\(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [388]

Optimal result

Integrand size = 31, antiderivative size = 353 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (3 a A-A b+a B-3 b B) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

-2/3*(4*A*a*b-B*a^2-3*B*b^2)*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)/b^2/(a+b)^(3/2)/d+2/3*(3*A*a-A*b+B*a-3*B*b)*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)/b/(a+ 
b)^(3/2)/d-2/3*(A*b-B*a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3 
*(4*A*a*b-B*a^2-3*B*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 14.72 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.71 \[ \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^2(c+d x) (A+B \sec (c+d x)) \left (-\frac {2 \left (-4 a A b+a^2 B+3 b^2 B\right ) \sin (c+d x)}{3 b \left (-a^2+b^2\right )^2}+\frac {2 \left (A b^2 \sin (c+d x)-a b B \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {2 \left (-5 a^2 A b \sin (c+d x)+A b^3 \sin (c+d x)+2 a^3 B \sin (c+d x)+2 a b^2 B \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (B+A \cos (c+d x)) (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (A+B \sec (c+d x)) \left (2 (a+b) \left (-4 a A b+a^2 B+3 b^2 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (3 a A+A b-a B-3 b B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-4 a A b+a^2 B+3 b^2 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b \left (a^2-b^2\right )^2 d (B+A \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \] Input:

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

Output:

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((-2*(-4*a*A*b 
 + a^2*B + 3*b^2*B)*Sin[c + d*x])/(3*b*(-a^2 + b^2)^2) + (2*(A*b^2*Sin[c + 
 d*x] - a*b*B*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (2 
*(-5*a^2*A*b*Sin[c + d*x] + A*b^3*Sin[c + d*x] + 2*a^3*B*Sin[c + d*x] + 2* 
a*b^2*B*Sin[c + d*x]))/(3*a*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*(B + 
A*Cos[c + d*x])*(a + b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*Se 
c[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x] 
)*(2*(a + b)*(-4*a*A*b + a^2*B + 3*b^2*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d 
*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3*a*A + A*b - a*B 
- 3*b*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/( 
(a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/( 
a + b)] + (-4*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*S 
ec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*(a^2 - b^2)^2*d*(B + A*Cos[c + d 
*x])*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4491, 27, 3042, 4491, 27, 3042, 4493, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 A a^2-4 b B a+A b^2+\left (-B a^2+4 A b a-3 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\left (a^2 (-B)+4 a A b-3 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} (a (3 A+B)-b (A+3 B)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a^2-b^2}-\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {2 (a-b) \sqrt {a+b} (a (3 A+B)-b (A+3 B)) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a^2-b^2}-\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 \left (a^2-b^2\right )}-\frac {2 (A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

Input:

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

Output:

(-2*(A*b - a*B)*Tan[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) 
 + (((-2*(a - b)*Sqrt[a + b]*(4*a*A*b - a^2*B - 3*b^2*B)*Cot[c + d*x]*Elli 
pticE[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^2*d) + (2*(a - b)*Sqrt[a + b]*(a*(3*A + B) - b*(A + 3*B))*Cot[c + d*x]*E 
llipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sq 
rt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))] 
)/(b*d))/(a^2 - b^2) - (2*(4*a*A*b - a^2*B - 3*b^2*B)*Tan[c + d*x])/((a^2 
- b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*(a^2 - b^2))
 

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 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 4491

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e 
 + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 
/((m + 1)*(a^2 - b^2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp 
[(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free 
Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m 
, -1]
 

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 4493

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2042\) vs. \(2(323)=646\).

Time = 25.91 (sec) , antiderivative size = 2043, normalized size of antiderivative = 5.79

method	result	size
default	\(\text {Expression too large to display}\)	\(2043\)
parts	\(\text {Expression too large to display}\)	\(2098\)

Input:

int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

-2/3/d/(a-b)^2/(a+b)^2/b*(B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2 
)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^4*EllipticE(-csc(d*x+c)+cot(d*x+c),( 
(a-b)/(a+b))^(1/2))*(cos(d*x+c)^3+2*cos(d*x+c)^2+cos(d*x+c))+sin(d*x+c)*co 
s(d*x+c)*(-cos(d*x+c)-1)*A*b^4-4*A*a^3*b*cos(d*x+c)^2*sin(d*x+c)-2*B*a^3*b 
*cos(d*x+c)^2*sin(d*x+c)-3*B*b^4*cos(d*x+c)*sin(d*x+c)+B*a^4*cos(d*x+c)^2* 
sin(d*x+c)+(4*cos(d*x+c)^3+11*cos(d*x+c)^2+10*cos(d*x+c)+3)*A*(1/(a+b)*(b+ 
a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2* 
b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(cos(d*x+c)^3+6* 
cos(d*x+c)^2+9*cos(d*x+c)+4)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^( 
1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^3*EllipticF(-csc(d*x+c)+cot(d*x 
+c),((a-b)/(a+b))^(1/2))+(-4*cos(d*x+c)^3-9*cos(d*x+c)^2-6*cos(d*x+c)-1)*B 
*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c) 
))^(1/2)*a^2*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-3 
*cos(d*x+c)^3-10*cos(d*x+c)^2-11*cos(d*x+c)-4)*B*(1/(a+b)*(b+a*cos(d*x+c)) 
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^3*EllipticF(- 
csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-4*cos(d*x+c)^3-12*cos(d*x+c)^ 
2-12*cos(d*x+c)-4)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos( 
d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a- 
b)/(a+b))^(1/2))+(-4*cos(d*x+c)^2-8*cos(d*x+c)-4)*A*(1/(a+b)*(b+a*cos(d*x+ 
c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^3*Ellip...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((A + B*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**(5/2), x)
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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