\(\int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [751]

Optimal result

Integrand size = 27, antiderivative size = 517 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 b^2 \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 \left (6 a^2 A b-a A b^2-3 A b^3-a^3 C+3 a^2 b C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) b (a+b)^{3/2} d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^3 d}+\frac {2 \left (A b^2+a^2 C\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \] Output:

-2/3*(3*A*b^4-a^4*C-a^2*b^2*(7*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)/a^2/b^2/(a+b)^(1/2)/(a^2-b^2)/d-2/3*(6*A*a^ 
2*b-A*a*b^2-3*A*b^3-C*a^3+3*C*a^2*b)*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)/a^2/(a-b)/b/(a+b)^(3/2)/d-2*A*(a+b)^(1/2)*cot 
(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((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^ 
3/d+2/3*(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-2/3* 
(3*A*b^4-a^4*C-a^2*b^2*(7*A+3*C))*tan(d*x+c)/a^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. \(1715\) vs. \(2(517)=1034\).

Time = 17.36 (sec) , antiderivative size = 1715, normalized size of antiderivative = 3.32 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(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)*((4*(-7*a^2*A* 
b^2 + 3*A*b^4 - a^4*C - 3*a^2*b^2*C)*Sin[c + d*x])/(3*a^2*b*(-a^2 + b^2)^2 
) - (4*(A*b^3*Sin[c + d*x] + a^2*b*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)*(b 
+ a*Cos[c + d*x])^2) + (8*(4*a^2*A*b^2*Sin[c + d*x] - 2*A*b^4*Sin[c + d*x] 
 + a^4*C*Sin[c + d*x] + a^2*b^2*C*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(b + 
 a*Cos[c + d*x]))))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x]) 
^(5/2)) - (4*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + 
d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*Sqrt[(a + b - a*Tan[(c + d*x)/ 
2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]*(7*a^3*A*b^2*Tan[(c 
 + d*x)/2] + 7*a^2*A*b^3*Tan[(c + d*x)/2] - 3*a*A*b^4*Tan[(c + d*x)/2] - 3 
*A*b^5*Tan[(c + d*x)/2] + a^5*C*Tan[(c + d*x)/2] + a^4*b*C*Tan[(c + d*x)/2 
] + 3*a^3*b^2*C*Tan[(c + d*x)/2] + 3*a^2*b^3*C*Tan[(c + d*x)/2] - 14*a^3*A 
*b^2*Tan[(c + d*x)/2]^3 + 6*a*A*b^4*Tan[(c + d*x)/2]^3 - 2*a^5*C*Tan[(c + 
d*x)/2]^3 - 6*a^3*b^2*C*Tan[(c + d*x)/2]^3 + 7*a^3*A*b^2*Tan[(c + d*x)/2]^ 
5 - 7*a^2*A*b^3*Tan[(c + d*x)/2]^5 - 3*a*A*b^4*Tan[(c + d*x)/2]^5 + 3*A*b^ 
5*Tan[(c + d*x)/2]^5 + a^5*C*Tan[(c + d*x)/2]^5 - a^4*b*C*Tan[(c + d*x)/2] 
^5 + 3*a^3*b^2*C*Tan[(c + d*x)/2]^5 - 3*a^2*b^3*C*Tan[(c + d*x)/2]^5 + 6*a 
^4*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - 
Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2 
]^2)/(a + b)] - 12*a^2*A*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (...
 

Rubi [A] (verified)

Time = 2.10 (sec) , antiderivative size = 538, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {3042, 4549, 27, 3042, 4548, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4549

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

\(\displaystyle \frac {\frac {\left (a^4 (-C)-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) \left (a^3 C-3 a^2 b (2 A+C)+a A b^2+3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 A \sqrt {a+b} \left (a^2-b^2\right )^2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 \left (a^4 (-C)-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x)}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

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

Output:

(2*(A*b^2 + a^2*C)*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^( 
3/2)) + (((-2*(a - b)*Sqrt[a + b]*(3*A*b^4 - a^4*C - a^2*b^2*(7*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)]*Sqrt[-((b*(1 + Sec[c + d* 
x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(a*A*b^2 + 3*A*b^3 + a^3* 
C - 3*a^2*b*(2*A + 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)]*Sq 
rt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (6*A*Sqrt[a + b]*(a^2 - b^2 
)^2*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqr 
t[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b 
*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/(a*(a^2 - b^2)) - (2*(3*A*b^4 - a^4 
*C - a^2*b^2*(7*A + 3*C))*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c 
+ d*x]]))/(3*a*(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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   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, C}, x] && NeQ[a^2 - b^2, 0]
 

rule 4548

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - 
a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 
 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2))   Int[(a + b*Csc[e + f*x])^( 
m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x 
] + (A*b^2 - a*b*B + a^2*C)*(m + 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 4549

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3083\) vs. \(2(478)=956\).

Time = 29.90 (sec) , antiderivative size = 3084, normalized size of antiderivative = 5.97

method	result	size
default	\(\text {Expression too large to display}\)	\(3084\)
parts	\(\text {Expression too large to display}\)	\(3108\)

Input:

int((A+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/a^2*(sin(d*x+c)*cos(d*x+c)*(8*cos(d*x+c)-6)*A*a^3* 
b^3-7*A*a^4*b^2*cos(d*x+c)^2*sin(d*x+c)+(3*cos(d*x+c)^3+9*cos(d*x+c)^2+9*c 
os(d*x+c)+3)*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)*a*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b) 
)^(1/2))-C*a^6*cos(d*x+c)^2*sin(d*x+c)-3*A*b^6*cos(d*x+c)*sin(d*x+c)+sin(d 
*x+c)*cos(d*x+c)*(3*cos(d*x+c)+7)*a^2*A*b^4+sin(d*x+c)*cos(d*x+c)*(-4*cos( 
d*x+c)+2)*A*b^5*a+sin(d*x+c)*cos(d*x+c)*(-3*cos(d*x+c)+1)*a^4*b^2*C+sin(d* 
x+c)*cos(d*x+c)*(2*cos(d*x+c)-4)*C*a^3*b^3+(3*cos(d*x+c)^2+6*cos(d*x+c)+3) 
*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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-6*c 
os(d*x+c)^2-12*cos(d*x+c)-6)*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*EllipticPi(-csc(d*x+c)+cot(d*x+ 
c),-1,((a-b)/(a+b))^(1/2))+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*EllipticE(-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*C*a^2*b^4 
*cos(d*x+c)*sin(d*x+c)+2*C*a^5*b*cos(d*x+c)^2*sin(d*x+c)+(-cos(d*x+c)^3-3* 
cos(d*x+c)^2-3*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)*a^5*b*EllipticE(-csc(d*x+c)+cot(d*x 
+c),((a-b)/(a+b))^(1/2))+(-3*cos(d*x+c)^3-7*cos(d*x+c)^2-5*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...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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