\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [718]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [F]
Sympy [F]
Maxima [F(-1)]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 454 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (21 A+11 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C+6 a^2 b C-21 b^3 (9 A+7 C)+3 a b^2 (21 A+13 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac {2 a \left (63 A b^2+8 a^2 C+39 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{315 b^2 d}+\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b^2 d}-\frac {8 a C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{9 b d} \] Output:

-2/315*(a-b)*(a+b)^(1/2)*(8*a^4*C+21*b^4*(9*A+7*C)+3*a^2*b^2*(21*A+11*C))* 
cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2 
))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d-2/ 
315*(a-b)*(a+b)^(1/2)*(8*a^3*C+6*a^2*b*C-21*b^3*(9*A+7*C)+3*a*b^2*(21*A+13 
*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^3 
/d+2/315*a*(63*A*b^2+8*C*a^2+39*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b 
^2/d+2/315*(8*C*a^2+7*b^2*(9*A+7*C))*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^2 
/d-8/63*a*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b^2/d+2/9*C*sec(d*x+c)*(a+b* 
sec(d*x+c))^(5/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 19.75 (sec) , antiderivative size = 685, normalized size of antiderivative = 1.51 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (2 (a+b) \left (8 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (21 A+11 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b (a+b) \left (8 a^3 C-6 a^2 b C+21 b^3 (9 A+7 C)+3 a b^2 (21 A+13 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (8 a^4 C+21 b^4 (9 A+7 C)+3 a^2 b^2 (21 A+11 C)\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{315 b^3 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {7}{2}}(c+d x)}+\frac {\cos ^3(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (63 a^2 A b^2+189 A b^4+8 a^4 C+33 a^2 b^2 C+147 b^4 C\right ) \sin (c+d x)}{315 b^3}+\frac {4 \sec ^2(c+d x) \left (63 A b^2 \sin (c+d x)+3 a^2 C \sin (c+d x)+49 b^2 C \sin (c+d x)\right )}{315 b}+\frac {8 \sec (c+d x) \left (63 a A b^2 \sin (c+d x)-2 a^3 C \sin (c+d x)+44 a b^2 C \sin (c+d x)\right )}{315 b^2}+\frac {40}{63} a C \sec ^2(c+d x) \tan (c+d x)+\frac {4}{9} b C \sec ^3(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x)) (A+2 C+A \cos (2 c+2 d x))} \] Input:

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

Output:

(-4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(A + 
C*Sec[c + d*x]^2)*(2*(a + b)*(8*a^4*C + 21*b^4*(9*A + 7*C) + 3*a^2*b^2*(21 
*A + 11*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x] 
)/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b 
)/(a + b)] - 2*b*(a + b)*(8*a^3*C - 6*a^2*b*C + 21*b^3*(9*A + 7*C) + 3*a*b 
^2*(21*A + 13*C))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c 
+ d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], 
(a - b)/(a + b)] + (8*a^4*C + 21*b^4*(9*A + 7*C) + 3*a^2*b^2*(21*A + 11*C) 
)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/ 
(315*b^3*d*(b + a*Cos[c + d*x])^2*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[ 
(c + d*x)/2]^2]*Sec[c + d*x]^(7/2)) + (Cos[c + d*x]^3*(a + b*Sec[c + d*x]) 
^(3/2)*(A + C*Sec[c + d*x]^2)*((4*(63*a^2*A*b^2 + 189*A*b^4 + 8*a^4*C + 33 
*a^2*b^2*C + 147*b^4*C)*Sin[c + d*x])/(315*b^3) + (4*Sec[c + d*x]^2*(63*A* 
b^2*Sin[c + d*x] + 3*a^2*C*Sin[c + d*x] + 49*b^2*C*Sin[c + d*x]))/(315*b) 
+ (8*Sec[c + d*x]*(63*a*A*b^2*Sin[c + d*x] - 2*a^3*C*Sin[c + d*x] + 44*a*b 
^2*C*Sin[c + d*x]))/(315*b^2) + (40*a*C*Sec[c + d*x]^2*Tan[c + d*x])/63 + 
(4*b*C*Sec[c + d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])*(A + 2*C + 
 A*Cos[2*c + 2*d*x]))
 

Rubi [A] (verified)

Time = 2.01 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.486, Rules used = {3042, 4581, 27, 3042, 4570, 27, 3042, 4490, 27, 3042, 4490, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4581

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (-4 a C \sec ^2(c+d x)+b (9 A+7 C) \sec (c+d x)+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (-4 a C \sec ^2(c+d x)+b (9 A+7 C) \sec (c+d x)+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (-4 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (9 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a C\right )dx}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {2 \int -\frac {1}{2} \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (6 a b C-\left (8 C a^2+7 b^2 (9 A+7 C)\right ) \sec (c+d x)\right )dx}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (6 a b C-\left (8 C a^2+7 b^2 (9 A+7 C)\right ) \sec (c+d x)\right )dx}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (6 a b C+\left (-8 C a^2-7 b^2 (9 A+7 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {-\frac {\frac {2}{5} \int -\frac {3}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 C a^2+63 A b^2+49 b^2 C\right )+a \left (8 C a^2+63 A b^2+39 b^2 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {3}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b \left (-2 C a^2+63 A b^2+49 b^2 C\right )+a \left (8 C a^2+63 A b^2+39 b^2 C\right ) \sec (c+d x)\right )dx-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {3}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b \left (-2 C a^2+63 A b^2+49 b^2 C\right )+a \left (8 C a^2+63 A b^2+39 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {-\frac {-\frac {3}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (2 a b \left (126 A b^2+\left (a^2+93 b^2\right ) C\right )+\left (8 C a^4+3 b^2 (21 A+11 C) a^2+21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {3}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (2 a b \left (126 A b^2+\left (a^2+93 b^2\right ) C\right )+\left (8 C a^4+3 b^2 (21 A+11 C) a^2+21 b^4 (9 A+7 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {3}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a b \left (126 A b^2+\left (a^2+93 b^2\right ) C\right )+\left (8 C a^4+3 b^2 (21 A+11 C) a^2+21 b^4 (9 A+7 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {-\frac {-\frac {3}{5} \left (\frac {1}{3} \left (\left (8 a^4 C+3 a^2 b^2 (21 A+11 C)+21 b^4 (9 A+7 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (8 a^3 C+6 a^2 b C+3 a b^2 (21 A+13 C)-21 b^3 (9 A+7 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {-\frac {3}{5} \left (\frac {1}{3} \left (\left (8 a^4 C+3 a^2 b^2 (21 A+11 C)+21 b^4 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (8 a^3 C+6 a^2 b C+3 a b^2 (21 A+13 C)-21 b^3 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {-\frac {-\frac {3}{5} \left (\frac {1}{3} \left (\left (8 a^4 C+3 a^2 b^2 (21 A+11 C)+21 b^4 (9 A+7 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C+6 a^2 b C+3 a b^2 (21 A+13 C)-21 b^3 (9 A+7 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {-\frac {-\frac {2 \left (8 a^2 C+7 b^2 (9 A+7 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}-\frac {3}{5} \left (\frac {2 a \left (8 a^2 C+63 A b^2+39 b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (-\frac {2 (a-b) \sqrt {a+b} \left (8 a^4 C+3 a^2 b^2 (21 A+11 C)+21 b^4 (9 A+7 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^3 C+6 a^2 b C+3 a b^2 (21 A+13 C)-21 b^3 (9 A+7 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )\right )}{7 b}-\frac {8 a C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}}{9 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{5/2}}{9 b d}\)

Input:

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

Output:

(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(9*b*d) + ((-8* 
a*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) - ((-2*(8*a^2*C + 7*b 
^2*(9*A + 7*C))*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) - (3*(((-2* 
(a - b)*Sqrt[a + b]*(8*a^4*C + 21*b^4*(9*A + 7*C) + 3*a^2*b^2*(21*A + 11*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]*(8*a^3*C + 6*a^2*b*C - 
 21*b^3*(9*A + 7*C) + 3*a*b^2*(21*A + 13*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))])/(b*d))/3 + (2 
*a*(63*A*b^2 + 8*a^2*C + 39*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/ 
(3*d)))/5)/(7*b))/(9*b)
 

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 4490
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[Csc[e + f*x]* 
(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* 
B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 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 4581
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(cs 
c[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Csc[e + f* 
x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/( 
b*(m + 3))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2 
) + A*(m + 3))*Csc[e + f*x] - 2*a*C*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, 
 b, e, f, A, C, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2248\) vs. \(2(416)=832\).

Time = 87.38 (sec) , antiderivative size = 2249, normalized size of antiderivative = 4.95

method result size
default \(\text {Expression too large to display}\) \(2249\)
parts \(\text {Expression too large to display}\) \(2260\)

Input:

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

Output:

2/315/d/b^3*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+ 
c)+b)*((33*cos(d*x+c)^2-cos(d*x+c)-1)*C*a^3*b^2*tan(d*x+c)+189*(cos(d*x+c) 
^2+cos(d*x+c)+1)*a*A*b^4*tan(d*x+c)+63*sin(d*x+c)*(3+2*cos(d*x+c))*a^2*A*b 
^3+7*(21*cos(d*x+c)^4+7*cos(d*x+c)^3+7*cos(d*x+c)^2+5*cos(d*x+c)+5)*C*b^5* 
tan(d*x+c)*sec(d*x+c)^3+8*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(1/(a+b)*(b+a*co 
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^5*Elli 
pticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(88*cos(d*x+c)^3+121*cos 
(d*x+c)^2+53*cos(d*x+c)+53)*C*a^2*b^3*tan(d*x+c)*sec(d*x+c)+(147*cos(d*x+c 
)^4+137*cos(d*x+c)^3+137*cos(d*x+c)^2+85*cos(d*x+c)+85)*C*b^4*a*tan(d*x+c) 
*sec(d*x+c)^2+189*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(cos(d*x+c)/(cos(d*x+c)+ 
1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*EllipticE(-c 
sc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+147*(cos(d*x+c)^2+2*cos(d*x+c)+1 
)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c 
)+1))^(1/2)*b^5*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+189* 
(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b 
)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5*EllipticF(-csc(d*x+c)+cot(d*x 
+c),((a-b)/(a+b))^(1/2))+147*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(cos(d*x+c)/ 
(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^5* 
EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+63*A*a^3*b^2*cos(d*x 
+c)*sin(d*x+c)+4*sin(d*x+c)*(1-cos(d*x+c))*a^4*b*C+2*(-cos(d*x+c)^2-2*c...
 

Fricas [F]

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

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

Output:

integral((C*b*sec(d*x + c)^5 + C*a*sec(d*x + c)^4 + A*b*sec(d*x + c)^3 + A 
*a*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**5,x)*b*c + int(sqrt(sec(c + d*x 
)*b + a)*sec(c + d*x)**4,x)*a*c + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x 
)**3,x)*a*b + int(sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**2,x)*a**2