\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\) [461]

Optimal result

Integrand size = 35, antiderivative size = 280 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \left (7 a^2 A b+8 A b^3-5 a^3 B-10 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{15 a^3 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^2 A+8 A b^2-10 a b B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b-5 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d \sqrt {\sec (c+d x)}} \] Output:

-2/15*(7*A*a^2*b+8*A*b^3-5*B*a^3-10*B*a*b^2)*((b+a*cos(d*x+c))/(a+b))^(1/2 
)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(a/(a+b))^(1/2))*sec(d*x+c)^(1/2)/ 
a^3/d/(a+b*sec(d*x+c))^(1/2)+2/15*(9*A*a^2+8*A*b^2-10*B*a*b)*EllipticE(sin 
(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a^3/d/((b+ 
a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)+2/5*A*(a+b*sec(d*x+c))^(1/2)*s 
in(d*x+c)/a/d/sec(d*x+c)^(3/2)-2/15*(4*A*b-5*B*a)*(a+b*sec(d*x+c))^(1/2)*s 
in(d*x+c)/a^2/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.67 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.71 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \left ((a+b) \left (9 a^2 A+8 A b^2-10 a b B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )+\left (-7 a^2 A b-8 A b^3+5 a^3 B+10 a b^2 B\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )+a (b+a \cos (c+d x)) (-4 A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)\right )}{15 a^3 d \sqrt {a+b \sec (c+d x)}} \] Input:

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

Output:

(2*Sqrt[Sec[c + d*x]]*((a + b)*(9*a^2*A + 8*A*b^2 - 10*a*b*B)*Sqrt[(b + a* 
Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*a)/(a + b)] + (-7*a^2*A*b 
 - 8*A*b^3 + 5*a^3*B + 10*a*b^2*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elli 
pticF[(c + d*x)/2, (2*a)/(a + b)] + a*(b + a*Cos[c + d*x])*(-4*A*b + 5*a*B 
 + 3*a*A*Cos[c + d*x])*Sin[c + d*x]))/(15*a^3*d*Sqrt[a + b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 2.11 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.543, Rules used = {3042, 4522, 27, 3042, 4592, 27, 3042, 4523, 3042, 4343, 3042, 3134, 3042, 3132, 4345, 3042, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4522

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \int \frac {-2 A b \sec ^2(c+d x)-3 a A \sec (c+d x)+4 A b-5 a B}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx}{5 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 A b \sec ^2(c+d x)-3 a A \sec (c+d x)+4 A b-5 a B}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx}{5 a}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {9 A a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {9 A a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {9 A a^2-10 b B a+(2 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{5 a}\)

\(\Big \downarrow \) 4523

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 4343

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\)

\(\Big \downarrow \) 4345

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (9 a^2 A-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \left (-5 a^3 B+7 a^2 A b-10 a b^2 B+8 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\)

Input:

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

Output:

(2*A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - ( 
-1/3*((-2*(7*a^2*A*b + 8*A*b^3 - 5*a^3*B - 10*a*b^2*B)*Sqrt[(b + a*Cos[c + 
 d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/ 
(a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2*A + 8*A*b^2 - 10*a*b*B)*Ellipti 
cE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(a*d*Sqrt[(b + a* 
Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]))/a + (2*(4*A*b - 5*a*B)*Sqrt[a 
+ b*Sec[c + d*x]]*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/(5*a)
 

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 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 4343

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] 
*(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S 
qrt[b + a*Sin[e + f*x]])   Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a 
, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4345

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/S 
qrt[a + b*Csc[e + f*x]])   Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[ 
{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4522

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

rule 4523

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

rule 4592

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1534\) vs. \(2(261)=522\).

Time = 30.92 (sec) , antiderivative size = 1535, normalized size of antiderivative = 5.48

method	result	size
default	\(\text {Expression too large to display}\)	\(1535\)
parts	\(\text {Expression too large to display}\)	\(1559\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

2/15/d/a^3/((a-b)/(a+b))^(1/2)*((9*cos(d*x+c)^2+18*cos(d*x+c)+9)*A*(1/(1+c 
os(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^3*Elli 
pticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(- 
9*cos(d*x+c)^2-18*cos(d*x+c)-9)*A*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*c 
os(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE(((a-b)/(a+b))^(1/2)*(csc( 
d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(8*cos(d*x+c)^2+16*cos(d*x+c)+8)* 
A*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2) 
*a*b^2*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b) 
)^(1/2))+(-8*cos(d*x+c)^2-16*cos(d*x+c)-8)*A*(1/(1+cos(d*x+c)))^(1/2)*(1/( 
a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^3*EllipticE(((a-b)/(a+b))^(1 
/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-10*cos(d*x+c)^2-20*cos 
(d*x+c)-10)*B*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d* 
x+c)))^(1/2)*a^2*b*EllipticE(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),( 
-(a+b)/(a-b))^(1/2))+(10*cos(d*x+c)^2+20*cos(d*x+c)+10)*B*(1/(1+cos(d*x+c) 
))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b^2*EllipticE(( 
(a-b)/(a+b))^(1/2)*(csc(d*x+c)-cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-9*cos(d 
*x+c)^2-18*cos(d*x+c)-9)*A*(1/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+ 
c))/(1+cos(d*x+c)))^(1/2)*a^3*EllipticF(((a-b)/(a+b))^(1/2)*(csc(d*x+c)-co 
t(d*x+c)),(-(a+b)/(a-b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*A*(1/(1+co 
s(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*b*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 520, normalized size of antiderivative = 1.86 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algo 
rithm="fricas")
 

Output:

1/45*(sqrt(2)*(-15*I*B*a^3 + 12*I*A*a^2*b - 20*I*B*a*b^2 + 16*I*A*b^3)*sqr 
t(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/ 
a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt(2)*(15*I* 
B*a^3 - 12*I*A*a^2*b + 20*I*B*a*b^2 - 16*I*A*b^3)*sqrt(a)*weierstrassPInve 
rse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x 
 + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*sqrt(2)*(-9*I*A*a^3 + 10*I*B*a^2* 
b - 8*I*A*a*b^2)*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9 
*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9 
*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) 
 - 3*sqrt(2)*(9*I*A*a^3 - 10*I*B*a^2*b + 8*I*A*a*b^2)*sqrt(a)*weierstrassZ 
eta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInve 
rse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x 
 + c) - 3*I*a*sin(d*x + c) + 2*b)/a)) + 6*(3*A*a^3*cos(d*x + c)^2 + (5*B*a 
^3 - 4*A*a^2*b)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sin( 
d*x + c)/sqrt(cos(d*x + c)))/(a^4*d)
 

Sympy [F(-1)]

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

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+b*sec(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F]

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

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algo 
rithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(5/2 
)),x)
 

Output:

int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(5/2 
)), x)
 

Reduce [F]

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

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a))/sec(c + d*x)**3,x)