3.14.25 \(\int \frac {\sqrt {\cos (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [1325]

3.14.25.1 Optimal result

Integrand size = 43, antiderivative size = 257 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (3 A b^2-a b B-a^2 (2 A-C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (3 A b^3+2 a^3 B-a b^2 B-a^2 b (4 A+C)\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{a^3 \left (a^2-b^2\right ) d}-\frac {\left (3 A b^4+3 a^3 b B-a b^3 B-a^4 C-a^2 b^2 (5 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^3 (a-b) (a+b)^2 d}+\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{a \left (a^2-b^2\right ) d (b+a \cos (c+d x))} \]

output

-(3*A*b^2-B*a*b-a^2*(2*A-C))*(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))/a^2/(a^2-b^2)/d+(3*A*b^3+2*B*a^3- 
B*a*b^2-a^2*b*(4*A+C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell 
ipticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/(a^2-b^2)/d-(3*A*b^4+3*B*a^3*b-B*a* 
b^3-a^4*C-a^2*b^2*(5*A+C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c) 
*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/a^3/(a-b)/(a+b)^2/d+(A*b 
^2-a*(B*b-C*a))*sin(d*x+c)*cos(d*x+c)^(1/2)/a/(a^2-b^2)/d/(b+a*cos(d*x+c))
 

3.14.25.2 Mathematica [A] (verified)

Time = 4.18 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 \left (A b^2+a (-b B+a C)\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (b+a \cos (c+d x))}+\frac {\frac {2 \left (-A b^2-a b B+a^2 (2 A+C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {8 (A b-a B+b C) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {2 \left (-3 A b^2+a b B+a^2 (2 A-C)\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a^2 b \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 a d} \]

input

Integrate[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a 
+ b*Sec[c + d*x])^2,x]
 

output

((4*(A*b^2 + a*(-(b*B) + a*C))*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^ 
2)*(b + a*Cos[c + d*x])) + ((2*(-(A*b^2) - a*b*B + a^2*(2*A + C))*Elliptic 
Pi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) - (8*(A*b - a*B + b*C)*((a + b) 
*EllipticF[(c + d*x)/2, 2] - b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])) 
/(a + b) + (2*(-3*A*b^2 + a*b*B + a^2*(2*A - C))*(-2*a*b*EllipticE[ArcSin[ 
Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]] 
], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt[Cos[c + d*x]]], -1]) 
*Sin[c + d*x])/(a^2*b*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/(4*a*d)
 

3.14.25.3 Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 253, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4600, 3042, 3526, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4600

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3526

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\frac {\left (2 a^3 B-a^2 b (4 A+C)-a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a}-\frac {\left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+C)-a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (2 a^3 B-a^2 b (4 A+C)-a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (a^4 (-C)+3 a^3 b B-a^2 b^2 (5 A+C)-a b^3 B+3 A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{a}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-\left (a^2 (2 A-C)\right )-a b B+3 A b^2\right )}{a d}}{2 a \left (a^2-b^2\right )}+\frac {\sin (c+d x) \sqrt {\cos (c+d x)} \left (A b^2-a (b B-a C)\right )}{a d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}\)

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

input

Int[(Sqrt[Cos[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Se 
c[c + d*x])^2,x]
 

output

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

3.14.25.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3526

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - 
d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2))   Int[(a + b*Sin[e + f*x])^(m - 
 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* 
d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 
1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x 
] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f 
*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d 
, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

rule 4600

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (B_.)*sec[(e_.) + (f_.)*(x_)] + (C_.)*sec[(e_.) + (f_.) 
*(x_)]^2), x_Symbol] :> Simp[d^(m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[ 
e + f*x])^(n - m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]
 

3.14.25.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(855\) vs. \(2(333)=666\).

Time = 3.57 (sec) , antiderivative size = 856, normalized size of antiderivative = 3.33

method	result	size
default	\(\text {Expression too large to display}\)	\(856\)

input

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^2,x, 
method=_RETURNVERBOSE)
 

output

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/a^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)^(1/2) 
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*A*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1 
/2))+a*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-B*a*EllipticF(cos(1/2*d*x+1 
/2*c),2^(1/2)))-2/a^2*(3*A*b^2-2*B*a*b+C*a^2)/(a^2-a*b)*(sin(1/2*d*x+1/2*c 
)^2)^(1/2)*(-2*cos(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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2 
*b*(A*b^2-B*a*b+C*a^2)/a^3*(a^2/b/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-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*a-a+b)-1 
/2/(a+b)/b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x 
+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*a/b/(a^2-b^2)*(sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^ 
2-b^2)/(a^2-a*b)*a^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi( 
cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+3/2*b/(a^2-b^2)/(a^2-a*b)*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)/(-2*sin(1/2*d*x+...
 

3.14.25.5 Fricas [F(-1)]

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

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^2,x, algorithm="fricas")
 

output

Timed out
 

3.14.25.6 Sympy [F]

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

input

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

output

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sqrt(cos(c + d*x))/(a + 
b*sec(c + d*x))**2, x)
 

3.14.25.7 Maxima [F]

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

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^2,x, algorithm="maxima")
 

output

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(b*se 
c(d*x + c) + a)^2, x)
 

3.14.25.8 Giac [F]

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

input

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*cos(d*x+c)^(1/2)/(a+b*sec(d*x+c) 
)^2,x, algorithm="giac")
 

output

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(cos(d*x + c))/(b*se 
c(d*x + c) + a)^2, x)
 

3.14.25.9 Mupad [F(-1)]

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

input

int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co 
s(c + d*x))^2,x)
 

output

int((cos(c + d*x)^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co 
s(c + d*x))^2, x)