3.6.2 \(\int \sec (c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [502]

3.6.2.1 Optimal result

Integrand size = 41, antiderivative size = 184 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {64 a^3 (21 A+15 B+13 C) \tan (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (21 A+15 B+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 a (21 A+15 B+13 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 d}+\frac {2 (9 B-2 C) (a+a \sec (c+d x))^{5/2} \tan (c+d x)}{63 d}+\frac {2 C (a+a \sec (c+d x))^{7/2} \tan (c+d x)}{9 a d} \]

output

2/105*a*(21*A+15*B+13*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/d+2/63*(9*B-2*C 
)*(a+a*sec(d*x+c))^(5/2)*tan(d*x+c)/d+2/9*C*(a+a*sec(d*x+c))^(7/2)*tan(d*x 
+c)/a/d+64/315*a^3*(21*A+15*B+13*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+16 
/315*a^2*(21*A+15*B+13*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/d
 

3.6.2.2 Mathematica [A] (verified)

Time = 2.49 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.57 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a^3 \left (903 A+690 B+584 C+(294 A+345 B+292 C) \sec (c+d x)+3 (21 A+60 B+73 C) \sec ^2(c+d x)+5 (9 B+26 C) \sec ^3(c+d x)+35 C \sec ^4(c+d x)\right ) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \]

input

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

output

(2*a^3*(903*A + 690*B + 584*C + (294*A + 345*B + 292*C)*Sec[c + d*x] + 3*( 
21*A + 60*B + 73*C)*Sec[c + d*x]^2 + 5*(9*B + 26*C)*Sec[c + d*x]^3 + 35*C* 
Sec[c + d*x]^4)*Tan[c + d*x])/(315*d*Sqrt[a*(1 + Sec[c + d*x])])
 

3.6.2.3 Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {3042, 4570, 27, 3042, 4489, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4570

\(\displaystyle \frac {2 \int \frac {1}{2} \sec (c+d x) (\sec (c+d x) a+a)^{5/2} (a (9 A+7 C)+a (9 B-2 C) \sec (c+d x))dx}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \sec (c+d x) (\sec (c+d x) a+a)^{5/2} (a (9 A+7 C)+a (9 B-2 C) \sec (c+d x))dx}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (a (9 A+7 C)+a (9 B-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \int \sec (c+d x) (\sec (c+d x) a+a)^{5/2}dx+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}dx+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 4280

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \left (\frac {8}{5} a \int \sec (c+d x) (\sec (c+d x) a+a)^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \left (\frac {8}{5} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}dx+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 4280

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

\(\Big \downarrow \) 4279

\(\displaystyle \frac {\frac {3}{7} a (21 A+15 B+13 C) \left (\frac {8}{5} a \left (\frac {8 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}\right )+\frac {2 a \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d}\right )+\frac {2 a (9 B-2 C) \tan (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d}}{9 a}+\frac {2 C \tan (c+d x) (a \sec (c+d x)+a)^{7/2}}{9 a d}\)

input

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

output

(2*C*(a + a*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*a*d) + ((2*a*(9*B - 2*C)* 
(a + a*Sec[c + d*x])^(5/2)*Tan[c + d*x])/(7*d) + (3*a*(21*A + 15*B + 13*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]]*Ta 
n[c + d*x])/(3*d)))/5))/7)/(9*a)
 

3.6.2.3.1 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 4489

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

rule 4570

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

3.6.2.4 Maple [A] (verified)

Time = 73.12 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.90

method	result	size
default	\(\frac {2 a^{2} \left (903 A \cos \left (d x +c \right )^{4}+690 B \cos \left (d x +c \right )^{4}+584 C \cos \left (d x +c \right )^{4}+294 A \cos \left (d x +c \right )^{3}+345 B \cos \left (d x +c \right )^{3}+292 C \cos \left (d x +c \right )^{3}+63 A \cos \left (d x +c \right )^{2}+180 B \cos \left (d x +c \right )^{2}+219 C \cos \left (d x +c \right )^{2}+45 B \cos \left (d x +c \right )+130 C \cos \left (d x +c \right )+35 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\)	\(166\)
parts	\(\frac {2 A \,a^{2} \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (43 \sin \left (d x +c \right )+14 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \,a^{2} \left (46 \cos \left (d x +c \right )^{3}+23 \cos \left (d x +c \right )^{2}+12 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{21 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \,a^{2} \left (584 \cos \left (d x +c \right )^{4}+292 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+130 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\)	\(224\)

input

int(sec(d*x+c)*(a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,me 
thod=_RETURNVERBOSE)
 

output

2/315*a^2/d*(903*A*cos(d*x+c)^4+690*B*cos(d*x+c)^4+584*C*cos(d*x+c)^4+294* 
A*cos(d*x+c)^3+345*B*cos(d*x+c)^3+292*C*cos(d*x+c)^3+63*A*cos(d*x+c)^2+180 
*B*cos(d*x+c)^2+219*C*cos(d*x+c)^2+45*B*cos(d*x+c)+130*C*cos(d*x+c)+35*C)* 
(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^3
 

3.6.2.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.78 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (903 \, A + 690 \, B + 584 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + {\left (294 \, A + 345 \, B + 292 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 60 \, B + 73 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, B + 26 \, C\right )} a^{2} \cos \left (d x + c\right ) + 35 \, C a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]

input

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

output

2/315*((903*A + 690*B + 584*C)*a^2*cos(d*x + c)^4 + (294*A + 345*B + 292*C 
)*a^2*cos(d*x + c)^3 + 3*(21*A + 60*B + 73*C)*a^2*cos(d*x + c)^2 + 5*(9*B 
+ 26*C)*a^2*cos(d*x + c) + 35*C*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c 
))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
 

3.6.2.6 Sympy [F]

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

input

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

output

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

3.6.2.7 Maxima [F(-1)]

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

input

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

output

Timed out
 

3.6.2.8 Giac [F]

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

input

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

output

sage0*x
 

3.6.2.9 Mupad [B] (verification not implemented)

Time = 27.14 (sec) , antiderivative size = 885, normalized size of antiderivative = 4.81 \[ \int \sec (c+d x) (a+a \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

input

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

output

(((A*a^2*2i)/d - (a^2*exp(c*1i + d*x*1i)*(903*A + 690*B + 584*C)*2i)/(315* 
d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2))/(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)*((a^2*(3*A + 4*B + 4*C)*10i)/(7*d) - (a^2*(11 
*A + 10*B + 20*C)*2i)/(7*d) - (a^2*(5*A + 2*B)*2i)/(7*d) + (a^2*(9*A - 16* 
C)*2i)/(63*d)) + (A*a^2*2i)/(7*d) - (a^2*(5*A + 2*B - 16*C)*2i)/(7*d) - (a 
^2*(11*A + 10*B + 4*C)*2i)/(7*d) + (a^2*(15*A + 20*B + 36*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*(189 
*A + 240*B + 292*C)*2i)/(315*d) - (a^2*(5*A + 2*B)*2i)/(3*d)) + (A*a^2*2i) 
/(3*d) - (a^2*(9*A + 10*B + 4*C)*2i)/(3*d)))/((exp(c*1i + d*x*1i) + 1)*(ex 
p(c*2i + d*x*2i) + 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)*((A*a^2*2i)/(9*d) - (a^2*(6*A + 5*B + 2* 
C)*4i)/(9*d) - (a^2*(10*A + 11*B + 10*C)*4i)/(9*d) + (a^2*(13*A + 15*B + 2 
0*C)*4i)/(9*d) + (a^2*(5*A + 2*B)*2i)/(9*d)) - (A*a^2*2i)/(9*d) + (a^2*(6* 
A + 5*B + 2*C)*4i)/(9*d) + (a^2*(10*A + 11*B + 10*C)*4i)/(9*d) - (a^2*(13* 
A + 15*B + 20*C)*4i)/(9*d) - (a^2*(5*A + 2*B)*2i)/(9*d)))/((exp(c*1i + d*x 
*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((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*A + 9*B + 10*C)* 
4i)/(5*d) + (a^2*(24*B - 21*A + 32*C)*2i)/(105*d) - (a^2*(5*A + 2*B)*2i...