\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [958]

Optimal result

Integrand size = 43, antiderivative size = 429 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (56 a^2 b B+63 b^3 B-48 a^3 C-2 a b^2 (35 A+22 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^5 d}+\frac {2 \sqrt {a+b} \left (48 a^3 C-4 a^2 b (14 B+3 C)+2 a b^2 (35 A+7 B+22 C)+b^3 (35 A-63 B+25 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^4 d}+\frac {2 \left (35 A b^2-28 a b B+24 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b^3 d}+\frac {2 (7 b B-6 a C) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{35 b^2 d}+\frac {2 C \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{7 b d} \] Output:

-2/105*(a-b)*(a+b)^(1/2)*(56*B*a^2*b+63*B*b^3-48*a^3*C-2*a*b^2*(35*A+22*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^5/d+ 
2/105*(a+b)^(1/2)*(48*a^3*C-4*a^2*b*(14*B+3*C)+2*a*b^2*(35*A+7*B+22*C)+b^3 
*(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))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a 
-b))^(1/2)/b^4/d+2/105*(35*A*b^2-28*B*a*b+24*C*a^2+25*C*b^2)*(a+b*sec(d*x+ 
c))^(1/2)*tan(d*x+c)/b^3/d+2/35*(7*B*b-6*C*a)*sec(d*x+c)*(a+b*sec(d*x+c))^ 
(1/2)*tan(d*x+c)/b^2/d+2/7*C*sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c 
)/b/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3811\) vs. \(2(429)=858\).

Time = 25.09 (sec) , antiderivative size = 3811, normalized size of antiderivative = 8.88 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \] Input:

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

Output:

(Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) 
*((4*(-70*a*A*b^2 + 56*a^2*b*B + 63*b^3*B - 48*a^3*C - 44*a*b^2*C)*Sin[c + 
 d*x])/(105*b^4) + (4*Sec[c + d*x]^2*(7*b*B*Sin[c + d*x] - 6*a*C*Sin[c + d 
*x]))/(35*b^2) + (4*Sec[c + d*x]*(35*A*b^2*Sin[c + d*x] - 28*a*b*B*Sin[c + 
 d*x] + 24*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b^3) + (4*C*S 
ec[c + d*x]^2*Tan[c + d*x])/(7*b)))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos 
[2*c + 2*d*x])*Sqrt[a + b*Sec[c + d*x]]) + (4*((4*a*A)/(3*b*Sqrt[b + a*Cos 
[c + d*x]]*Sqrt[Sec[c + d*x]]) - (6*B)/(5*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Se 
c[c + d*x]]) - (16*a^2*B)/(15*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d* 
x]]) + (32*a^3*C)/(35*b^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + ( 
88*a*C)/(105*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*Sqrt[Se 
c[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) + (4*a^2*A*Sqrt[Sec[c + d*x]])/( 
3*b^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^3*B*Sqrt[Sec[c + d*x]])/(15*b^3*Sq 
rt[b + a*Cos[c + d*x]]) - (14*a*B*Sqrt[Sec[c + d*x]])/(15*b*Sqrt[b + a*Cos 
[c + d*x]]) + (10*C*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + a*Cos[c + d*x]]) + (3 
2*a^4*C*Sqrt[Sec[c + d*x]])/(35*b^4*Sqrt[b + a*Cos[c + d*x]]) + (64*a^2*C* 
Sqrt[Sec[c + d*x]])/(105*b^2*Sqrt[b + a*Cos[c + d*x]]) + (4*a^2*A*Cos[2*(c 
 + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^3*B* 
Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(15*b^3*Sqrt[b + a*Cos[c + d*x]]) - ( 
6*a*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*b*Sqrt[b + a*Cos[c + d*x]...
 

Rubi [A] (verified)

Time = 1.84 (sec) , antiderivative size = 449, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4590, 27, 3042, 4580, 27, 3042, 4570, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4590

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4580

\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (\left (24 C a^2-28 b B a+35 A b^2+25 b^2 C\right ) \sec ^2(c+d x)+b (21 b B+2 a C) \sec (c+d x)+2 a (7 b B-6 a C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (\left (24 C a^2-28 b B a+35 A b^2+25 b^2 C\right ) \sec ^2(c+d x)+b (21 b B+2 a C) \sec (c+d x)+2 a (7 b B-6 a C)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (24 C a^2-28 b B a+35 A b^2+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (21 b B+2 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (7 b B-6 a C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {\frac {2 \int \frac {\sec (c+d x) \left (b \left (-12 C a^2+14 b B a+35 A b^2+25 b^2 C\right )+\left (-48 C a^3+56 b B a^2-2 b^2 (35 A+22 C) a+63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (-12 C a^2+14 b B a+35 A b^2+25 b^2 C\right )+\left (-48 C a^3+56 b B a^2-2 b^2 (35 A+22 C) a+63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-12 C a^2+14 b B a+35 A b^2+25 b^2 C\right )+\left (-48 C a^3+56 b B a^2-2 b^2 (35 A+22 C) a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\frac {\left (48 a^3 C-4 a^2 b (14 B+3 C)+2 a b^2 (35 A+7 B+22 C)+b^3 (35 A-63 B+25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (-48 a^3 C+56 a^2 b B-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (48 a^3 C-4 a^2 b (14 B+3 C)+2 a b^2 (35 A+7 B+22 C)+b^3 (35 A-63 B+25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (-48 a^3 C+56 a^2 b B-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {\left (-48 a^3 C+56 a^2 b B-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (48 a^3 C-4 a^2 b (14 B+3 C)+2 a b^2 (35 A+7 B+22 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}+\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {2 \tan (c+d x) \left (24 a^2 C-28 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 b d}+\frac {\frac {2 \sqrt {a+b} \cot (c+d x) \left (48 a^3 C-4 a^2 b (14 B+3 C)+2 a b^2 (35 A+7 B+22 C)+b^3 (35 A-63 B+25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-48 a^3 C+56 a^2 b B-2 a b^2 (35 A+22 C)+63 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}}{5 b}+\frac {2 (7 b B-6 a C) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)}}{7 b d}\)

Input:

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

Output:

(2*C*Sec[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(7*b*d) + ((2*( 
7*b*B - 6*a*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) 
 + (((-2*(a - b)*Sqrt[a + b]*(56*a^2*b*B + 63*b^3*B - 48*a^3*C - 2*a*b^2*( 
35*A + 22*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*Sqrt[a + b]*(48*a^3*C - 4*a^2*b 
*(14*B + 3*C) + 2*a*b^2*(35*A + 7*B + 22*C) + b^3*(35*A - 63*B + 25*C))*Co 
t[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*b) + (2*(35*A*b^2 - 28*a*b*B + 24*a^2*C + 25*b^2*C 
)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*d))/(5*b))/(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 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 4580

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ 
(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(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 - b*B* 
(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & 
& NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]
 

rule 4590

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2342\) vs. \(2(395)=790\).

Time = 110.04 (sec) , antiderivative size = 2343, normalized size of antiderivative = 5.46

method	result	size
default	\(\text {Expression too large to display}\)	\(2343\)
parts	\(\text {Expression too large to display}\)	\(2394\)

Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, 
method=_RETURNVERBOSE)
 

Output:

-2/105/d/b^4*(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)*(48*(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^4*EllipticE(-csc(d*x+ 
c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+35*(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^4*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+25*(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^4*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b) 
/(a+b))^(1/2))+(-25*cos(d*x+c)^3+19*cos(d*x+c)^2+3*cos(d*x+c)+3)*C*a*b^3*t 
an(d*x+c)*sec(d*x+c)+70*A*a^2*b^2*cos(d*x+c)*sin(d*x+c)+24*sin(d*x+c)*(1-c 
os(d*x+c))*a^3*b*C+63*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(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*b^3*Ellip 
ticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+70*(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))+48*(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^3*b*EllipticE(-csc(d*x+c) 
+cot(d*x+c),((a-b)/(a+b))^(1/2))+44*(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^2*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+44*(cos...
 

Fricas [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1 
/2),x, algorithm="fricas")
 

Output:

integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4 + A*sec(d*x + c)^3)/sqrt(b*s 
ec(d*x + c) + a), x)
 

Sympy [F]

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

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))* 
*(1/2),x)
 

Output:

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

Maxima [F(-1)]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1 
/2),x, algorithm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1 
/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + 
 d*x))^(1/2)),x)
 

Output:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + 
 d*x))^(1/2)), x)
 

Reduce [F]

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

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x)
 

Output:

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