3.1.87 \(\int \frac {1}{(a+a \sec (c+d x))^5} \, dx\) [87]

3.1.87.1 Optimal result

Integrand size = 12, antiderivative size = 144 \[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=\frac {x}{a^5}-\frac {\tan (c+d x)}{9 d (a+a \sec (c+d x))^5}-\frac {13 \tan (c+d x)}{63 a d (a+a \sec (c+d x))^4}-\frac {34 \tan (c+d x)}{105 a^2 d (a+a \sec (c+d x))^3}-\frac {173 \tan (c+d x)}{315 a^3 d (a+a \sec (c+d x))^2}-\frac {488 \tan (c+d x)}{315 d \left (a^5+a^5 \sec (c+d x)\right )} \]

output

x/a^5-1/9*tan(d*x+c)/d/(a+a*sec(d*x+c))^5-13/63*tan(d*x+c)/a/d/(a+a*sec(d* 
x+c))^4-34/105*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^3-173/315*tan(d*x+c)/a^3/ 
d/(a+a*sec(d*x+c))^2-488/315*tan(d*x+c)/d/(a^5+a^5*sec(d*x+c))
 

3.1.87.2 Mathematica [A] (verified)

Time = 4.87 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.94 \[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^9\left (\frac {1}{2} (c+d x)\right ) \left (39690 d x \cos \left (\frac {d x}{2}\right )+39690 d x \cos \left (c+\frac {d x}{2}\right )+26460 d x \cos \left (c+\frac {3 d x}{2}\right )+26460 d x \cos \left (2 c+\frac {3 d x}{2}\right )+11340 d x \cos \left (2 c+\frac {5 d x}{2}\right )+11340 d x \cos \left (3 c+\frac {5 d x}{2}\right )+2835 d x \cos \left (3 c+\frac {7 d x}{2}\right )+2835 d x \cos \left (4 c+\frac {7 d x}{2}\right )+315 d x \cos \left (4 c+\frac {9 d x}{2}\right )+315 d x \cos \left (5 c+\frac {9 d x}{2}\right )-116676 \sin \left (\frac {d x}{2}\right )+100800 \sin \left (c+\frac {d x}{2}\right )-88284 \sin \left (c+\frac {3 d x}{2}\right )+56700 \sin \left (2 c+\frac {3 d x}{2}\right )-43236 \sin \left (2 c+\frac {5 d x}{2}\right )+18900 \sin \left (3 c+\frac {5 d x}{2}\right )-12384 \sin \left (3 c+\frac {7 d x}{2}\right )+3150 \sin \left (4 c+\frac {7 d x}{2}\right )-1726 \sin \left (4 c+\frac {9 d x}{2}\right )\right )}{161280 a^5 d} \]

input

Integrate[(a + a*Sec[c + d*x])^(-5),x]
 

output

(Sec[c/2]*Sec[(c + d*x)/2]^9*(39690*d*x*Cos[(d*x)/2] + 39690*d*x*Cos[c + ( 
d*x)/2] + 26460*d*x*Cos[c + (3*d*x)/2] + 26460*d*x*Cos[2*c + (3*d*x)/2] + 
11340*d*x*Cos[2*c + (5*d*x)/2] + 11340*d*x*Cos[3*c + (5*d*x)/2] + 2835*d*x 
*Cos[3*c + (7*d*x)/2] + 2835*d*x*Cos[4*c + (7*d*x)/2] + 315*d*x*Cos[4*c + 
(9*d*x)/2] + 315*d*x*Cos[5*c + (9*d*x)/2] - 116676*Sin[(d*x)/2] + 100800*S 
in[c + (d*x)/2] - 88284*Sin[c + (3*d*x)/2] + 56700*Sin[2*c + (3*d*x)/2] - 
43236*Sin[2*c + (5*d*x)/2] + 18900*Sin[3*c + (5*d*x)/2] - 12384*Sin[3*c + 
(7*d*x)/2] + 3150*Sin[4*c + (7*d*x)/2] - 1726*Sin[4*c + (9*d*x)/2]))/(1612 
80*a^5*d)
 

3.1.87.3 Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.19, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.333, Rules used = {3042, 4264, 25, 3042, 4410, 27, 3042, 4410, 25, 3042, 4410, 25, 3042, 4407, 3042, 4281}

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.

\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^5} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^5}dx\)

\(\Big \downarrow \) 4264

\(\displaystyle -\frac {\int -\frac {9 a-4 a \sec (c+d x)}{(\sec (c+d x) a+a)^4}dx}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {9 a-4 a \sec (c+d x)}{(\sec (c+d x) a+a)^4}dx}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {9 a-4 a \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4}dx}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 4410

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {21 a^2-13 a^2 \sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

\(\displaystyle \frac {\frac {3 \left (-\frac {\int -\frac {105 a^3-68 a^3 \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {105 a^3-68 a^3 \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {105 a^3-68 a^3 \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 4410

\(\displaystyle \frac {\frac {3 \left (\frac {-\frac {\int -\frac {315 a^4-173 a^4 \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {315 a^4-173 a^4 \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {315 a^4-173 a^4 \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {\frac {3 \left (\frac {\frac {315 a^3 x-488 a^4 \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3 \left (\frac {\frac {315 a^3 x-488 a^4 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {3 \left (\frac {\frac {315 a^3 x-\frac {488 a^4 \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {173 a^3 \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {34 a^2 \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\right )}{7 a^2}-\frac {13 a \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}}{9 a^2}-\frac {\tan (c+d x)}{9 d (a \sec (c+d x)+a)^5}\)

input

Int[(a + a*Sec[c + d*x])^(-5),x]
 

output

-1/9*Tan[c + d*x]/(d*(a + a*Sec[c + d*x])^5) + ((-13*a*Tan[c + d*x])/(7*d* 
(a + a*Sec[c + d*x])^4) + (3*((-34*a^2*Tan[c + d*x])/(5*d*(a + a*Sec[c + d 
*x])^3) + ((-173*a^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + (315*a^3 
*x - (488*a^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2)))/( 
7*a^2))/(9*a^2)
 

3.1.87.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 4264

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-Cot[c 
+ d*x])*((a + b*Csc[c + d*x])^n/(d*(2*n + 1))), x] + Simp[1/(a^2*(2*n + 1)) 
   Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]), 
 x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && Int 
egerQ[2*n]
 

rule 4281

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

rule 4407

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4410

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

3.1.87.4 Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.53

method	result	size
parallelrisch	\(\frac {-35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}+270 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}-1008 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+2730 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+5040 d x -9765 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5040 d \,a^{5}}\)	\(77\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+32 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{5}}\)	\(85\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{9}+\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {26 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+32 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d \,a^{5}}\)	\(85\)
norman	\(\frac {\frac {x}{a}-\frac {31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 a d}+\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{56 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{144 a d}}{a^{4}}\)	\(104\)
risch	\(\frac {x}{a^{5}}-\frac {2 i \left (1575 \,{\mathrm e}^{8 i \left (d x +c \right )}+9450 \,{\mathrm e}^{7 i \left (d x +c \right )}+28350 \,{\mathrm e}^{6 i \left (d x +c \right )}+50400 \,{\mathrm e}^{5 i \left (d x +c \right )}+58338 \,{\mathrm e}^{4 i \left (d x +c \right )}+44142 \,{\mathrm e}^{3 i \left (d x +c \right )}+21618 \,{\mathrm e}^{2 i \left (d x +c \right )}+6192 \,{\mathrm e}^{i \left (d x +c \right )}+863\right )}{315 d \,a^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{9}}\)	\(119\)

input

int(1/(a+a*sec(d*x+c))^5,x,method=_RETURNVERBOSE)
 

output

1/5040*(-35*tan(1/2*d*x+1/2*c)^9+270*tan(1/2*d*x+1/2*c)^7-1008*tan(1/2*d*x 
+1/2*c)^5+2730*tan(1/2*d*x+1/2*c)^3+5040*d*x-9765*tan(1/2*d*x+1/2*c))/d/a^ 
5
 

3.1.87.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.31 \[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=\frac {315 \, d x \cos \left (d x + c\right )^{5} + 1575 \, d x \cos \left (d x + c\right )^{4} + 3150 \, d x \cos \left (d x + c\right )^{3} + 3150 \, d x \cos \left (d x + c\right )^{2} + 1575 \, d x \cos \left (d x + c\right ) + 315 \, d x - {\left (863 \, \cos \left (d x + c\right )^{4} + 2740 \, \cos \left (d x + c\right )^{3} + 3549 \, \cos \left (d x + c\right )^{2} + 2125 \, \cos \left (d x + c\right ) + 488\right )} \sin \left (d x + c\right )}{315 \, {\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \]

input

integrate(1/(a+a*sec(d*x+c))^5,x, algorithm="fricas")
 

output

1/315*(315*d*x*cos(d*x + c)^5 + 1575*d*x*cos(d*x + c)^4 + 3150*d*x*cos(d*x 
 + c)^3 + 3150*d*x*cos(d*x + c)^2 + 1575*d*x*cos(d*x + c) + 315*d*x - (863 
*cos(d*x + c)^4 + 2740*cos(d*x + c)^3 + 3549*cos(d*x + c)^2 + 2125*cos(d*x 
 + c) + 488)*sin(d*x + c))/(a^5*d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 
+ 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c) 
 + a^5*d)
 

3.1.87.6 Sympy [F]

\[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=\frac {\int \frac {1}{\sec ^{5}{\left (c + d x \right )} + 5 \sec ^{4}{\left (c + d x \right )} + 10 \sec ^{3}{\left (c + d x \right )} + 10 \sec ^{2}{\left (c + d x \right )} + 5 \sec {\left (c + d x \right )} + 1}\, dx}{a^{5}} \]

input

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

output

Integral(1/(sec(c + d*x)**5 + 5*sec(c + d*x)**4 + 10*sec(c + d*x)**3 + 10* 
sec(c + d*x)**2 + 5*sec(c + d*x) + 1), x)/a**5
 

3.1.87.7 Maxima [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=-\frac {\frac {\frac {9765 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2730 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1008 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {270 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac {10080 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{5}}}{5040 \, d} \]

input

integrate(1/(a+a*sec(d*x+c))^5,x, algorithm="maxima")
 

output

-1/5040*((9765*sin(d*x + c)/(cos(d*x + c) + 1) - 2730*sin(d*x + c)^3/(cos( 
d*x + c) + 1)^3 + 1008*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 270*sin(d*x + 
 c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^5 - 
 10080*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^5)/d
 

3.1.87.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=\frac {\frac {5040 \, {\left (d x + c\right )}}{a^{5}} - \frac {35 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 270 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1008 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2730 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9765 \, a^{40} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{45}}}{5040 \, d} \]

input

integrate(1/(a+a*sec(d*x+c))^5,x, algorithm="giac")
 

output

1/5040*(5040*(d*x + c)/a^5 - (35*a^40*tan(1/2*d*x + 1/2*c)^9 - 270*a^40*ta 
n(1/2*d*x + 1/2*c)^7 + 1008*a^40*tan(1/2*d*x + 1/2*c)^5 - 2730*a^40*tan(1/ 
2*d*x + 1/2*c)^3 + 9765*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d
 

3.1.87.9 Mupad [B] (verification not implemented)

Time = 13.44 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+a \sec (c+d x))^5} \, dx=\frac {x}{a^5}-\frac {\frac {863\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{315}-\frac {356\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{315}+\frac {169\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{420}-\frac {41\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{504}+\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{144}}{a^5\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9} \]

input

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

output

x/a^5 - (sin(c/2 + (d*x)/2)/144 - (41*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x) 
/2))/504 + (169*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2))/420 - (356*cos(c/ 
2 + (d*x)/2)^6*sin(c/2 + (d*x)/2))/315 + (863*cos(c/2 + (d*x)/2)^8*sin(c/2 
 + (d*x)/2))/315)/(a^5*d*cos(c/2 + (d*x)/2)^9)