\(\int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx\) [175]

Optimal result

Integrand size = 26, antiderivative size = 152 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}-\frac {A \tan (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {7 A \tan (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}} \] Output:

2*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(5/2)/d-23/16*A*ar 
ctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))*2^(1/2)/a^(5/2 
)/d-1/2*A*tan(d*x+c)/d/(a-a*sec(d*x+c))^(5/2)-7/8*A*tan(d*x+c)/a/d/(a-a*se 
c(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {A \left (16 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) (-1+\sec (c+d x))^2+\sqrt {1+\sec (c+d x)} (-11+7 \sec (c+d x))-46 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) \sec ^2(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right ) \tan (c+d x)}{8 a^2 d (-1+\sec (c+d x))^2 \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \] Input:

Integrate[(A + A*Sec[c + d*x])/(a - a*Sec[c + d*x])^(5/2),x]
 

Output:

(A*(16*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*(-1 + Sec[c + d*x])^2 + Sqrt[1 + Se 
c[c + d*x]]*(-11 + 7*Sec[c + d*x]) - 46*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d 
*x]]/Sqrt[2]]*Sec[c + d*x]^2*Sin[(c + d*x)/2]^4)*Tan[c + d*x])/(8*a^2*d*(- 
1 + Sec[c + d*x])^2*Sqrt[1 + Sec[c + d*x]]*Sqrt[a - a*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.38, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4392, 3042, 4375, 373, 402, 27, 397, 216}

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.

\(\displaystyle \int \frac {A \sec (c+d x)+A}{(a-a \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4392

\(\displaystyle -a A \int \frac {\tan ^2(c+d x)}{(a-a \sec (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle -a A \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx\)

\(\Big \downarrow \) 4375

\(\displaystyle \frac {2 A \int \frac {\tan ^2(c+d x)}{(a-a \sec (c+d x)) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1\right ) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^3}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a d}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {2 A \left (-\frac {\int \frac {1-\frac {3 a \tan ^2(c+d x)}{a-a \sec (c+d x)}}{\left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1\right ) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a}-\frac {\tan (c+d x)}{4 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}\right )}{a d}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {2 A \left (-\frac {\frac {\int \frac {a \left (9-\frac {7 a \tan ^2(c+d x)}{a-a \sec (c+d x)}\right )}{\left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1\right ) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a}+\frac {7 \tan (c+d x)}{4 \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}}{4 a}-\frac {\tan (c+d x)}{4 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}\right )}{a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \left (-\frac {\frac {1}{4} \int \frac {9-\frac {7 a \tan ^2(c+d x)}{a-a \sec (c+d x)}}{\left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1\right ) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )+\frac {7 \tan (c+d x)}{4 \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}}{4 a}-\frac {\tan (c+d x)}{4 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}\right )}{a d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {2 A \left (-\frac {\frac {1}{4} \left (16 \int \frac {1}{\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )-23 \int \frac {1}{\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )\right )+\frac {7 \tan (c+d x)}{4 \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}}{4 a}-\frac {\tan (c+d x)}{4 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}\right )}{a d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 A \left (-\frac {\frac {1}{4} \left (\frac {23 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a}}-\frac {16 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a}}\right )+\frac {7 \tan (c+d x)}{4 \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}}{4 a}-\frac {\tan (c+d x)}{4 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}\right )}{a d}\)

Input:

Int[(A + A*Sec[c + d*x])/(a - a*Sec[c + d*x])^(5/2),x]
 

Output:

(2*A*(-1/4*Tan[c + d*x]/(a*Sqrt[a - a*Sec[c + d*x]]*(2 + (a*Tan[c + d*x]^2 
)/(a - a*Sec[c + d*x]))^2) - (((-16*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - 
 a*Sec[c + d*x]]])/Sqrt[a] + (23*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sq 
rt[a - a*Sec[c + d*x]])])/(Sqrt[2]*Sqrt[a]))/4 + (7*Tan[c + d*x])/(4*Sqrt[ 
a - a*Sec[c + d*x]]*(2 + (a*Tan[c + d*x]^2)/(a - a*Sec[c + d*x]))))/(4*a)) 
)/(a*d)
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

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

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4375

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d)   Subst[Int[x^m*((2 + a*x^2 
)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] 
]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I 
ntegerQ[n - 1/2]
 

rule 4392

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m   Int[Cot[e + f*x]^(2*m)*( 
c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E 
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !( 
IntegerQ[n] && GtQ[m - n, 0])
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(795\) vs. \(2(127)=254\).

Time = 2.51 (sec) , antiderivative size = 796, normalized size of antiderivative = 5.24

method	result	size
default	\(\text {Expression too large to display}\)	\(796\)
parts	\(\text {Expression too large to display}\)	\(796\)

Input:

int((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

A*(-1/64/d/(-a/(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/((2* 
cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)/a^2*(arctanh(2^(1/ 
2)*cos(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/( 
cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*(128*cot(1/2*d*x+1/2*c)-128*csc(1/2*d*x+1/ 
2*c))+2^(1/2)*arctanh((2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1)/((2* 
cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*(-43*cot(1/2*d*x+ 
1/2*c)+43*csc(1/2*d*x+1/2*c))+2^(1/2)*ln(2*(((2*cos(1/2*d*x+1/2*c)^2-1)/(c 
os(1/2*d*x+1/2*c)+1)^2)^(1/2)*cos(1/2*d*x+1/2*c)+((2*cos(1/2*d*x+1/2*c)^2- 
1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)-2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/ 
2*c)+1))*(43*cot(1/2*d*x+1/2*c)-43*csc(1/2*d*x+1/2*c))+2^(1/2)*((2*cos(1/2 
*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(30*cot(1/2*d*x+1/2*c)^3- 
26*cot(1/2*d*x+1/2*c)*csc(1/2*d*x+1/2*c)^2))-1/64/d*2^(1/2)/(-a/(2*cos(1/2 
*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/((2*cos(1/2*d*x+1/2*c)^2-1)/( 
cos(1/2*d*x+1/2*c)+1)^2)^(1/2)/a^2*(arctanh((2*cos(1/2*d*x+1/2*c)-1)/(cos( 
1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1 
/2))*(-3*cot(1/2*d*x+1/2*c)+3*csc(1/2*d*x+1/2*c))+ln(2*(((2*cos(1/2*d*x+1/ 
2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*cos(1/2*d*x+1/2*c)+((2*cos(1/2*d 
*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)-2*cos(1/2*d*x+1/2*c)-1)/(co 
s(1/2*d*x+1/2*c)+1))*(3*cot(1/2*d*x+1/2*c)-3*csc(1/2*d*x+1/2*c))+((2*cos(1 
/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*(14*cot(1/2*d*x+1/2*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (127) = 254\).

Time = 0.10 (sec) , antiderivative size = 590, normalized size of antiderivative = 3.88 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="fricas")
 

Output:

[-1/32*(23*sqrt(2)*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(-a)*log( 
(2*sqrt(2)*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - 
 a)/cos(d*x + c)) + (3*a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c) - 
1)*sin(d*x + c)))*sin(d*x + c) + 32*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + 
 A)*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d 
*x + c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c) + a)*sin(d*x + c))/sin(d*x 
+ c))*sin(d*x + c) - 4*(11*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 - 7*A*cos 
(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^3*d*cos(d*x + c)^2 
 - 2*a^3*d*cos(d*x + c) + a^3*d)*sin(d*x + c)), 1/16*(23*sqrt(2)*(A*cos(d* 
x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + 
c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c) - 
32*(A*cos(d*x + c)^2 - 2*A*cos(d*x + c) + A)*sqrt(a)*arctan(sqrt((a*cos(d* 
x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin(d*x + c 
) + 2*(11*A*cos(d*x + c)^3 + 4*A*cos(d*x + c)^2 - 7*A*cos(d*x + c))*sqrt(( 
a*cos(d*x + c) - a)/cos(d*x + c)))/((a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d* 
x + c) + a^3*d)*sin(d*x + c))]
 

Sympy [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=A \left (\int \frac {\sec {\left (c + d x \right )}}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {1}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(5/2),x)
 

Output:

A*(Integral(sec(c + d*x)/(a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x)**2 - 
 2*a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a**2*sqrt(-a*sec(c + d*x) 
 + a)), x) + Integral(1/(a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x)**2 - 
2*a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a**2*sqrt(-a*sec(c + d*x) 
+ a)), x))
 

Maxima [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\int { \frac {A \sec \left (d x + c\right ) + A}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="maxima")
 

Output:

integrate((A*sec(d*x + c) + A)/(-a*sec(d*x + c) + a)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {23 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {32 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {\sqrt {2} {\left (9 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 7 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{16 \, d} \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="giac")
 

Output:

1/16*(23*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(5 
/2) - 32*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/ 
a^(5/2) - sqrt(2)*(9*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 7*sqrt(a*tan 
(1/2*d*x + 1/2*c)^2 - a)*A*a)/(a^4*tan(1/2*d*x + 1/2*c)^4))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {A}{\cos \left (c+d\,x\right )}}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(5/2),x)
 

Output:

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}-3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )-1}d x +\int \frac {\sqrt {-\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{3}-3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )-1}d x \right )}{a^{2}} \] Input:

int((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x)
 

Output:

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