\(\int \frac {\sqrt {a+a \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [582]

Optimal result

Integrand size = 45, antiderivative size = 120 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {2 a (A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {2 A \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a \left ((2 A+3 B+A \cos (c+d x)) \sqrt {1-\sec (c+d x)}-3 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) \sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{3 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2 
))/Sec[c + d*x]^(3/2),x]
 

Output:

(2*a*((2*A + 3*B + A*Cos[c + d*x])*Sqrt[1 - Sec[c + d*x]] - 3*C*ArcSin[Sqr 
t[Sec[c + d*x]]]*Sqrt[Sec[c + d*x]])*Tan[c + d*x])/(3*d*Sqrt[-((-1 + Sec[c 
 + d*x])*Sec[c + d*x])]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3042, 4574, 27, 3042, 4503, 3042, 4288, 222}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4574

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4503

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {2 a^2 (A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {6 a C \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {6 a^{3/2} C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {2 a^2 (A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}+\frac {2 A \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\)

Input:

Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec 
[c + d*x]^(3/2),x]
 

Output:

(2*A*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + ((6 
*a^(3/2)*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + ( 
2*a^2*(A + 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x 
]]))/(3*a)
 

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

rule 4503

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co 
t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp 
[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n)   Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ 
e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a 
*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && LtQ[n, 0]
 

rule 4574

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e 
 + f*x])^n/(f*n)), x] - Simp[1/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[ 
e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] 
, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
&&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
 

Maple [A] (warning: unable to verify)

Time = 4.59 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.58

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (3+3 \sec \left (d x +c \right )\right )+\sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (3+3 \sec \left (d x +c \right )\right )+A \left (4 \sin \left (d x +c \right )+8 \tan \left (d x +c \right )\right )+12 B \tan \left (d x +c \right )\right )}{6 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(189\)
parts	\(\frac {A \left (2 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (3 \cos \left (d x +c \right )+3\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}-\frac {2 B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{d \sqrt {\sec \left (d x +c \right )}}-\frac {C \left (\arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \cos \left (d x +c \right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(222\)

Input:

int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 
),x,method=_RETURNVERBOSE)
 

Output:

1/6/d*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*(2^(1/2)*C* 
(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(cos(d 
*x+c)+1))^(1/2))*(3+3*sec(d*x+c))+2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*arct 
an(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*(3+3*sec(d*x+ 
c))+A*(4*sin(d*x+c)+8*tan(d*x+c))+12*B*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.96 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\left [\frac {3 \, {\left (C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (A \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {3 \, {\left (C \cos \left (d x + c\right ) + C\right )} \sqrt {-a} \arctan \left (\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (A \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{3 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(3/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(C*cos(d*x + c) + C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + 
 c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + 
 a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + 
 cos(d*x + c)^2)) + 4*(A*cos(d*x + c)^2 + (2*A + 3*B)*cos(d*x + c))*sqrt(( 
a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos( 
d*x + c) + d), 1/3*(3*(C*cos(d*x + c) + C)*sqrt(-a)*arctan(1/2*(cos(d*x + 
c)^2 - 2*cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a 
*sqrt(cos(d*x + c))*sin(d*x + c))) + 2*(A*cos(d*x + c)^2 + (2*A + 3*B)*cos 
(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d 
*x + c)))/(d*cos(d*x + c) + d)]
 

Sympy [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right )}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate((a+a*sec(d*x+c))**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x 
+c)**(3/2),x)
 

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)** 
2)/sec(c + d*x)**(3/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (102) = 204\).

Time = 0.28 (sec) , antiderivative size = 373, normalized size of antiderivative = 3.11 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(3/2),x, algorithm="maxima")
 

Output:

1/6*(sqrt(2)*(3*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) 
))*sin(3/2*d*x + 3/2*c) - 3*cos(3/2*d*x + 3/2*c)*sin(2/3*arctan2(sin(3/2*d 
*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(3/2*d*x + 3/2*c) + 3*sin(1/3*a 
rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*A*sqrt(a) + 12*sqrt(2 
)*B*sqrt(a)*sin(1/2*d*x + 1/2*c) + 3*C*sqrt(a)*(log(2*cos(1/2*d*x + 1/2*c) 
^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2) 
*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x 
+ 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2* 
c) + 2) + log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt 
(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - log(2*cos 
(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1 
/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (102) = 204\).

Time = 1.14 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.92 \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {3 \, C a^{\frac {3}{2}} \log \left (\frac {{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{{\left | a \right |}} + \frac {2 \, {\left (3 \, \sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + {\left (\sqrt {2} A a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 3 \, \sqrt {2} B a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}}{3 \, d} \] Input:

integrate((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c 
)^(3/2),x, algorithm="giac")
 

Output:

1/3*(3*C*a^(3/2)*log(abs(2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2* 
d*x + 1/2*c)^2 + a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(a)*tan(1/2*d* 
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 4*sqrt(2)*abs(a) - 6* 
a))*sgn(cos(d*x + c))/abs(a) + 2*(3*sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 3*sq 
rt(2)*B*a^2*sgn(cos(d*x + c)) + (sqrt(2)*A*a^2*sgn(cos(d*x + c)) + 3*sqrt( 
2)*B*a^2*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/( 
a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 
1/cos(c + d*x))^(3/2),x)
 

Output:

int(((a + a/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/( 
1/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )}d x \right ) b +\left (\int \sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}d x \right ) c \right ) \] Input:

int((a+a*sec(d*x+c))^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2 
),x)
 

Output:

sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x)**2,x 
)*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1))/sec(c + d*x),x)*b + 
int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1),x)*c)