\(\int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [936]

Optimal result

Integrand size = 43, antiderivative size = 413 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 (a-b) \sqrt {a+b} \left (14 a^2 b B-63 b^3 B-8 a^3 C-a b^2 (35 A+19 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^4 d}-\frac {2 (a-b) \sqrt {a+b} \left (35 A b^2-b^2 (63 B-25 C)+8 a^2 C-a (14 b B-6 b C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{105 b^3 d}+\frac {2 \left (35 A b^2-14 a b B+8 a^2 C+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{105 b^2 d}+\frac {2 (7 b B-4 a C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{35 b^2 d}+\frac {2 C \sec (c+d x) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{7 b d} \] Output:

2/105*(a-b)*(a+b)^(1/2)*(14*B*a^2*b-63*B*b^3-8*a^3*C-a*b^2*(35*A+19*C))*co 
t(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-2/10 
5*(a-b)*(a+b)^(1/2)*(35*A*b^2-b^2*(63*B-25*C)+8*C*a^2-a*(14*B*b-6*C*b))*co 
t(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+2/10 
5*(35*A*b^2-14*B*a*b+8*C*a^2+25*C*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b 
^2/d+2/35*(7*B*b-4*C*a)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b^2/d+2/7*C*sec( 
d*x+c)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3706\) vs. \(2(413)=826\).

Time = 23.92 (sec) , antiderivative size = 3706, normalized size of antiderivative = 8.97 \[ \int \sec ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:

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

Output:

(Cos[c + d*x]^2*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d 
*x]^2)*((4*(35*a*A*b^2 - 14*a^2*b*B + 63*b^3*B + 8*a^3*C + 19*a*b^2*C)*Sin 
[c + d*x])/(105*b^3) + (4*Sec[c + d*x]^2*(7*b*B*Sin[c + d*x] + a*C*Sin[c + 
 d*x]))/(35*b) + (4*Sec[c + d*x]*(35*A*b^2*Sin[c + d*x] + 7*a*b*B*Sin[c + 
d*x] - 4*a^2*C*Sin[c + d*x] + 25*b^2*C*Sin[c + d*x]))/(105*b^2) + (4*C*Sec 
[c + d*x]^2*Tan[c + d*x])/7))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 
 2*d*x])) - (4*((-2*a*A)/(3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + 
 (4*a^2*B)/(15*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (6*b*B)/(5 
*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (38*a*C)/(105*Sqrt[b + a*C 
os[c + d*x]]*Sqrt[Sec[c + d*x]]) - (16*a^3*C)/(105*b^2*Sqrt[b + a*Cos[c + 
d*x]]*Sqrt[Sec[c + d*x]]) - (2*a^2*A*Sqrt[Sec[c + d*x]])/(3*b*Sqrt[b + a*C 
os[c + d*x]]) + (2*A*b*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]) - 
(4*a*B*Sqrt[Sec[c + d*x]])/(15*Sqrt[b + a*Cos[c + d*x]]) + (4*a^3*B*Sqrt[S 
ec[c + d*x]])/(15*b^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^4*C*Sqrt[Sec[c + d 
*x]])/(105*b^3*Sqrt[b + a*Cos[c + d*x]]) - (34*a^2*C*Sqrt[Sec[c + d*x]])/( 
105*b*Sqrt[b + a*Cos[c + d*x]]) + (10*b*C*Sqrt[Sec[c + d*x]])/(21*Sqrt[b + 
 a*Cos[c + d*x]]) - (2*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b*Sqr 
t[b + a*Cos[c + d*x]]) - (6*a*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*Sq 
rt[b + a*Cos[c + d*x]]) + (4*a^3*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(1 
5*b^2*Sqrt[b + a*Cos[c + d*x]]) - (16*a^4*C*Cos[2*(c + d*x)]*Sqrt[Sec[c...
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4580, 27, 3042, 4570, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4580

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {2 \int \frac {1}{2} \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b (21 b B-2 a C)+\left (8 C a^2-14 b B a+35 A b^2+25 b^2 C\right ) \sec (c+d x)\right )dx}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (b (21 b B-2 a C)+\left (8 C a^2-14 b B a+35 A b^2+25 b^2 C\right ) \sec (c+d x)\right )dx}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )} \left (b (21 b B-2 a C)+\left (8 C a^2-14 b B a+35 A b^2+25 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {\frac {2}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^2+49 b B a+35 A b^2+25 b^2 C\right )-\left (-8 C a^3+14 b B a^2-b^2 (35 A+19 C) a-63 b^3 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {1}{3} \int \frac {\sec (c+d x) \left (b \left (2 C a^2+49 b B a+35 A b^2+25 b^2 C\right )-\left (-8 C a^3+14 b B a^2-b^2 (35 A+19 C) a-63 b^3 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (2 C a^2+49 b B a+35 A b^2+25 b^2 C\right )+\left (8 C a^3-14 b B a^2+b^2 (35 A+19 C) a+63 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\frac {1}{3} \left (-\left ((a-b) \left (8 a^2 C-a (14 b B-6 b C)+35 A b^2-b^2 (63 B-25 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )-\left (-8 a^3 C+14 a^2 b B-a b^2 (35 A+19 C)-63 b^3 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {1}{3} \left (-\left ((a-b) \left (8 a^2 C-a (14 b B-6 b C)+35 A b^2-b^2 (63 B-25 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\left (-8 a^3 C+14 a^2 b B-a b^2 (35 A+19 C)-63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {1}{3} \left (-\left (-8 a^3 C+14 a^2 b B-a b^2 (35 A+19 C)-63 b^3 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (8 a^2 C-a (14 b B-6 b C)+35 A b^2-b^2 (63 B-25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {2 \tan (c+d x) \left (8 a^2 C-14 a b B+35 A b^2+25 b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{3 d}+\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (-8 a^3 C+14 a^2 b B-a b^2 (35 A+19 C)-63 b^3 B\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (8 a^2 C-a (14 b B-6 b C)+35 A b^2-b^2 (63 B-25 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )}{5 b}+\frac {2 (7 b B-4 a C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}}{7 b}+\frac {2 C \tan (c+d x) \sec (c+d x) (a+b \sec (c+d x))^{3/2}}{7 b d}\)

Input:

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

Output:

(2*C*Sec[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(7*b*d) + ((2*( 
7*b*B - 4*a*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*b*d) + (((2*(a 
- b)*Sqrt[a + b]*(14*a^2*b*B - 63*b^3*B - 8*a^3*C - a*b^2*(35*A + 19*C))*C 
ot[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]*(35*A*b^2 - b^2*(63*B - 25 
*C) + 8*a^2*C - a*(14*b*B - 6*b*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 - 14*a*b*B + 8*a^2*C + 25*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/ 
(3*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 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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2338\) vs. \(2(379)=758\).

Time = 84.38 (sec) , antiderivative size = 2339, normalized size of antiderivative = 5.66

method	result	size
default	\(\text {Expression too large to display}\)	\(2339\)
parts	\(\text {Expression too large to display}\)	\(2384\)

Input:

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

Output:

2/105/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)*(8*(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+44*cos(d*x+c)^2+18*cos(d*x+c)+18)*C*a*b^3* 
tan(d*x+c)*sec(d*x+c)+35*A*a^2*b^2*cos(d*x+c)*sin(d*x+c)+35*sin(d*x+c)*(co 
s(d*x+c)+2)*a*A*b^3+4*sin(d*x+c)*(1-cos(d*x+c))*a^3*b*C+(19*cos(d*x+c)^2-c 
os(d*x+c)-1)*C*a^2*b^2*tan(d*x+c)+5*(5*cos(d*x+c)^3+5*cos(d*x+c)^2+3*cos(d 
*x+c)+3)*C*b^4*tan(d*x+c)*sec(d*x+c)^2+63*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B* 
(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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+63*(-cos 
(d*x+c)^2-2*cos(d*x+c)-1)*B*(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))+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)*a*b^3*Ellip 
ticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(cos(d*x+c)^2+2*cos(...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Maxima [F]

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

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(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)*sqrt(b*sec(d*x + c) + a) 
*sec(d*x + c)^2, x)
 

Giac [F]

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

integrate(sec(d*x+c)^2*(a+b*sec(d*x+c))^(1/2)*(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)*sqrt(b*sec(d*x + c) + a) 
*sec(d*x + c)^2, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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