\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx\) [616]

Optimal result

Integrand size = 45, antiderivative size = 161 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-3 B-C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A-B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/4*(7*A-3*B-C)*arctanh(1/2*a^(1/2)*sec(d*x+c)^(1/2)*sin(d*x+c)*2^(1/2)/( 
a+a*sec(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/d-1/2*(A-B+C)*sec(d*x+c)^(1/2)*sin( 
d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/2*(5*A-B+C)*sec(d*x+c)^(1/2)*sin(d*x+c)/ 
a/d/(a+a*sec(d*x+c))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(512\) vs. \(2(161)=322\).

Time = 2.46 (sec) , antiderivative size = 512, normalized size of antiderivative = 3.18 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\frac {10 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-2 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+2 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+8 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} \sin (c+d x)+7 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-3 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+7 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-3 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)+2 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) (1+\sec (c+d x)) \tan (c+d x)+2 C \arcsin \left (\sqrt {\sec (c+d x)}\right ) (1+\sec (c+d x)) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(10*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] - 2*B*Sqrt[1 
- Sec[c + d*x]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 2*C*Sqrt[1 - Sec[c + d*x 
]]*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 8*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c 
+ d*x])]*Sin[c + d*x] + 7*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sq 
rt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 3*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[ 
c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - Sqrt[2]*C*ArcTan[(Sqrt[2 
]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] + 7*Sqrt[2]*A*A 
rcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Ta 
n[c + d*x] - 3*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[ 
c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] - Sqrt[2]*C*ArcTan[(Sqrt[2]*Sqrt[Sec[ 
c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] + 2*C*ArcSin[ 
Sqrt[1 - Sec[c + d*x]]]*(1 + Sec[c + d*x])*Tan[c + d*x] + 2*C*ArcSin[Sqrt[ 
Sec[c + d*x]]]*(1 + Sec[c + d*x])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x] 
]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {3042, 4572, 27, 3042, 4501, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (5 A-B+C)-2 a (A-B-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {2 a (5 A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a (7 A-3 B-C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (5 A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-a (7 A-3 B-C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {2 a (7 A-3 B-C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a (5 A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a (5 A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {\sqrt {2} \sqrt {a} (7 A-3 B-C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((A - B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) 
^(3/2)) + (-((Sqrt[2]*Sqrt[a]*(7*A - 3*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*(5*A 
- B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4 
*a^2)
 

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 3042

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

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 4501

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

rule 4572

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(298\) vs. \(2(136)=272\).

Time = 4.15 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.86

method	result	size
default	\(\frac {\left (\left (7 \cos \left (d x +c \right )^{2}+14 \cos \left (d x +c \right )+7\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, A \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, B \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\left (8 \cos \left (d x +c \right )+10\right ) \sin \left (d x +c \right ) A -2 B \sin \left (d x +c \right )+2 C \sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{4 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\sec \left (d x +c \right )}}\)	\(299\)
parts	\(\frac {A \left (\frac {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}}{4}+\frac {7 \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}}{4}+\frac {9 \csc \left (d x +c \right )}{4}-\frac {9 \cot \left (d x +c \right )}{4}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{2} \sqrt {\sec \left (d x +c \right )}}+\frac {B \left (\left (3 \cos \left (d x +c \right )+3\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{4 d \,a^{2} \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {C \left (\left (\cos \left (d x +c \right )+1\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \cos \left (d x +c \right )^{2}}{4 d \,a^{2} \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(381\)

Input:

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

Output:

1/4/d/a^2*((7*cos(d*x+c)^2+14*cos(d*x+c)+7)*(-2/(cos(d*x+c)+1))^(1/2)*A*ar 
ctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))+(-3*c 
os(d*x+c)^2-6*cos(d*x+c)-3)*(-2/(cos(d*x+c)+1))^(1/2)*B*arctan(1/2*2^(1/2) 
/(-1/(cos(d*x+c)+1))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))+(-cos(d*x+c)^2-2*cos( 
d*x+c)-1)*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)/(-1/(cos(d*x+c)+1 
))^(1/2)*(-csc(d*x+c)+cot(d*x+c)))+(8*cos(d*x+c)+10)*sin(d*x+c)*A-2*B*sin( 
d*x+c)+2*C*sin(d*x+c))*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c) 
+1)/sec(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.80 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right ) + 7 \, A - 3 \, B - C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \frac {4 \, {\left (4 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A - B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {\sqrt {2} {\left ({\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A - 3 \, B - C\right )} \cos \left (d x + c\right ) + 7 \, A - 3 \, B - C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (4 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A - B + C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

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

Output:

[-1/8*(sqrt(2)*((7*A - 3*B - C)*cos(d*x + c)^2 + 2*(7*A - 3*B - C)*cos(d*x 
 + c) + 7*A - 3*B - C)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)* 
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 
2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*(4*A*co 
s(d*x + c)^2 + (5*A - B + C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d 
*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d* 
cos(d*x + c) + a^2*d), 1/4*(sqrt(2)*((7*A - 3*B - C)*cos(d*x + c)^2 + 2*(7 
*A - 3*B - C)*cos(d*x + c) + 7*A - 3*B - C)*sqrt(-a)*arctan(sqrt(2)*sqrt(- 
a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + 
 c))) + 2*(4*A*cos(d*x + c)^2 + (5*A - B + C)*cos(d*x + c))*sqrt((a*cos(d* 
x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x 
+ c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 23927 vs. \(2 (136) = 272\).

Time = 0.91 (sec) , antiderivative size = 23927, normalized size of antiderivative = 148.61 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*((4*(7*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) - 7*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) - 8*sin(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3/ 
2*c)^4 + 63*(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) - 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))*cos(1/2*d*x + 1/2*c)^4 + 4*(7*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) - 7* 
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) - 8*sin(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c)^4 + 70*(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) - l 
og(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))*cos(1/2*d*x + 1/2*c)^2*sin(1/2*d*x + 1/2*c)^2 + 7*(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) - log(c 
os(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))*sin(1/2*d*x + 1/2*c)^4 - 8*sin(1/2*d*x + 1/2*c)^5 + 28*(7*(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) - l 
og(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))*cos(1/2*d*x + 1/2*c) - 8*cos(1/2*d*x + 1/2*c)*sin(1/2*d*x + 1/2*c) 
)*cos(3/2*d*x + 3/2*c)^3 + 4*(21*(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) - log(cos(1/2*d*x + 1/2*c)^2 ...
 

Giac [A] (verification not implemented)

Time = 2.51 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {{\left (\frac {\sqrt {2} {\left (A a^{2} - B a^{2} + C a^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3}} + \frac {\sqrt {2} {\left (9 \, A a^{2} - B a^{2} + C a^{2}\right )}}{a^{3}}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} + \frac {\sqrt {2} {\left (7 \, A - 3 \, B - C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}}}{4 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

1/4*((sqrt(2)*(A*a^2 - B*a^2 + C*a^2)*tan(1/2*d*x + 1/2*c)^2/a^3 + sqrt(2) 
*(9*A*a^2 - B*a^2 + C*a^2)/a^3)*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 
1/2*c)^2 + a) + sqrt(2)*(7*A - 3*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)))/a^(3/2))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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