\(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [736]

Optimal result

Integrand size = 35, antiderivative size = 320 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 d}-\frac {2 \sqrt {a+b} \left (8 a^2 C-2 a b C+3 b^2 (5 A+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}-\frac {8 a C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 C \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d} \] Output:

-2/15*(a-b)*(a+b)^(1/2)*(8*C*a^2+3*b^2*(5*A+3*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^4/d-2/15*(a+b)^(1/2)*(8*C*a^2- 
2*C*a*b+3*b^2*(5*A+3*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^3/d-8/15*a*C*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/ 
5*C*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 14.90 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {4 \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (15 A b^2+8 a^2 C+9 b^2 C\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-2 b \left (15 A b^2+\left (8 a^2+2 a b+9 b^2\right ) C\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+\left (15 A b^2+8 a^2 C+9 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+(b+a \cos (c+d x)) \sqrt {\sec (c+d x)} \left (\left (15 A b^2+8 a^2 C+9 b^2 C\right ) \sin (c+d x)+b C (-4 a+3 b \sec (c+d x)) \tan (c+d x)\right )\right )}{15 b^3 d (A+2 C+A \cos (2 (c+d x))) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \] Input:

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

Output:

(4*(A + C*Sec[c + d*x]^2)*(-((Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a 
+ b)*(15*A*b^2 + 8*a^2*C + 9*b^2*C)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a 
 - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a 
 + b)*(1 + Sec[c + d*x]))] - 2*b*(15*A*b^2 + (8*a^2 + 2*a*b + 9*b^2)*C)*El 
lipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + d*x]) 
^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + (15*A*b^2 
 + 8*a^2*C + 9*b^2*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2 
*Tan[(c + d*x)/2]))/Sqrt[Sec[(c + d*x)/2]^2]) + (b + a*Cos[c + d*x])*Sqrt[ 
Sec[c + d*x]]*((15*A*b^2 + 8*a^2*C + 9*b^2*C)*Sin[c + d*x] + b*C*(-4*a + 3 
*b*Sec[c + d*x])*Tan[c + d*x])))/(15*b^3*d*(A + 2*C + A*Cos[2*(c + d*x)])* 
Sec[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4581, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4581

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (2 a b C+\left (8 C a^2+3 b^2 (5 A+3 C)\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (2 a b C+\left (8 C a^2+3 b^2 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a b C+\left (8 C a^2+3 b^2 (5 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-\left (8 a^2 C-2 a b C+3 b^2 (5 A+3 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (8 a^2 C-2 a b C+3 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (8 a^2 C-2 a b C+3 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {-\frac {2 \sqrt {a+b} \left (8 a^2 C-2 a b C+3 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}-\frac {8 a C \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 C \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

Input:

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

Output:

(2*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]*(8*a^2*C + 3*b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[Ar 
cSin[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]*(8*a^2*C - 2*a*b*C + 3*b^2*(5*A + 3*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*b) - (8*a*C*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*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 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 4581

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(cs 
c[(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*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, 
 b, e, f, A, 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. \(1202\) vs. \(2(290)=580\).

Time = 58.32 (sec) , antiderivative size = 1203, normalized size of antiderivative = 3.76

method	result	size
default	\(\text {Expression too large to display}\)	\(1203\)
parts	\(\text {Expression too large to display}\)	\(1225\)

Input:

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

Output:

2/15/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)*(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))+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^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/ 
(a+b))^(1/2))+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^2*b*EllipticE(-c 
sc(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*(c 
os(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)*b^3*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*co 
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3*Elli 
pticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+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))/(co 
s(d*x+c)+1))^(1/2)*a^2*b*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))...
 

Fricas [F]

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

integrate(sec(d*x+c)^2*(A+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)^4 + A*sec(d*x + c)^2)/sqrt(b*sec(d*x + c) + a), x 
)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^2*(A+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)**4)/(sec(c + d*x)*b + a),x)*c + 
 int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x)**2)/(sec(c + d*x)*b + a),x)*a