\(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [256]

Optimal result

Integrand size = 25, antiderivative size = 177 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {11 \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac {7 \sin (c+d x)}{6 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {19 \sqrt {\sec (c+d x)} \sin (c+d x)}{6 a d \sqrt {a+a \sec (c+d x)}} \] Output:

11/4*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*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d* 
x+c))^(3/2)+7/6*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)-19/ 
6*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 = 0.62 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\frac {-33 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+\sqrt {1-\sec (c+d x)} (4 \sin (c+d x)-(12+19 \sec (c+d x)) \tan (c+d x))}{6 d \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (a (1+\sec (c+d x)))^{3/2}} \] Input:

Integrate[1/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)),x]
 

Output:

(-33*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*C 
os[(c + d*x)/2]^2*Sec[c + d*x]^(5/2)*Sin[c + d*x] + Sqrt[1 - Sec[c + d*x]] 
*(4*Sin[c + d*x] - (12 + 19*Sec[c + d*x])*Tan[c + d*x]))/(6*d*Sqrt[-((-1 + 
 Sec[c + d*x])*Sec[c + d*x])]*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4304, 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.

\(\displaystyle \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4304

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4510

\(\displaystyle \frac {\frac {2 \int -\frac {19 a^2-14 a^2 \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {19 a^2-14 a^2 \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {19 a^2-14 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {38 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-33 a^2 \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {38 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-33 a^2 \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {66 a^2 \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {38 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {14 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {38 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {33 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{4 a^2}-\frac {\sin (c+d x)}{2 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}\)

Input:

Int[1/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)),x]
 

Output:

-1/2*Sin[c + d*x]/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((14 
*a*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((-33 
*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2 
]*Sqrt[a + a*Sec[c + d*x]])])/d + (38*a^2*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 4304

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc 
[e + f*x])^n/(f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1))   Int[(a + b*Csc[e 
 + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e 
+ f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ 
[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
 

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]
 

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (8 \cos \left (d x +c \right )^{2}-24 \cos \left (d x +c \right )-38\right ) \tan \left (d x +c \right )+\arctan \left (\frac {\sqrt {2}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (-33 \cos \left (d x +c \right )-66-33 \sec \left (d x +c \right )\right )\right )}{12 d \,a^{2} \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\)	\(145\)

Input:

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

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)*((8*cos(d*x+c)^2-24*cos(d*x+c)-38)*tan(d*x+c)+arctan(1/2*2^(1/2)* 
(cot(d*x+c)-csc(d*x+c))/(-1/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/ 
2)*(-33*cos(d*x+c)-66-33*sec(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.25 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {33 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 12 \, \cos \left (d x + c\right )^{2} - 19 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac {33 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (4 \, \cos \left (d x + c\right )^{3} - 12 \, \cos \left (d x + c\right )^{2} - 19 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \] Input:

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

Output:

[1/24*(33*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log(-(a*co 
s(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s 
qrt(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*cos(d*x + c)^3 - 12*cos(d*x + c)^2 - 19*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*(33*s 
qrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*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*cos(d*x + c)^3 - 12*cos(d*x + c)^2 - 19*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]

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

integrate(1/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**(3/2),x)
 

Output:

Integral(1/((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. 33960 vs. \(2 (146) = 292\).

Time = 0.60 (sec) , antiderivative size = 33960, normalized size of antiderivative = 191.86 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="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*arc 
tan2(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*(co 
s(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*arct 
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(5/3*arctan2(sin(3...
 

Giac [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left ({\left (3 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 46 \, \sqrt {2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 27 \, \sqrt {2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} + \frac {33 \, \sqrt {2} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{a^{\frac {3}{2}}}}{12 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

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

Output:

-1/12*(((3*sqrt(2)*tan(1/2*d*x + 1/2*c)^2 + 46*sqrt(2))*tan(1/2*d*x + 1/2* 
c)^2 + 27*sqrt(2))*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/ 
2) + 33*sqrt(2)*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))/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(1/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**2