\(\int \sqrt {a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [938]

Optimal result

Integrand size = 35, antiderivative size = 366 \[ \int \sqrt {a+b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} (3 b B+a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 \sqrt {a+b} (3 A b+(a-b) (3 B-C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 d} \] Output:

-2/3*(a-b)*(a+b)^(1/2)*(3*B*b+C*a)*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^2/d+2/3*(a+b)^(1/2)*(3*A*b+(a-b)*(3*B-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/d-2*A*(a+ 
b)^(1/2)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a, 
((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)/d+2/3*C*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(785\) vs. \(2(366)=732\).

Time = 17.53 (sec) , antiderivative size = 785, normalized size of antiderivative = 2.14 \[ \int \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 {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \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 ) \left (-12 a A b \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 {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+2 (a+b) (3 b B+a C) \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 ) \sec ^2\left (\frac {1}{2} (c+d x)\right )-2 b (a (-3 A+3 B+C)+b (3 A+3 B+C)) \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 ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+3 a b B \tan \left (\frac {1}{2} (c+d x)\right )+3 b^2 B \tan \left (\frac {1}{2} (c+d x)\right )+a^2 C \tan \left (\frac {1}{2} (c+d x)\right )+a b C \tan \left (\frac {1}{2} (c+d x)\right )-12 a A b \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 {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-6 a b B \tan ^3\left (\frac {1}{2} (c+d x)\right )-2 a^2 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+3 a b B \tan ^5\left (\frac {1}{2} (c+d x)\right )-3 b^2 B \tan ^5\left (\frac {1}{2} (c+d x)\right )+a^2 C \tan ^5\left (\frac {1}{2} (c+d x)\right )-a b C \tan ^5\left (\frac {1}{2} (c+d x)\right )\right )}{3 b d (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 (3 b B+a C) \sin (c+d x)}{3 b}+\frac {4}{3} C \tan (c+d x)\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))} \] Input:

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

Output:

(-4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(A + B* 
Sec[c + d*x] + C*Sec[c + d*x]^2)*(-12*a*A*b*Sqrt[Cos[c + d*x]/(1 + Cos[c + 
 d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticPi 
[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*(a + b)*(3*b*B + a*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)]*S 
ec[(c + d*x)/2]^2 - 2*b*(a*(-3*A + 3*B + C) + b*(3*A + 3*B + 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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sec[(c + d 
*x)/2]^2 + 3*a*b*B*Tan[(c + d*x)/2] + 3*b^2*B*Tan[(c + d*x)/2] + a^2*C*Tan 
[(c + d*x)/2] + a*b*C*Tan[(c + d*x)/2] - 12*a*A*b*Sqrt[Cos[c + d*x]/(1 + C 
os[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Elli 
pticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2 - 
 6*a*b*B*Tan[(c + d*x)/2]^3 - 2*a^2*C*Tan[(c + d*x)/2]^3 + 3*a*b*B*Tan[(c 
+ d*x)/2]^5 - 3*b^2*B*Tan[(c + d*x)/2]^5 + a^2*C*Tan[(c + d*x)/2]^5 - a*b* 
C*Tan[(c + d*x)/2]^5))/(3*b*d*(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*(3*b*B + a*C)*Sin[c + d*x])/(3*b) + (4*C*Tan[c + d*x])/3))/(d 
*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x]))
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{3} \left (((a-b) (3 B-C)+3 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{3} \left ((a C+3 b B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) ((a-b) (3 B-C)+3 A b) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {6 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{3} \left (\frac {2 \sqrt {a+b} \cot (c+d x) ((a-b) (3 B-C)+3 A b) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {6 A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}-\frac {2 (a-b) \sqrt {a+b} (a C+3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}\right )+\frac {2 C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\)

Input:

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

Output:

((-2*(a - b)*Sqrt[a + b]*(3*b*B + a*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]*(3*A*b + (a - b)*(3*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) - (6*A*Sqrt[a + b] 
*Cot[c + d*x]*EllipticPi[(a + b)/a, 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))])/d)/3 + (2*C*Sqrt[a + b*Sec[c + d*x]]*Tan[c + 
d*x])/(3*d)
 

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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4544

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

rule 4546

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C 
)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C   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, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1149\) vs. \(2(331)=662\).

Time = 40.95 (sec) , antiderivative size = 1150, normalized size of antiderivative = 3.14

method	result	size
parts	\(\text {Expression too large to display}\)	\(1150\)
default	\(\text {Expression too large to display}\)	\(1237\)

Input:

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

Output:

2*A/d*(EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*a-EllipticF(- 
csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*b-2*a*EllipticPi(-csc(d*x+c)+co 
t(d*x+c),-1,((a-b)/(a+b))^(1/2)))*(cos(d*x+c)+1)*(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*sec(d*x+c))^ 
(1/2)/(b+a*cos(d*x+c))-1/2*B/d/(b+a*cos(d*x+c))*(cos(d*x+c)+1)^2*((-(1-cos 
(d*x+c))^3*csc(d*x+c)^3+csc(d*x+c)-cot(d*x+c))*a+((1-cos(d*x+c))^3*csc(d*x 
+c)^3+csc(d*x+c)-cot(d*x+c))*b-2*(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)*EllipticF(-csc(d*x+c)+cot(d*x+c), 
((a-b)/(a+b))^(1/2))*a-2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*c 
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/( 
a+b))^(1/2))*b+2*(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)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1 
/2))*a+2*(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)*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))*b)* 
((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(a+b*sec(d*x+c))^(1/2)*sec(d*x+c)-2/3*C/ 
d/b*(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)*(( 
-cos(d*x+c)^2-2*cos(d*x+c)-1)*(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*EllipticE(-csc(d*x+c)+cot(d*x+c) 
,((a-b)/(a+b))^(1/2))+(-cos(d*x+c)^2-2*cos(d*x+c)-1)*(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*Ellipt...
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

integrate((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) 
, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Reduce [F]

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

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