\(\int \frac {\sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1035]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 45, antiderivative size = 273 \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{15 a^2 d \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (2 A b^2-5 a b B-3 a^2 (3 A+5 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 (A b+5 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt {\sec (c+d x)}} \] Output:

-2/15*(a^2-b^2)*(2*A*b-5*B*a)*((b+a*cos(d*x+c))/(a+b))^(1/2)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2)*(a/(a+b))^(1/2))*sec(d*x+c)^(1/2)/a^2/d/(a+b*sec( 
d*x+c))^(1/2)-2/15*(2*A*b^2-5*B*a*b-3*a^2*(3*A+5*C))*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^2/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)*sin(d*x+c 
)/d/sec(d*x+c)^(3/2)+2/15*(A*b+5*B*a)*(a+b*sec(d*x+c))^(1/2)*sin(d*x+c)/a/ 
d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 6.88 (sec) , antiderivative size = 3426, normalized size of antiderivative = 12.55 \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(9* 
a^2*A - 2*A*b^2 + 5*a*b*B + 15*a^2*C)*Cot[c])/(15*a^2*d) + (4*(A*b + 5*a*B 
)*Cos[d*x]*Sin[c])/(15*a*d) + (2*A*Cos[2*d*x]*Sin[2*c])/(5*d) + (4*(A*b + 
5*a*B)*Cos[c]*Sin[d*x])/(15*a*d) + (2*A*Cos[2*c]*Sin[2*d*x])/(5*d)))/((A + 
 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(5/2)) - (28*A* 
b*AppellF1[1/2, 1/2, 1/2, 3/2, (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Si 
n[d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(1 + (b*Csc[c])/(a*Sqrt[1 
+ Cot[c]^2]))), (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[ 
Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(-1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))) 
]*Csc[c]*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)* 
Sec[d*x - ArcTan[Cot[c]]]*Sqrt[(a*Sqrt[1 + Cot[c]^2] - a*Sqrt[1 + Cot[c]^2 
]*Sin[d*x - ArcTan[Cot[c]]])/(a*Sqrt[1 + Cot[c]^2] - b*Csc[c])]*Sqrt[(a*Sq 
rt[1 + Cot[c]^2] + a*Sqrt[1 + Cot[c]^2]*Sin[d*x - ArcTan[Cot[c]]])/(a*Sqrt 
[1 + Cot[c]^2] + b*Csc[c])]*Sqrt[b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - 
 ArcTan[Cot[c]]]])/(15*a*d*Sqrt[b + a*Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + 
 d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*Sec[c + d*x]^(5/2)) - (4*B* 
AppellF1[1/2, 1/2, 1/2, 3/2, (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[ 
d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(1 + (b*Csc[c])/(a*Sqrt[1 + 
Cot[c]^2]))), (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Co 
t[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(-1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))...
 

Rubi [A] (verified)

Time = 2.11 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.03, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4582, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4582

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {b (2 A+5 C) \sec ^2(c+d x)+(3 a A+5 b B+5 a C) \sec (c+d x)+A b+5 a B}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {b (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+A b+5 a B}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4592

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sec (c+d x) a+2 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \sec (c+d x) a+2 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-3 (3 A+5 C) a^2-5 b B a-(7 A b+15 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4523

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4343

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {2 (5 a B+A b) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (-3 a^2 (3 A+5 C)-5 a b B+2 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {\left (a^2-b^2\right ) (2 A b-5 a B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}\right )+\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 4345

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

(2*A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (-1 
/3*((2*(a^2 - b^2)*(2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elli 
pticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(a*d*Sqrt[a + b*Sec[ 
c + d*x]]) + (2*(2*A*b^2 - 5*a*b*B - 3*a^2*(3*A + 5*C))*EllipticE[(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*(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
 

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 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 4582
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*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(d*n)   Int[(a + b*Csc[e + f*x])^(m - 1)*(d* 
Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs 
c[e + f*x] - b*(C*n + A*(m + n + 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] && GtQ[m, 0] && LeQ[n, -1]
 

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. \(2029\) vs. \(2(254)=508\).

Time = 22.56 (sec) , antiderivative size = 2030, normalized size of antiderivative = 7.44

method result size
default \(\text {Expression too large to display}\) \(2030\)
parts \(\text {Expression too large to display}\) \(2096\)

Input:

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

Output:

2/15/d/((a-b)/(a+b))^(1/2)/a^2*((-9*cos(d*x+c)^2-18*cos(d*x+c)-9)*A*(1/(co 
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*Ell 
ipticE(((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/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a* 
cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticE(((a-b)/(a+b))^(1/2)*(-cs 
c(d*x+c)+cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2) 
*A*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2 
)*a*b^2*EllipticE(((a-b)/(a+b))^(1/2)*(-csc(d*x+c)+cot(d*x+c)),(-(a+b)/(a- 
b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*A*(1/(cos(d*x+c)+1))^(1/2)*(1/ 
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(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))+(-5*cos(d*x+c)^2-10*co 
s(d*x+c)-5)*B*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+ 
c)+1))^(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))+(5*cos(d*x+c)^2+10*cos(d*x+c)+5)*B*(1/(cos(d*x+c)+1) 
)^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^2*EllipticE((( 
a-b)/(a+b))^(1/2)*(-csc(d*x+c)+cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(-15*cos( 
d*x+c)^2-30*cos(d*x+c)-15)*C*(1/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d* 
x+c))/(cos(d*x+c)+1))^(1/2)*a^3*EllipticE(((a-b)/(a+b))^(1/2)*(-csc(d*x+c) 
+cot(d*x+c)),(-(a+b)/(a-b))^(1/2))+(15*cos(d*x+c)^2+30*cos(d*x+c)+15)*C*(1 
/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 539, normalized size of antiderivative = 1.97 \[ \int \frac {\sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

1/45*(sqrt(2)*(-15*I*B*a^3 - 3*I*(A + 5*C)*a^2*b + 10*I*B*a*b^2 - 4*I*A*b^ 
3)*sqrt(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 + 3*I*(A + 5*C)*a^2*b - 10*I*B*a*b^2 + 4*I*A*b^3)*sqrt(a)*weie 
rstrassPInverse(-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)*(-3*I*(3*A + 
5*C)*a^3 - 5*I*B*a^2*b + 2*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*si 
n(d*x + c) + 2*b)/a)) - 3*sqrt(2)*(3*I*(3*A + 5*C)*a^3 + 5*I*B*a^2*b - 2*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)) + 6*(3 
*A*a^3*cos(d*x + c)^2 + (5*B*a^3 + 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^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a))/sec(c + d*x)**3,x)*a + i 
nt((sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a))/sec(c + d*x)**2,x)*b + in 
t((sqrt(sec(c + d*x))*sqrt(sec(c + d*x)*b + a))/sec(c + d*x),x)*c