\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx\) [254]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 230 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {2 (A-7 B) \sin (c+d x)}{35 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 (31 A-7 B) \sin (c+d x)}{105 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 (43 A-91 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}} \] Output:

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+2/7*A*sin(d*x+c)/d/sec(d*x+c)^(5/2)/(a+a*sec 
(d*x+c))^(1/2)-2/35*(A-7*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c)) 
^(1/2)+2/105*(31*A-7*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/ 
2)-2/105*(43*A-91*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.66 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {-\frac {2 \left (-15 A+3 (A-7 B) \sec (c+d x)+(-31 A+7 B) \sec ^2(c+d x)+(43 A-91 B) \sec ^3(c+d x)\right ) \sin (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}-\frac {105 \sqrt {2} (A-B) \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{\sqrt {1-\sec (c+d x)}}}{105 d \sqrt {a (1+\sec (c+d x))}} \] Input:

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x] 
]),x]
 

Output:

((-2*(-15*A + 3*(A - 7*B)*Sec[c + d*x] + (-31*A + 7*B)*Sec[c + d*x]^2 + (4 
3*A - 91*B)*Sec[c + d*x]^3)*Sin[c + d*x])/Sec[c + d*x]^(5/2) - (105*Sqrt[2 
]*(A - B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[ 
c + d*x])/Sqrt[1 - Sec[c + d*x]])/(105*d*Sqrt[a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4510, 27, 3042, 4510, 27, 3042, 4510, 27, 3042, 4501, 3042, 4295, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2} \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {2 \int -\frac {a (A-7 B)-6 a A \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{7 a}+\frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-7 B)-6 a A \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a (A-7 B)-6 a A \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{7 a}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^2 (31 A-7 B)-4 a^2 (A-7 B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}+\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}}{7 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (31 A-7 B)-4 a^2 (A-7 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {\sec (c+d x) a+a}}dx}{5 a}}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^2 (31 A-7 B)-4 a^2 (A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}}{7 a}\)

\(\Big \downarrow \) 4510

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 \int -\frac {a^3 (43 A-91 B)-2 a^3 (31 A-7 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}+\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}}{5 a}}{7 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (43 A-91 B)-2 a^3 (31 A-7 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\int \frac {a^3 (43 A-91 B)-2 a^3 (31 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{7 a}\)

\(\Big \downarrow \) 4501

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-105 a^3 (A-B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx}{3 a}}{5 a}}{7 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-105 a^3 (A-B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}}{5 a}}{7 a}\)

\(\Big \downarrow \) 4295

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {210 a^3 (A-B) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}}{3 a}}{5 a}}{7 a}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {2 A \sin (c+d x)}{7 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a (A-7 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^2 (31 A-7 B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {\frac {2 a^3 (43 A-91 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d \sqrt {a \sec (c+d x)+a}}-\frac {105 \sqrt {2} a^{5/2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{3 a}}{5 a}}{7 a}\)

Input:

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(7/2)*Sqrt[a + a*Sec[c + d*x]]),x]
 

Output:

(2*A*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) - ((2 
*a*(A - 7*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x] 
]) - ((2*a^2*(31*A - 7*B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a 
*Sec[c + d*x]]) - ((-105*Sqrt[2]*a^(5/2)*(A - B)*ArcTanh[(Sqrt[a]*Sqrt[Sec 
[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/d + (2*a^3*( 
43*A - 91*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) 
)/(3*a))/(5*a))/(7*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 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 4501
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[(a*A*m 
 - b*B*n)/(b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] 
, x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a 
^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]
 

rule 4510
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e 
 + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d 
*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B* 
n - A*b*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, 
 m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]
 
Maple [A] (verified)

Time = 1.52 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.99

method result size
default \(\frac {\left (\left (105 \cos \left (d x +c \right )+105\right ) A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (-105 \cos \left (d x +c \right )-105\right ) B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (30 \cos \left (d x +c \right )^{3}-6 \cos \left (d x +c \right )^{2}+62 \cos \left (d x +c \right )-86\right ) \sin \left (d x +c \right ) A +\left (42 \cos \left (d x +c \right )^{2}-14 \cos \left (d x +c \right )+182\right ) \sin \left (d x +c \right ) B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{105 d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}}\) \(228\)
parts \(\frac {A \left (\left (30 \cos \left (d x +c \right )^{3}-6 \cos \left (d x +c \right )^{2}+62 \cos \left (d x +c \right )-86\right ) \sin \left (d x +c \right )+105 \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{105 d a \left (1+\cos \left (d x +c \right )\right ) \sqrt {\sec \left (d x +c \right )}}+\frac {B \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (6 \sin \left (d x +c \right )-2 \tan \left (d x +c \right )+26 \sec \left (d x +c \right ) \tan \left (d x +c \right )+\arctan \left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-15 \sec \left (d x +c \right )-15 \sec \left (d x +c \right )^{2}\right )\right )}{15 d a \left (1+\cos \left (d x +c \right )\right ) \sec \left (d x +c \right )^{\frac {5}{2}}}\) \(277\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/105/d/a*((105*cos(d*x+c)+105)*A*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^( 
1/2)/(-1/(1+cos(d*x+c)))^(1/2)*(csc(d*x+c)-cot(d*x+c)))+(-105*cos(d*x+c)-1 
05)*B*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*2^(1/2)/(-1/(1+cos(d*x+c)))^(1/ 
2)*(csc(d*x+c)-cot(d*x+c)))+(30*cos(d*x+c)^3-6*cos(d*x+c)^2+62*cos(d*x+c)- 
86)*sin(d*x+c)*A+(42*cos(d*x+c)^2-14*cos(d*x+c)+182)*sin(d*x+c)*B)*(a*(1+s 
ec(d*x+c)))^(1/2)/(1+cos(d*x+c))/sec(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.83 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\left [-\frac {\frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} - \frac {4 \, {\left (15 \, A \cos \left (d x + c\right )^{4} - 3 \, {\left (A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (31 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 91 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{210 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, -\frac {105 \, \sqrt {2} {\left ({\left (A - B\right )} a \cos \left (d x + c\right ) + {\left (A - B\right )} a\right )} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \sqrt {\cos \left (d x + c\right )}}{\sin \left (d x + c\right )}\right ) - \frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{4} - 3 \, {\left (A - 7 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (31 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (43 \, A - 91 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="fricas")
 

Output:

[-1/210*(105*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*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*(15*A*cos(d*x + c)^4 - 3*(A - 7*B)*cos(d*x + c)^3 + 
(31*A - 7*B)*cos(d*x + c)^2 - (43*A - 91*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*d*cos(d*x + c) 
 + a*d), -1/105*(105*sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*sqrt(-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)) - 2*(15*A*cos(d*x + c)^4 - 3*(A - 7*B)*cos(d*x 
 + c)^3 + (31*A - 7*B)*cos(d*x + c)^2 - (43*A - 91*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*d*co 
s(d*x + c) + a*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(7/2)/(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. 805 vs. \(2 (195) = 390\).

Time = 0.31 (sec) , antiderivative size = 805, normalized size of antiderivative = 3.50 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="maxima")
 

Output:

-1/840*(sqrt(2)*(525*cos(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7 
/2*c)))*sin(7/2*d*x + 7/2*c) - 175*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), c 
os(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) + 21*cos(2/7*arctan2(sin(7/2*d* 
x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 525*cos(7/2*d*x 
+ 7/2*c)*sin(6/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 17 
5*cos(7/2*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 
 7/2*c))) - 21*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2(sin(7/2*d*x + 7/2*c), 
cos(7/2*d*x + 7/2*c))) - 420*log(cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos 
(7/2*d*x + 7/2*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x 
+ 7/2*c)))^2 + 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c 
))) + 1) + 420*log(cos(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2 
*c)))^2 + sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))^2 - 
 2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 1) - 30* 
sin(7/2*d*x + 7/2*c) + 21*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d* 
x + 7/2*c))) - 175*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2 
*c))) + 525*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))* 
A/sqrt(a) - 14*sqrt(2)*(60*cos(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d 
*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 5*cos(2/5*arctan2(sin(5/2*d*x + 5/2*c 
), cos(5/2*d*x + 5/2*c)))*sin(5/2*d*x + 5/2*c) - 60*cos(5/2*d*x + 5/2*c)*s 
in(4/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 5*cos(5/2...
 

Giac [A] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.83 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=-\frac {\frac {105 \, \sqrt {2} {\left (A - B\right )} \log \left ({\left | -\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {a}} - \frac {2 \, {\left (105 \, \sqrt {2} B a^{3} - {\left ({\left (\sqrt {2} {\left (92 \, A a^{3} - 119 \, B a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 7 \, \sqrt {2} {\left (16 \, A a^{3} - 37 \, B a^{3}\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {2} {\left (4 \, A a^{3} - 7 \, B a^{3}\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {7}{2}}}}{105 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(1/2),x, algo 
rithm="giac")
 

Output:

-1/105*(105*sqrt(2)*(A - 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)))/sqrt(a) - 2*(105*sqrt(2)*B*a^3 - ((sqrt(2)* 
(92*A*a^3 - 119*B*a^3)*tan(1/2*d*x + 1/2*c)^2 + 7*sqrt(2)*(16*A*a^3 - 37*B 
*a^3))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(4*A*a^3 - 7*B*a^3))*tan(1/2*d* 
x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(7/2))/( 
d*sgn(cos(d*x + c)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}} \,d x \] Input:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(7/2 
)),x)
 

Output:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(1/2)*(1/cos(c + d*x))^(7/2 
)), x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{5}+\sec \left (d x +c \right )^{4}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{4}+\sec \left (d x +c \right )^{3}}d x \right ) b \right )}{a} \] Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(7/2)/(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)**5 
 + sec(c + d*x)**4),x)*a + int((sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)) 
/(sec(c + d*x)**4 + sec(c + d*x)**3),x)*b))/a