\(\int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\) [241]

Optimal result

Integrand size = 35, antiderivative size = 180 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^{5/2} (38 A+25 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}+\frac {a^3 (54 A+49 B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{24 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (2 A+3 B) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d} \] Output:

1/8*a^(5/2)*(38*A+25*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2)) 
/d+1/24*a^3*(54*A+49*B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/ 
2)+1/4*a^2*(2*A+3*B)*sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(1/2)*sin(d*x+c)/d+ 
1/3*a*B*sec(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d
 

Mathematica [A] (warning: unable to verify)

Time = 0.60 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.13 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (75 B \arcsin \left (\sqrt {1-\sec (c+d x)}\right )-114 A \arcsin \left (\sqrt {\sec (c+d x)}\right )+12 A \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+34 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {3}{2}}(c+d x)+8 B \sqrt {1-\sec (c+d x)} \sec ^{\frac {5}{2}}(c+d x)+66 A \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}+75 B \sqrt {-((-1+\sec (c+d x)) \sec (c+d x))}\right ) \tan (c+d x)}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(a^3*(75*B*ArcSin[Sqrt[1 - Sec[c + d*x]]] - 114*A*ArcSin[Sqrt[Sec[c + d*x] 
]] + 12*A*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(3/2) + 34*B*Sqrt[1 - Sec[c 
+ d*x]]*Sec[c + d*x]^(3/2) + 8*B*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^(5/2) 
 + 66*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] + 75*B*Sqrt[-((-1 + Sec[ 
c + d*x])*Sec[c + d*x])])*Tan[c + d*x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[ 
a*(1 + Sec[c + d*x])])
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{3} \int \frac {1}{2} \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} (a (6 A+B)+3 a (2 A+3 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^{3/2} (a (6 A+B)+3 a (2 A+3 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (a (6 A+B)+3 a (2 A+3 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4506

\(\displaystyle \frac {1}{6} \left (\frac {1}{2} \int \frac {1}{2} \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left ((30 A+13 B) a^2+(54 A+49 B) \sec (c+d x) a^2\right )dx+\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a} \left ((30 A+13 B) a^2+(54 A+49 B) \sec (c+d x) a^2\right )dx+\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((30 A+13 B) a^2+(54 A+49 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4504

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^2 (38 A+25 B) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx+\frac {a^3 (54 A+49 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {3}{2} a^2 (38 A+25 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {a^3 (54 A+49 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )+\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 4288

\(\displaystyle \frac {1}{6} \left (\frac {1}{4} \left (\frac {a^3 (54 A+49 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}-\frac {3 a^2 (38 A+25 B) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}\right )+\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{6} \left (\frac {3 a^2 (2 A+3 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}+\frac {1}{4} \left (\frac {3 a^{5/2} (38 A+25 B) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^3 (54 A+49 B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}\right )\right )+\frac {a B \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d}\)

Input:

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

Output:

(a*B*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d) + ( 
(3*a^2*(2*A + 3*B)*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x 
])/(2*d) + ((3*a^(5/2)*(38*A + 25*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a 
 + a*Sec[c + d*x]]])/d + (a^3*(54*A + 49*B)*Sec[c + d*x]^(3/2)*Sin[c + d*x 
])/(d*Sqrt[a + a*Sec[c + d*x]]))/4)/6
 

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]
 

Maple [A] (verified)

Time = 4.32 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.72

method	result	size
default	\(\frac {a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (114 A \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+75 B \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+114 A \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+75 B \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+6 \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (11 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right ) A +\sqrt {2}\, B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (75 \sin \left (d x +c \right )+34 \tan \left (d x +c \right )+8 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )\right )}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(309\)
parts	\(\frac {A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {\sec \left (d x +c \right )}\, \left (19 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+19 \cos \left (d x +c \right ) \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (11 \sin \left (d x +c \right )+2 \tan \left (d x +c \right )\right )\right )}{8 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}+\frac {B \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sec \left (d x +c \right )^{\frac {3}{2}} \left (75 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )-1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+75 \cos \left (d x +c \right )^{2} \arctan \left (\frac {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}{2 \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\right )+\left (75 \cos \left (d x +c \right )^{2}+34 \cos \left (d x +c \right )+8\right ) \sqrt {2}\, \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )}{48 d \left (1+\cos \left (d x +c \right )\right ) \sqrt {-\frac {1}{1+\cos \left (d x +c \right )}}}\)	\(357\)

Input:

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

Output:

1/48/d*a^2*(a*(1+sec(d*x+c)))^(1/2)*sec(d*x+c)^(1/2)/(1+cos(d*x+c))/(-1/(1 
+cos(d*x+c)))^(1/2)*(114*A*cos(d*x+c)*arctan(1/2*(cot(d*x+c)-csc(d*x+c)-1) 
/(-1/(1+cos(d*x+c)))^(1/2))+75*B*cos(d*x+c)*arctan(1/2*(cot(d*x+c)-csc(d*x 
+c)-1)/(-1/(1+cos(d*x+c)))^(1/2))+114*A*cos(d*x+c)*arctan(1/2*(cot(d*x+c)- 
csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+75*B*cos(d*x+c)*arctan(1/2*(cot(d 
*x+c)-csc(d*x+c)+1)/(-1/(1+cos(d*x+c)))^(1/2))+6*2^(1/2)*(-2/(1+cos(d*x+c) 
))^(1/2)*(11*sin(d*x+c)+2*tan(d*x+c))*A+2^(1/2)*B*(-2/(1+cos(d*x+c)))^(1/2 
)*(75*sin(d*x+c)+34*tan(d*x+c)+8*sec(d*x+c)*tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 475, normalized size of antiderivative = 2.64 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\left [\frac {3 \, {\left ({\left (38 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (38 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\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 (22 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 17 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, B 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 )}}}{96 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac {3 \, {\left ({\left (38 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (38 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )^{2}\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 (22 \, A + 25 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (6 \, A + 17 \, B\right )} a^{2} \cos \left (d x + c\right ) + 8 \, B 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 )}}}{48 \, {\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \] Input:

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

Output:

[1/96*(3*((38*A + 25*B)*a^2*cos(d*x + c)^3 + (38*A + 25*B)*a^2*cos(d*x + c 
)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^ 
2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d* 
x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(3 
*(22*A + 25*B)*a^2*cos(d*x + c)^2 + 2*(6*A + 17*B)*a^2*cos(d*x + c) + 8*B* 
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)^3 + d*cos(d*x + c)^2), 1/48*(3*((38*A + 25*B)*a^2*cos( 
d*x + c)^3 + (38*A + 25*B)*a^2*cos(d*x + c)^2)*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*(22*A + 25*B)*a^2*cos(d*x + 
c)^2 + 2*(6*A + 17*B)*a^2*cos(d*x + c) + 8*B*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)^3 + d*cos 
(d*x + c)^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6297 vs. \(2 (154) = 308\).

Time = 3.20 (sec) , antiderivative size = 6297, normalized size of antiderivative = 34.98 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/96*(6*(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*sqr 
t(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)*c 
os(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))*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*c)...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 783 vs. \(2 (154) = 308\).

Time = 3.98 (sec) , antiderivative size = 783, normalized size of antiderivative = 4.35 \[ \int \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Too large to display} \] Input:

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

Output:

1/48*(3*(38*A*a^(5/2)*sgn(cos(d*x + c)) + 25*B*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*(38*A*a^(5/2)*sgn(cos(d*x + c)) + 25*B*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*(114*sqrt(2)*(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^(7/2)*sg 
n(cos(d*x + c)) + 75*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/ 
2*d*x + 1/2*c)^2 + a))^10*B*a^(7/2)*sgn(cos(d*x + c)) - 1710*sqrt(2)*(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^(9/2) 
*sgn(cos(d*x + c)) - 1125*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t 
an(1/2*d*x + 1/2*c)^2 + a))^8*B*a^(9/2)*sgn(cos(d*x + c)) + 6804*sqrt(2)*( 
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^( 
11/2)*sgn(cos(d*x + c)) + 6174*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr 
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*B*a^(11/2)*sgn(cos(d*x + c)) - 4284*sqr 
t(2)*(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^(13/2)*sgn(cos(d*x + c)) - 4314*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) 
 - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a^(13/2)*sgn(cos(d*x + c)) + 85 
8*sqrt(2)*(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^(15/2)*sgn(cos(d*x + c)) + 807*sqrt(2)*(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^(15/2)*sgn(cos(d*x + c...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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