3.10.61 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [961]

3.10.61.1 Optimal result

Integrand size = 35, antiderivative size = 317 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} C \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^2 d}+\frac {2 \sqrt {a+b} (B-C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d} \]

output

-2*(a-b)*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^2/d+2*(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))*(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*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))*(a+b)^(1/2)*(b*(1-sec 
(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a/d
 

3.10.61.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(758\) vs. \(2(317)=634\).

Time = 20.61 (sec) , antiderivative size = 758, normalized size of antiderivative = 2.39 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {4 C \cos (c+d x) (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sin (c+d x)}{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)}}-\frac {4 \sqrt {b+a \cos (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (a C \tan \left (\frac {1}{2} (c+d x)\right )+b C \tan \left (\frac {1}{2} (c+d x)\right )-2 a C \tan ^3\left (\frac {1}{2} (c+d x)\right )+a C \tan ^5\left (\frac {1}{2} (c+d x)\right )-b C \tan ^5\left (\frac {1}{2} (c+d x)\right )-2 A b \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 A b \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 ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+(a+b) C E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (A-B-C) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{b d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}} \]

input

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

output

(4*C*Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x 
]^2)*Sin[c + d*x])/(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*Sqrt[b + a*Cos[c + d*x]]*(A + B*Sec[c + d*x 
] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(a*C*Tan[(c + d* 
x)/2] + b*C*Tan[(c + d*x)/2] - 2*a*C*Tan[(c + d*x)/2]^3 + a*C*Tan[(c + d*x 
)/2]^5 - b*C*Tan[(c + d*x)/2]^5 - 2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x 
)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[( 
c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*A*b*EllipticPi[-1, ArcS 
in[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c 
+ d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a 
 + b)] + (a + b)*C*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sq 
rt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c 
 + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + b*(A - B - C)*EllipticF[Ar 
cSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + 
 Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/ 
2]^2)/(a + b)]))/(b*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Se 
c[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*(1 + Tan[(c + d*x)/2]^2)^(3/2)*S 
qrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d* 
x)/2]^2)])
 

3.10.61.3 Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4546

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4409

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

input

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

output

(-2*(a - b)*Sqrt[a + b]*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]*(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) - (2*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))])/(a*d 
)
 

3.10.61.3.1 Defintions of rubi rules used

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 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]
 

3.10.61.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1077\) vs. \(2(290)=580\).

Time = 16.92 (sec) , antiderivative size = 1078, normalized size of antiderivative = 3.40

method	result	size
parts	\(\text {Expression too large to display}\)	\(1078\)
default	\(\text {Expression too large to display}\)	\(1589\)

input

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

output

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

3.10.61.5 Fricas [F]

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

input

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

3.10.61.6 Sympy [F]

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

input

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

3.10.61.7 Maxima [F]

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

input

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

3.10.61.8 Giac [F]

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

input

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

3.10.61.9 Mupad [F(-1)]

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

input

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

output

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