3.3.63 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [263]

3.3.63.1 Optimal result
3.3.63.2 Mathematica [B] (verified)
3.3.63.3 Rubi [A] (verified)
3.3.63.4 Maple [B] (warning: unable to verify)
3.3.63.5 Fricas [B] (verification not implemented)
3.3.63.6 Sympy [F(-1)]
3.3.63.7 Maxima [B] (verification not implemented)
3.3.63.8 Giac [F]
3.3.63.9 Mupad [F(-1)]

3.3.63.1 Optimal result

Integrand size = 35, antiderivative size = 194 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A-43 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(3 A-11 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]

output
2*B*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(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*(3*A-11*B)*sec( 
d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/32*(3*A-43*B)*arctanh 
(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a 
^(5/2)/d*2^(1/2)
 
3.3.63.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(570\) vs. \(2(194)=388\).

Time = 3.94 (sec) , antiderivative size = 570, normalized size of antiderivative = 2.94 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {16 (3 A-11 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+16 (3 A-43 B) \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+6 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)-22 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+14 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-30 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)-3 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+43 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-6 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)+86 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-3 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)+43 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec ^2(c+d x) \tan (c+d x)}{32 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{5/2}} \]

input
Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^( 
5/2),x]
 
output
(16*(3*A - 11*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*(3*A - 43*B)*ArcSin[Sqrt[Sec[c + d*x]]]*Cos[ 
(c + d*x)/2]^5*Sec[c + d*x]^3*Sin[(c + d*x)/2] + 6*A*Sqrt[1 - Sec[c + d*x] 
]*Sec[c + d*x]^(3/2)*Sin[c + d*x] - 22*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d* 
x]^(3/2)*Sin[c + d*x] + 14*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin 
[c + d*x] - 30*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d*x] - 
3*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Ta 
n[c + d*x] + 43*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec 
[c + d*x]]]*Tan[c + d*x] - 6*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]]) 
/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c + d*x] + 86*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] - 3*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] + 43*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 
 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))
 
3.3.63.3 Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.06, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4507, 27, 3042, 4507, 27, 3042, 4511, 3042, 4288, 222, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((3 A-11 B) a^2+32 B \sec (c+d x) a^2\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left ((3 A-11 B) a^2+32 B \sec (c+d x) a^2\right )}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left ((3 A-11 B) a^2+32 B \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4511

\(\displaystyle \frac {\frac {a^2 (3 A-43 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx+32 a B \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (3 A-43 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+32 a B \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {a^2 (3 A-43 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {64 a B \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {a^2 (3 A-43 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {64 a^{3/2} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {\frac {64 a^{3/2} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^2 (3 A-43 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\sqrt {2} a^{3/2} (3 A-43 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {64 a^{3/2} B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}+\frac {a (3 A-11 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\)

input
Int[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x 
]
 
output
((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) 
 + (((64*a^(3/2)*B*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]] 
])/d + (Sqrt[2]*a^(3/2)*(3*A - 43*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*S 
in[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/(4*a^2) + (a*(3*A - 1 
1*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)))/(8 
*a^2)
 

3.3.63.3.1 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 222
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4288
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
+ (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)]   Subst[Int[1/Sqrt[1 
+ x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a 
, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
 

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 4507
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b 
- a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 
2*m + 1))), x] - Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)* 
(d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m 
 - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, 
A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G 
tQ[n, 0]
 

rule 4511
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)/b   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b 
 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, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0]
 
3.3.63.4 Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(553\) vs. \(2(163)=326\).

Time = 8.24 (sec) , antiderivative size = 554, normalized size of antiderivative = 2.86

method result size
default \(\frac {\left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (-2 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+2 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-5 A \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+16 B \sqrt {2}\, \arctan \left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right ) \sqrt {2}}{2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+16 B \sqrt {2}\, \arctan \left (\frac {\left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \sqrt {2}}{2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+13 B \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 A \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )-43 B \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 a^{3} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{2} \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\) \(554\)
parts \(\frac {A \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{3} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (-2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-5 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{2} \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}-\frac {B \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {7}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{4} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+16 \sqrt {2}\, \arctan \left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right ) \sqrt {2}}{2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+16 \sqrt {2}\, \arctan \left (\frac {\left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \sqrt {2}}{2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+13 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-43 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 d \,a^{3} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right )^{3} \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\) \(716\)

input
int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x,method=_RET 
URNVERBOSE)
 
output
1/32/a^3/d*(-((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/((1-cos(d*x+c))^2*csc(d*x+c 
)^2-1))^(5/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^3*(-2*a/((1-cos(d*x+c))^2* 
csc(d*x+c)^2-1))^(1/2)*(-2*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-c 
os(d*x+c))^3*csc(d*x+c)^3+2*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1- 
cos(d*x+c))^3*csc(d*x+c)^3-5*A*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(- 
cot(d*x+c)+csc(d*x+c))+16*B*2^(1/2)*arctan(1/2*(csc(d*x+c)-cot(d*x+c)+1)*2 
^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+16*B*2^(1/2)*arctan(1/2*( 
-cot(d*x+c)+csc(d*x+c)-1)*2^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2) 
)+13*B*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))+3 
*A*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+ 
c)))-43*B*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+c 
sc(d*x+c))))/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^2/(-(1-cos(d*x+c))^2*csc(d* 
x+c)^2-1)^(1/2)
 
3.3.63.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (163) = 326\).

Time = 0.35 (sec) , antiderivative size = 749, normalized size of antiderivative = 3.86 \[ \int \frac {\sec ^{\frac {5}{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 (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, 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 ) - 32 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \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 )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) - \frac {4 \, {\left ({\left (3 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 15 \, 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 (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, 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 ) - 32 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\right )} \sqrt {-a} \arctan \left (\frac {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 )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) - \frac {2 \, {\left ({\left (3 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 15 \, 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)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="fricas")
 
output
[-1/64*(sqrt(2)*((3*A - 43*B)*cos(d*x + c)^3 + 3*(3*A - 43*B)*cos(d*x + c) 
^2 + 3*(3*A - 43*B)*cos(d*x + c) + 3*A - 43*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)) - 32*(B*cos(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c 
) + B)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c 
)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin( 
d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4* 
((3*A - 11*B)*cos(d*x + c)^2 + (7*A - 15*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), -1/32*(sqrt 
(2)*((3*A - 43*B)*cos(d*x + c)^3 + 3*(3*A - 43*B)*cos(d*x + c)^2 + 3*(3*A 
- 43*B)*cos(d*x + c) + 3*A - 43*B)*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))) - 3 
2*(B*cos(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B)*sqrt(-a)* 
arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c 
))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) - 2*((3*A - 11* 
B)*cos(d*x + c)^2 + (7*A - 15*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/c 
os(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)]
 
3.3.63.6 Sympy [F(-1)]

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

input
integrate(sec(d*x+c)**(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(5/2),x)
 
output
Timed out
 
3.3.63.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 89320 vs. \(2 (163) = 326\).

Time = 15.02 (sec) , antiderivative size = 89320, normalized size of antiderivative = 460.41 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="maxima")
 
output
1/32*((512*((2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x + 5/2*c) + cos 
(5/2*d*x + 5/2*c)*sin(4*d*x + 4*c) + 2*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3* 
c) + (2*cos(2*d*x + 2*c) + cos(d*x + c))*sin(5/2*d*x + 5/2*c) + cos(4*d*x 
+ 4*c)*sin(5/2*d*x + 5/2*c) + 2*cos(3*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos 
(5*d*x + 5*c)^2 + 2560*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2*d*x 
+ 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2*c)* 
sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos(2*d* 
x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c)*sin 
(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3*d*x 
 + 3*c)*sin(5/2*d*x + 5/2*c))*cos(8/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/ 
2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos(5/2* 
d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 5/2 
*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10*cos( 
2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x + 5*c) 
*sin(5/2*d*x + 5/2*c) - 5*cos(4*d*x + 4*c)*sin(5/2*d*x + 5/2*c) - 10*cos(3 
*d*x + 3*c)*sin(5/2*d*x + 5/2*c))*cos(6/5*arctan2(sin(5/2*d*x + 5/2*c), co 
s(5/2*d*x + 5/2*c)))^2 + 10240*(5*(2*sin(2*d*x + 2*c) + sin(d*x + c))*cos( 
5/2*d*x + 5/2*c) + cos(5/2*d*x + 5/2*c)*sin(5*d*x + 5*c) + 5*cos(5/2*d*x + 
 5/2*c)*sin(4*d*x + 4*c) + 10*cos(5/2*d*x + 5/2*c)*sin(3*d*x + 3*c) - (10* 
cos(2*d*x + 2*c) + 5*cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c) - cos(5*d*x...
 
3.3.63.8 Giac [F]

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

input
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algo 
rithm="giac")
 
output
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*sec(d*x + c) + a)^(5/ 
2), x)
 
3.3.63.9 Mupad [F(-1)]

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

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