\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) [420]

Optimal result

Integrand size = 33, antiderivative size = 196 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 \left (a^2 A+3 A b^2-3 a b B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^3 d}-\frac {2 b^2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^3 (a+b) d}+\frac {2 A \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}} \] Output:

-2*(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^2/d+2/3*(A*a^2+3*A*b^2-3*B*a*b)*cos(d*x+c)^(1/2)*InverseJacob 
iAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^3/d-2*b^2*(A*b-B*a)*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^3/(a+b)/d+2/3*A*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 4.84 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.42 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {2 \csc (c+d x) \left (3 a A b-3 a^2 B-3 a A b \sec ^2(c+d x)+3 a^2 B \sec ^2(c+d x)+a^2 A \sin (c+d x) \tan (c+d x)-3 a (-A b+a B) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+a (a A-3 A b+3 a B) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+3 A b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-3 a b B \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{3 a^3 d \sec ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

(2*Csc[c + d*x]*(3*a*A*b - 3*a^2*B - 3*a*A*b*Sec[c + d*x]^2 + 3*a^2*B*Sec[ 
c + d*x]^2 + a^2*A*Sin[c + d*x]*Tan[c + d*x] - 3*a*(-(A*b) + a*B)*Elliptic 
E[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] 
 + a*(a*A - 3*A*b + 3*a*B)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[ 
Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 3*A*b^2*EllipticPi[-(b/a), ArcSin[Sq 
rt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 3*a*b*B* 
EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt 
[-Tan[c + d*x]^2]))/(3*a^3*d*Sec[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4522, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 A \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {3 b^2 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 \left (a^2 A-3 a b B+3 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{3 a}\)

\(\Big \downarrow \) 4336

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

-1/3*(((6*a*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[ 
Sec[c + d*x]])/d - (2*(a^2*A + 3*A*b^2 - 3*a*b*B)*Sqrt[Cos[c + d*x]]*Ellip 
ticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (6*b^2*(A*b - a*B)*Sqrt[ 
Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] 
)/(a^2*(a + b)*d))/a + (2*A*Sin[c + d*x])/(3*a*d*Sqrt[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 4522

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim 
p[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* 
n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* 
x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] 
 && NeQ[a^2 - b^2, 0] && LeQ[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. \(821\) vs. \(2(183)=366\).

Time = 6.02 (sec) , antiderivative size = 822, normalized size of antiderivative = 4.19

method	result	size
default	\(\text {Expression too large to display}\)	\(822\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c)),x,method=_RETURNVER 
BOSE)
                                                                                    
                                                                                    
 

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral((A + B*sec(c + d*x))/((a + b*sec(c + d*x))*sec(c + d*x)**(3/2)), 
x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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