\(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [607]

Optimal result

Integrand size = 45, antiderivative size = 141 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {(2 B-C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d \sqrt {a+a \sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.71 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left (C \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-2 (B-C) \arcsin \left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} (A-B+C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

((C*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 2*(B - C)*ArcSin[Sqrt[Sec[c + d*x]]] 
- Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + 
 d*x]]] + C*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Tan[c + d*x])/(d*Sq 
rt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4576, 27, 3042, 4511, 3042, 4288, 222, 4295, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {2 \sqrt {a} (2 B-C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a (A-B+C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 4295

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

rule 4511

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

rule 4576

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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs 
c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Cs 
c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* 
B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m 
, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && 
NeQ[m + n + 1, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(388\) vs. \(2(120)=240\).

Time = 6.54 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.76

method	result	size
default	\(-\frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 A \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-2 B \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right ) \sqrt {2}+2 C \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+2 B \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 B \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-C \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-C \cos \left (d x +c \right ) \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right )}{2 d a \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(389\)
parts	\(\frac {A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{d a \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}-\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{\frac {3}{2}} \left (\arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}-\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 )}{d a \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (2 \sqrt {2}\, \cos \left (d x +c \right )^{3} \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{2}-\cos \left (d x +c \right )^{3} \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 )-\cos \left (d x +c \right )^{3} \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 )}{2 d a \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(489\)

Input:

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

Output:

-1/2/d/a*sec(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2)*(2*A*2^(1/2)*arctan(1/2 
*2^(1/2)*(csc(d*x+c)-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-2*B 
*2^(1/2)*arctan(1/2*2^(1/2)*(csc(d*x+c)-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1 
/2))*cos(d*x+c)-C*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*2^(1/2)+2*C*2^(1/2) 
*arctan(1/2*2^(1/2)*(csc(d*x+c)-cot(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))*cos 
(d*x+c)+2*B*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(cos(d*x+ 
c)+1))^(1/2))+2*B*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(co 
s(d*x+c)+1))^(1/2))-C*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1 
/(cos(d*x+c)+1))^(1/2))-C*cos(d*x+c)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1) 
/(-1/(cos(d*x+c)+1))^(1/2)))/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.78 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1442 vs. \(2 (120) = 240\).

Time = 0.40 (sec) , antiderivative size = 1442, normalized size of antiderivative = 10.23 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(2*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si 
n(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x 
 + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*A/sqrt(a) - 2*(sqrt(2)*log(cos( 
1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arctan2(sin(d*x + c), 
 cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 1) - 
sqrt(2)*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + sin(1/2*arcta 
n2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*arctan2(sin(d*x + c), cos(d* 
x + c))) + 1) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*s 
in(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2( 
sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos 
(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 
2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arcta 
n2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), 
cos(d*x + c))) + 2) - log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 
 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*ar 
ctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c 
), cos(d*x + c))) + 2) + log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)) 
)^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2 
*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x 
+ c), cos(d*x + c))) + 2))*B/sqrt(a) - (4*sqrt(2)*cos(3/2*arctan2(sin(d...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (120) = 240\).

Time = 1.78 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.30 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {\sqrt {2} {\left (A \sqrt {a} - B \sqrt {a} + C \sqrt {a}\right )} \log \left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{a} - \frac {{\left (2 \, B - C\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right )}{\sqrt {a}} + \frac {{\left (2 \, B - C\right )} \log \left ({\left | {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right )}{\sqrt {a}} - \frac {4 \, \sqrt {2} {\left (3 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} C \sqrt {a} - C a^{\frac {3}{2}}\right )}}{{\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} - 6 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a + a^{2}}}{2 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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