\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [960]

Optimal result

Integrand size = 41, antiderivative size = 267 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 (a-b) \sqrt {a+b} (3 b B-2 a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 d}+\frac {2 \sqrt {a+b} (3 A b-b (3 B-C)+2 a C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^2 d}+\frac {2 C \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b d} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 15.11 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.76 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {4 \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (a+b) (-3 b B+2 a C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (3 A b-2 a C+b (3 B+C)) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-(3 b B-2 a C) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {\cos (c+d x) (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 (3 b B-2 a C) \sin (c+d x)}{3 b^2}+\frac {4 C \tan (c+d x)}{3 b}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {a+b \sec (c+d x)}} \] Input:

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

Output:

(4*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d 
*x]^2)*(2*(a + b)*(-3*b*B + 2*a*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*S 
qrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Ta 
n[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(3*A*b - 2*a*C + b*(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)] - (3* 
b*B - 2*a*C)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + 
 d*x)/2]))/(3*b^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt 
[Sec[(c + d*x)/2]^2]*Sec[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) + (Cos[c 
 + d*x]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*( 
3*b*B - 2*a*C)*Sin[c + d*x])/(3*b^2) + (4*C*Tan[c + d*x])/(3*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]])
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

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

\(\Big \downarrow \) 4492

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

Input:

Int[(Sec[c + d*x]*(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]*(3*b*B - 2*a*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqr 
t[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 - b*(3*B - C) + 2*a*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*b) + (2*C* 
Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b*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 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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(943\) vs. \(2(241)=482\).

Time = 59.42 (sec) , antiderivative size = 944, normalized size of antiderivative = 3.54

method	result	size
default	\(\text {Expression too large to display}\)	\(944\)
parts	\(\text {Expression too large to display}\)	\(981\)

Input:

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

Output:

-2/3/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)*((-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*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*EllipticE(-csc(d*x+c) 
+cot(d*x+c),((a-b)/(a+b))^(1/2))+(-3*cos(d*x+c)^2-6*cos(d*x+c)-3)*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 
)*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(2*cos(d*x+c)^ 
2+4*cos(d*x+c)+2)*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^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a 
+b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*C*(1/(a+b)*(b+a*cos(d*x+c))/(c 
os(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b*EllipticE(-csc(d 
*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(3*cos(d*x+c)^2+6*cos(d*x+c)+3)*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^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+(3*cos(d*x 
+c)^2+6*cos(d*x+c)+3)*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(c 
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b 
)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*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*EllipticF(- 
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)*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^2*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))-3*B*...
 

Fricas [F]

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

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

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

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)*(A+B*sec(d*x+c)+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)**3)/(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)*b 
+ int((sqrt(sec(c + d*x)*b + a)*sec(c + d*x))/(sec(c + d*x)*b + a),x)*a