3.6.26 \(\int \frac {1}{(3+5 \sec (c+d x))^4} \, dx\) [526]

3.6.26.1 Optimal result

Integrand size = 12, antiderivative size = 106 \[ \int \frac {1}{(3+5 \sec (c+d x))^4} \, dx=\frac {21553 x}{2654208}+\frac {11215 \arctan \left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{1327104 d}-\frac {25 \tan (c+d x)}{144 d (3+5 \sec (c+d x))^3}-\frac {25 \tan (c+d x)}{4608 d (3+5 \sec (c+d x))^2}-\frac {16925 \tan (c+d x)}{221184 d (3+5 \sec (c+d x))} \]

output

21553/2654208*x+11215/1327104*arctan(sin(d*x+c)/(3+cos(d*x+c)))/d-25/144*t 
an(d*x+c)/d/(3+5*sec(d*x+c))^3-25/4608*tan(d*x+c)/d/(3+5*sec(d*x+c))^2-169 
25/221184*tan(d*x+c)/d/(3+5*sec(d*x+c))
 

3.6.26.2 Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(3+5 \sec (c+d x))^4} \, dx=\frac {6307840 c+6307840 d x+8036352 (c+d x) \cos (c+d x)+22430 \arctan \left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right ) (5+3 \cos (c+d x))^3+2211840 c \cos (2 (c+d x))+2211840 d x \cos (2 (c+d x))+221184 c \cos (3 (c+d x))+221184 d x \cos (3 (c+d x))-5660475 \sin (c+d x)-3082500 \sin (2 (c+d x))-582975 \sin (3 (c+d x))}{2654208 d (5+3 \cos (c+d x))^3} \]

input

Integrate[(3 + 5*Sec[c + d*x])^(-4),x]
 

output

(6307840*c + 6307840*d*x + 8036352*(c + d*x)*Cos[c + d*x] + 22430*ArcTan[2 
*Cot[(c + d*x)/2]]*(5 + 3*Cos[c + d*x])^3 + 2211840*c*Cos[2*(c + d*x)] + 2 
211840*d*x*Cos[2*(c + d*x)] + 221184*c*Cos[3*(c + d*x)] + 221184*d*x*Cos[3 
*(c + d*x)] - 5660475*Sin[c + d*x] - 3082500*Sin[2*(c + d*x)] - 582975*Sin 
[3*(c + d*x)])/(2654208*d*(5 + 3*Cos[c + d*x])^3)
 

3.6.26.3 Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.24, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {3042, 4272, 3042, 4548, 27, 3042, 4548, 3042, 4407, 3042, 4318, 3042, 3136}

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.

\(\displaystyle \int \frac {1}{(5 \sec (c+d x)+3)^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (5 \csc \left (c+d x+\frac {\pi }{2}\right )+3\right )^4}dx\)

\(\Big \downarrow \) 4272

\(\displaystyle \frac {1}{144} \int \frac {-50 \sec ^2(c+d x)+45 \sec (c+d x)+48}{(5 \sec (c+d x)+3)^3}dx-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{144} \int \frac {-50 \csc \left (c+d x+\frac {\pi }{2}\right )^2+45 \csc \left (c+d x+\frac {\pi }{2}\right )+48}{\left (5 \csc \left (c+d x+\frac {\pi }{2}\right )+3\right )^3}dx-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 4548

\(\displaystyle \frac {1}{144} \left (\frac {1}{96} \int \frac {3 \left (-25 \sec ^2(c+d x)-290 \sec (c+d x)+512\right )}{(5 \sec (c+d x)+3)^2}dx-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \int \frac {-25 \sec ^2(c+d x)-290 \sec (c+d x)+512}{(5 \sec (c+d x)+3)^2}dx-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \int \frac {-25 \csc \left (c+d x+\frac {\pi }{2}\right )^2-290 \csc \left (c+d x+\frac {\pi }{2}\right )+512}{\left (5 \csc \left (c+d x+\frac {\pi }{2}\right )+3\right )^2}dx-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 4548

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \int \frac {9915 \sec (c+d x)+8192}{5 \sec (c+d x)+3}dx-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \int \frac {9915 \csc \left (c+d x+\frac {\pi }{2}\right )+8192}{5 \csc \left (c+d x+\frac {\pi }{2}\right )+3}dx-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \left (\frac {8192 x}{3}-\frac {11215}{3} \int \frac {\sec (c+d x)}{5 \sec (c+d x)+3}dx\right )-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \left (\frac {8192 x}{3}-\frac {11215}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{5 \csc \left (c+d x+\frac {\pi }{2}\right )+3}dx\right )-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \left (\frac {8192 x}{3}-\frac {2243}{3} \int \frac {1}{\frac {3}{5} \cos (c+d x)+1}dx\right )-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \left (\frac {8192 x}{3}-\frac {2243}{3} \int \frac {1}{\frac {3}{5} \sin \left (c+d x+\frac {\pi }{2}\right )+1}dx\right )-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

\(\Big \downarrow \) 3136

\(\displaystyle \frac {1}{144} \left (\frac {1}{32} \left (\frac {1}{48} \left (\frac {8192 x}{3}-\frac {2243}{3} \left (\frac {5 x}{4}-\frac {5 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{2 d}\right )\right )-\frac {16925 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{32 d (5 \sec (c+d x)+3)^2}\right )-\frac {25 \tan (c+d x)}{144 d (5 \sec (c+d x)+3)^3}\)

input

Int[(3 + 5*Sec[c + d*x])^(-4),x]
 

output

(-25*Tan[c + d*x])/(144*d*(3 + 5*Sec[c + d*x])^3) + ((-25*Tan[c + d*x])/(3 
2*d*(3 + 5*Sec[c + d*x])^2) + (((8192*x)/3 - (2243*((5*x)/4 - (5*ArcTan[Si 
n[c + d*x]/(3 + Cos[c + d*x])])/(2*d)))/3)/48 - (16925*Tan[c + d*x])/(48*d 
*(3 + 5*Sec[c + d*x])))/32)/144
 

3.6.26.3.1 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 3136

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[ 
a^2 - b^2, 2]}, Simp[x/q, x] + Simp[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q 
 + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2, 0] && 
 PosQ[a]
 

rule 4272

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

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[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 4548

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

3.6.26.4 Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{81}+\frac {-\frac {25925 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{221184}-\frac {3575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6912}-\frac {17675 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{13824}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{3}}-\frac {11215 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1327104}}{d}\)	\(87\)
default	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{81}+\frac {-\frac {25925 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{221184}-\frac {3575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6912}-\frac {17675 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{13824}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{3}}-\frac {11215 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1327104}}{d}\)	\(87\)
risch	\(\frac {x}{81}-\frac {25 i \left (164835 \,{\mathrm e}^{5 i \left (d x +c \right )}+931257 \,{\mathrm e}^{4 i \left (d x +c \right )}+1995070 \,{\mathrm e}^{3 i \left (d x +c \right )}+1610514 \,{\mathrm e}^{2 i \left (d x +c \right )}+534735 \,{\mathrm e}^{i \left (d x +c \right )}+69957\right )}{995328 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )^{3}}-\frac {11215 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3\right )}{2654208 d}+\frac {11215 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{3}\right )}{2654208 d}\)	\(130\)
parallelrisch	\(\frac {11215 i \left (770+27 \cos \left (3 d x +3 c \right )+981 \cos \left (d x +c \right )+270 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 i\right )+11215 i \left (-770-27 \cos \left (3 d x +3 c \right )-981 \cos \left (d x +c \right )-270 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 i\right )[phone] d x \cos \left (d x +c \right )+8847360 d x \cos \left (2 d x +2 c \right )+884736 d x \cos \left (3 d x +3 c \right )[phone] d x -22641900 \sin \left (d x +c \right )-12330000 \sin \left (2 d x +2 c \right )-2331900 \sin \left (3 d x +3 c \right )}{2654208 d \left (770+27 \cos \left (3 d x +3 c \right )+981 \cos \left (d x +c \right )+270 \cos \left (2 d x +2 c \right )\right )}\)	\(207\)

input

int(1/(3+5*sec(d*x+c))^4,x,method=_RETURNVERBOSE)
 

output

1/d*(2/81*arctan(tan(1/2*d*x+1/2*c))+5/648*(-15555/1024*tan(1/2*d*x+1/2*c) 
^5-2145/32*tan(1/2*d*x+1/2*c)^3-10605/64*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+ 
1/2*c)^2+4)^3-11215/1327104*arctan(1/2*tan(1/2*d*x+1/2*c)))
 

3.6.26.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.50 \[ \int \frac {1}{(3+5 \sec (c+d x))^4} \, dx=\frac {884736 \, d x \cos \left (d x + c\right )^{3} + 4423680 \, d x \cos \left (d x + c\right )^{2} + 7372800 \, d x \cos \left (d x + c\right ) + 4096000 \, d x + 11215 \, {\left (27 \, \cos \left (d x + c\right )^{3} + 135 \, \cos \left (d x + c\right )^{2} + 225 \, \cos \left (d x + c\right ) + 125\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) - 300 \, {\left (7773 \, \cos \left (d x + c\right )^{2} + 20550 \, \cos \left (d x + c\right ) + 16925\right )} \sin \left (d x + c\right )}{2654208 \, {\left (27 \, d \cos \left (d x + c\right )^{3} + 135 \, d \cos \left (d x + c\right )^{2} + 225 \, d \cos \left (d x + c\right ) + 125 \, d\right )}} \]

input

integrate(1/(3+5*sec(d*x+c))^4,x, algorithm="fricas")
 

output

1/2654208*(884736*d*x*cos(d*x + c)^3 + 4423680*d*x*cos(d*x + c)^2 + 737280 
0*d*x*cos(d*x + c) + 4096000*d*x + 11215*(27*cos(d*x + c)^3 + 135*cos(d*x 
+ c)^2 + 225*cos(d*x + c) + 125)*arctan(1/4*(5*cos(d*x + c) + 3)/sin(d*x + 
 c)) - 300*(7773*cos(d*x + c)^2 + 20550*cos(d*x + c) + 16925)*sin(d*x + c) 
)/(27*d*cos(d*x + c)^3 + 135*d*cos(d*x + c)^2 + 225*d*cos(d*x + c) + 125*d 
)
 

3.6.26.6 Sympy [F]

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

input

integrate(1/(3+5*sec(d*x+c))**4,x)
 

output

Integral((5*sec(c + d*x) + 3)**(-4), x)
 

3.6.26.7 Maxima [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.61 \[ \int \frac {1}{(3+5 \sec (c+d x))^4} \, dx=-\frac {\frac {150 \, {\left (\frac {11312 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4576 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {1037 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{\frac {48 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {12 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 64} - 32768 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 11215 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{1327104 \, d} \]

input

integrate(1/(3+5*sec(d*x+c))^4,x, algorithm="maxima")
 

output

-1/1327104*(150*(11312*sin(d*x + c)/(cos(d*x + c) + 1) + 4576*sin(d*x + c) 
^3/(cos(d*x + c) + 1)^3 + 1037*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(48*si 
n(d*x + c)^2/(cos(d*x + c) + 1)^2 + 12*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 
 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 64) - 32768*arctan(sin(d*x + c)/( 
cos(d*x + c) + 1)) + 11215*arctan(1/2*sin(d*x + c)/(cos(d*x + c) + 1)))/d
 

3.6.26.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(3+5 \sec (c+d x))^4} \, dx=\frac {21553 \, d x + 21553 \, c - \frac {300 \, {\left (1037 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4576 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11312 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4\right )}^{3}} + 22430 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{2654208 \, d} \]

input

integrate(1/(3+5*sec(d*x+c))^4,x, algorithm="giac")
 

output

1/2654208*(21553*d*x + 21553*c - 300*(1037*tan(1/2*d*x + 1/2*c)^5 + 4576*t 
an(1/2*d*x + 1/2*c)^3 + 11312*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^ 
2 + 4)^3 + 22430*arctan(sin(d*x + c)/(cos(d*x + c) + 3)))/d
 

3.6.26.9 Mupad [B] (verification not implemented)

Time = 14.43 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(3+5 \sec (c+d x))^4} \, dx=\frac {x}{81}-\frac {11215\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{1327104\,d}-\frac {\frac {25925\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{221184}+\frac {3575\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6912}+\frac {17675\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{13824}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+64\right )} \]

input

int(1/(5/cos(c + d*x) + 3)^4,x)
 

output

x/81 - (11215*atan(tan(c/2 + (d*x)/2)/2))/(1327104*d) - ((17675*tan(c/2 + 
(d*x)/2))/13824 + (3575*tan(c/2 + (d*x)/2)^3)/6912 + (25925*tan(c/2 + (d*x 
)/2)^5)/221184)/(d*(48*tan(c/2 + (d*x)/2)^2 + 12*tan(c/2 + (d*x)/2)^4 + ta 
n(c/2 + (d*x)/2)^6 + 64))