\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [970]

Optimal result

Integrand size = 43, antiderivative size = 549 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (8 a^4 b B-15 a^2 b^3 B+3 b^5 B-2 a^3 b^2 (A-14 C)+2 a b^4 (3 A-4 C)-16 a^5 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^5 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)-16 a^4 C+b^4 (3 A-3 B+C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 a \left (4 A b^4+a \left (3 a^2 b B-7 b^3 B-6 a^3 C+10 a b^2 C\right )\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2-a b B+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \] Output:

-2/3*(8*B*a^4*b-15*B*a^2*b^3+3*B*b^5-2*a^3*b^2*(A-14*C)+2*a*b^4*(3*A-4*C)- 
16*a^5*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)/(a^2-b^2)/d-2/3*(a^3*b*(8*B-12*C)-2*a^2*b^2*(A-3*B-8*C)- 
3*a*b^3*(A+3*B-3*C)-16*a^4*C+b^4*(3*A-3*B+C))*cot(d*x+c)*EllipticF((a+b*se 
c(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)/(a^2-b^2)/d-2/3*(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))^(3/ 
2)-2/3*a*(4*A*b^4+a*(3*B*a^2*b-7*B*b^3-6*C*a^3+10*C*a*b^2))*tan(d*x+c)/b^3 
/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)+2/3*(A*b^2-B*a*b+2*C*a^2-C*b^2)*(a+b 
*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(5489\) vs. \(2(549)=1098\).

Time = 29.11 (sec) , antiderivative size = 5489, normalized size of antiderivative = 10.00 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \] 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])^(5/2),x]
 

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 2.56 (sec) , antiderivative size = 562, normalized size of antiderivative = 1.02, 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, 4578, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4586

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4578

\(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (3 b \left (a^2-b^2\right ) \left (2 C a^2-b B a+A b^2-b^2 C\right ) \sec ^2(c+d x)+\left (-12 C a^5+6 b B a^4+22 b^2 C a^3-13 b^3 B a^2+2 b^4 (2 A-3 C) a+3 b^5 B\right ) \sec (c+d x)+b \left (4 A b^4+a \left (-6 C a^3+3 b B a^2+10 b^2 C a-7 b^3 B\right )\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}+\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\sec (c+d x) \left (3 b \left (a^2-b^2\right ) \left (2 C a^2-b B a+A b^2-b^2 C\right ) \sec ^2(c+d x)+\left (-12 C a^5+6 b B a^4+22 b^2 C a^3-13 b^3 B a^2+2 b^4 (2 A-3 C) a+3 b^5 B\right ) \sec (c+d x)+b \left (4 A b^4+a \left (-6 C a^3+3 b B a^2+10 b^2 C a-7 b^3 B\right )\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 b \left (a^2-b^2\right ) \left (2 C a^2-b B a+A b^2-b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (-12 C a^5+6 b B a^4+22 b^2 C a^3-13 b^3 B a^2+2 b^4 (2 A-3 C) a+3 b^5 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b \left (4 A b^4+a \left (-6 C a^3+3 b B a^2+10 b^2 C a-7 b^3 B\right )\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4570

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {2 \int \frac {3 \sec (c+d x) \left (\left (-4 C a^4+2 b B a^3+b^2 (A+7 C) a^2-6 b^3 B a+b^4 (3 A+C)\right ) b^2+\left (-16 C a^5+8 b B a^4-2 b^2 (A-14 C) a^3-15 b^3 B a^2+2 b^4 (3 A-4 C) a+3 b^5 B\right ) \sec (c+d x) b\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\sec (c+d x) \left (\left (-4 C a^4+2 b B a^3+b^2 (A+7 C) a^2-6 b^3 B a+b^4 (3 A+C)\right ) b^2+\left (-16 C a^5+8 b B a^4-2 b^2 (A-14 C) a^3-15 b^3 B a^2+2 b^4 (3 A-4 C) a+3 b^5 B\right ) \sec (c+d x) b\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b}+\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (-4 C a^4+2 b B a^3+b^2 (A+7 C) a^2-6 b^3 B a+b^4 (3 A+C)\right ) b^2+\left (-16 C a^5+8 b B a^4-2 b^2 (A-14 C) a^3-15 b^3 B a^2+2 b^4 (3 A-4 C) a+3 b^5 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) b\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4493

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (-16 a^5 C+8 a^4 b B-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) \left (-16 a^4 C+a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{b}+\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (-16 a^5 C+8 a^4 b B-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) \left (-16 a^4 C+a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4319

\(\displaystyle -\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (-16 a^5 C+8 a^4 b B-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-16 a^4 C+a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{b}+\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4492

\(\displaystyle -\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {\frac {2 a \tan (c+d x) \left (a \left (-6 a^3 C+3 a^2 b B+10 a b^2 C-7 b^3 B\right )+4 A b^4\right )}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\frac {2 \left (a^2-b^2\right ) \tan (c+d x) \left (2 a^2 C-a b B+A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{d}+\frac {-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-16 a^4 C+a^3 b (8 B-12 C)-2 a^2 b^2 (A-3 B-8 C)-3 a b^3 (A+3 B-3 C)+b^4 (3 A-3 B+C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-16 a^5 C+8 a^4 b B-2 a^3 b^2 (A-14 C)-15 a^2 b^3 B+2 a b^4 (3 A-4 C)+3 b^5 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}}{b}}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}\)

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])^(5/2),x]
 

Output:

(-2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)* 
d*(a + b*Sec[c + d*x])^(3/2)) - ((2*a*(4*A*b^4 + a*(3*a^2*b*B - 7*b^3*B - 
6*a^3*C + 10*a*b^2*C))*Tan[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + 
 d*x]]) - (((-2*(a - b)*Sqrt[a + b]*(8*a^4*b*B - 15*a^2*b^3*B + 3*b^5*B - 
2*a^3*b^2*(A - 14*C) + 2*a*b^4*(3*A - 4*C) - 16*a^5*C)*Cot[c + d*x]*Ellipt 
icE[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]*(a^3*b*(8*B - 12*C) - 2*a^2*b^2*(A - 3*B - 8*C 
) - 3*a*b^3*(A + 3*B - 3*C) - 16*a^4*C + b^4*(3*A - 3*B + 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) 
)])/d)/b + (2*(a^2 - b^2)*(A*b^2 - a*b*B + 2*a^2*C - b^2*C)*Sqrt[a + b*Sec 
[c + d*x]]*Tan[c + d*x])/d)/(b^2*(a^2 - b^2)))/(3*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 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 
]
 

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. \(4724\) vs. \(2(515)=1030\).

Time = 86.32 (sec) , antiderivative size = 4725, normalized size of antiderivative = 8.61

method	result	size
default	\(\text {Expression too large to display}\)	\(4725\)
parts	\(\text {Expression too large to display}\)	\(4809\)

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))^(5/2),x, 
method=_RETURNVERBOSE)
 

Output:

-2/3/d/(a-b)^2/(a+b)^2/b^4*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)/(a^2*cos( 
d*x+c)^2+2*a*b*cos(d*x+c)+b^2)*(sin(d*x+c)*cos(d*x+c)*(-cos(d*x+c)+3)*A*a^ 
4*b^3-8*B*a^6*b*cos(d*x+c)^2*sin(d*x+c)-6*B*a*b^6*cos(d*x+c)*sin(d*x+c)+2* 
A*a^5*b^2*cos(d*x+c)^2*sin(d*x+c)+6*A*a*b^6*cos(d*x+c)*sin(d*x+c)+(13*cos( 
d*x+c)^3-43*cos(d*x+c)^2-cos(d*x+c)-1)*C*a^4*b^3*tan(d*x+c)+(-cos(d*x+c)^3 
+15*cos(d*x+c)^2+2*cos(d*x+c)+2)*C*a^2*b^5*tan(d*x+c)+2*sin(d*x+c)*cos(d*x 
+c)*(-1-3*cos(d*x+c))*A*a^3*b^4+2*sin(d*x+c)*cos(d*x+c)*(11-4*cos(d*x+c))* 
a^3*b^4*B+sin(d*x+c)*(15*cos(d*x+c)^2+5*cos(d*x+c)-3)*B*a^4*b^3+3*sin(d*x+ 
c)*(-cos(d*x+c)^2-3*cos(d*x+c)+2)*a^2*b^5*B+2*sin(d*x+c)*(3-14*cos(d*x+c)^ 
2-5*cos(d*x+c))*a^5*b^2*C+4*sin(d*x+c)*(-3+2*cos(d*x+c)^2+4*cos(d*x+c))*C* 
a^3*b^4+2*sin(d*x+c)*(-cos(d*x+c)+3)*C*a*b^6+sin(d*x+c)*cos(d*x+c)*(5*cos( 
d*x+c)-7)*a^2*A*b^5+(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^7*EllipticF( 
-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*sin(d*x+c)*cos(d*x+c)*(-cos( 
d*x+c)+3)*C*a^6*b+4*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)-3)*a^5*b^2*B+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^7*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)/(cos(d*x+ 
c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^7*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...
 

Fricas [F]

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

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

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

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

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] 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))^(5 
/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

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

Mupad [F(-1)]

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

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

Reduce [F]

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

int(sec(d*x+c)^3*(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)**5)/(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)**4)/(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)**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)*a