\(\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+C \sec ^2(c+d x)) \, dx\) [719]

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

Optimal result

Integrand size = 33, antiderivative size = 374 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 a (a-b) \sqrt {a+b} \left (70 A b^2-3 a^2 C+41 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (105 a A b-35 A b^2+6 a^2 C+57 a b C-25 b^2 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}-\frac {4 a C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \] Output:

-4/105*a*(a-b)*(a+b)^(1/2)*(70*A*b^2-3*C*a^2+41*C*b^2)*cot(d*x+c)*Elliptic 
E((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+2/105*(a-b)*(a+b)^(1/ 
2)*(105*A*a*b-35*A*b^2+6*C*a^2+57*C*a*b-25*C*b^2)*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^2/d-2/105*(6*C*a^2-5*b^2*(7*A+ 
5*C))*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d-4/35*a*C*(a+b*sec(d*x+c))^(3/2 
)*tan(d*x+c)/b/d+2/7*C*(a+b*sec(d*x+c))^(5/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 16.34 (sec) , antiderivative size = 574, normalized size of antiderivative = 1.53 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (2 a (a+b) \left (-70 A b^2+3 a^2 C-41 b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+b (a+b) \left (-6 a^2 C+5 b^2 (7 A+5 C)+3 a b (35 A+19 C)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \left (-70 A b^2+3 a^2 C-41 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{105 b^2 d (b+a \cos (c+d x))^2 (A+2 C+A \cos (2 c+2 d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {7}{2}}(c+d x)}+\frac {\cos ^3(c+d x) (a+b \sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {8 a \left (-70 A b^2+3 a^2 C-41 b^2 C\right ) \sin (c+d x)}{105 b^2}+\frac {4 \sec (c+d x) \left (35 A b^2 \sin (c+d x)+3 a^2 C \sin (c+d x)+25 b^2 C \sin (c+d x)\right )}{105 b}+\frac {32}{35} a C \sec (c+d x) \tan (c+d x)+\frac {4}{7} b C \sec ^2(c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x)) (A+2 C+A \cos (2 c+2 d x))} \] Input:

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

Output:

(8*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*(a + b)*(-70*A*b^2 + 3*a^2*C - 41*b^2*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)] + b*(a + b)* 
(-6*a^2*C + 5*b^2*(7*A + 5*C) + 3*a*b*(35*A + 19*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]))]*E 
llipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(-70*A*b^2 + 3*a^2 
*C - 41*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c 
 + d*x)/2]))/(105*b^2*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)*((-8*a*(-70*A*b^2 + 3*a^2*C - 4 
1*b^2*C)*Sin[c + d*x])/(105*b^2) + (4*Sec[c + d*x]*(35*A*b^2*Sin[c + d*x] 
+ 3*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b) + (32*a*C*Sec[c + 
 d*x]*Tan[c + d*x])/35 + (4*b*C*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a 
*Cos[c + d*x])*(A + 2*C + A*Cos[2*c + 2*d*x]))
 

Rubi [A] (verified)

Time = 1.54 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4571, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4571

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (a b (35 A+19 C)-\left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sec (c+d x)\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (a b (35 A+19 C)-\left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \sec (c+d x)\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (a b (35 A+19 C)+\left (5 b^2 (7 A+5 C)-6 a^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (3 (35 A+17 C) a^2+5 b^2 (7 A+5 C)\right )+2 a \left (-3 C a^2+70 A b^2+41 b^2 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (3 (35 A+17 C) a^2+5 b^2 (7 A+5 C)\right )+2 a \left (-3 C a^2+70 A b^2+41 b^2 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (3 (35 A+17 C) a^2+5 b^2 (7 A+5 C)\right )+2 a \left (-3 C a^2+70 A b^2+41 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+2 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (2 a \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 C+105 a A b+57 a b C-35 A b^2-25 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {4 a (a-b) \sqrt {a+b} \left (-3 a^2 C+70 A b^2+41 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )-\frac {2 \left (6 a^2 C-5 b^2 (7 A+5 C)\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )-\frac {4 a C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

Input:

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

Output:

(2*C*(a + b*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*b*d) + ((-4*a*C*(a + b*Se 
c[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-4*a*(a - b)*Sqrt[a + b]*(70*A* 
b^2 - 3*a^2*C + 41*b^2*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] 
*(105*a*A*b - 35*A*b^2 + 6*a^2*C + 57*a*b*C - 25*b^2*C)*Cot[c + d*x]*Ellip 
ticF[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*(6*a^2*C - 5*b^2*(7*A + 5*C))*Sqrt[a + b*Sec[c + d*x]]*Tan[c + 
 d*x])/(3*d))/5)/(7*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 4571
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ 
(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x] 
*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   In 
t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1617\) vs. \(2(340)=680\).

Time = 73.54 (sec) , antiderivative size = 1618, normalized size of antiderivative = 4.33

method result size
default \(\text {Expression too large to display}\) \(1618\)
parts \(\text {Expression too large to display}\) \(1634\)

Input:

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

Output:

2/105/d/b^2*(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)*(140*(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)*a^2*b^2*EllipticE(-csc( 
d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+140*(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)*a*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+6*(-c 
os(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1 
/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4*EllipticE(-csc(d*x+c)+cot(d*x+c) 
,((a-b)/(a+b))^(1/2))+6*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*C*(1/(a+b)*(b+a*cos 
(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*Ell 
ipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+82*(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))/(c 
os(d*x+c)+1))^(1/2)*a^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b)) 
^(1/2))+82*(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)*a*b^3*EllipticE(-csc(d* 
x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+105*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A* 
(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1) 
)^(1/2)*a^2*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+140* 
(-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)*a*b^3*EllipticF(-csc(d*x+c)+co...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate(sec(d*x+c)*(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), 
 x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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