\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [748]

Optimal result

Integrand size = 35, antiderivative size = 488 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {4 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^5 \sqrt {a+b} \left (a^2-b^2\right ) d}+\frac {2 \left (2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)+16 a^4 C+12 a^3 b C-b^4 (3 A+C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d} \] Output:

4/3*a*(a^2*b^2*(A-14*C)-b^4*(3*A-4*C)+8*a^4*C)*cot(d*x+c)*EllipticE((a+b*s 
ec(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*(2* 
a^2*b^2*(A-8*C)+3*a*b^3*(A-3*C)+16*a^4*C+12*a^3*b*C-b^4*(3*A+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)/( 
a^2-b^2)/d-2/3*(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*se 
c(d*x+c))^(3/2)-4/3*a*(2*A*b^4-3*C*a^4+5*C*a^2*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+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. \(4050\) vs. \(2(488)=976\).

Time = 23.68 (sec) , antiderivative size = 4050, normalized size of antiderivative = 8.30 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[(Sec[c + d*x]^3*(A + 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 + C*Sec[c + d*x]^2)*((-8*a*(a^2*A* 
b^2 - 3*A*b^4 + 8*a^4*C - 14*a^2*b^2*C + 4*b^4*C)*Sin[c + d*x])/(3*b^4*(a^ 
2 - b^2)^2) - (4*(a*A*b^2*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(3*b^2*(-a^2 
 + b^2)*(b + a*Cos[c + d*x])^2) - (4*(-(a^3*A*b^2*Sin[c + d*x]) + 5*a*A*b^ 
4*Sin[c + d*x] - 7*a^5*C*Sin[c + d*x] + 11*a^3*b^2*C*Sin[c + d*x]))/(3*b^3 
*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (4*C*Tan[c + d*x])/(3*b^3)))/(d*(A 
 + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)) + (8*(b + a*Cos[c 
 + d*x])^2*((4*a^3*A)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Se 
c[c + d*x]]) - (4*a*A*b)/((-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec 
[c + d*x]]) + (32*a^5*C)/(3*b^3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]*Sq 
rt[Sec[c + d*x]]) - (56*a^3*C)/(3*b*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x] 
]*Sqrt[Sec[c + d*x]]) + (16*a*b*C)/(3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d* 
x]]*Sqrt[Sec[c + d*x]]) - (10*a^2*A*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)^2* 
Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*A*Sqrt[Sec[c + d*x]])/(3*b^2*(-a^2 + b^ 
2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/((-a^2 + b^2 
)^2*Sqrt[b + a*Cos[c + d*x]]) + (10*a^2*C*Sqrt[Sec[c + d*x]])/((-a^2 + b^2 
)^2*Sqrt[b + a*Cos[c + d*x]]) + (32*a^6*C*Sqrt[Sec[c + d*x]])/(3*b^4*(-a^2 
 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (64*a^4*C*Sqrt[Sec[c + d*x]])/(3*b^2 
*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) + (2*b^2*C*Sqrt[Sec[c + d*x]])/( 
3*(-a^2 + b^2)^2*Sqrt[b + a*Cos[c + d*x]]) - (4*a^2*A*Cos[2*(c + d*x)]*...
 

Rubi [A] (verified)

Time = 2.30 (sec) , antiderivative size = 510, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4587, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4587

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4578

\(\displaystyle -\frac {\frac {2 \int -\frac {\sec (c+d x) \left (3 b \left (a^2-b^2\right ) \left (2 C a^2+A b^2-b^2 C\right ) \sec ^2(c+d x)+2 a \left (2 A b^4-\left (6 a^4-11 b^2 a^2+3 b^4\right ) C\right ) \sec (c+d x)+2 b \left (-3 C a^4+5 b^2 C a^2+2 A b^4\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}+\frac {4 a \left (-3 a^4 C+5 a^2 b^2 C+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {4 a \left (-3 a^4 C+5 a^2 b^2 C+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\sec (c+d x) \left (3 b \left (a^2-b^2\right ) \left (2 C a^2+A b^2-b^2 C\right ) \sec ^2(c+d x)+2 a \left (2 A b^4-\left (6 a^4-11 b^2 a^2+3 b^4\right ) C\right ) \sec (c+d x)+2 b \left (-3 C a^4+5 b^2 C a^2+2 A b^4\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {4 a \left (-3 a^4 C+5 a^2 b^2 C+2 A b^4\right ) \tan (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 b \left (a^2-b^2\right ) \left (2 C a^2+A b^2-b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+2 a \left (2 A b^4-\left (6 a^4-11 b^2 a^2+3 b^4\right ) C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 b \left (-3 C a^4+5 b^2 C a^2+2 A b^4\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b^2 \left (a^2-b^2\right )}}{3 b \left (a^2-b^2\right )}-\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

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

Output:

(-2*(A*b^2 + a^2*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)) - ((4*a*(2*A*b^4 - 3*a^4*C + 5*a^2*b^2*C)*Tan[c + d* 
x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (-(((-4*a*(a - b)*Sqrt[ 
a + b]*(a^2*b^2*(A - 14*C) - b^4*(3*A - 4*C) + 8*a^4*C)*Cot[c + d*x]*Ellip 
ticE[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*Sqrt[a + b]*(2*a^3*b^2*(A - 14*C) - 2*a*b^4*(3*A - 4*C) + 16*a^5* 
C - 4*a^4*b*C + b^5*(3*A + C) + a^2*b^3*(A + 7*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 + 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 4587

Int[((A_.) + 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^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^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(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, 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. \(3104\) vs. \(2(454)=908\).

Time = 63.85 (sec) , antiderivative size = 3105, normalized size of antiderivative = 6.36

method	result	size
default	\(\text {Expression too large to display}\)	\(3105\)
parts	\(\text {Expression too large to display}\)	\(3147\)

Input:

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETUR 
NVERBOSE)
 

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+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*(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^7*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1 
/2))+2*sin(d*x+c)*cos(d*x+c)*(3*cos(d*x+c)+1)*A*a^3*b^4+8*sin(d*x+c)*cos(d 
*x+c)*(cos(d*x+c)-3)*C*b*a^6+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^ 
7*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+16*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^7*E 
llipticE(-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))-2*A*a^5*b^2*cos(d*x+c)^2*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)*(-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^6-6*A*a*b^6*cos(d*x+c)*sin(d*x+ 
c)-16*C*a^7*cos(d*x+c)^2*sin(d*x+c)+C*b^7*(sin(d*x+c)+tan(d*x+c))+4*(cos(d 
*x+c)^3+6*cos(d*x+c)^2+9*cos(d*x+c)+4)*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^5*b^2*EllipticF(-csc(d* 
x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+4*(1-7*cos(d*x+c)^3-13*cos(d*x+c)^...
 

Fricas [F]

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

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algori 
thm="fricas")
 

Output:

integral((C*sec(d*x + c)^5 + 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]

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

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

Output:

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(5/ 
2), x)
 

Maxima [F(-1)]

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

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algori 
thm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^3*(A+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)**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