\(\int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [287]

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 37, antiderivative size = 201 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {(11 A+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {(7 A+3 C) \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {(19 A+3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}} \] Output:

1/4*(11*A+3*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+C)*sin(d*x+c)/d/sec(d*x+c)^( 
1/2)/(a+a*sec(d*x+c))^(3/2)+1/6*(7*A+3*C)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/ 
(a+a*sec(d*x+c))^(1/2)-1/6*(19*A+3*C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a+a 
*sec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 1.61 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.85 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {-6 \sqrt {2} (11 A+3 C) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-2 (17 A+3 C+12 A \cos (c+d x)-2 A \cos (2 (c+d x))) \sqrt {1-\sec (c+d x)} \sec (c+d x) \tan (c+d x)}{12 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

(-6*Sqrt[2]*(11*A + 3*C)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[ 
c + d*x]]]*Cos[(c + d*x)/2]^2*Sec[c + d*x]^(5/2)*Sin[c + d*x] - 2*(17*A + 
3*C + 12*A*Cos[c + d*x] - 2*A*Cos[2*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sec 
[c + d*x]*Tan[c + d*x])/(12*d*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])]*(a 
*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 4573, 27, 3042, 4510, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {2 \int -\frac {a^2 (19 A+3 C)-2 a^2 (7 A+3 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A+3 C)-2 a^2 (7 A+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (19 A+3 C)-2 a^2 (7 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A+3 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-3 a^2 (11 A+3 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {6 a^2 (11 A+3 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {2 a (7 A+3 C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (19 A+3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 \sqrt {2} a^{3/2} (11 A+3 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {(A+C) \sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

Input:

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

Output:

-1/2*((A + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/ 
2)) + ((2*a*(7*A + 3*C)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*S 
ec[c + d*x]]) - ((-3*Sqrt[2]*a^(3/2)*(11*A + 3*C)*ArcTanh[(Sqrt[a]*Sqrt[Se 
c[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^2* 
(19*A + 3*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) 
)/(3*a))/(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 4510
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[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 

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 [A] (verified)

Time = 1.97 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.12

method result size
default \(-\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (A \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (33 \cos \left (d x +c \right )+66+33 \sec \left (d x +c \right )\right )+C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \left (9 \cos \left (d x +c \right )+18+9 \sec \left (d x +c \right )\right )+\left (-8 \cos \left (d x +c \right )^{2}+24 \cos \left (d x +c \right )+38\right ) A \tan \left (d x +c \right )+6 C \tan \left (d x +c \right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) \(225\)
parts \(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-8 \cos \left (d x +c \right )^{2}+24 \cos \left (d x +c \right )+38\right ) \tan \left (d x +c \right )+\arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (33 \cos \left (d x +c \right )+66+33 \sec \left (d x +c \right )\right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}-\frac {C \left (\left (-3 \cos \left (d x +c \right )-3\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\sin \left (d x +c \right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\right ) \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{4 d \,a^{2} \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(277\)

Input:

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

Output:

-1/12/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^2+2*cos(d*x+c)+1)/sec(d*x 
+c)^(3/2)*(A*(-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)))*(33*cos(d*x+c)+66+33*sec(d*x+c))+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)))*(9*cos(d*x+c)+18+9*sec(d*x+c))+(-8*cos(d*x+c)^2+24*cos 
(d*x+c)+38)*A*tan(d*x+c)+6*C*tan(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.27 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A + 3 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} - {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} \cos \left (d x + c\right ) + 11 \, A + 3 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (4 \, A \cos \left (d x + c\right )^{3} - 12 \, A \cos \left (d x + c\right )^{2} - {\left (19 \, A + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

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

Output:

[1/24*(3*sqrt(2)*((11*A + 3*C)*cos(d*x + c)^2 + 2*(11*A + 3*C)*cos(d*x + c 
) + 11*A + 3*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*cos(d*x + 
 c)^3 - 12*A*cos(d*x + c)^2 - (19*A + 3*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), -1/12*(3*sqrt(2)*((11*A + 3*C)*cos(d* 
x + c)^2 + 2*(11*A + 3*C)*cos(d*x + c) + 11*A + 3*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)^3 - 12*A*cos(d*x + c)^2 - (19*A + 3*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]

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

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34992 vs. \(2 (170) = 340\).

Time = 0.67 (sec) , antiderivative size = 34992, normalized size of antiderivative = 174.09 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/12*((4*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin 
(3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*ar 
ctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(7/3*arctan2(sin(3/ 
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 64*(cos(3*d*x + 3*c)^2*sin(3/2* 
d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3* 
c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d 
*x + 3/2*c))))*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) 
)^4 + 4*sin(3/2*d*x + 3/2*c)^5 + 4*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c 
) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin( 
3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c) 
)))*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^4 + 64*(c 
os(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin(3/2*d*x + 
3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arctan2(sin(3 
/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(5/3*arctan2(sin(3/2*d*x + 3/2 
*c), cos(3/2*d*x + 3/2*c)))^4 + 4*(2*cos(3*d*x + 3*c)^2*cos(3/2*d*x + 3/2* 
c)*sin(3/2*d*x + 3/2*c) + 2*cos(3/2*d*x + 3/2*c)*sin(3*d*x + 3*c)^2*sin(3/ 
2*d*x + 3/2*c) + 2*cos(3*d*x + 3*c)*cos(3/2*d*x + 3/2*c)*sin(3/2*d*x + 3/2 
*c) + 8*(cos(3*d*x + 3*c)^2*sin(3/2*d*x + 3/2*c) + sin(3*d*x + 3*c)^2*sin( 
3/2*d*x + 3/2*c) - 9*(cos(3*d*x + 3*c)^2 + sin(3*d*x + 3*c)^2)*sin(1/3*arc 
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(5/3*arctan2(sin(...
 

Giac [A] (verification not implemented)

Time = 2.42 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.88 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {3 \, {\left (11 \, \sqrt {2} A + 3 \, \sqrt {2} C\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}} + \frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} A a + \sqrt {2} C a\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} + \frac {2 \, {\left (23 \, \sqrt {2} A a + 3 \, \sqrt {2} C a\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {3 \, {\left (9 \, \sqrt {2} A a + \sqrt {2} C a\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}}}{12 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

-1/12*(3*(11*sqrt(2)*A + 3*sqrt(2)*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) + ((3*(sqrt(2)*A*a + sqrt 
(2)*C*a)*tan(1/2*d*x + 1/2*c)^2/a + 2*(23*sqrt(2)*A*a + 3*sqrt(2)*C*a)/a)* 
tan(1/2*d*x + 1/2*c)^2 + 3*(9*sqrt(2)*A*a + sqrt(2)*C*a)/a)*tan(1/2*d*x + 
1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}+2 \sec \left (d x +c \right )^{3}+\sec \left (d x +c \right )^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) c \right )}{a^{2}} \] Input:

int((A+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(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)**4 
 + 2*sec(c + d*x)**3 + sec(c + d*x)**2),x)*a + int((sqrt(sec(c + d*x))*sqr 
t(sec(c + d*x) + 1))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*c))/a**2