3.3.56 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [256]

3.3.56.1 Optimal result
3.3.56.2 Mathematica [B] (verified)
3.3.56.3 Rubi [A] (verified)
3.3.56.4 Maple [B] (verified)
3.3.56.5 Fricas [A] (verification not implemented)
3.3.56.6 Sympy [F(-1)]
3.3.56.7 Maxima [B] (verification not implemented)
3.3.56.8 Giac [F]
3.3.56.9 Mupad [F(-1)]

3.3.56.1 Optimal result

Integrand size = 35, antiderivative size = 197 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(2 A-3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {(5 A-9 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(A-3 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \]

output
(2*A-3*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d+1/2 
*(A-B)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/4*(5*A-9*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^(3/2)/d*2^(1/2)-1/2*(A-3*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(a+a*s 
ec(d*x+c))^(1/2)
 
3.3.56.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(429\) vs. \(2(197)=394\).

Time = 1.31 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.18 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {-8 (A-3 B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )-8 (5 A-9 B) \arcsin \left (\sqrt {\sec (c+d x)}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \sin \left (\frac {1}{2} (c+d x)\right )-2 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+6 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)+4 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)+5 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)-9 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)+5 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)-9 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \sec (c+d x) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

input
Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^( 
3/2),x]
 
output
(-8*(A - 3*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^3*Sec[c + d* 
x]^2*Sin[(c + d*x)/2] - 8*(5*A - 9*B)*ArcSin[Sqrt[Sec[c + d*x]]]*Cos[(c + 
d*x)/2]^3*Sec[c + d*x]^2*Sin[(c + d*x)/2] - 2*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] + 4*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2)*Sin[c + d* 
x] + 5*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x] 
]]*Tan[c + d*x] - 9*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - 
 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]*Tan[c + d*x] - 9*Sqrt[2]*B*ArcTa 
n[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Sec[c + d*x]*Tan[c 
+ d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))
 
3.3.56.3 Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04, 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, 4509, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4507

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4511

\(\displaystyle \frac {-\frac {a^2 (5 A-9 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-2 a (2 A-3 B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {a^2 (5 A-9 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-2 a (2 A-3 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {-\frac {a^2 (5 A-9 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {4 a (2 A-3 B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {-\frac {a^2 (5 A-9 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^{3/2} (2 A-3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {-\frac {-\frac {2 a^2 (5 A-9 B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {4 a^{3/2} (2 A-3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {\frac {\sqrt {2} a^{3/2} (5 A-9 B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {4 a^{3/2} (2 A-3 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{a}-\frac {2 a (A-3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}+\frac {(A-B) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\)

input
Int[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(3/2),x 
]
 
output
((A - B)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) 
 + (-(((-4*a^(3/2)*(2*A - 3*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*S 
ec[c + d*x]]])/d + (Sqrt[2]*a^(3/2)*(5*A - 9*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[ 
c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/a) - (2*a* 
(A - 3*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/( 
4*a^2)
 

3.3.56.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4509
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), 
 x] + Simp[d/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 
 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] 
&& GtQ[n, 1]
 

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]
 
3.3.56.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(770\) vs. \(2(166)=332\).

Time = 7.41 (sec) , antiderivative size = 771, normalized size of antiderivative = 3.91

method result size
default \(-\frac {\sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (-5 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, \cos \left (d x +c \right )^{4}+9 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, \cos \left (d x +c \right )^{4}+4 A \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{4}+4 A \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-5 A \cos \left (d x +c \right )^{3} \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 A \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{3}-6 B \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{4}-6 B \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+9 B \cos \left (d x +c \right )^{3} \sqrt {2}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-6 B \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{3}+4 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+4 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-6 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-6 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-4 B \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\right )}{4 a^{2} d \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(771\)
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 {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 )-2 \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 )-\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 )+5 \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 )}{4 d \,a^{2} \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 \sec \left (d x +c \right )^{\frac {7}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (9 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, \cos \left (d x +c \right )^{5}+6 \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{5}-6 \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{5}+9 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {2}}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \sqrt {2}\, \cos \left (d x +c \right )^{4}-6 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{4} \sin \left (d x +c \right )+6 \arctan \left (\frac {\cos \left (d x +c \right )-\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{4}-6 \arctan \left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{4}-4 \sin \left (d x +c \right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )^{3}\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}}}\) \(784\)

input
int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x,method=_RET 
URNVERBOSE)
 
output
-1/4/a^2/d*sec(d*x+c)^(5/2)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)^2/(-1/ 
(cos(d*x+c)+1))^(1/2)*(-5*A*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/( 
-1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)^4+9*B*arctan(1/2*sin(d*x+c)*2 
^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)^4+4*A* 
arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^( 
1/2))*cos(d*x+c)^4+4*A*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/( 
cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-5*A*cos(d*x+c)^3*2^(1/2)*arctan(1 
/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+2*A*sin(d* 
x+c)*(-1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3-6*B*arctan(1/2*(-cos(d*x+c)+si 
n(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^4-6*B*cos 
(d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x 
+c)+1))^(1/2))+9*B*cos(d*x+c)^3*2^(1/2)*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos 
(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-6*B*sin(d*x+c)*(-1/(cos(d*x+c)+1))^( 
1/2)*cos(d*x+c)^3+4*A*cos(d*x+c)^3*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/( 
cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+4*A*cos(d*x+c)^3*arctan(1/2*(cos( 
d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-6*B*cos(d*x 
+c)^3*arctan(1/2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c) 
+1))^(1/2))-6*B*cos(d*x+c)^3*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x 
+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-4*B*cos(d*x+c)^2*sin(d*x+c)*(-1/(cos(d*x 
+c)+1))^(1/2))
 
3.3.56.5 Fricas [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 669, normalized size of antiderivative = 3.40 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 9 \, 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 ) + 2 \, {\left ({\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 3 \, 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 (A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, B\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 )}}}{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 {\sqrt {2} {\left ({\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - 9 \, B\right )} \cos \left (d x + c\right ) + 5 \, A - 9 \, 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 ) + 2 \, {\left ({\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A - 3 \, 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 (A - 3 \, B\right )} \cos \left (d x + c\right ) - 2 \, B\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 )}}}{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(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="fricas")
 
output
[-1/8*(sqrt(2)*((5*A - 9*B)*cos(d*x + c)^2 + 2*(5*A - 9*B)*cos(d*x + c) + 
5*A - 9*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)) + 2*((2*A - 3*B)*cos(d* 
x + c)^2 + 2*(2*A - 3*B)*cos(d*x + c) + 2*A - 3*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*((A - 3*B)*cos(d*x + c) - 2*B) 
*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), 1/4*(sqrt(2)*((5*A 
- 9*B)*cos(d*x + c)^2 + 2*(5*A - 9*B)*cos(d*x + c) + 5*A - 9*B)*sqrt(-a)*a 
rctan(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*((2*A - 3*B)*cos(d*x + c)^2 + 2*(2*A - 3*B)* 
cos(d*x + c) + 2*A - 3*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*c 
os(d*x + c) - 2*a)) - 2*((A - 3*B)*cos(d*x + c) - 2*B)*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)]
 
3.3.56.6 Sympy [F(-1)]

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

input
integrate(sec(d*x+c)**(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)
 
output
Timed out
 
3.3.56.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7057 vs. \(2 (166) = 332\).

Time = 0.75 (sec) , antiderivative size = 7057, normalized size of antiderivative = 35.82 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="maxima")
 
output
1/4*((4*(sin(2*d*x + 2*c) + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x 
+ 2*c))))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqrt(2 
)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c)))^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(2*d*x + 2*c)*si 
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*sqrt(2)*sin(1/2*arc 
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*(sqrt(2)*cos(2*d*x + 2*c) 
+ sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqrt(2 
)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^ 
2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqr 
t(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 2*(sqrt(2 
)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d 
*x + 2*c)))^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(2*d*x + 2*c)*si 
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*sqrt(2)*sin(1/2*arc 
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*(sqrt(2)*cos(2*d*x + 2*c) 
+ sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqrt(2 
)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos( 
2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^ 
2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*sqr 
t(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 2*(sqr...
 
3.3.56.8 Giac [F]

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

input
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algo 
rithm="giac")
 
output
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*sec(d*x + c) + a)^(3/ 
2), x)
 
3.3.56.9 Mupad [F(-1)]

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

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