\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [386]

Optimal result

Integrand size = 33, antiderivative size = 417 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (2 a^3 A b-6 a A b^3-8 a^4 B+15 a^2 b^2 B-3 b^4 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^4 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 b (A-3 B)-3 b^3 (A-B)-8 a^3 B+3 a b^2 (A+3 B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

2/3*(2*A*a^3*b-6*A*a*b^3-8*B*a^4+15*B*a^2*b^2-3*B*b^4)*cot(d*x+c)*Elliptic 
E((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^4/(a+b)^(3/2)/d+2/3 
*(2*a^2*b*(A-3*B)-3*b^3*(A-B)-8*B*a^3+3*a*b^2*(A+3*B))*cot(d*x+c)*Elliptic 
F((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)/b^3/(a+b)^(1/2)/(a^2-b^2)/d 
-2/3*a^2*(A*b-B*a)*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)+2/3*a 
*(2*A*a^2*b-6*A*b^3-5*B*a^3+9*B*a*b^2)*tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*s 
ec(d*x+c))^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 19.42 (sec) , antiderivative size = 641, normalized size of antiderivative = 1.54 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (\frac {2 \left (-2 a^3 A b+6 a A b^3+8 a^4 B-15 a^2 b^2 B+3 b^4 B\right ) \sin (c+d x)}{3 b^3 \left (-a^2+b^2\right )^2}-\frac {2 \left (a A b \sin (c+d x)-a^2 B \sin (c+d x)\right )}{3 b \left (-a^2+b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 \left (-a^3 A b \sin (c+d x)+5 a A b^3 \sin (c+d x)+4 a^4 B \sin (c+d x)-8 a^2 b^2 B \sin (c+d x)\right )}{3 b^2 \left (-a^2+b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}-\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-2 a^3 A b+6 a A b^3+8 a^4 B-15 a^2 b^2 B+3 b^4 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \left (3 a b^2 (A-3 B)+8 a^3 B+3 b^3 (A+B)-2 a^2 b (A+3 B)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-2 a^3 A b+6 a A b^3+8 a^4 B-15 a^2 b^2 B+3 b^4 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \] Input:

Integrate[(Sec[c + d*x]^3*(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]^3*((2*(-2*a^3*A*b + 6*a*A*b^3 + 8*a^4 
*B - 15*a^2*b^2*B + 3*b^4*B)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)^2) - (2*(a* 
A*b*Sin[c + d*x] - a^2*B*Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(b + a*Cos[c + d 
*x])^2) - (2*(-(a^3*A*b*Sin[c + d*x]) + 5*a*A*b^3*Sin[c + d*x] + 4*a^4*B*S 
in[c + d*x] - 8*a^2*b^2*B*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)^2*(b + a*Cos[ 
c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) - (2*(b + a*Cos[c + d*x])^2*Se 
c[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-2*a^3* 
A*b + 6*a*A*b^3 + 8*a^4*B - 15*a^2*b^2*B + 3*b^4*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]))]*El 
lipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(a + b)*(3*a*b^2* 
(A - 3*B) + 8*a^3*B + 3*b^3*(A + B) - 2*a^2*b*(A + 3*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)] + (-2*a^3*A*b + 6* 
a*A*b^3 + 8*a^4*B - 15*a^2*b^2*B + 3*b^4*B)*Cos[c + d*x]*(b + a*Cos[c + d* 
x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b^3*(a^2 - b^2)^2*d*Sqrt[Sec[ 
(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.63 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4516, 27, 3042, 4568, 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.

\(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4516

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

\(\displaystyle \frac {\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int -\frac {\sec (c+d x) \left (b^2 \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right )-b \left (-8 B a^4+2 A b a^3+15 b^2 B a^2-6 A b^3 a-3 b^4 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (b^2 \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right )-b \left (-8 B a^4+2 A b a^3+15 b^2 B a^2-6 A b^3 a-3 b^4 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b^2 \left (2 B a^3+A b a^2-6 b^2 B a+3 A b^3\right )-b \left (-8 B a^4+2 A b a^3+15 b^2 B a^2-6 A b^3 a-3 b^4 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {b (a-b) \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-b \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b (a-b) \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-b \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {2 a \left (-5 a^3 B+2 a^2 A b+9 a b^2 B-6 A b^3\right ) \tan (c+d x)}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {\frac {2 (a-b) \sqrt {a+b} \left (-8 a^3 B+2 a^2 b (A-3 B)+3 a b^2 (A+3 B)-3 b^3 (A-B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} \left (-8 a^4 B+2 a^3 A b+15 a^2 b^2 B-6 a A b^3-3 b^4 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{b \left (a^2-b^2\right )}}{3 b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \tan (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

Input:

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(5/2),x]
 

Output:

(-2*a^2*(A*b - a*B)*Tan[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Sec[c + d*x] 
)^(3/2)) + (((2*(a - b)*Sqrt[a + b]*(2*a^3*A*b - 6*a*A*b^3 - 8*a^4*B + 15* 
a^2*b^2*B - 3*b^4*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)*Sqrt[a + b]*(2*a^2 
*b*(A - 3*B) - 3*b^3*(A - B) - 8*a^3*B + 3*a*b^2*(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))] 
)/d)/(b*(a^2 - b^2)) + (2*a*(2*a^2*A*b - 6*A*b^3 - 5*a^3*B + 9*a*b^2*B)*Ta 
n[c + d*x])/((a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*b^2*(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 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]
 

rule 4516

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-a^2)*(A*b - a*B) 
*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x 
] + Simp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x] 
)^(m + 1)*Simp[a*b*(A*b - a*B)*(m + 1) - (A*b - a*B)*(a^2 + b^2*(m + 1))*Cs 
c[e + f*x] + b*B*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1 
]
 

rule 4568

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] 
)^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) 
 Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m 
+ 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], 
x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 
2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2907\) vs. \(2(387)=774\).

Time = 45.32 (sec) , antiderivative size = 2908, normalized size of antiderivative = 6.97

method	result	size
default	\(\text {Expression too large to display}\)	\(2908\)
parts	\(\text {Expression too large to display}\)	\(2956\)

Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/3/d/(a-b)^2/(a+b)^2/b^3*(3*(-cos(d*x+c)^2-3*cos(d*x+c)+2)*sin(d*x+c)*B* 
a^2*b^4+(15*cos(d*x+c)^2+5*cos(d*x+c)-3)*sin(d*x+c)*B*a^4*b^2+2*A*a^5*b*co 
s(d*x+c)^2*sin(d*x+c)+6*A*a*b^5*cos(d*x+c)*sin(d*x+c)-6*B*a*b^5*cos(d*x+c) 
*sin(d*x+c)+sin(d*x+c)*cos(d*x+c)*(3-cos(d*x+c))*A*a^4*b^2+sin(d*x+c)*cos( 
d*x+c)*(5*cos(d*x+c)-7)*A*a^2*b^4+8*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)*a^6*EllipticE(-csc(d*x+c)+co 
t(d*x+c),((a-b)/(a+b))^(1/2))*(-cos(d*x+c)^3-2*cos(d*x+c)^2-cos(d*x+c))+2* 
sin(d*x+c)*cos(d*x+c)*(11-4*cos(d*x+c))*B*a^3*b^3+3*(cos(d*x+c)^2+2*cos(d* 
x+c)+1)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+c 
os(d*x+c)))^(1/2)*b^6*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2) 
)+3*(cos(d*x+c)^2+2*cos(d*x+c)+1)*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)*b^6*EllipticF(-csc(d*x+c)+cot( 
d*x+c),((a-b)/(a+b))^(1/2))+3*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*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)*b^6 
*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+2*sin(d*x+c)*cos(d* 
x+c)*(-1-3*cos(d*x+c))*A*a^3*b^3+4*sin(d*x+c)*cos(d*x+c)*(-3+cos(d*x+c))*B 
*a^5*b+(cos(d*x+c)^3-3*cos(d*x+c)-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)*a^3*b^3*EllipticF(-csc(d*x+ 
c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(6*cos(d*x+c)^3+13*cos(d*x+c)^2+8*cos(d 
*x+c)+1)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/...
 

Fricas [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x, algorith 
m="fricas")
 

Output:

integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*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]

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

integrate(sec(d*x+c)**3*(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)**3/(a + b*sec(c + d*x))**(5/2), 
 x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x, algorith 
m="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x, algorith 
m="giac")
 

Output:

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/2), 
x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)),x)
 

Output:

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)), x)
 

Reduce [F]

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

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