\(\int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [269]

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

Optimal result

Integrand size = 37, antiderivative size = 218 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{5/2} (304 A+163 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^3 (432 A+299 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{192 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (16 A+17 C) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{32 d}+\frac {5 a C \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{24 d}+\frac {C \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{4 d} \] Output:

1/64*a^(5/2)*(304*A+163*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/ 
2))/d+1/192*a^3*(432*A+299*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c 
))^(1/2)+1/32*a^2*(16*A+17*C)*sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*sin( 
d*x+c)/d+5/24*a*C*sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+1/4 
*C*sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.53 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.06 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (489 C \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-912 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+96 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+326 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+184 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+48 C \sqrt {1-\sec (c+d x)} \sec ^{\frac {7}{2}}(c+d x)+528 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+489 C \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{192 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(489*C*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 912*A*ArcSin[Sqrt[Sec[c + d*x 
]]] + 96*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 326*C*Sqrt[1 - Sec[ 
c + d*x]]*Sec[c + d*x]^(3/2) + 184*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^( 
5/2) + 48*C*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(7/2) + 528*A*Sqrt[-((-1 + 
 Sec[c + d*x])*Sec[c + d*x])] + 489*C*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d 
*x])])*Tan[c + d*x])/(192*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x 
])])
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 4577, 27, 3042, 4506, 27, 3042, 4506, 27, 3042, 4504, 3042, 4288, 222}

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.

\(\displaystyle \int \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4577

\(\displaystyle \frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{5/2} (a (8 A+C)+5 a C \sec (c+d x))dx}{4 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {1}{3} \int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} \left ((48 A+11 C) a^2+3 (16 A+17 C) \sec (c+d x) a^2\right )dx+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} \left ((48 A+11 C) a^2+3 (16 A+17 C) \sec (c+d x) a^2\right )dx+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left ((48 A+11 C) a^2+3 (16 A+17 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{2} \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left (5 (48 A+19 C) a^3+(432 A+299 C) \sec (c+d x) a^3\right )dx+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left (5 (48 A+19 C) a^3+(432 A+299 C) \sec (c+d x) a^3\right )dx+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (5 (48 A+19 C) a^3+(432 A+299 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (304 A+163 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a^4 (432 A+299 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^3 (304 A+163 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^4 (432 A+299 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {\frac {1}{6} \left (\frac {1}{4} \left (\frac {a^4 (432 A+299 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 a^3 (304 A+163 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {\frac {5 a^2 C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}+\frac {1}{6} \left (\frac {3 a^3 (16 A+17 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{7/2} (304 A+163 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^4 (432 A+299 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )}{8 a}+\frac {C \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}}{4 d}\)

Input:

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

Output:

(C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(4*d) + ((5 
*a^2*C*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + 
 ((3*a^3*(16*A + 17*C)*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + 
 d*x])/(2*d) + ((3*a^(7/2)*(304*A + 163*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/ 
Sqrt[a + a*Sec[c + d*x]]])/d + (a^4*(432*A + 299*C)*Sec[c + d*x]^(3/2)*Sin 
[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/4)/6)/(8*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 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 4504
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) 
 + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C 
ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] 
 + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1))   Int[Sqrt[a + b*Csc[e + f* 
x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ 
[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && 
!LtQ[n, 0]
 

rule 4506
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* 
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), 
 x] + Simp[1/(d*(m + n))   Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] 
)^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - 
 a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]
 

rule 4577
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C) 
*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + n + 1))), 
x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n 
*Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
 b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && 
!LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
 
Maple [A] (verified)

Time = 2.58 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.48

method result size
default \(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (912 A \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+489 C \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+912 A \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+489 C \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )+48 \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (11 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right ) A +\left (489 \cos \left (d x +c \right )^{3}+326 \cos \left (d x +c \right )^{2}+184 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, C \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}\right )}{384 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(322\)
parts \(\frac {A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (19 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-19 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )+\sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \left (11 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )\right )}{8 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}+\frac {C \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {5}{2}} \left (489 \cos \left (d x +c \right )^{3} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right )-489 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{3}+\left (489 \cos \left (d x +c \right )^{3}+326 \cos \left (d x +c \right )^{2}+184 \cos \left (d x +c \right )+48\right ) \sqrt {2}\, \sqrt {-\frac {2}{\cos \left (d x +c \right )+1}}\, \tan \left (d x +c \right )\right )}{384 d \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {1}{\cos \left (d x +c \right )+1}}}\) \(367\)

Input:

int(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/384/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(1/2)/(cos(d*x+c)+1)/(-1/( 
cos(d*x+c)+1))^(1/2)*(912*A*cos(d*x+c)*arctan(1/2/(-1/(cos(d*x+c)+1))^(1/2 
)*(cot(d*x+c)-csc(d*x+c)+1))+489*C*cos(d*x+c)*arctan(1/2/(-1/(cos(d*x+c)+1 
))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1))+912*A*cos(d*x+c)*arctan(1/2*(cot(d*x+c 
)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+489*C*cos(d*x+c)*arctan(1/2*(co 
t(d*x+c)-csc(d*x+c)-1)/(-1/(cos(d*x+c)+1))^(1/2))+48*2^(1/2)*(-2/(cos(d*x+ 
c)+1))^(1/2)*(11*sin(d*x+c)+2*tan(d*x+c))*A+(489*cos(d*x+c)^3+326*cos(d*x+ 
c)^2+184*cos(d*x+c)+48)*2^(1/2)*C*(-2/(cos(d*x+c)+1))^(1/2)*tan(d*x+c)*sec 
(d*x+c)^2)
 

Fricas [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 503, normalized size of antiderivative = 2.31 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\left [\frac {3 \, {\left ({\left (304 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (304 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\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 (3 \, {\left (176 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 184 \, C a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\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 )}}}{768 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}, \frac {3 \, {\left ({\left (304 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (304 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{3}\right )} \sqrt {-a} \arctan \left (\frac {{\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 )}}}{2 \, a \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (3 \, {\left (176 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 2 \, {\left (48 \, A + 163 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 184 \, C a^{2} \cos \left (d x + c\right ) + 48 \, C a^{2}\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 )}}}{384 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}}\right ] \] Input:

integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, al 
gorithm="fricas")
 

Output:

[1/768*(3*((304*A + 163*C)*a^2*cos(d*x + c)^4 + (304*A + 163*C)*a^2*cos(d* 
x + c)^3)*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))*s 
in(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 
 4*(3*(176*A + 163*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 163*C)*a^2*cos(d*x + 
c)^2 + 184*C*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d* 
x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 + d*cos(d*x + c 
)^3), 1/384*(3*((304*A + 163*C)*a^2*cos(d*x + c)^4 + (304*A + 163*C)*a^2*c 
os(d*x + c)^3)*sqrt(-a)*arctan(1/2*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt( 
-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))/(a*sqrt(cos(d*x + c))*sin(d*x 
+ c))) + 2*(3*(176*A + 163*C)*a^2*cos(d*x + c)^3 + 2*(48*A + 163*C)*a^2*co 
s(d*x + c)^2 + 184*C*a^2*cos(d*x + c) + 48*C*a^2)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^4 + d*cos 
(d*x + c)^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6687 vs. \(2 (186) = 372\).

Time = 3.26 (sec) , antiderivative size = 6687, normalized size of antiderivative = 30.67 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, al 
gorithm="maxima")
 

Output:

-1/768*(48*(88*sqrt(2)*a^2*cos(7/2*d*x + 7/2*c)*sin(2*d*x + 2*c) - 56*sqrt 
(2)*a^2*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) - 28*sqrt(2)*a^2*sin(3/2*d*x 
 + 3/2*c) + 44*sqrt(2)*a^2*sin(1/2*d*x + 1/2*c) - 19*(a^2*log(2*cos(1/2*d* 
x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 
 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 
2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin( 
1/2*d*x + 1/2*c) + 2) + a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 
 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c 
) + 2) - a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*s 
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(4*d 
*x + 4*c)^2 - 76*(a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c 
)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) 
 - a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2) 
*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a^2*log(2*co 
s(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 
1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a^2*log(2*cos(1/2*d*x + 1/2 
*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt 
(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c)^2 - 19*a^2*log(2*cos(1/2*d 
*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) 
+ 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 19*a^2*log(2*cos(1/2*d*x + 1/2*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 942 vs. \(2 (186) = 372\).

Time = 4.53 (sec) , antiderivative size = 942, normalized size of antiderivative = 4.32 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate(sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x, al 
gorithm="giac")
 

Output:

1/384*(3*(304*A*a^(5/2)*sgn(cos(d*x + c)) + 163*C*a^(5/2)*sgn(cos(d*x + c) 
))*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))) - 3*(304*A*a^(5/2)*sgn(cos(d*x + c)) + 163*C* 
a^(5/2)*sgn(cos(d*x + c)))*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))) + 4*sqrt(2)*(912*(sqr 
t(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*a^(7/ 
2)*sgn(cos(d*x + c)) + 489*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2* 
d*x + 1/2*c)^2 + a))^14*C*a^(7/2)*sgn(cos(d*x + c)) - 19152*(sqrt(a)*tan(1 
/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*a^(9/2)*sgn(cos 
(d*x + c)) - 10269*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/ 
2*c)^2 + a))^12*C*a^(9/2)*sgn(cos(d*x + c)) + 137424*(sqrt(a)*tan(1/2*d*x 
+ 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*a^(11/2)*sgn(cos(d*x + 
 c)) + 69885*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 
 + a))^10*C*a^(11/2)*sgn(cos(d*x + c)) - 374544*(sqrt(a)*tan(1/2*d*x + 1/2 
*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^8*A*a^(13/2)*sgn(cos(d*x + c)) - 
 259233*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) 
)^8*C*a^(13/2)*sgn(cos(d*x + c)) + 266928*(sqrt(a)*tan(1/2*d*x + 1/2*c) - 
sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*a^(15/2)*sgn(cos(d*x + c)) + 20997 
9*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C* 
a^(15/2)*sgn(cos(d*x + c)) - 75888*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*a**2*(int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)** 
4,x)*c + 2*int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**3,x 
)*c + int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*a + 
 int(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2,x)*c + 2*in 
t(sqrt(sec(c + d*x))*sqrt(sec(c + d*x) + 1)*sec(c + d*x),x)*a + int(sqrt(s 
ec(c + d*x))*sqrt(sec(c + d*x) + 1),x)*a)