3.10.44 \(\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [944]

3.10.44.1 Optimal result
3.10.44.2 Mathematica [B] (warning: unable to verify)
3.10.44.3 Rubi [A] (verified)
3.10.44.4 Maple [B] (verified)
3.10.44.5 Fricas [F]
3.10.44.6 Sympy [F]
3.10.44.7 Maxima [F]
3.10.44.8 Giac [F]
3.10.44.9 Mupad [F(-1)]

3.10.44.1 Optimal result

Integrand size = 41, antiderivative size = 406 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (21 a^2 b B+63 b^3 B-6 a^3 C+2 a b^2 (70 A+41 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 (a-b) \sqrt {a+b} \left (6 a^2 C+3 a b (35 A-7 B+19 C)-b^2 (35 A-63 B+25 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^2 d}+\frac {2 \left (35 A b^2+21 a b B-6 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b d}+\frac {2 (7 b B-2 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b d}+\frac {2 C (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{7 b d} \]

output
-2/105*(a-b)*(21*B*a^2*b+63*B*b^3-6*a^3*C+2*a*b^2*(70*A+41*C))*cot(d*x+c)* 
EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(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)*(6*C*a^2+3*a*b*(35*A-7*B+19*C)-b^2*(35*A-63*B+25*C))*cot(d*x+c) 
*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(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+ 
2/35*(7*B*b-2*C*a)*(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+2/105*(35*A*b^2+21*B*a*b-6*C*a^2+25*C*b^2)*(a+b 
*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
 
3.10.44.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3724\) vs. \(2(406)=812\).

Time = 24.21 (sec) , antiderivative size = 3724, normalized size of antiderivative = 9.17 \[ \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \]

input
Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C* 
Sec[c + d*x]^2),x]
 
output
(Cos[c + d*x]^3*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + 
 d*x]^2)*((-4*(-140*a*A*b^2 - 21*a^2*b*B - 63*b^3*B + 6*a^3*C - 82*a*b^2*C 
)*Sin[c + d*x])/(105*b^2) + (4*Sec[c + d*x]^2*(7*b*B*Sin[c + d*x] + 8*a*C* 
Sin[c + d*x]))/35 + (4*Sec[c + d*x]*(35*A*b^2*Sin[c + d*x] + 42*a*b*B*Sin[ 
c + d*x] + 3*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b) + (4*b*C 
*Sec[c + d*x]^2*Tan[c + d*x])/7))/(d*(b + a*Cos[c + d*x])*(A + 2*C + 2*B*C 
os[c + d*x] + A*Cos[2*c + 2*d*x])) + (4*((-8*a*A*b)/(3*Sqrt[b + a*Cos[c + 
d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^2*B)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec 
[c + d*x]]) - (6*b^2*B)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 
(4*a^3*C)/(35*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (164*a*b*C) 
/(105*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^2*A*Sqrt[Sec[c + 
 d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) + (2*A*b^2*Sqrt[Sec[c + d*x]])/(3*Sqr 
t[b + a*Cos[c + d*x]]) - (2*a^3*B*Sqrt[Sec[c + d*x]])/(5*b*Sqrt[b + a*Cos[ 
c + d*x]]) + (2*a*b*B*Sqrt[Sec[c + d*x]])/(5*Sqrt[b + a*Cos[c + d*x]]) - ( 
62*a^2*C*Sqrt[Sec[c + d*x]])/(105*Sqrt[b + a*Cos[c + d*x]]) + (4*a^4*C*Sqr 
t[Sec[c + d*x]])/(35*b^2*Sqrt[b + a*Cos[c + d*x]]) + (10*b^2*C*Sqrt[Sec[c 
+ d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) - (8*a^2*A*Cos[2*(c + d*x)]*Sqrt[Se 
c[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) - (2*a^3*B*Cos[2*(c + d*x)]*Sqrt 
[Sec[c + d*x]])/(5*b*Sqrt[b + a*Cos[c + d*x]]) - (6*a*b*B*Cos[2*(c + d*x)] 
*Sqrt[Sec[c + d*x]])/(5*Sqrt[b + a*Cos[c + d*x]]) - (164*a^2*C*Cos[2*(c...
 
3.10.44.3 Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {2}{5} \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b (35 a A+21 b B+19 a C)+\left (-6 C a^2+21 b B a+35 A b^2+25 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b (35 a A+21 b B+19 a C)+\left (-6 C a^2+21 b B a+35 A b^2+25 b^2 C\right ) \sec (c+d x)\right )dx+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (35 a A+21 b B+19 a C)+\left (-6 C a^2+21 b B a+35 A b^2+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (3 (35 A+17 C) a^2+84 b B a+5 b^2 (7 A+5 C)\right )+\left (-6 C a^3+21 b B a^2+2 b^2 (70 A+41 C) a+63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (3 (35 A+17 C) a^2+84 b B a+5 b^2 (7 A+5 C)\right )+\left (-6 C a^3+21 b B a^2+2 b^2 (70 A+41 C) a+63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (3 (35 A+17 C) a^2+84 b B a+5 b^2 (7 A+5 C)\right )+\left (-6 C a^3+21 b B a^2+2 b^2 (70 A+41 C) a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2 C+3 a b (35 A-7 B+19 C)-b^2 (35 A-63 B+25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (-6 a^3 C+21 a^2 b B+2 a b^2 (70 A+41 C)+63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left ((a-b) \left (6 a^2 C+3 a b (35 A-7 B+19 C)-b^2 (35 A-63 B+25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (-6 a^3 C+21 a^2 b B+2 a b^2 (70 A+41 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{5} \left (\frac {1}{3} \left (\left (-6 a^3 C+21 a^2 b B+2 a b^2 (70 A+41 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (6 a^2 C+3 a b (35 A-7 B+19 C)-b^2 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{5} \left (\frac {2 \tan (c+d x) \left (-6 a^2 C+21 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (6 a^2 C+3 a b (35 A-7 B+19 C)-b^2 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-6 a^3 C+21 a^2 b B+2 a b^2 (70 A+41 C)+63 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )\right )+\frac {2 (7 b B-2 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}}{7 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{7 b d}\)

input
Int[Sec[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + 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) + ((2*(7*b*B - 2*a*C 
)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a + 
b]*(21*a^2*b*B + 63*b^3*B - 6*a^3*C + 2*a*b^2*(70*A + 41*C))*Cot[c + d*x]* 
EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*S 
qrt[(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]*(6*a^2*C + 3*a*b*(35*A - 7*B + 19*C) - 
 b^2*(35*A - 63*B + 25*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*(35*A*b^2 + 21*a* 
b*B - 6*a^2*C + 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/5) 
/(7*b)
 

3.10.44.3.1 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]
 
3.10.44.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5703\) vs. \(2(372)=744\).

Time = 34.22 (sec) , antiderivative size = 5704, normalized size of antiderivative = 14.05

method result size
parts \(\text {Expression too large to display}\) \(5704\)
default \(\text {Expression too large to display}\) \(5771\)

input
int(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 
output
result too large to display
 
3.10.44.5 Fricas [F]

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

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="fricas")
 
output
integral((C*b*sec(d*x + c)^4 + (C*a + B*b)*sec(d*x + c)^3 + A*a*sec(d*x + 
c) + (B*a + A*b)*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
 
3.10.44.6 Sympy [F]

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

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)* 
*2),x)
 
output
Integral((a + b*sec(c + d*x))**(3/2)*(A + B*sec(c + d*x) + C*sec(c + d*x)* 
*2)*sec(c + d*x), x)
 
3.10.44.7 Maxima [F]

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

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="maxima")
 
output
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/ 
2)*sec(d*x + c), x)
 
3.10.44.8 Giac [F]

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

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2 
),x, algorithm="giac")
 
output
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/ 
2)*sec(d*x + c), x)
 
3.10.44.9 Mupad [F(-1)]

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

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