3.12.58 \(\int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx\) [1158]

3.12.58.1 Optimal result
3.12.58.2 Mathematica [A] (verified)
3.12.58.3 Rubi [A] (verified)
3.12.58.4 Maple [A] (verified)
3.12.58.5 Fricas [A] (verification not implemented)
3.12.58.6 Sympy [F]
3.12.58.7 Maxima [B] (verification not implemented)
3.12.58.8 Giac [F]
3.12.58.9 Mupad [F(-1)]

3.12.58.1 Optimal result

Integrand size = 37, antiderivative size = 173 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=-\frac {C \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {\sqrt {2} (A+C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{\sqrt {a} d}+\frac {C \sin (c+d x)}{d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \]

output
-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)/d/a^(1/2)+(A+C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1 
/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)*cos(d*x+c)^(1/2)*sec(d*x+c)^(1 
/2)/d/a^(1/2)+C*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)
 
3.12.58.2 Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.10 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {\cos (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (C \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+2 C \arcsin \left (\sqrt {\sec (c+d x)}\right )-\sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )-\sqrt {2} C \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \sin (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

input
Integrate[(A + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d* 
x]]),x]
 
output
(Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(C*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 
 2*C*ArcSin[Sqrt[Sec[c + d*x]]] - Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d 
*x]])/Sqrt[1 - Sec[c + d*x]]] - Sqrt[2]*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])]) 
*Sin[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 
3.12.58.3 Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 4753, 3042, 4577, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4753

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4577

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4288

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

\(\Big \downarrow \) 222

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

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

input
Int[(A + C*Sec[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]),x 
]
 
output
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((-2*Sqrt[a]*C*ArcSinh[(Sqrt[a]*Tan 
[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*Sqrt[2]*Sqrt[a]*(A + C)*ArcTa 
nh[(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]]))
 

3.12.58.3.1 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 4577
Int[((A_.) + 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*Csc[e + f*x])^n/(f*(m + n + 1))), 
x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n 
*Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
 b, d, e, f, A, 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 
]
 
3.12.58.4 Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.59

method result size
default \(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (2 A \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+2 C \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-C \cos \left (d x +c \right ) \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-C \cos \left (d x +c \right ) \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-2 C \sin \left (d x +c \right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\right )}{2 d \left (1+\cos \left (d x +c \right )\right ) a \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\cos \left (d x +c \right )}}\) \(275\)

input
int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2),x,method=_R 
ETURNVERBOSE)
 
output
-1/2/d*(a*(1+sec(d*x+c)))^(1/2)*(2*A*cos(d*x+c)*2^(1/2)*arctan(1/2*sin(d*x 
+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))+2*C*cos(d*x+c)*2^(1/ 
2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(1+cos(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2)) 
-C*cos(d*x+c)*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(1+cos(d*x+c))/(-1/(1+ 
cos(d*x+c)))^(1/2))-C*cos(d*x+c)*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(1+c 
os(d*x+c))/(-1/(1+cos(d*x+c)))^(1/2))-2*C*sin(d*x+c)*(-1/(1+cos(d*x+c)))^( 
1/2))/(1+cos(d*x+c))/a/(-1/(1+cos(d*x+c)))^(1/2)/cos(d*x+c)^(1/2)
 
3.12.58.5 Fricas [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 542, normalized size of antiderivative = 3.13 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\left [\frac {4 \, C \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 ) + {\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} + 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {2 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A + C\right )} a \cos \left (d x + c\right )\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \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 )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{4 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}, -\frac {2 \, \sqrt {2} {\left ({\left (A + C\right )} a \cos \left (d x + c\right )^{2} + {\left (A + C\right )} a \cos \left (d x + c\right )\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - 2 \, C \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 ) + {\left (C \cos \left (d x + c\right )^{2} + C \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {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 )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{2 \, {\left (a d \cos \left (d x + c\right )^{2} + a d \cos \left (d x + c\right )\right )}}\right ] \]

input
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2),x, al 
gorithm="fricas")
 
output
[1/4*(4*C*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d 
*x + c) + (C*cos(d*x + c)^2 + C*cos(d*x + c))*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)*s 
qrt(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)) + 2*sqrt(2)*((A + C)*a*cos(d*x + c)^2 + (A + C)*a*cos( 
d*x + c))*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)^2 + a*d*cos 
(d*x + c)), -1/2*(2*sqrt(2)*((A + C)*a*cos(d*x + c)^2 + (A + C)*a*cos(d*x 
+ c))*sqrt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq 
rt(-1/a)*sqrt(cos(d*x + c))/sin(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) + (C*cos(d*x + c)^2 + C*cos( 
d*x + c))*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)^2 + a*d*cos(d*x + c))]
 
3.12.58.6 Sympy [F]

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

input
integrate((A+C*sec(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+a*sec(d*x+c))**(1/2),x)
 
output
Integral((A + C*sec(c + d*x)**2)/(sqrt(a*(sec(c + d*x) + 1))*sqrt(cos(c + 
d*x))), x)
 
3.12.58.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18526 vs. \(2 (144) = 288\).

Time = 0.62 (sec) , antiderivative size = 18526, normalized size of antiderivative = 107.09 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \]

input
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2),x, al 
gorithm="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*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) 
 - 2*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) - 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)*co 
s(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*s 
in(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))*cos(d*x + c)^4 + (2*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) - 2*sqrt(2)*log(co 
s(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(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*co 
s(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*s 
in(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(...
 
3.12.58.8 Giac [F]

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

input
integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2),x, al 
gorithm="giac")
 
output
integrate((C*sec(d*x + c)^2 + A)/(sqrt(a*sec(d*x + c) + a)*sqrt(cos(d*x + 
c))), x)
 
3.12.58.9 Mupad [F(-1)]

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

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