Integrand size = 12, antiderivative size = 69 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=-\frac {99 x}{19652}+\frac {5 \log (3 \cos (c+d x)+5 \sin (c+d x))}{19652 d}-\frac {5}{68 d (3+5 \tan (c+d x))^2}-\frac {15}{578 d (3+5 \tan (c+d x))} \]
-99/19652*x+5/19652*ln(3*cos(d*x+c)+5*sin(d*x+c))/d-5/68/d/(3+5*tan(d*x+c) )^2-15/578/d/(3+5*tan(d*x+c))
Result contains complex when optimal does not.
Time = 0.64 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{39304}+\frac {i}{39304}\right ) \left ((47+52 i) \log (i-\tan (c+d x))-(52+47 i) \log (i+\tan (c+d x))+(5-5 i) \left (\log (3+5 \tan (c+d x))-\frac {85 (7+6 \tan (c+d x))}{(3+5 \tan (c+d x))^2}\right )\right )}{d} \]
((1/39304 + I/39304)*((47 + 52*I)*Log[I - Tan[c + d*x]] - (52 + 47*I)*Log[ I + Tan[c + d*x]] + (5 - 5*I)*(Log[3 + 5*Tan[c + d*x]] - (85*(7 + 6*Tan[c + d*x]))/(3 + 5*Tan[c + d*x])^2)))/d
Time = 0.50 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3964, 3042, 4012, 27, 3042, 4014, 3042, 4013}
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.
| \(\displaystyle \int \frac {1}{(5 \tan (c+d x)+3)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(5 \tan (c+d x)+3)^3}dx\) |
| \(\Big \downarrow \) 3964 |
| \(\displaystyle \frac {1}{34} \int \frac {3-5 \tan (c+d x)}{(5 \tan (c+d x)+3)^2}dx-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \int \frac {3-5 \tan (c+d x)}{(5 \tan (c+d x)+3)^2}dx-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{34} \int -\frac {2 (15 \tan (c+d x)+8)}{5 \tan (c+d x)+3}dx-\frac {15}{17 d (5 \tan (c+d x)+3)}\right )-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{34} \left (-\frac {1}{17} \int \frac {15 \tan (c+d x)+8}{5 \tan (c+d x)+3}dx-\frac {15}{17 d (5 \tan (c+d x)+3)}\right )-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (-\frac {1}{17} \int \frac {15 \tan (c+d x)+8}{5 \tan (c+d x)+3}dx-\frac {15}{17 d (5 \tan (c+d x)+3)}\right )-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {5}{34} \int \frac {5-3 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {99 x}{34}\right )-\frac {15}{17 d (5 \tan (c+d x)+3)}\right )-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {5}{34} \int \frac {5-3 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {99 x}{34}\right )-\frac {15}{17 d (5 \tan (c+d x)+3)}\right )-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
| \(\Big \downarrow \) 4013 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {5 \log (5 \sin (c+d x)+3 \cos (c+d x))}{34 d}-\frac {99 x}{34}\right )-\frac {15}{17 d (5 \tan (c+d x)+3)}\right )-\frac {5}{68 d (5 \tan (c+d x)+3)^2}\) |
-5/(68*d*(3 + 5*Tan[c + d*x])^2) + (((-99*x)/34 + (5*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(34*d))/17 - 15/(17*d*(3 + 5*Tan[c + d*x])))/34
3.5.97.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* (x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si n[e + f*x], x]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Time = 0.30 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {-\frac {5}{68 \left (3+5 \tan \left (d x +c \right )\right )^{2}}-\frac {15}{578 \left (3+5 \tan \left (d x +c \right )\right )}+\frac {5 \ln \left (3+5 \tan \left (d x +c \right )\right )}{19652}-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{39304}-\frac {99 \arctan \left (\tan \left (d x +c \right )\right )}{19652}}{d}\) | \(69\) |
| default | \(\frac {-\frac {5}{68 \left (3+5 \tan \left (d x +c \right )\right )^{2}}-\frac {15}{578 \left (3+5 \tan \left (d x +c \right )\right )}+\frac {5 \ln \left (3+5 \tan \left (d x +c \right )\right )}{19652}-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{39304}-\frac {99 \arctan \left (\tan \left (d x +c \right )\right )}{19652}}{d}\) | \(69\) |
| risch | \(-\frac {99 x}{19652}-\frac {5 i x}{19652}-\frac {5 i c}{9826 d}+\frac {\left (-\frac {2375}{2083112}-\frac {3675 i}{2083112}\right ) \left (1802 \,{\mathrm e}^{2 i \left (d x +c \right )}+27+1575 i\right )}{d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}-8+15 i\right )^{2}}+\frac {5 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {8}{17}+\frac {15 i}{17}\right )}{19652 d}\) | \(74\) |
| norman | \(\frac {-\frac {891 x}{19652}-\frac {1485 x \tan \left (d x +c \right )}{9826}-\frac {2475 x \left (\tan ^{2}\left (d x +c \right )\right )}{19652}-\frac {175}{1156 d}-\frac {75 \tan \left (d x +c \right )}{578 d}}{\left (3+5 \tan \left (d x +c \right )\right )^{2}}+\frac {5 \ln \left (3+5 \tan \left (d x +c \right )\right )}{19652 d}-\frac {5 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{39304 d}\) | \(87\) |
| parallelrisch | \(\frac {-123750 \left (\tan ^{2}\left (d x +c \right )\right ) x d -148750+6250 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )-3125 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )-148500 \tan \left (d x +c \right ) x d +7500 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )-3750 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )-44550 d x +2250 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right )-1125 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )-127500 \tan \left (d x +c \right )}{982600 d \left (3+5 \tan \left (d x +c \right )\right )^{2}}\) | \(154\) |
1/d*(-5/68/(3+5*tan(d*x+c))^2-15/578/(3+5*tan(d*x+c))+5/19652*ln(3+5*tan(d *x+c))-5/39304*ln(1+tan(d*x+c)^2)-99/19652*arctan(tan(d*x+c)))
Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.74 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=-\frac {50 \, {\left (99 \, d x - 25\right )} \tan \left (d x + c\right )^{2} + 1782 \, d x - 5 \, {\left (25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9\right )} \log \left (\frac {25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9}{\tan \left (d x + c\right )^{2} + 1}\right ) + 180 \, {\left (33 \, d x + 20\right )} \tan \left (d x + c\right ) + 5500}{39304 \, {\left (25 \, d \tan \left (d x + c\right )^{2} + 30 \, d \tan \left (d x + c\right ) + 9 \, d\right )}} \]
-1/39304*(50*(99*d*x - 25)*tan(d*x + c)^2 + 1782*d*x - 5*(25*tan(d*x + c)^ 2 + 30*tan(d*x + c) + 9)*log((25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9)/(ta n(d*x + c)^2 + 1)) + 180*(33*d*x + 20)*tan(d*x + c) + 5500)/(25*d*tan(d*x + c)^2 + 30*d*tan(d*x + c) + 9*d)
Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (60) = 120\).
Time = 0.38 (sec) , antiderivative size = 442, normalized size of antiderivative = 6.41 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=\begin {cases} - \frac {4950 d x \tan ^{2}{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {5940 d x \tan {\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {1782 d x}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} + \frac {250 \log {\left (5 \tan {\left (c + d x \right )} + 3 \right )} \tan ^{2}{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} + \frac {300 \log {\left (5 \tan {\left (c + d x \right )} + 3 \right )} \tan {\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} + \frac {90 \log {\left (5 \tan {\left (c + d x \right )} + 3 \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {125 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan ^{2}{\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {150 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {45 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {5100 \tan {\left (c + d x \right )}}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} - \frac {5950}{982600 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 353736 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (5 \tan {\left (c \right )} + 3\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((-4950*d*x*tan(c + d*x)**2/(982600*d*tan(c + d*x)**2 + 1179120*d *tan(c + d*x) + 353736*d) - 5940*d*x*tan(c + d*x)/(982600*d*tan(c + d*x)** 2 + 1179120*d*tan(c + d*x) + 353736*d) - 1782*d*x/(982600*d*tan(c + d*x)** 2 + 1179120*d*tan(c + d*x) + 353736*d) + 250*log(5*tan(c + d*x) + 3)*tan(c + d*x)**2/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) + 300*log(5*tan(c + d*x) + 3)*tan(c + d*x)/(982600*d*tan(c + d*x)**2 + 117 9120*d*tan(c + d*x) + 353736*d) + 90*log(5*tan(c + d*x) + 3)/(982600*d*tan (c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 125*log(tan(c + d*x)** 2 + 1)*tan(c + d*x)**2/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 150*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 45*log(tan(c + d*x)**2 + 1) /(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 5100*tan (c + d*x)/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d) - 5950/(982600*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 353736*d), Ne(d , 0)), (x/(5*tan(c) + 3)**3, True))
Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=-\frac {198 \, d x + 198 \, c + \frac {850 \, {\left (6 \, \tan \left (d x + c\right ) + 7\right )}}{25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9} + 5 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 10 \, \log \left (5 \, \tan \left (d x + c\right ) + 3\right )}{39304 \, d} \]
-1/39304*(198*d*x + 198*c + 850*(6*tan(d*x + c) + 7)/(25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9) + 5*log(tan(d*x + c)^2 + 1) - 10*log(5*tan(d*x + c) + 3))/d
Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=-\frac {198 \, d x + 198 \, c + \frac {5 \, {\left (75 \, \tan \left (d x + c\right )^{2} + 1110 \, \tan \left (d x + c\right ) + 1217\right )}}{{\left (5 \, \tan \left (d x + c\right ) + 3\right )}^{2}} + 5 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 10 \, \log \left ({\left | 5 \, \tan \left (d x + c\right ) + 3 \right |}\right )}{39304 \, d} \]
-1/39304*(198*d*x + 198*c + 5*(75*tan(d*x + c)^2 + 1110*tan(d*x + c) + 121 7)/(5*tan(d*x + c) + 3)^2 + 5*log(tan(d*x + c)^2 + 1) - 10*log(abs(5*tan(d *x + c) + 3)))/d
Time = 4.44 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(3+5 \tan (c+d x))^3} \, dx=\frac {5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {3}{5}\right )}{19652\,d}-\frac {\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{578}+\frac {7}{1156}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+\frac {6\,\mathrm {tan}\left (c+d\,x\right )}{5}+\frac {9}{25}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {5}{39304}+\frac {99}{39304}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-\frac {5}{39304}-\frac {99}{39304}{}\mathrm {i}\right )}{d} \]