3.6.25 \(\int \frac {1}{(3+5 \sec (c+d x))^3} \, dx\) [525]

3.6.25.1 Optimal result

Integrand size = 12, antiderivative size = 81 \[ \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx=-\frac {1007 x}{55296}+\frac {3055 \arctan \left (\frac {\sin (c+d x)}{3+\cos (c+d x)}\right )}{27648 d}-\frac {25 \tan (c+d x)}{96 d (3+5 \sec (c+d x))^2}-\frac {125 \tan (c+d x)}{4608 d (3+5 \sec (c+d x))} \]

output

-1007/55296*x+3055/27648*arctan(sin(d*x+c)/(3+cos(d*x+c)))/d-25/96*tan(d*x 
+c)/d/(3+5*sec(d*x+c))^2-125/4608*tan(d*x+c)/d/(3+5*sec(d*x+c))
 

3.6.25.2 Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.33 \[ \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx=\frac {30208 c+30208 d x+30720 (c+d x) \cos (c+d x)+3055 \arctan \left (2 \cot \left (\frac {1}{2} (c+d x)\right )\right ) (5+3 \cos (c+d x))^2+4608 c \cos (2 (c+d x))+4608 d x \cos (2 (c+d x))-3750 \sin (c+d x)-4725 \sin (2 (c+d x))}{27648 d (5+3 \cos (c+d x))^2} \]

input

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

output

(30208*c + 30208*d*x + 30720*(c + d*x)*Cos[c + d*x] + 3055*ArcTan[2*Cot[(c 
 + d*x)/2]]*(5 + 3*Cos[c + d*x])^2 + 4608*c*Cos[2*(c + d*x)] + 4608*d*x*Co 
s[2*(c + d*x)] - 3750*Sin[c + d*x] - 4725*Sin[2*(c + d*x)])/(27648*d*(5 + 
3*Cos[c + d*x])^2)
 

3.6.25.3 Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 4272, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

\(\displaystyle \frac {1}{96} \int \frac {-25 \sec ^2(c+d x)+30 \sec (c+d x)+32}{(5 \sec (c+d x)+3)^2}dx-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{96} \int \frac {-25 \csc \left (c+d x+\frac {\pi }{2}\right )^2+30 \csc \left (c+d x+\frac {\pi }{2}\right )+32}{\left (5 \csc \left (c+d x+\frac {\pi }{2}\right )+3\right )^2}dx-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 4548

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \int \frac {512-165 \sec (c+d x)}{5 \sec (c+d x)+3}dx-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \int \frac {512-165 \csc \left (c+d x+\frac {\pi }{2}\right )}{5 \csc \left (c+d x+\frac {\pi }{2}\right )+3}dx-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \left (\frac {512 x}{3}-\frac {3055}{3} \int \frac {\sec (c+d x)}{5 \sec (c+d x)+3}dx\right )-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \left (\frac {512 x}{3}-\frac {3055}{3} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{5 \csc \left (c+d x+\frac {\pi }{2}\right )+3}dx\right )-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \left (\frac {512 x}{3}-\frac {611}{3} \int \frac {1}{\frac {3}{5} \cos (c+d x)+1}dx\right )-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \left (\frac {512 x}{3}-\frac {611}{3} \int \frac {1}{\frac {3}{5} \sin \left (c+d x+\frac {\pi }{2}\right )+1}dx\right )-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

\(\Big \downarrow \) 3136

\(\displaystyle \frac {1}{96} \left (\frac {1}{48} \left (\frac {512 x}{3}-\frac {611}{3} \left (\frac {5 x}{4}-\frac {5 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)+3}\right )}{2 d}\right )\right )-\frac {125 \tan (c+d x)}{48 d (5 \sec (c+d x)+3)}\right )-\frac {25 \tan (c+d x)}{96 d (5 \sec (c+d x)+3)^2}\)

input

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

output

(-25*Tan[c + d*x])/(96*d*(3 + 5*Sec[c + d*x])^2) + (((512*x)/3 - (611*((5* 
x)/4 - (5*ArcTan[Sin[c + d*x]/(3 + Cos[c + d*x])])/(2*d)))/3)/48 - (125*Ta 
n[c + d*x])/(48*d*(3 + 5*Sec[c + d*x])))/96
 

3.6.25.3.1 Defintions of rubi rules used

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]
 

3.6.25.4 Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {-\frac {5 \left (-\frac {285 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128}+\frac {165 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}\right )}{108 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{2}}-\frac {3055 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{27648}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{27}}{d}\)	\(74\)
default	\(\frac {-\frac {5 \left (-\frac {285 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128}+\frac {165 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}\right )}{108 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{2}}-\frac {3055 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{27648}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{27}}{d}\)	\(74\)
risch	\(\frac {x}{27}-\frac {25 i \left (185 \,{\mathrm e}^{3 i \left (d x +c \right )}+413 \,{\mathrm e}^{2 i \left (d x +c \right )}+235 \,{\mathrm e}^{i \left (d x +c \right )}+63\right )}{2304 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )^{2}}+\frac {3055 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {1}{3}\right )}{55296 d}-\frac {3055 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}+3\right )}{55296 d}\)	\(108\)
parallelrisch	\(\frac {3055 i \left (59+9 \cos \left (2 d x +2 c \right )+60 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 i\right )+3055 i \left (-59-9 \cos \left (2 d x +2 c \right )-60 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 i\right )+122880 d x \cos \left (d x +c \right )+18432 d x \cos \left (2 d x +2 c \right )+120832 d x -15000 \sin \left (d x +c \right )-18900 \sin \left (2 d x +2 c \right )}{55296 d \left (59+9 \cos \left (2 d x +2 c \right )+60 \cos \left (d x +c \right )\right )}\)	\(150\)

input

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

output

1/d*(-5/108*(-285/128*tan(1/2*d*x+1/2*c)^3+165/32*tan(1/2*d*x+1/2*c))/(tan 
(1/2*d*x+1/2*c)^2+4)^2-3055/27648*arctan(1/2*tan(1/2*d*x+1/2*c))+2/27*arct 
an(tan(1/2*d*x+1/2*c)))
 

3.6.25.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx=\frac {18432 \, d x \cos \left (d x + c\right )^{2} + 61440 \, d x \cos \left (d x + c\right ) + 51200 \, d x + 3055 \, {\left (9 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 25\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) + 3}{4 \, \sin \left (d x + c\right )}\right ) - 300 \, {\left (63 \, \cos \left (d x + c\right ) + 25\right )} \sin \left (d x + c\right )}{55296 \, {\left (9 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 25 \, d\right )}} \]

input

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

output

1/55296*(18432*d*x*cos(d*x + c)^2 + 61440*d*x*cos(d*x + c) + 51200*d*x + 3 
055*(9*cos(d*x + c)^2 + 30*cos(d*x + c) + 25)*arctan(1/4*(5*cos(d*x + c) + 
 3)/sin(d*x + c)) - 300*(63*cos(d*x + c) + 25)*sin(d*x + c))/(9*d*cos(d*x 
+ c)^2 + 30*d*cos(d*x + c) + 25*d)
 

3.6.25.6 Sympy [F]

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

input

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

output

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

3.6.25.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.62 \[ \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx=-\frac {\frac {150 \, {\left (\frac {44 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {19 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 16} - 2048 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 3055 \, \arctan \left (\frac {\sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}{27648 \, d} \]

input

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

output

-1/27648*(150*(44*sin(d*x + c)/(cos(d*x + c) + 1) - 19*sin(d*x + c)^3/(cos 
(d*x + c) + 1)^3)/(8*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/ 
(cos(d*x + c) + 1)^4 + 16) - 2048*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) 
+ 3055*arctan(1/2*sin(d*x + c)/(cos(d*x + c) + 1)))/d
 

3.6.25.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx=-\frac {1007 \, d x + 1007 \, c - \frac {300 \, {\left (19 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 4\right )}^{2}} - 6110 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 3}\right )}{55296 \, d} \]

input

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

output

-1/55296*(1007*d*x + 1007*c - 300*(19*tan(1/2*d*x + 1/2*c)^3 - 44*tan(1/2* 
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 4)^2 - 6110*arctan(sin(d*x + c)/(c 
os(d*x + c) + 3)))/d
 

3.6.25.9 Mupad [B] (verification not implemented)

Time = 14.55 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(3+5 \sec (c+d x))^3} \, dx=\frac {x}{27}-\frac {3055\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{27648\,d}-\frac {\frac {275\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{1152}-\frac {475\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4608}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \]

input

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

output

x/27 - (3055*atan(tan(c/2 + (d*x)/2)/2))/(27648*d) - ((275*tan(c/2 + (d*x) 
/2))/1152 - (475*tan(c/2 + (d*x)/2)^3)/4608)/(d*(8*tan(c/2 + (d*x)/2)^2 + 
tan(c/2 + (d*x)/2)^4 + 16))