\(\int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [381]

Optimal result

Integrand size = 25, antiderivative size = 376 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 (A b-a B) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a b \sqrt {a+b} d}-\frac {2 (A b-a B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a b \sqrt {a+b} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 d}+\frac {2 b (A b-a B) \tan (c+d x)}{a \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \] Output:

2*(A*b-B*a)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b) 
/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1 
/2)/a/b/(a+b)^(1/2)/d-2*(A*b-B*a)*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1 
/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1 
+sec(d*x+c))/(a-b))^(1/2)/a/b/(a+b)^(1/2)/d-2*A*(a+b)^(1/2)*cot(d*x+c)*Ell 
ipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(b 
*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d+2*b*(A* 
b-B*a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1138\) vs. \(2(376)=752\).

Time = 12.82 (sec) , antiderivative size = 1138, normalized size of antiderivative = 3.03 \[ \int \frac {A+B \sec (c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

((b + a*Cos[c + d*x])^2*Sec[c + d*x]*(A + B*Sec[c + d*x])*((2*(-(A*b) + a* 
B)*Sin[c + d*x])/(a*(a^2 - b^2)) - (2*(-(A*b^2*Sin[c + d*x]) + a*b*B*Sin[c 
 + d*x]))/(a*(a^2 - b^2)*(b + a*Cos[c + d*x]))))/(d*(B + A*Cos[c + d*x])*( 
a + b*Sec[c + d*x])^(3/2)) - (2*(b + a*Cos[c + d*x])^(3/2)*Sqrt[Sec[c + d* 
x]]*(A + B*Sec[c + d*x])*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d 
*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(-(a*A*b*Tan[(c + d*x)/2]) - A*b^2*Tan 
[(c + d*x)/2] + a^2*B*Tan[(c + d*x)/2] + a*b*B*Tan[(c + d*x)/2] + 2*a*A*b* 
Tan[(c + d*x)/2]^3 - 2*a^2*B*Tan[(c + d*x)/2]^3 - a*A*b*Tan[(c + d*x)/2]^5 
 + A*b^2*Tan[(c + d*x)/2]^5 + a^2*B*Tan[(c + d*x)/2]^5 - a*b*B*Tan[(c + d* 
x)/2]^5 - 2*a^2*A*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b) 
]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[ 
(c + d*x)/2]^2)/(a + b)] + 2*A*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]] 
, (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d 
*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a^2*A*EllipticPi[-1, ArcSin[ 
Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d 
*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + 
b)] + 2*A*b^2*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Ta 
n[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x 
)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(-(A*b) + a*B)*EllipticE 
[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2...
 

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4411, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4411

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

\(\displaystyle \frac {-(a-b) (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 A \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {-b (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 A \sqrt {a+b} \left (a^2-b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} (A b-a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}+\frac {2 b (A b-a B) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4492

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

Input:

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

Output:

((2*(a - b)*Sqrt[a + b]*(A*b - a*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + 
 b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]) 
)/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (2*(a - b)*Sqr 
t[a + b]*(A*b - a*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x] 
]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt 
[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (2*A*Sqrt[a + b]*(a^2 - b^2)* 
Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a 
+ b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 
+ Sec[c + d*x]))/(a - b))])/(a*d))/(a*(a^2 - b^2)) + (2*b*(A*b - a*B)*Tan[ 
c + d*x])/(a*(a^2 - b^2)*d*Sqrt[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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

rule 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4409

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

rule 4411

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[b*(b*c - a*d)*Cot[e + f*x]*((a + b*Csc[e + f 
*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2) 
)   Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - 
 a*d)*(m + 1))*Csc[e + f*x] + b*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && N 
eQ[a^2 - b^2, 0] && IntegerQ[2*m]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   Int[Csc[e + f*x]*(( 
1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, 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. \(1069\) vs. \(2(347)=694\).

Time = 13.94 (sec) , antiderivative size = 1070, normalized size of antiderivative = 2.85

method	result	size
default	\(\text {Expression too large to display}\)	\(1070\)
parts	\(\text {Expression too large to display}\)	\(1134\)

Input:

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

Output:

2/d/a/(a+b)/(a-b)*((-(1-cos(d*x+c))^3*csc(d*x+c)^3-cot(d*x+c)+csc(d*x+c))* 
A*a*b+(-(1-cos(d*x+c))^3*csc(d*x+c)^3-cot(d*x+c)+csc(d*x+c))*B*a*b+((1-cos 
(d*x+c))^3*csc(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*b^2*A+((1-cos(d*x+c))^3*csc 
(d*x+c)^3-csc(d*x+c)+cot(d*x+c))*a^2*B-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 
2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+c 
ot(d*x+c),((a-b)/(a+b))^(1/2))*a^2-2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*( 
1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d 
*x+c),((a-b)/(a+b))^(1/2))*a*b+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a 
+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c 
),((a-b)/(a+b))^(1/2))*a*b+2*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)* 
(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticE(-csc(d*x+c)+cot(d*x+c),(( 
a-b)/(a+b))^(1/2))*b^2+4*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a 
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(( 
a-b)/(a+b))^(1/2))*a^2-4*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a 
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(( 
a-b)/(a+b))^(1/2))*b^2+2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a 
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b) 
/(a+b))^(1/2))*a^2+2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos 
(d*x+c))/(1+cos(d*x+c)))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+ 
b))^(1/2))*a*b-2*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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