\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx\) [1277]

Optimal result

Integrand size = 45, antiderivative size = 235 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {(8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 \sqrt {a} d}-\frac {\sqrt {2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {C \sin (c+d x)}{2 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {(4 B-C) \sin (c+d x)}{4 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(8*A-4*B+7*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))*cos(d 
*x+c)^(1/2)*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*sec(d*x+c))^(1/2))*cos(d*x+c)^(1 
/2)*sec(d*x+c)^(1/2)/a^(1/2)/d+1/2*C*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*se 
c(d*x+c))^(1/2)+1/4*(4*B-C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c)) 
^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 0.53 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left ((4 B-C) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-8 (A-B+C) \arcsin \left (\sqrt {\sec (c+d x)}\right )+4 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )-4 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+4 \sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+2 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+4 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}-C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*((4*B - C)*ArcSin[Sqrt[1 - Sec[c + 
d*x]]] - 8*(A - B + C)*ArcSin[Sqrt[Sec[c + d*x]]] + 4*Sqrt[2]*A*ArcTan[(Sq 
rt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] - 4*Sqrt[2]*B*ArcTan[(Sq 
rt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] + 4*Sqrt[2]*C*ArcTan[(Sq 
rt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]] + 2*C*Sqrt[1 - Sec[c + d 
*x]]*Sec[c + d*x]^(3/2) + 4*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] - 
C*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])])*Sin[c + d*x])/(4*d*Sqrt[1 - S 
ec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.49 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4753, 3042, 4576, 27, 3042, 4509, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4576

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((4 B-C) a^2+(8 A-4 B+7 C) \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((4 B-C) a^2+(8 A-4 B+7 C) \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((4 B-C) a^2+(8 A-4 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 4511

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a (8 A-4 B+7 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx-8 a^2 (A-B+C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a (8 A-4 B+7 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-8 a^2 (A-B+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 4288

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {-8 a^2 (A-B+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a (8 A-4 B+7 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 222

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^{3/2} (8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-8 a^2 (A-B+C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 4295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {16 a^2 (A-B+C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^{3/2} (8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^{3/2} (8 A-4 B+7 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {8 \sqrt {2} a^{3/2} (A-B+C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 B-C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((C*Sec[c + d*x]^(5/2)*Sin[c + d*x]) 
/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (((2*a^(3/2)*(8*A - 4*B + 7*C)*ArcSinh[( 
Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (8*Sqrt[2]*a^(3/2)*(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) + (a*(4*B - C)*Sec[c + d*x]^(3/2)*Sin[c + 
d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*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 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 4509

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

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]
 

rule 4753

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a 
+ b*x])^m*(c*Sec[a + b*x])^m   Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[u, x 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(531\) vs. \(2(196)=392\).

Time = 3.18 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.26

method	result	size
default	\(\frac {\left (-8 A \cos \left (d x +c \right )^{2} \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 )+8 B \cos \left (d x +c \right )^{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 ) \sqrt {2}-8 C \cos \left (d x +c \right )^{2} \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 )+8 A \cos \left (d x +c \right )^{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 )-4 B \cos \left (d x +c \right )^{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 )+7 C \cos \left (d x +c \right )^{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 )+8 A \cos \left (d x +c \right )^{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 )-4 B \cos \left (d x +c \right )^{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 )+7 C \cos \left (d x +c \right )^{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 )+4 B \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+\sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )+2\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{8 d \cos \left (d x +c \right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )+1\right ) a \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(532\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 645, normalized size of antiderivative = 2.74 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(4*((4*B - C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x 
+ c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A - 4*B + 7*C)*cos(d*x + c)^3 
+ (8*A - 4*B + 7*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt 
(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d* 
x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x 
 + c)^2)) + 8*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a*cos(d* 
x + c)^2)*log(-(cos(d*x + c)^2 + 2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d 
*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(co 
s(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos 
(d*x + c)^2), 1/8*(8*sqrt(2)*((A - B + C)*a*cos(d*x + c)^3 + (A - B + C)*a 
*cos(d*x + c)^2)*sqrt(-1/a)*arctan(1/2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/c 
os(d*x + c))*sqrt(-1/a)*sqrt(cos(d*x + c))*sin(d*x + c)/(cos(d*x + c) + 1) 
) + 2*((4*B - C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c 
))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A - 4*B + 7*C)*cos(d*x + c)^3 + ( 
8*A - 4*B + 7*C)*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d* 
x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^ 
2 - a*cos(d*x + c) - 2*a)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3334 vs. \(2 (196) = 392\).

Time = 0.41 (sec) , antiderivative size = 3334, normalized size of antiderivative = 14.19 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/16*(8*(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*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sin(1/2*ar 
ctan2(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*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) + 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) - 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)*s 
in(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + log(2*cos(1/2*arctan2(s 
in(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))*A/sqrt(a) + 4*(4*sqrt 
(2)*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x + 2*c) 
- 4*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(2*d*x 
 + 2*c) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 570 vs. \(2 (196) = 392\).

Time = 165.94 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.43 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/8*(4*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*sgn(cos(d*x + c))) 
+ (8*A - 4*B + 7*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)*sgn(cos(d*x + c))) 
- (8*A - 4*B + 7*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)*sgn(cos(d*x + c))) 
+ 4*sqrt(2)*(12*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c 
)^2 + a))^6*B*sqrt(a) - 17*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2* 
d*x + 1/2*c)^2 + a))^6*C*sqrt(a) - 76*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt 
(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(3/2) + 57*(sqrt(a)*tan(1/2*d*x + 1/ 
2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*a^(3/2) + 36*(sqrt(a)*tan(1 
/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(5/2) - 19*(sq 
rt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^(5/ 
2) - 4*B*a^(7/2) + 3*C*a^(7/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t 
an(1/2*d*x + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t 
an(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

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

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

Reduce [F]

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

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

Output:

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