\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [971]

Optimal result

Integrand size = 43, antiderivative size = 449 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (2 a^3 b B-6 a b^3 B+3 b^4 (A-C)-8 a^4 C+a^2 b^2 (A+15 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 \left (2 a^2 b (B-3 C)-3 b^3 (A+B-C)-8 a^3 C+a b^2 (A+3 B+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (3 A b^4+2 a^3 b B-6 a b^3 B-5 a^4 C+a^2 b^2 (A+9 C)\right ) \tan (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

2/3*(2*B*a^3*b-6*B*a*b^3+3*b^4*(A-C)-8*a^4*C+a^2*b^2*(A+15*C))*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-se 
c(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/(a+b)^(1/2)/(a^ 
2-b^2)/d+2/3*(2*a^2*b*(B-3*C)-3*b^3*(A+B-C)-8*a^3*C+a*b^2*(A+3*B+9*C))*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)/b^3/(a+b)^( 
1/2)/(a^2-b^2)/d+2/3*a*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+b 
*sec(d*x+c))^(3/2)+2/3*(3*A*b^4+2*B*a^3*b-6*B*a*b^3-5*a^4*C+a^2*b^2*(A+9*C 
))*tan(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4504\) vs. \(2(449)=898\).

Time = 24.26 (sec) , antiderivative size = 4504, normalized size of antiderivative = 10.03 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

((b + a*Cos[c + d*x])^3*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 
2)*((-4*(a^2*A*b^2 + 3*A*b^4 + 2*a^3*b*B - 6*a*b^3*B - 8*a^4*C + 15*a^2*b^ 
2*C - 3*b^4*C)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)^2) + (4*(A*b^2*Sin[c + d* 
x] - a*b*B*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(3*b*(-a^2 + b^2)*(b + a*Co 
s[c + d*x])^2) + (4*(2*a^2*A*b^2*Sin[c + d*x] + 2*A*b^4*Sin[c + d*x] + a^3 
*b*B*Sin[c + d*x] - 5*a*b^3*B*Sin[c + d*x] - 4*a^4*C*Sin[c + d*x] + 8*a^2* 
b^2*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)^2*(b + a*Cos[c + d*x]))))/(d*(A + 
 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) 
- (4*(b + a*Cos[c + d*x])^2*((2*a^2*A)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c 
+ d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*b^2)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + 
 d*x]]*Sqrt[Sec[c + d*x]]) + (4*a^3*B)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[ 
c + d*x]]*Sqrt[Sec[c + d*x]]) - (4*a*b*B)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c 
 + d*x]]*Sqrt[Sec[c + d*x]]) + (10*a^2*C)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c 
 + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^4*C)/(3*b^2*(-a^2 + b^2)^2*Sqrt[b + a 
*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*b^2*C)/((-a^2 + b^2)^2*Sqrt[b + a* 
Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*a^3*A*Sqrt[Sec[c + d*x]])/(3*b*(-a^ 
2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (2*a*A*b*Sqrt[Sec[c + d*x]])/(3*(-a 
^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (10*a^2*B*Sqrt[Sec[c + d*x]])/(3*( 
-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*B*Sqrt[Sec[c + d*x]])/(3* 
b^2*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*b^2*B*Sqrt[Sec[c + d*...
 

Rubi [A] (verified)

Time = 1.76 (sec) , antiderivative size = 461, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4578, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4578

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

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

Output:

(2*a*(A*b^2 - a*(b*B - a*C))*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*b*B - 6*a*b^3*B + 3* 
b^4*(A - C) - 8*a^4*C + a^2*b^2*(A + 15*C))*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*(B - 3*C) - 3*b^3*(A + B - C) - 8*a^3*C + a*b^2* 
(A + 3*B + 9*C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sq 
rt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-(( 
b*(1 + Sec[c + d*x]))/(a - b))])/d)/(b*(a^2 - b^2))) - (2*(3*A*b^4 + 2*a^3 
*b*B - 6*a*b^3*B - 5*a^4*C + a^2*b^2*(A + 9*C))*Tan[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 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]
 

rule 4578

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4016\) vs. \(2(419)=838\).

Time = 62.32 (sec) , antiderivative size = 4017, normalized size of antiderivative = 8.95

method	result	size
default	\(\text {Expression too large to display}\)	\(4017\)
parts	\(\text {Expression too large to display}\)	\(4130\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((C*sec(d*x + c)^4 + B*sec(d*x + c)^3 + A*sec(d*x + c)^2)*sqrt(b*s 
ec(d*x + c) + a)/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*se 
c(d*x + c) + a^3), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{4}}{\sec \left (d x +c \right )^{3} b^{3}+3 \sec \left (d x +c \right )^{2} a \,b^{2}+3 \sec \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{3} b^{3}+3 \sec \left (d x +c \right )^{2} a \,b^{2}+3 \sec \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right ) b +a}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3} b^{3}+3 \sec \left (d x +c \right )^{2} a \,b^{2}+3 \sec \left (d x +c \right ) a^{2} b +a^{3}}d x \right ) a \] Input:

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

Output:

int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**4)/(sec(c + d*x)**3*b**3 + 3*s 
ec(c + d*x)**2*a*b**2 + 3*sec(c + d*x)*a**2*b + a**3),x)*c + int((sqrt(sec 
(c + d*x)*b + a)*sec(c + d*x)**3)/(sec(c + d*x)**3*b**3 + 3*sec(c + d*x)** 
2*a*b**2 + 3*sec(c + d*x)*a**2*b + a**3),x)*b + int((sqrt(sec(c + d*x)*b + 
 a)*sec(c + d*x)**2)/(sec(c + d*x)**3*b**3 + 3*sec(c + d*x)**2*a*b**2 + 3* 
sec(c + d*x)*a**2*b + a**3),x)*a