\(\int \sec (c+d x) \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [937]

Optimal result

Integrand size = 41, antiderivative size = 324 \[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (3 b^2 (5 A+3 C)+a (5 b B-2 a C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}+\frac {2 (a-b) \sqrt {a+b} (15 A b-5 b B+2 a C+9 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^2 d}+\frac {2 (5 b B-2 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac {2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d} \] Output:

-2/15*(a-b)*(a+b)^(1/2)*(3*b^2*(5*A+3*C)+a*(5*B*b-2*C*a))*cot(d*x+c)*Ellip 
ticE((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+2/15*(a-b)*(a+b)^( 
1/2)*(15*A*b-5*B*b+2*C*a+9*C*b)*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+s 
ec(d*x+c))/(a-b))^(1/2)/b^2/d+2/15*(5*B*b-2*C*a)*(a+b*sec(d*x+c))^(1/2)*ta 
n(d*x+c)/b/d+2/5*C*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/b/d
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 18.17 (sec) , antiderivative size = 579, normalized size of antiderivative = 1.79 \[ \int \sec (c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \sqrt {2} \sqrt {\frac {\cos (c+d x)}{(1+\cos (c+d x))^2}} \sqrt {\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((a+b) \left (-\left (\left (15 A b^2+5 a b B-2 a^2 C+9 b^2 C\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )\right )+b (15 A b+5 b B-2 a C+9 b C) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}} \sec (c+d x)-\left (15 A b^2+5 a b B-2 a^2 C+9 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{15 b^2 d \sqrt {\frac {1}{1+\cos (c+d x)}} (b+a \cos (c+d x)) (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \sec ^{\frac {5}{2}}(c+d x)}+\frac {\cos ^2(c+d x) \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (15 A b^2+5 a b B-2 a^2 C+9 b^2 C\right ) \sin (c+d x)}{15 b^2}+\frac {4 \sec (c+d x) (5 b B \sin (c+d x)+a C \sin (c+d x))}{15 b}+\frac {4}{5} C \sec (c+d x) \tan (c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))} \] Input:

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

Output:

(4*Sqrt[2]*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])^2]*Sqrt[Cos[c + d*x]*Sec[( 
c + d*x)/2]^2]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*Sqrt[a + b*Sec[c + 
d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((a + b)*(-((15*A*b^2 + 5*a* 
b*B - 2*a^2*C + 9*b^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + 
b)]) + b*(15*A*b + 5*b*B - 2*a*C + 9*b*C)*EllipticF[ArcSin[Tan[(c + d*x)/2 
]], (a - b)/(a + b)])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(3/2)*Sqrt[((b + a 
*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)]*Sec[c + d*x] - (15*A*b^2 + 5*a 
*b*B - 2*a^2*C + 9*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/ 
2]^4*Tan[(c + d*x)/2]))/(15*b^2*d*Sqrt[(1 + Cos[c + d*x])^(-1)]*(b + a*Cos 
[c + d*x])*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(Sec[(c + d*x 
)/2]^2)^(3/2)*Sec[c + d*x]^(5/2)) + (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*(15*A*b^2 + 5*a*b*B - 2*a^2 
*C + 9*b^2*C)*Sin[c + d*x])/(15*b^2) + (4*Sec[c + d*x]*(5*b*B*Sin[c + d*x] 
 + a*C*Sin[c + d*x]))/(15*b) + (4*C*Sec[c + d*x]*Tan[c + d*x])/5))/(d*(A + 
 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {2}{3} \int \frac {\sec (c+d x) \left (b (15 a A+5 b B+7 a C)+\left (3 (5 A+3 C) b^2+a (5 b B-2 a C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {\sec (c+d x) \left (b (15 a A+5 b B+7 a C)+\left (3 (5 A+3 C) b^2+a (5 b B-2 a C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (15 a A+5 b B+7 a C)+\left (3 (5 A+3 C) b^2+a (5 b B-2 a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {1}{3} \left (\left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+(a-b) (2 a C+15 A b-5 b B+9 b C) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) (2 a C+15 A b-5 b B+9 b C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {1}{3} \left (\left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) (2 a C+15 A b-5 b B+9 b C) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {1}{3} \left (\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) (2 a C+15 A b-5 b B+9 b C) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (a (5 b B-2 a C)+3 b^2 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {2 (5 b B-2 a C) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}}{5 b}+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}\)

Input:

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

Output:

(2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*b*d) + (((-2*(a - b)*Sqrt 
[a + b]*(3*b^2*(5*A + 3*C) + a*(5*b*B - 2*a*C))*Cot[c + d*x]*EllipticE[Arc 
Sin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - S 
ec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + 
(2*(a - b)*Sqrt[a + b]*(15*A*b - 5*b*B + 2*a*C + 9*b*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*(5*b*B - 2*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/ 
(5*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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1770\) vs. \(2(294)=588\).

Time = 60.14 (sec) , antiderivative size = 1771, normalized size of antiderivative = 5.47

method	result	size
default	\(\text {Expression too large to display}\)	\(1771\)
parts	\(\text {Expression too large to display}\)	\(1818\)

Input:

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

Output:

2/15/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)*(15*(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*b^2*EllipticE(-csc(d*x+ 
c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+15*(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)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+5*(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)*a^2*b*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b 
)/(a+b))^(1/2))+5*(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^2*EllipticE( 
-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+2*(-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*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+2*( 
-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*EllipticE(-csc(d*x+c)+cot(d* 
x+c),((a-b)/(a+b))^(1/2))+9*(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*b^2* 
EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(cos(d*x+c)^2+2*co 
s(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)*b^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b)...
 

Fricas [F]

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

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

Sympy [F]

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

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

Reduce [F]

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

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