\(\int (a+a \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [377]

Optimal result

Integrand size = 34, antiderivative size = 138 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (7 B+5 C) \tan (c+d x)}{105 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (7 B+5 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{105 d}+\frac {2 a (7 B+5 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{35 d}+\frac {2 C (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{7 d} \] Output:

64/105*a^3*(7*B+5*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16/105*a^2*(7*B+5 
*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/35*a*(7*B+5*C)*(a+a*sec(d*x+c))^ 
(3/2)*tan(d*x+c)/d+2/7*C*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.57 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 \left (301 B+230 C+(98 B+115 C) \sec (c+d x)+3 (7 B+20 C) \sec ^2(c+d x)+15 C \sec ^3(c+d x)\right ) \tan (c+d x)}{105 d \sqrt {a (1+\sec (c+d x))}} \] Input:

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

Output:

(2*a^3*(301*B + 230*C + (98*B + 115*C)*Sec[c + d*x] + 3*(7*B + 20*C)*Sec[c 
 + d*x]^2 + 15*C*Sec[c + d*x]^3)*Tan[c + d*x])/(105*d*Sqrt[a*(1 + Sec[c + 
d*x])])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 4542, 27, 3042, 4280, 3042, 4280, 3042, 4279}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4542

\(\displaystyle \frac {2 \int \frac {1}{2} a (7 B+5 C) \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx}{7 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} (7 B+5 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} (7 B+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 4280

\(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4280

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4279

\(\displaystyle \frac {1}{7} (7 B+5 C) \left (\frac {8}{5} a \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}\)

Input:

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

Output:

(2*C*(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + ((7*B + 5*C)*((2*a*( 
a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (8*a*((8*a^2*Tan[c + d*x]) 
/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d* 
x])/(3*d)))/5))/7
 

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4279

Int[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free 
Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
 

rule 4280

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[(-b)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] 
+ Simp[a*((2*m - 1)/m)   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m - 1), x], 
 x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && Intege 
rQ[2*m]
 

rule 4542

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot 
[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(b*(m + 1))   I 
nt[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + 
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
 &&  !LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 12.90 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {2 a^{2} \left (\cos \left (d x +c \right ) \left (301 \cos \left (d x +c \right )^{2}+98 \cos \left (d x +c \right )+21\right ) B +\left (230 \cos \left (d x +c \right )^{3}+115 \cos \left (d x +c \right )^{2}+60 \cos \left (d x +c \right )+15\right ) C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{105 d \left (\cos \left (d x +c \right )+1\right )}\)	\(106\)
parts	\(\frac {B \left (86 \sin \left (d x +c \right )+28 \tan \left (d x +c \right )+6 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right ) a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \left (15 \cos \left (d x +c \right )+15\right )}+\frac {2 C \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}\)	\(140\)

Input:

int((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNV 
ERBOSE)
 

Output:

2/105/d*a^2*(cos(d*x+c)*(301*cos(d*x+c)^2+98*cos(d*x+c)+21)*B+(230*cos(d*x 
+c)^3+115*cos(d*x+c)^2+60*cos(d*x+c)+15)*C)*(a*(1+sec(d*x+c)))^(1/2)/(cos( 
d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (301 \, B + 230 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (98 \, B + 115 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B + 20 \, C\right )} a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{4} + d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorith 
m="fricas")
 

Output:

2/105*((301*B + 230*C)*a^2*cos(d*x + c)^3 + (98*B + 115*C)*a^2*cos(d*x + c 
)^2 + 3*(7*B + 20*C)*a^2*cos(d*x + c) + 15*C*a^2)*sqrt((a*cos(d*x + c) + a 
)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^4 + d*cos(d*x + c)^3)
 

Sympy [F]

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

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

Output:

Integral((a*(sec(c + d*x) + 1))**(5/2)*(B + C*sec(c + d*x))*sec(c + d*x), 
x)
 

Maxima [F]

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

integrate((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorith 
m="maxima")
 

Output:

-2/105*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) 
^(1/4)*(7*(15*B*a^2*sin(6*d*x + 6*c) + 5*(17*B + 10*C)*a^2*sin(4*d*x + 4*c 
) + (113*B + 100*C)*a^2*sin(2*d*x + 2*c))*cos(7/2*arctan2(sin(2*d*x + 2*c) 
, cos(2*d*x + 2*c) + 1)) - (105*B*a^2*cos(6*d*x + 6*c) + 35*(17*B + 10*C)* 
a^2*cos(4*d*x + 4*c) + 7*(113*B + 100*C)*a^2*cos(2*d*x + 2*c) + (301*B + 2 
30*C)*a^2)*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt( 
a) - 105*(((5*B + 2*C)*a^2*d*cos(2*d*x + 2*c)^4 + (5*B + 2*C)*a^2*d*sin(2* 
d*x + 2*c)^4 + 4*(5*B + 2*C)*a^2*d*cos(2*d*x + 2*c)^3 + 6*(5*B + 2*C)*a^2* 
d*cos(2*d*x + 2*c)^2 + 4*(5*B + 2*C)*a^2*d*cos(2*d*x + 2*c) + (5*B + 2*C)* 
a^2*d + 2*((5*B + 2*C)*a^2*d*cos(2*d*x + 2*c)^2 + 2*(5*B + 2*C)*a^2*d*cos( 
2*d*x + 2*c) + (5*B + 2*C)*a^2*d)*sin(2*d*x + 2*c)^2)*integrate((((cos(6*d 
*x + 6*c)*cos(2*d*x + 2*c) + 2*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d 
*x + 2*c)^2 + sin(6*d*x + 6*c)*sin(2*d*x + 2*c) + 2*sin(4*d*x + 4*c)*sin(2 
*d*x + 2*c) + sin(2*d*x + 2*c)^2)*cos(7/2*arctan2(sin(2*d*x + 2*c), cos(2* 
d*x + 2*c))) + (cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin 
(4*d*x + 4*c) - cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin 
(2*d*x + 2*c))*sin(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(5 
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*s 
in(6*d*x + 6*c) + 2*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(6*d*x + 6*c)*s 
in(2*d*x + 2*c) - 2*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*cos(7/2*arctan2(...
 

Giac [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.57 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left ({\left (4 \, {\left (2 \, \sqrt {2} {\left (7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, \sqrt {2} {\left (7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, \sqrt {2} {\left (7 \, B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 5 \, C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, \sqrt {2} {\left (B a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + C a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{105 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} d} \] Input:

integrate((a+a*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorith 
m="giac")
 

Output:

8/105*((4*(2*sqrt(2)*(7*B*a^6*sgn(cos(d*x + c)) + 5*C*a^6*sgn(cos(d*x + c) 
))*tan(1/2*d*x + 1/2*c)^2 - 7*sqrt(2)*(7*B*a^6*sgn(cos(d*x + c)) + 5*C*a^6 
*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(7*B*a^6*sgn(cos( 
d*x + c)) + 5*C*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 105*sqrt( 
2)*(B*a^6*sgn(cos(d*x + c)) + C*a^6*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2* 
c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*d 
)
 

Mupad [B] (verification not implemented)

Time = 15.58 (sec) , antiderivative size = 590, normalized size of antiderivative = 4.28 \[ \int (a+a \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((B*a^2*2i) 
/d - (a^2*exp(c*1i + d*x*1i)*(301*B + 230*C)*2i)/(105*d)))/(exp(c*1i + d*x 
*1i) + 1) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2) 
*(exp(c*1i + d*x*1i)*((B*a^2*2i)/(7*d) + (a^2*(3*B + 4*C)*10i)/(7*d) - (a^ 
2*(5*B + 2*C)*2i)/(7*d) - (a^2*(11*B + 10*C)*2i)/(7*d)) + (B*a^2*2i)/(7*d) 
 + (a^2*(3*B + 4*C)*10i)/(7*d) - (a^2*(5*B + 2*C)*2i)/(7*d) - (a^2*(11*B + 
 10*C)*2i)/(7*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^3) - 
 ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i 
+ d*x*1i)*((a^2*(5*B + 2*C)*2i)/(5*d) - (a^2*(5*B + 9*C)*4i)/(5*d) + (a^2* 
(7*B - 8*C)*2i)/(35*d)) - (B*a^2*2i)/(5*d) - (a^2*(B + 2*C)*2i)/d + (a^2*( 
B + C)*4i)/d))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) + ((a 
 + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d* 
x*1i)*((a^2*(5*B + 2*C)*2i)/(3*d) - (a^2*(63*B + 80*C)*2i)/(105*d)) - (B*a 
^2*2i)/(3*d) + (a^2*(9*B + 10*C)*2i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(ex 
p(c*2i + d*x*2i) + 1))
 

Reduce [F]

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

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

Output:

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