Integrand size = 12, antiderivative size = 95 \[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=\frac {x}{25}+\frac {123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}-\frac {123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{1600 d}+\frac {9 \tan (c+d x)}{80 d (5+3 \sec (c+d x))} \]
1/25*x+123/1600*ln(2*cos(1/2*d*x+1/2*c)-sin(1/2*d*x+1/2*c))/d-123/1600*ln( 2*cos(1/2*d*x+1/2*c)+sin(1/2*d*x+1/2*c))/d+9/80*tan(d*x+c)/d/(5+3*sec(d*x+ c))
Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=\frac {5 \cos (c+d x) \left (64 (c+d x)+123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 \left (64 c+64 d x+123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-123 \log \left (2 \cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+60 \sin (c+d x)\right )}{1600 d (3+5 \cos (c+d x))} \]
(5*Cos[c + d*x]*(64*(c + d*x) + 123*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x) /2]] - 123*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 3*(64*c + 64*d*x + 123*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 123*Log[2*Cos[(c + d*x) /2] + Sin[(c + d*x)/2]] + 60*Sin[c + d*x]))/(1600*d*(3 + 5*Cos[c + d*x]))
Time = 0.45 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.61, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 4272, 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.
| \(\displaystyle \int \frac {1}{(3 \sec (c+d x)+5)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (3 \csc \left (c+d x+\frac {\pi }{2}\right )+5\right )^2}dx\) |
| \(\Big \downarrow \) 4272 |
| \(\displaystyle \frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}-\frac {1}{80} \int -\frac {16-15 \sec (c+d x)}{3 \sec (c+d x)+5}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{80} \int \frac {16-15 \sec (c+d x)}{3 \sec (c+d x)+5}dx+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{80} \int \frac {16-15 \csc \left (c+d x+\frac {\pi }{2}\right )}{3 \csc \left (c+d x+\frac {\pi }{2}\right )+5}dx+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {1}{80} \left (\frac {16 x}{5}-\frac {123}{5} \int \frac {\sec (c+d x)}{3 \sec (c+d x)+5}dx\right )+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{80} \left (\frac {16 x}{5}-\frac {123}{5} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{3 \csc \left (c+d x+\frac {\pi }{2}\right )+5}dx\right )+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {1}{80} \left (\frac {16 x}{5}-\frac {41}{5} \int \frac {1}{\frac {5}{3} \cos (c+d x)+1}dx\right )+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{80} \left (\frac {16 x}{5}-\frac {41}{5} \int \frac {1}{\frac {5}{3} \sin \left (c+d x+\frac {\pi }{2}\right )+1}dx\right )+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {1}{80} \left (\frac {16 x}{5}-\frac {82 \int \frac {1}{\frac {8}{3}-\frac {2}{3} \tan ^2\left (\frac {1}{2} (c+d x)\right )}d\tan \left (\frac {1}{2} (c+d x)\right )}{5 d}\right )+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{80} \left (\frac {16 x}{5}-\frac {123 \text {arctanh}\left (\frac {1}{2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{10 d}\right )+\frac {9 \tan (c+d x)}{80 d (3 \sec (c+d x)+5)}\) |
((16*x)/5 - (123*ArcTanh[Tan[(c + d*x)/2]/2])/(10*d))/80 + (9*Tan[c + d*x] )/(80*d*(5 + 3*Sec[c + d*x]))
3.6.28.3.1 Defintions of rubi rules used
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])
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]
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]
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]
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]
Time = 0.39 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{25}-\frac {9}{160 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{1600}-\frac {9}{160 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{1600}}{d}\) | \(76\) |
| default | \(\frac {\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{25}-\frac {9}{160 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}-\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{1600}-\frac {9}{160 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}+\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{1600}}{d}\) | \(76\) |
| norman | \(\frac {-\frac {4 x}{25}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{80 d}+\frac {x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{25}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-4}+\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{1600 d}-\frac {123 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{1600 d}\) | \(84\) |
| risch | \(\frac {x}{25}+\frac {9 i \left (3 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}{200 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}-\frac {123 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}+\frac {4 i}{5}\right )}{1600 d}+\frac {123 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {3}{5}-\frac {4 i}{5}\right )}{1600 d}\) | \(88\) |
| parallelrisch | \(\frac {320 d x \cos \left (d x +c \right )+615 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right ) \cos \left (d x +c \right )-615 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right ) \cos \left (d x +c \right )+192 d x +180 \sin \left (d x +c \right )+369 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )-369 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{8000 d \cos \left (d x +c \right )+4800 d}\) | \(108\) |
1/d*(2/25*arctan(tan(1/2*d*x+1/2*c))-9/160/(tan(1/2*d*x+1/2*c)+2)-123/1600 *ln(tan(1/2*d*x+1/2*c)+2)-9/160/(tan(1/2*d*x+1/2*c)-2)+123/1600*ln(tan(1/2 *d*x+1/2*c)-2))
Time = 0.27 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=\frac {640 \, d x \cos \left (d x + c\right ) + 384 \, d x - 123 \, {\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 123 \, {\left (5 \, \cos \left (d x + c\right ) + 3\right )} \log \left (\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 360 \, \sin \left (d x + c\right )}{3200 \, {\left (5 \, d \cos \left (d x + c\right ) + 3 \, d\right )}} \]
1/3200*(640*d*x*cos(d*x + c) + 384*d*x - 123*(5*cos(d*x + c) + 3)*log(3/2* cos(d*x + c) + 2*sin(d*x + c) + 5/2) + 123*(5*cos(d*x + c) + 3)*log(3/2*co s(d*x + c) - 2*sin(d*x + c) + 5/2) + 360*sin(d*x + c))/(5*d*cos(d*x + c) + 3*d)
\[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=\int \frac {1}{\left (3 \sec {\left (c + d x \right )} + 5\right )^{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.17 \[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=-\frac {\frac {180 \, \sin \left (d x + c\right )}{{\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 4\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - 128 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right ) + 123 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 123 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{1600 \, d} \]
-1/1600*(180*sin(d*x + c)/((sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4)*(cos( d*x + c) + 1)) - 128*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) + 123*log(sin (d*x + c)/(cos(d*x + c) + 1) + 2) - 123*log(sin(d*x + c)/(cos(d*x + c) + 1 ) - 2))/d
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=\frac {64 \, d x + 64 \, c - \frac {180 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 4} - 123 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right ) + 123 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{1600 \, d} \]
1/1600*(64*d*x + 64*c - 180*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 4) - 123*log(abs(tan(1/2*d*x + 1/2*c) + 2)) + 123*log(abs(tan(1/2*d*x + 1 /2*c) - 2)))/d
Time = 14.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(5+3 \sec (c+d x))^2} \, dx=\frac {x}{25}-\frac {\frac {123\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}\right )}{800}+\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{80\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\right )}}{d} \]