\(\int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [264]

Optimal result

Integrand size = 35, antiderivative size = 156 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(5 A+3 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \] Output:

1/32*(5*A+3*B)*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^(5/2)/d-1/4*(A-B)*sec(d*x+c)^(5/2)*sin(d*x+ 
c)/d/(a+a*sec(d*x+c))^(5/2)+1/16*(5*A+3*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d 
/(a+a*sec(d*x+c))^(3/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(570\) vs. \(2(156)=312\).

Time = 3.74 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.65 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {16 (5 A+3 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+16 (5 A+3 B) \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+10 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+6 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+2 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+14 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-5 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-3 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-10 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-6 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-5 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)-3 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)}{32 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \] Input:

Integrate[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^( 
5/2),x]
 

Output:

(16*(5*A + 3*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^5*Sec[c + 
d*x]^3*Sin[(c + d*x)/2] + 16*(5*A + 3*B)*ArcSin[Sqrt[Sec[c + d*x]]]*Cos[(c 
 + d*x)/2]^5*Sec[c + d*x]^3*Sin[(c + d*x)/2] + 10*A*Sqrt[1 - Sec[c + d*x]] 
*Sec[c + d*x]^(3/2)*Sin[c + d*x] + 6*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x] 
^(3/2)*Sin[c + d*x] + 2*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c 
+ d*x] + 14*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] - 5*S 
qrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c 
 + d*x] - 3*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + 
 d*x]]]*Tan[c + d*x] - 10*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sq 
rt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] - 6*Sqrt[2]*B*ArcTan[(Sqrt 
[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] 
- 5*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]* 
Sec[c + d*x]^2*Tan[c + d*x] - 3*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x 
]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]^2*Tan[c + d*x])/(32*d*Sqrt[1 - Se 
c[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4500, 3042, 4297, 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.

\(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\)

\(\Big \downarrow \) 4500

\(\displaystyle \frac {(5 A+3 B) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4297

\(\displaystyle \frac {(5 A+3 B) \left (\frac {\int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{4 a}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4295

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {(5 A+3 B) \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\right )}{8 a}-\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

Input:

Int[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x 
]
 

Output:

-1/4*((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/ 
2)) + ((5*A + 3*B)*(ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqr 
t[2]*Sqrt[a + a*Sec[c + d*x]])]/(2*Sqrt[2]*a^(3/2)*d) + (Sec[c + d*x]^(3/2 
)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2))))/(8*a)
 

Defintions of rubi rules used

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 4297

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

rule 4500

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))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 
 1))), x] + Simp[(a*A*m + b*B*(m + 1))/(a^2*(2*m + 1))   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, n} 
, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && LeQ[ 
m, -1]
 

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.69

method	result	size
default	\(\frac {\left (\left (-5 \cos \left (d x +c \right )^{2}-10 \cos \left (d x +c \right )-5\right ) A \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-3 \cos \left (d x +c \right )^{2}-6 \cos \left (d x +c \right )-3\right ) B \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (5 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+\left (3 \cos \left (d x +c \right )+7\right ) \sin \left (d x +c \right ) B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \sec \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(263\)
parts	\(\frac {A \left (\left (5 \cos \left (d x +c \right )^{2}+10 \cos \left (d x +c \right )+5\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (5 \cos \left (d x +c \right )+1\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B \left (\left (3 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+3\right ) \arctan \left (\frac {\sqrt {2}\, \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (3 \cos \left (d x +c \right )+7\right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \cos \left (d x +c \right )^{3} \sqrt {2}\, \sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{32 d \,a^{3} \left (\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(346\)

Input:

int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/32/d/a^3*((-5*cos(d*x+c)^2-10*cos(d*x+c)-5)*A*arctan(1/2*2^(1/2)/(-1/(1+ 
cos(d*x+c)))^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(-3*cos(d*x+c)^2-6*cos(d*x+c)- 
3)*B*arctan(1/2*2^(1/2)/(-1/(1+cos(d*x+c)))^(1/2)*(csc(d*x+c)-cot(d*x+c))) 
+(5*cos(d*x+c)+1)*sin(d*x+c)*A*(-2/(1+cos(d*x+c)))^(1/2)+(3*cos(d*x+c)+7)* 
sin(d*x+c)*B*(-2/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2*2^(1/2)*sec(d*x+c)^(3 
/2)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)^3+3*cos(d*x+c)^2+3*cos(d*x+c)+1)/ 
(-1/(1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 498, normalized size of antiderivative = 3.19 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B\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 ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A + 7 \, B\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 )}}}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B\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 ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (A + 7 \, B\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 )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \] Input:

integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="fricas")
 

Output:

[1/64*(sqrt(2)*((5*A + 3*B)*cos(d*x + c)^3 + 3*(5*A + 3*B)*cos(d*x + c)^2 
+ 3*(5*A + 3*B)*cos(d*x + c) + 5*A + 3*B)*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*((5*A + 3*B)*cos(d*x + c)^2 + (A + 7*B)*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^3*d*co 
s(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/ 
32*(sqrt(2)*((5*A + 3*B)*cos(d*x + c)^3 + 3*(5*A + 3*B)*cos(d*x + c)^2 + 3 
*(5*A + 3*B)*cos(d*x + c) + 5*A + 3*B)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sq 
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) 
 - 2*((5*A + 3*B)*cos(d*x + c)^2 + (A + 7*B)*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^3*d*cos(d*x + 
 c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate(sec(d*x+c)**(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(5/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 87207 vs. \(2 (131) = 262\).

Time = 14.31 (sec) , antiderivative size = 87207, normalized size of antiderivative = 559.02 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="maxima")
 

Output:

1/32*((4*(3*sin(3/2*d*x + 3/2*c) + 5*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), 
 cos(3/2*d*x + 3/2*c))) - 3*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* 
d*x + 3/2*c))) - 5*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2 
*c))))*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 40*( 
2*sin(3*d*x + 3*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 
 3/2*c))) + 2*sin(2/3*arctan2(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))) + 24*(2*sin 
(3*d*x + 3*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* 
c))) + 2*sin(2/3*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))) + 24*(3*sin(3/2* 
d*x + 3/2*c) - 5*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c 
))))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*(3* 
sin(3/2*d*x + 3/2*c) - 5*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x 
 + 3/2*c))))*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) 
+ 5*(16*cos(3*d*x + 3*c)^2 + 2*(4*cos(3*d*x + 3*c) + 6*cos(4/3*arctan2(sin 
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*cos(2/3*arctan2(sin(3/2*d*x 
+ 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c) 
, cos(3/2*d*x + 3/2*c))) + cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d 
*x + 3/2*c)))^2 + 12*(4*cos(3*d*x + 3*c) + 4*cos(2/3*arctan2(sin(3/2*d*x + 
 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*...
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:Warning, need to choose a branch for the root of a poly 
nomial with parameters. This might be wrong.Non regular value [0,0] was di 
scarded and
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x))^(5/ 
2),x)
 

Output:

int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + a/cos(c + d*x))^(5/ 
2), x)
 

Reduce [F]

\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a \right )}{a^{3}} \] Input:

int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2)/ 
(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*b + int((sqr 
t(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x))/(sec(c + d*x)**3 + 3* 
sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a))/a**3