I*B)*cos(d*x + c) + sqrt(2)*(9*I*A + I*B))*weierstrassZeta(-4, 0, wei 
erstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-9*I 
*A - I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A - I*B)*cos(d*x + c)^2 + 3*sqr 
t(2)*(-9*I*A - I*B)*cos(d*x + c) + sqrt(2)*(-9*I*A - I*B))*weierstrassZeta 
(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3* 
(9*A + B)*cos(d*x + c)^3 + 2*(18*A + 7*B)*cos(d*x + c)^2 + 5*(3*A + B)*cos 
(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*x + c)^3 + 3*a^3* 
d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

Sympy [F(-1)]

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

integrate(sec(d*x+c)**(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-1)]

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

integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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