\(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\) [248]

Optimal result

Integrand size = 35, antiderivative size = 190 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {(4 A-7 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 \sqrt {a} d}+\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {(4 A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {B \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {a+a \sec (c+d x)}} \] Output:

-1/4*(4*A-7*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(1/2)/ 
d+2^(1/2)*(A-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))/a^(1/2)/d+1/4*(4*A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/ 
(a+a*sec(d*x+c))^(1/2)+1/2*B*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c) 
)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 0.46 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {\left ((4 A-B) \arcsin \left (\sqrt {1-\sec (c+d x)}\right )+8 (A-B) \arcsin \left (\sqrt {\sec (c+d x)}\right )-4 \sqrt {2} A \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+4 \sqrt {2} B \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )+2 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+4 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}-B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(((4*A - B)*ArcSin[Sqrt[1 - Sec[c + d*x]]] + 8*(A - B)*ArcSin[Sqrt[Sec[c + 
 d*x]]] - 4*Sqrt[2]*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + 
 d*x]]] + 4*Sqrt[2]*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + 
 d*x]]] + 2*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 4*A*Sqrt[-((-1 + 
 Sec[c + d*x])*Sec[c + d*x])] - B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x]) 
])*Tan[c + d*x])/(4*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 4509, 27, 3042, 4509, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4509

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^2 (4 A-B)-a^2 (4 A-7 B) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^2 (4 A-B)-a^2 (4 A-7 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^2 (4 A-B)-a^2 (4 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4511

\(\displaystyle \frac {\frac {8 a^2 (A-B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-a (4 A-7 B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {8 a^2 (A-B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-a (4 A-7 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {8 a^2 (A-B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {2 a (4 A-7 B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {8 a^2 (A-B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^{3/2} (4 A-7 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {\frac {-\frac {16 a^2 (A-B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {2 a^{3/2} (4 A-7 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {8 \sqrt {2} a^{3/2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^{3/2} (4 A-7 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}+\frac {a (4 A-B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a}+\frac {B \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{2 d \sqrt {a \sec (c+d x)+a}}\)

Input:

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

Output:

(B*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*Sqrt[a + a*Sec[c + d*x]]) + (((-2 
*a^(3/2)*(4*A - 7*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x 
]]])/d + (8*Sqrt[2]*a^(3/2)*(A - B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Si 
n[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d)/(2*a) + (a*(4*A - B)*S 
ec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/(4*a)
 

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 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]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(159)=318\).

Time = 2.20 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.11

method	result	size
default	\(-\frac {\sec \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (8 A \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{3}-8 B \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{3}-4 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+7 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4 A \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+7 B \cos \left (d x +c \right )^{3} \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )-4 A \sqrt {2}\, \cos \left (d x +c \right )^{2} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )-2\right ) \sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )\right )}{8 d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(401\)
parts	\(-\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (2 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{3}-\sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{2}+\cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{3}\right )}{2 d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {7}{2}} \left (8 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{4}+7 \cos \left (d x +c \right )^{4} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+7 \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right )^{4}+\sin \left (d x +c \right ) \left (-\cos \left (d x +c \right )+2\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )^{2}\right )}{8 d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(451\)

Input:

int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/8/d/a*sec(d*x+c)^(5/2)*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(-1/(1+c 
os(d*x+c)))^(1/2)*(8*A*2^(1/2)*arctan(1/2*2^(1/2)/(-1/(1+cos(d*x+c)))^(1/2 
)*(csc(d*x+c)-cot(d*x+c)))*cos(d*x+c)^3-8*B*2^(1/2)*arctan(1/2*2^(1/2)/(-1 
/(1+cos(d*x+c)))^(1/2)*(csc(d*x+c)-cot(d*x+c)))*cos(d*x+c)^3-4*A*cos(d*x+c 
)^3*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))+7*B*c 
os(d*x+c)^3*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)/(-1/(1+cos(d*x+c)))^(1/2 
))-4*A*cos(d*x+c)^3*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(1+cos(d*x+c 
)))^(1/2))+7*B*cos(d*x+c)^3*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)+1)/(-1/(1+c 
os(d*x+c)))^(1/2))-4*A*2^(1/2)*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)*sin( 
d*x+c)+sin(d*x+c)*(cos(d*x+c)-2)*2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d 
*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 614, normalized size of antiderivative = 3.23 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="fricas")
 

Output:

[-1/16*(((4*A - 7*B)*cos(d*x + c)^2 + (4*A - 7*B)*cos(d*x + c))*sqrt(a)*lo 
g((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)) + 8*sqrt(2)*((A - B)* 
a*cos(d*x + c)^2 + (A - B)*a*cos(d*x + c))*log(-(cos(d*x + c)^2 + 2*sqrt(2 
)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/ 
sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt( 
a) - 4*((4*A - 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*d*cos(d*x + c)^2 + a*d*cos(d*x + c 
)), -1/8*(8*sqrt(2)*((A - B)*a*cos(d*x + c)^2 + (A - B)*a*cos(d*x + c))*sq 
rt(-1/a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a) 
*sqrt(cos(d*x + c))/sin(d*x + c)) + ((4*A - 7*B)*cos(d*x + c)^2 + (4*A - 7 
*B)*cos(d*x + c))*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sq 
rt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d 
*x + c))) - 2*((4*A - 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*d*cos(d*x + c)^2 + a*d*cos( 
d*x + c))]
 

Sympy [F(-1)]

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

integrate(sec(d*x+c)**(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(1/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2524 vs. \(2 (159) = 318\).

Time = 0.34 (sec) , antiderivative size = 2524, normalized size of antiderivative = 13.28 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="maxima")
 

Output:

-1/16*(4*(4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x 
 + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x 
 + 2*c) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 
1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan 
2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), 
 cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 
 2) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*l 
og(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(si 
n(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos 
(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) 
+ (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2 
*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d* 
x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x 
 + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - (c 
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos 
(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + 
c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c 
))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 2*(sqr 
t(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2)*cos(2*d*x 
 + 2*c) + sqrt(2))*log(cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 +...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (159) = 318\).

Time = 2.40 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.80 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="giac")
 

Output:

-1/8*(4*sqrt(2)*(A*sqrt(a) - B*sqrt(a))*log((sqrt(a)*tan(1/2*d*x + 1/2*c) 
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/a + (4*A - 7*B)*log(abs((sqrt(a)* 
tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - a*(2*sqrt(2 
) + 3)))/sqrt(a) - (4*A - 7*B)*log(abs((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr 
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + a*(2*sqrt(2) - 3)))/sqrt(a) - 4*sqrt( 
2)*(12*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)) 
^6*A*sqrt(a) - 17*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2 
*c)^2 + a))^6*B*sqrt(a) - 76*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/ 
2*d*x + 1/2*c)^2 + a))^4*A*a^(3/2) + 57*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sq 
rt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(3/2) + 36*(sqrt(a)*tan(1/2*d*x + 
1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*a^(5/2) - 19*(sqrt(a)*tan 
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*a^(5/2) - 4*A* 
a^(7/2) + 3*B*a^(7/2))/((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x 
 + 1/2*c)^2 + a))^4 - 6*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x 
 + 1/2*c)^2 + a))^2*a + a^2)^2)/(d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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