\(\int \frac {\sqrt {\sec (c+d x)} (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [233]

Optimal result

Integrand size = 35, antiderivative size = 152 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a d}-\frac {(A+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:

-(A+3*C)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c) 
^(1/2)/a/d+(A-C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*s 
ec(d*x+c)^(1/2)/a/d+(A+3*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d-(A+C)*sec(d*x+ 
c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.59 (sec) , antiderivative size = 776, normalized size of antiderivative = 5.11 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2* 
I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-3*Sqrt[1 + E^( 
(2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2 
, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E 
^(I*d*x)*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (Sqrt[2]*C 
*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d* 
x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + 
 d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4 
, -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(d*E^(I*d*x)*(A 
+ 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*A*Cos[c/2 + (d*x)/2 
]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + C* 
Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x 
]]*(a + a*Sec[c + d*x])) - (2*C*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Cs 
c[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sin[c])/( 
d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])) 
+ (Cos[c/2 + (d*x)/2]^2*(A + C*Sec[c + d*x]^2)*((2*(A + 3*C)*Cos[d*x]*Csc[ 
c/2]*Sec[c/2])/d - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[ 
(d*x)/2]))/d - (4*(A + C)*Tan[c/2])/d))/((A + 2*C + A*Cos[2*c + 2*d*x])*Sq 
rt[Sec[c + d*x]]*(a + a*Sec[c + d*x]))
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4573, 27, 3042, 4274, 3042, 4255, 3042, 4258, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 a (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+a (A+3 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}\)

Input:

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

Output:

-(((A + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))) + (( 
2*a*(A - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x] 
])/d + a*(A + 3*C)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[ 
Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/(2*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 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4573

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) 
*(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*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*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, 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. \(315\) vs. \(2(143)=286\).

Time = 1.64 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.08

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +3 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +5 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(316\)

Input:

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

Output:

-1/a*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-cos(1/2*d 
*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2* 
d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*d* 
x+1/2*c),2^(1/2))+A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-C*EllipticF(cos( 
1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(-2*s 
in(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+3*C)*sin(1/2*d*x+1/2*c) 
^4+(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+5*C)*sin(1/2*d* 
x+1/2*c)^2)/cos(1/2*d*x+1/2*c)/sin(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^ 
2-1)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left ({\left (A + 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/2*((sqrt(2)*(-I*A + I*C)*cos(d*x + c) + sqrt(2)*(-I*A + I*C))*weierstras 
sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - I*C)*cos 
(d*x + c) + sqrt(2)*(I*A - I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - 
 I*sin(d*x + c)) + (sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 
3*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I 
*sin(d*x + c))) + (sqrt(2)*(I*A + 3*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 3*I 
*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si 
n(d*x + c))) + 2*((A + 3*C)*cos(d*x + c) + 2*C)*sin(d*x + c)/sqrt(cos(d*x 
+ c)))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sqrt {\sec {\left (c + d x \right )}}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{\frac {5}{2}}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

(Integral(A*sqrt(sec(c + d*x))/(sec(c + d*x) + 1), x) + Integral(C*sec(c + 
 d*x)**(5/2)/(sec(c + d*x) + 1), x))/a
 

Maxima [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + A)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a), 
x)
 

Giac [F]

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

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

Output:

integrate((C*sec(d*x + c)^2 + A)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a), 
x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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