3.6.29 \(\int \frac {1}{(5+3 \sec (c+d x))^3} \, dx\) [529]

3.6.29.1 Optimal result

Integrand size = 12, antiderivative size = 120 \[ \int \frac {1}{(5+3 \sec (c+d x))^3} \, dx=\frac {x}{125}+\frac {8361 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{256000 d}-\frac {8361 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{256000 d}+\frac {9 \tan (c+d x)}{160 d (5+3 \sec (c+d x))^2}+\frac {963 \tan (c+d x)}{12800 d (5+3 \sec (c+d x))} \]

output

1/125*x+8361/256000*ln(2*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c))/d-8361/256 
000*ln(2*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d+9/160*tan(d*x+c)/d/(5+3* 
sec(d*x+c))^2+963/12800*tan(d*x+c)/d/(5+3*sec(d*x+c))
 

3.6.29.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(120)=240\).

Time = 0.37 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.01 \[ \int \frac {1}{(5+3 \sec (c+d x))^3} \, dx=\frac {88064 c+88064 d x+359523 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+60 \cos (c+d x) \left (2048 (c+d x)+8361 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8361 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+25 \cos (2 (c+d x)) \left (2048 (c+d x)+8361 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-8361 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-359523 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+115560 \sin (c+d x)+110700 \sin (2 (c+d x))}{512000 d (3+5 \cos (c+d x))^2} \]

input

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

output

(88064*c + 88064*d*x + 359523*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 
 60*Cos[c + d*x]*(2048*(c + d*x) + 8361*Log[2*Cos[(c + d*x)/2] - Sin[(c + 
d*x)/2]] - 8361*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 25*Cos[2*(c 
+ d*x)]*(2048*(c + d*x) + 8361*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] 
- 8361*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 359523*Log[2*Cos[(c + 
 d*x)/2] + Sin[(c + d*x)/2]] + 115560*Sin[c + d*x] + 110700*Sin[2*(c + d*x 
)])/(512000*d*(3 + 5*Cos[c + d*x])^2)
 

3.6.29.3 Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {3042, 4272, 25, 3042, 4548, 25, 3042, 4407, 3042, 4318, 3042, 3138, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

\(\displaystyle \frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}-\frac {1}{160} \int -\frac {9 \sec ^2(c+d x)-30 \sec (c+d x)+32}{(3 \sec (c+d x)+5)^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{160} \int \frac {9 \sec ^2(c+d x)-30 \sec (c+d x)+32}{(3 \sec (c+d x)+5)^2}dx+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{160} \int \frac {9 \csc \left (c+d x+\frac {\pi }{2}\right )^2-30 \csc \left (c+d x+\frac {\pi }{2}\right )+32}{\left (3 \csc \left (c+d x+\frac {\pi }{2}\right )+5\right )^2}dx+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 4548

\(\displaystyle \frac {1}{160} \left (\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}-\frac {1}{80} \int -\frac {512-1365 \sec (c+d x)}{3 \sec (c+d x)+5}dx\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \int \frac {512-1365 \sec (c+d x)}{3 \sec (c+d x)+5}dx+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \int \frac {512-1365 \csc \left (c+d x+\frac {\pi }{2}\right )}{3 \csc \left (c+d x+\frac {\pi }{2}\right )+5}dx+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \left (\frac {512 x}{5}-\frac {8361}{5} \int \frac {\sec (c+d x)}{3 \sec (c+d x)+5}dx\right )+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \left (\frac {512 x}{5}-\frac {8361}{5} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{3 \csc \left (c+d x+\frac {\pi }{2}\right )+5}dx\right )+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \left (\frac {512 x}{5}-\frac {2787}{5} \int \frac {1}{\frac {5}{3} \cos (c+d x)+1}dx\right )+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \left (\frac {512 x}{5}-\frac {2787}{5} \int \frac {1}{\frac {5}{3} \sin \left (c+d x+\frac {\pi }{2}\right )+1}dx\right )+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \left (\frac {512 x}{5}-\frac {5574 \int \frac {1}{\frac {8}{3}-\frac {2}{3} \tan ^2\left (\frac {1}{2} (c+d x)\right )}d\tan \left (\frac {1}{2} (c+d x)\right )}{5 d}\right )+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{160} \left (\frac {1}{80} \left (\frac {512 x}{5}-\frac {8361 \text {arctanh}\left (\frac {1}{2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{10 d}\right )+\frac {963 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\right )+\frac {9 \tan (c+d x)}{160 d (3 \sec (c+d x)+5)^2}\)

input

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

output

(9*Tan[c + d*x])/(160*d*(5 + 3*Sec[c + d*x])^2) + (((512*x)/5 - (8361*ArcT 
anh[Tan[(c + d*x)/2]/2])/(10*d))/80 + (963*Tan[c + d*x])/(80*d*(5 + 3*Sec[ 
c + d*x])))/160
 

3.6.29.3.1 Defintions of rubi rules used

rule 25

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

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

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]
 

3.6.29.4 Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {-\frac {27}{2560 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )^{2}}-\frac {1323}{25600 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {8361 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{256000}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{125}+\frac {27}{2560 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )^{2}}-\frac {1323}{25600 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {8361 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{256000}}{d}\)	\(106\)
default	\(\frac {-\frac {27}{2560 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )^{2}}-\frac {1323}{25600 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {8361 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{256000}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{125}+\frac {27}{2560 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )^{2}}-\frac {1323}{25600 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {8361 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{256000}}{d}\)	\(106\)
risch	\(\frac {x}{125}+\frac {27 i \left (695 \,{\mathrm e}^{3 i \left (d x +c \right )}+1763 \,{\mathrm e}^{2 i \left (d x +c \right )}+1765 \,{\mathrm e}^{i \left (d x +c \right )}+1025\right )}{32000 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )^{2}}+\frac {8361 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}-\frac {4 i}{5}\right )}{256000 d}-\frac {8361 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}+\frac {4 i}{5}\right )}{256000 d}\)	\(110\)
norman	\(\frac {\frac {16 x}{125}+\frac {1053 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3200 d}-\frac {1323 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12800 d}-\frac {8 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{125}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{125}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-4\right )^{2}}+\frac {8361 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{256000 d}-\frac {8361 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{256000 d}\)	\(114\)
parallelrisch	\(\frac {\left (501660 \cos \left (d x +c \right )+209025 \cos \left (2 d x +2 c \right )+359523\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )+\left (-501660 \cos \left (d x +c \right )-209025 \cos \left (2 d x +2 c \right )-359523\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )+122880 d x \cos \left (d x +c \right )+51200 d x \cos \left (2 d x +2 c \right )+88064 d x +115560 \sin \left (d x +c \right )+110700 \sin \left (2 d x +2 c \right )}{256000 d \left (43+25 \cos \left (2 d x +2 c \right )+60 \cos \left (d x +c \right )\right )}\)	\(144\)

input

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

output

1/d*(-27/2560/(tan(1/2*d*x+1/2*c)-2)^2-1323/25600/(tan(1/2*d*x+1/2*c)-2)+8 
361/256000*ln(tan(1/2*d*x+1/2*c)-2)+2/125*arctan(tan(1/2*d*x+1/2*c))+27/25 
60/(tan(1/2*d*x+1/2*c)+2)^2-1323/25600/(tan(1/2*d*x+1/2*c)+2)-8361/256000* 
ln(tan(1/2*d*x+1/2*c)+2))
 

3.6.29.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(5+3 \sec (c+d x))^3} \, dx=\frac {102400 \, d x \cos \left (d x + c\right )^{2} + 122880 \, d x \cos \left (d x + c\right ) + 36864 \, d x - 8361 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 8361 \, {\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 1080 \, {\left (205 \, \cos \left (d x + c\right ) + 107\right )} \sin \left (d x + c\right )}{512000 \, {\left (25 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 9 \, d\right )}} \]

input

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

output

1/512000*(102400*d*x*cos(d*x + c)^2 + 122880*d*x*cos(d*x + c) + 36864*d*x 
- 8361*(25*cos(d*x + c)^2 + 30*cos(d*x + c) + 9)*log(3/2*cos(d*x + c) + 2* 
sin(d*x + c) + 5/2) + 8361*(25*cos(d*x + c)^2 + 30*cos(d*x + c) + 9)*log(3 
/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2) + 1080*(205*cos(d*x + c) + 107)*si 
n(d*x + c))/(25*d*cos(d*x + c)^2 + 30*d*cos(d*x + c) + 9*d)
 

3.6.29.6 Sympy [F]

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

input

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

output

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

3.6.29.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.29 \[ \int \frac {1}{(5+3 \sec (c+d x))^3} \, dx=-\frac {\frac {540 \, {\left (\frac {156 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {49 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 16} - 4096 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 8361 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 8361 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{256000 \, d} \]

input

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

output

-1/256000*(540*(156*sin(d*x + c)/(cos(d*x + c) + 1) - 49*sin(d*x + c)^3/(c 
os(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) - 4096*arctan(sin(d*x + c)/(cos(d*x + c) + 1) 
) + 8361*log(sin(d*x + c)/(cos(d*x + c) + 1) + 2) - 8361*log(sin(d*x + c)/ 
(cos(d*x + c) + 1) - 2))/d
 

3.6.29.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(5+3 \sec (c+d x))^3} \, dx=\frac {2048 \, d x + 2048 \, c - \frac {540 \, {\left (49 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 156 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4\right )}^{2}} - 8361 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) + 8361 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{256000 \, d} \]

input

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

output

1/256000*(2048*d*x + 2048*c - 540*(49*tan(1/2*d*x + 1/2*c)^3 - 156*tan(1/2 
*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 4)^2 - 8361*log(abs(tan(1/2*d*x + 
 1/2*c) + 2)) + 8361*log(abs(tan(1/2*d*x + 1/2*c) - 2)))/d
 

3.6.29.9 Mupad [B] (verification not implemented)

Time = 14.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(5+3 \sec (c+d x))^3} \, dx=\frac {x}{125}-\frac {8361\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{128000\,d}+\frac {\frac {1053\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3200}-\frac {1323\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12800}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+16\right )} \]

input

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

output

x/125 - (8361*atanh(tan(c/2 + (d*x)/2)/2))/(128000*d) + ((1053*tan(c/2 + ( 
d*x)/2))/3200 - (1323*tan(c/2 + (d*x)/2)^3)/12800)/(d*(tan(c/2 + (d*x)/2)^ 
4 - 8*tan(c/2 + (d*x)/2)^2 + 16))