\(\int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [196]

Optimal result

Integrand size = 27, antiderivative size = 125 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {(5 A-3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}} \] Output:

2*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-1/4*(5*A-3 
*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/ 
a^(3/2)/d-1/2*(A+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.92 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.23 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\left ((A+C) (-1+\cos (c+d x))+8 A \arctan \left (\sqrt {-1+\sec (c+d x)}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}-\sqrt {2} (5 A-3 C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}\right ) \tan \left (\frac {1}{2} (c+d x)\right )}{2 a d (-1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

-1/2*(((A + C)*(-1 + Cos[c + d*x]) + 8*A*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*C 
os[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] - Sqrt[2]*(5*A - 3*C)*ArcTan[Sqr 
t[-1 + Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]])* 
Tan[(c + d*x)/2])/(a*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 4541, 27, 3042, 4408, 3042, 4261, 216, 4282, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4541

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4408

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4261

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 4282

\(\displaystyle \frac {\frac {2 a (5 A-3 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {8 \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A+C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {8 \sqrt {a} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} \sqrt {a} (5 A-3 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A+C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

Input:

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

Output:

((8*Sqrt[a]*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - 
 (Sqrt[2]*Sqrt[a]*(5*A - 3*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[ 
a + a*Sec[c + d*x]])])/d)/(4*a^2) - ((A + C)*Tan[c + d*x])/(2*d*(a + a*Sec 
[c + d*x])^(3/2))
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 3042

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

rule 4261

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

rule 4282

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

rule 4408

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

rule 4541

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*((a + b*Csc[e + 
 f*x])^m/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + 
 f*x])^(m + 1)*Simp[A*b*(2*m + 1) - a*(A*(m + 1) - C*m)*Csc[e + f*x], x], x 
], x] /; FreeQ[{a, b, e, f, A, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1 
)]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(244\) vs. \(2(104)=208\).

Time = 0.88 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.96

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (A \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+4 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+C \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-5 A \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 C \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{4 d \,a^{2}}\)	\(245\)
parts	\(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-4 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )-\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+5 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{4 d \,a^{2}}+\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+3 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{4 d \,a^{2}}\)	\(285\)

Input:

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

Output:

1/4/d/a^2*(a*(1+sec(d*x+c)))^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(A 
*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(csc(d*x+c)-cot(d*x+c))+4*A*2^(1/2)* 
arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(csc(d*x+c)-cot(d* 
x+c)))+C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(csc(d*x+c)-cot(d*x+c))-5*A* 
ln((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c))+3*C*ln((-2* 
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cot(d*x+c)+csc(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (104) = 208\).

Time = 1.59 (sec) , antiderivative size = 544, normalized size of antiderivative = 4.35 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {4 \, {\left (A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \sqrt {2} {\left ({\left (5 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 3 \, C\right )} \cos \left (d x + c\right ) + 5 \, A - 3 \, C\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 8 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\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 {2 \, {\left (A + C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - \sqrt {2} {\left ({\left (5 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 3 \, C\right )} \cos \left (d x + c\right ) + 5 \, A - 3 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 8 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\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+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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