\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [964]

Optimal result

Integrand size = 43, antiderivative size = 510 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^5 \sqrt {a+b} d}+\frac {2 \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \] Output:

2/15*(40*B*a^3*b-25*B*a*b^3-6*a^2*b^2*(5*A-4*C)-48*a^4*C+3*b^4*(5*A+3*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-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^5/(a+b 
)^(1/2)/d+2/15*(a^2*b*(40*B-36*C)-48*a^3*C-6*a*b^2*(5*A-5*B+2*C)-b^3*(15*A 
-5*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^4/(a+b)^(1/2)/d-2*(A*b^2-a*(B*b-C*a))*sec(d*x+c)^2*tan(d*x+c)/b/(a^2- 
b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/15*(20*B*a^2*b-5*B*b^3-3*a*b^2*(5*A-3*C)-2 
4*a^3*C)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(5*A*b^2-5* 
B*a*b+6*C*a^2-C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2 
-b^2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5013\) vs. \(2(510)=1020\).

Time = 28.27 (sec) , antiderivative size = 5013, normalized size of antiderivative = 9.83 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 2.37 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4586, 27, 3042, 4580, 27, 3042, 4570, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4586

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4580

\(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (\left (-24 C a^3+20 b B a^2-3 b^2 (5 A-3 C) a-5 b^3 B\right ) \sec ^2(c+d x)-b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2-5 b B a+5 A b^2-b^2 C\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\int \frac {\sec (c+d x) \left (\left (-24 C a^3+20 b B a^2-3 b^2 (5 A-3 C) a-5 b^3 B\right ) \sec ^2(c+d x)-b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \sec (c+d x)+2 a \left (6 C a^2-5 b B a+5 A b^2-b^2 C\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-24 C a^3+20 b B a^2-3 b^2 (5 A-3 C) a-5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (2 C a^2-5 b B a+5 A b^2+3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (6 C a^2-5 b B a+5 A b^2-b^2 C\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4570

\(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {\sec (c+d x) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 (5 A+C) a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3-6 b^2 (5 A-4 C) a^2-25 b^3 B a+3 b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\int \frac {\sec (c+d x) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 (5 A+C) a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3-6 b^2 (5 A-4 C) a^2-25 b^3 B a+3 b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 (5 A+C) a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3-6 b^2 (5 A-4 C) a^2-25 b^3 B a+3 b^4 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4493

\(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4319

\(\displaystyle -\frac {-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {\left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}}{5 b}-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\)

\(\Big \downarrow \) 4492

\(\displaystyle -\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}-\frac {-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}}{5 b}}{b \left (a^2-b^2\right )}\)

Input:

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

Output:

(-2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d* 
Sqrt[a + b*Sec[c + d*x]]) - ((-2*(5*A*b^2 - 5*a*b*B + 6*a^2*C - b^2*C)*Sec 
[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) - (-1/3*((-2*(a - 
 b)*Sqrt[a + b]*(40*a^3*b*B - 25*a*b^3*B - 6*a^2*b^2*(5*A - 4*C) - 48*a^4* 
C + 3*b^4*(5*A + 3*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)]*Sq 
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) - (2*(a - b)*Sqrt[a + b]*(a 
^2*b*(40*B - 36*C) - 48*a^3*C - 6*a*b^2*(5*A - 5*B + 2*C) - b^3*(15*A - 5* 
B + 9*C))*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))/b + (2*(20*a^2*b*B - 5*b^3*B - 3*a*b^2*(5 
*A - 3*C) - 24*a^3*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b 
))/(b*(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 4570

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[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

rule 4580

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[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 
1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3))   Int[Csc[e + f*x]*(a + b*Csc[e 
+ f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* 
(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & 
& NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 4586

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3113\) vs. \(2(478)=956\).

Time = 102.74 (sec) , antiderivative size = 3114, normalized size of antiderivative = 6.11

method	result	size
default	\(\text {Expression too large to display}\)	\(3114\)
parts	\(\text {Expression too large to display}\)	\(3163\)

Input:

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

Output:

-2/15/d/(a-b)/(a+b)/b^4*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c 
)+b*cos(d*x+c)+b)*(24*sin(d*x+c)*(cos(d*x+c)-1)*a^4*b*C+3*(-3*cos(d*x+c)^3 
+5*cos(d*x+c)^2-cos(d*x+c)-1)*C*a^2*b^3*tan(d*x+c)*sec(d*x+c)+48*(-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)*(c 
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b 
)/(a+b))^(1/2))+15*(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^5*EllipticE(- 
csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(cos(d*x+c)^2+2*cos(d*x+c)+1) 
*C*(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^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+15*(- 
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^5*EllipticF(-csc(d*x+c)+cot(d*x+c 
),((a-b)/(a+b))^(1/2))+3*(3*cos(d*x+c)^2-2*cos(d*x+c)-2)*C*a*b^4*tan(d*x+c 
)+15*A*a*b^4*cos(d*x+c)*sin(d*x+c)+6*(1+4*cos(d*x+c)^2+cos(d*x+c))*C*a^3*b 
^2*tan(d*x+c)+15*sin(d*x+c)*(cos(d*x+c)-1)*a^2*A*b^3+40*B*a^4*b*cos(d*x+c) 
*sin(d*x+c)+9*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(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^5*EllipticF(-csc( 
d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))-30*A*a^3*b^2*cos(d*x+c)*sin(d*x+c)+ 
25*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(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)*a*b^4*EllipticF(-csc(d*x+c)+...
 

Fricas [F]

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

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

Output:

integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(b*s 
ec(d*x + c) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5)/(sec(c + d*x)**2*b**2 + 2*s 
ec(c + d*x)*a*b + a**2),x)*c + int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)* 
*4)/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2),x)*b + int((sqrt(se 
c(c + d*x)*b + a)*sec(c + d*x)**3)/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)* 
a*b + a**2),x)*a