\(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [425]

Optimal result

Integrand size = 33, antiderivative size = 263 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {(A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a \left (a^2-b^2\right ) d}+\frac {\left (2 a^2 A-A b^2-a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}-\frac {\left (3 a^2 A b-A b^3-a^3 B-a b^2 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a-b) (a+b)^2 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

(A*b-B*a)*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/(a^2-b^2)/d+(2*A*a^2-A*b^2-B*a*b)*cos(d*x+c)^(1/2)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^2/(a^2-b^2)/d-(3*A*a^2*b-A*b^ 
3-B*a^3-B*a*b^2)*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b), 
2^(1/2))*sec(d*x+c)^(1/2)/a^2/(a-b)/(a+b)^2/d-(A*b-B*a)*sec(d*x+c)^(1/2)*s 
in(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 6.56 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.79 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x)) \sqrt {\sec (c+d x)} (A+B \sec (c+d x)) \left (\frac {4 a^2 (-A b+a B) \sin (c+d x)}{a^2-b^2}-\frac {\cos (c+d x) \cot (c+d x) (a+b \sec (c+d x)) \left (-2 a^2 (-A b+a B) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-8 a b (a A-b B) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+(A b-a B) \left (4 a b-4 a b \sec ^2(c+d x)+4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{(a-b) b (a+b)}\right )}{4 a^2 d (B+A \cos (c+d x)) (a+b \sec (c+d x))^2} \] Input:

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

Output:

((b + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x])*((4*a^2*(-(A 
*b) + a*B)*Sin[c + d*x])/(a^2 - b^2) - (Cos[c + d*x]*Cot[c + d*x]*(a + b*S 
ec[c + d*x])*(-2*a^2*(-(A*b) + a*B)*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], 
 -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*Sqrt[Sec[c + d* 
x]]*Sqrt[-Tan[c + d*x]^2] - 8*a*b*(a*A - b*B)*EllipticPi[-(b/a), ArcSin[Sq 
rt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + (A*b - a 
*B)*(4*a*b - 4*a*b*Sec[c + d*x]^2 + 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d* 
x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 2*a*(a - 2*b)*Ellipti 
cF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2 
] - 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + 
d*x]]*Sqrt[-Tan[c + d*x]^2] + 4*b^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + 
 d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/((a - b)*b*(a + b 
))))/(4*a^2*d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^2)
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4515, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4515

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 4336

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

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 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 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 4336

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]]   Int[ 
1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4515

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

rule 4594

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2)   Int[(d*Csc[e + 
f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2   Int[(a*A - (A*b - a 
*B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, 
B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(802\) vs. \(2(254)=508\).

Time = 4.44 (sec) , antiderivative size = 803, normalized size of antiderivative = 3.05

method	result	size
default	\(\text {Expression too large to display}\)	\(803\)

Input:

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

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/a^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))+ 
2/a*(2*A*b-B*a)/(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)*Ellipt 
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2*b*(A*b-B*a)/a^2*(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*a*cos(1/2*d*x+1/2*c)^2-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+sin(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+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipt 
icPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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