Integrand size = 12, antiderivative size = 88 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=-\frac {161 x}{334084}-\frac {60 \log (3 \cos (c+d x)+5 \sin (c+d x))}{83521 d}-\frac {5}{102 d (3+5 \tan (c+d x))^3}-\frac {15}{1156 d (3+5 \tan (c+d x))^2}-\frac {5}{19652 d (3+5 \tan (c+d x))} \]
-161/334084*x-60/83521*ln(3*cos(d*x+c)+5*sin(d*x+c))/d-5/102/d/(3+5*tan(d* x+c))^3-15/1156/d/(3+5*tan(d*x+c))^2-5/19652/d/(3+5*tan(d*x+c))
Result contains complex when optimal does not.
Time = 1.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=\frac {(720+483 i) \log (i-\tan (c+d x))+(720-483 i) \log (i+\tan (c+d x))-1440 \log (3+5 \tan (c+d x))-\frac {170 \left (1064+855 \tan (c+d x)+75 \tan ^2(c+d x)\right )}{(3+5 \tan (c+d x))^3}}{2004504 d} \]
((720 + 483*I)*Log[I - Tan[c + d*x]] + (720 - 483*I)*Log[I + Tan[c + d*x]] - 1440*Log[3 + 5*Tan[c + d*x]] - (170*(1064 + 855*Tan[c + d*x] + 75*Tan[c + d*x]^2))/(3 + 5*Tan[c + d*x])^3)/(2004504*d)
Time = 0.60 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3964, 3042, 4012, 27, 3042, 4012, 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.
| \(\displaystyle \int \frac {1}{(5 \tan (c+d x)+3)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(5 \tan (c+d x)+3)^4}dx\) |
| \(\Big \downarrow \) 3964 |
| \(\displaystyle \frac {1}{34} \int \frac {3-5 \tan (c+d x)}{(5 \tan (c+d x)+3)^3}dx-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \int \frac {3-5 \tan (c+d x)}{(5 \tan (c+d x)+3)^3}dx-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{34} \int -\frac {2 (15 \tan (c+d x)+8)}{(5 \tan (c+d x)+3)^2}dx-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{34} \left (-\frac {1}{17} \int \frac {15 \tan (c+d x)+8}{(5 \tan (c+d x)+3)^2}dx-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (-\frac {1}{17} \int \frac {15 \tan (c+d x)+8}{(5 \tan (c+d x)+3)^2}dx-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (-\frac {1}{34} \int \frac {5 \tan (c+d x)+99}{5 \tan (c+d x)+3}dx-\frac {5}{34 d (5 \tan (c+d x)+3)}\right )-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (-\frac {1}{34} \int \frac {5 \tan (c+d x)+99}{5 \tan (c+d x)+3}dx-\frac {5}{34 d (5 \tan (c+d x)+3)}\right )-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {1}{34} \left (-\frac {240}{17} \int \frac {5-3 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {161 x}{17}\right )-\frac {5}{34 d (5 \tan (c+d x)+3)}\right )-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {1}{34} \left (-\frac {240}{17} \int \frac {5-3 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {161 x}{17}\right )-\frac {5}{34 d (5 \tan (c+d x)+3)}\right )-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
| \(\Big \downarrow \) 4013 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {1}{34} \left (-\frac {240 \log (5 \sin (c+d x)+3 \cos (c+d x))}{17 d}-\frac {161 x}{17}\right )-\frac {5}{34 d (5 \tan (c+d x)+3)}\right )-\frac {15}{34 d (5 \tan (c+d x)+3)^2}\right )-\frac {5}{102 d (5 \tan (c+d x)+3)^3}\) |
-5/(102*d*(3 + 5*Tan[c + d*x])^3) + (-15/(34*d*(3 + 5*Tan[c + d*x])^2) + ( ((-161*x)/17 - (240*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(17*d))/34 - 5/( 34*d*(3 + 5*Tan[c + d*x])))/17)/34
3.5.98.3.1 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 = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {\frac {30 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{83521}-\frac {161 \arctan \left (\tan \left (d x +c \right )\right )}{334084}-\frac {5}{102 \left (3+5 \tan \left (d x +c \right )\right )^{3}}-\frac {15}{1156 \left (3+5 \tan \left (d x +c \right )\right )^{2}}-\frac {5}{19652 \left (3+5 \tan \left (d x +c \right )\right )}-\frac {60 \ln \left (3+5 \tan \left (d x +c \right )\right )}{83521}}{d}\) | \(83\) |
| default | \(\frac {\frac {30 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{83521}-\frac {161 \arctan \left (\tan \left (d x +c \right )\right )}{334084}-\frac {5}{102 \left (3+5 \tan \left (d x +c \right )\right )^{3}}-\frac {15}{1156 \left (3+5 \tan \left (d x +c \right )\right )^{2}}-\frac {5}{19652 \left (3+5 \tan \left (d x +c \right )\right )}-\frac {60 \ln \left (3+5 \tan \left (d x +c \right )\right )}{83521}}{d}\) | \(83\) |
| risch | \(-\frac {161 x}{334084}+\frac {60 i x}{83521}+\frac {120 i c}{83521 d}+\frac {\left (\frac {875}{75502984}-\frac {5825 i}{226508952}\right ) \left (391884 \,{\mathrm e}^{4 i \left (d x +c \right )}+531675 i {\mathrm e}^{2 i \left (d x +c \right )}+114393 \,{\mathrm e}^{2 i \left (d x +c \right )}-116591+67425 i\right )}{d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}-8+15 i\right )^{3}}-\frac {60 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {8}{17}+\frac {15 i}{17}\right )}{83521 d}\) | \(97\) |
| norman | \(\frac {-\frac {4347 x}{334084}-\frac {21735 x \tan \left (d x +c \right )}{334084}-\frac {36225 x \left (\tan ^{2}\left (d x +c \right )\right )}{334084}-\frac {20125 x \left (\tan ^{3}\left (d x +c \right )\right )}{334084}+\frac {166250 \left (\tan ^{3}\left (d x +c \right )\right )}{397953 d}+\frac {22325 \tan \left (d x +c \right )}{58956 d}+\frac {131875 \left (\tan ^{2}\left (d x +c \right )\right )}{176868 d}}{\left (3+5 \tan \left (d x +c \right )\right )^{3}}-\frac {60 \ln \left (3+5 \tan \left (d x +c \right )\right )}{83521 d}+\frac {30 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{83521 d}\) | \(119\) |
| parallelrisch | \(-\frac {1630125 \left (\tan ^{3}\left (d x +c \right )\right ) x d +2430000 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right ) \left (\tan ^{3}\left (d x +c \right )\right )-1215000 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{3}\left (d x +c \right )\right )+2934225 \left (\tan ^{2}\left (d x +c \right )\right ) x d +4374000 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )-2187000 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \left (\tan ^{2}\left (d x +c \right )\right )+1760535 \tan \left (d x +c \right ) x d -11305000 \left (\tan ^{3}\left (d x +c \right )\right )+2624400 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )-1312200 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )+352107 d x -20176875 \left (\tan ^{2}\left (d x +c \right )\right )+524880 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right )-262440 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )-10247175 \tan \left (d x +c \right )}{27060804 d \left (3+5 \tan \left (d x +c \right )\right )^{3}}\) | \(225\) |
1/d*(30/83521*ln(1+tan(d*x+c)^2)-161/334084*arctan(tan(d*x+c))-5/102/(3+5* tan(d*x+c))^3-15/1156/(3+5*tan(d*x+c))^2-5/19652/(3+5*tan(d*x+c))-60/83521 *ln(3+5*tan(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (78) = 156\).
Time = 0.25 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.78 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=-\frac {375 \, {\left (161 \, d x + 135\right )} \tan \left (d x + c\right )^{3} + 75 \, {\left (1449 \, d x + 1300\right )} \tan \left (d x + c\right )^{2} + 13041 \, d x + 360 \, {\left (125 \, \tan \left (d x + c\right )^{3} + 225 \, \tan \left (d x + c\right )^{2} + 135 \, \tan \left (d x + c\right ) + 27\right )} \log \left (\frac {25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9}{\tan \left (d x + c\right )^{2} + 1}\right ) + 45 \, {\left (1449 \, d x + 2830\right )} \tan \left (d x + c\right ) + 101375}{1002252 \, {\left (125 \, d \tan \left (d x + c\right )^{3} + 225 \, d \tan \left (d x + c\right )^{2} + 135 \, d \tan \left (d x + c\right ) + 27 \, d\right )}} \]
-1/1002252*(375*(161*d*x + 135)*tan(d*x + c)^3 + 75*(1449*d*x + 1300)*tan( d*x + c)^2 + 13041*d*x + 360*(125*tan(d*x + c)^3 + 225*tan(d*x + c)^2 + 13 5*tan(d*x + c) + 27)*log((25*tan(d*x + c)^2 + 30*tan(d*x + c) + 9)/(tan(d* x + c)^2 + 1)) + 45*(1449*d*x + 2830)*tan(d*x + c) + 101375)/(125*d*tan(d* x + c)^3 + 225*d*tan(d*x + c)^2 + 135*d*tan(d*x + c) + 27*d)
Leaf count of result is larger than twice the leaf count of optimal. 790 vs. \(2 (78) = 156\).
Time = 0.47 (sec) , antiderivative size = 790, normalized size of antiderivative = 8.98 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=\text {Too large to display} \]
Piecewise((-60375*d*x*tan(c + d*x)**3/(125281500*d*tan(c + d*x)**3 + 22550 6700*d*tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) - 108675*d *x*tan(c + d*x)**2/(125281500*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x) **2 + 135304020*d*tan(c + d*x) + 27060804*d) - 65205*d*x*tan(c + d*x)/(125 281500*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) - 13041*d*x/(125281500*d*tan(c + d*x)**3 + 225506700 *d*tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) - 90000*log(5* tan(c + d*x) + 3)*tan(c + d*x)**3/(125281500*d*tan(c + d*x)**3 + 225506700 *d*tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) - 162000*log(5 *tan(c + d*x) + 3)*tan(c + d*x)**2/(125281500*d*tan(c + d*x)**3 + 22550670 0*d*tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) - 97200*log(5 *tan(c + d*x) + 3)*tan(c + d*x)/(125281500*d*tan(c + d*x)**3 + 225506700*d *tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) - 19440*log(5*ta n(c + d*x) + 3)/(125281500*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)**2 + 135304020*d*tan(c + d*x) + 27060804*d) + 45000*log(tan(c + d*x)**2 + 1) *tan(c + d*x)**3/(125281500*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)** 2 + 135304020*d*tan(c + d*x) + 27060804*d) + 81000*log(tan(c + d*x)**2 + 1 )*tan(c + d*x)**2/(125281500*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)* *2 + 135304020*d*tan(c + d*x) + 27060804*d) + 48600*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(125281500*d*tan(c + d*x)**3 + 225506700*d*tan(c + d*x)...
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=-\frac {483 \, d x + 483 \, c + \frac {85 \, {\left (75 \, \tan \left (d x + c\right )^{2} + 855 \, \tan \left (d x + c\right ) + 1064\right )}}{125 \, \tan \left (d x + c\right )^{3} + 225 \, \tan \left (d x + c\right )^{2} + 135 \, \tan \left (d x + c\right ) + 27} - 360 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 720 \, \log \left (5 \, \tan \left (d x + c\right ) + 3\right )}{1002252 \, d} \]
-1/1002252*(483*d*x + 483*c + 85*(75*tan(d*x + c)^2 + 855*tan(d*x + c) + 1 064)/(125*tan(d*x + c)^3 + 225*tan(d*x + c)^2 + 135*tan(d*x + c) + 27) - 3 60*log(tan(d*x + c)^2 + 1) + 720*log(5*tan(d*x + c) + 3))/d
Time = 0.37 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=-\frac {483 \, d x + 483 \, c - \frac {25 \, {\left (6600 \, \tan \left (d x + c\right )^{3} + 11625 \, \tan \left (d x + c\right )^{2} + 4221 \, \tan \left (d x + c\right ) - 2192\right )}}{{\left (5 \, \tan \left (d x + c\right ) + 3\right )}^{3}} - 360 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 720 \, \log \left ({\left | 5 \, \tan \left (d x + c\right ) + 3 \right |}\right )}{1002252 \, d} \]
-1/1002252*(483*d*x + 483*c - 25*(6600*tan(d*x + c)^3 + 11625*tan(d*x + c) ^2 + 4221*tan(d*x + c) - 2192)/(5*tan(d*x + c) + 3)^3 - 360*log(tan(d*x + c)^2 + 1) + 720*log(abs(5*tan(d*x + c) + 3)))/d
Time = 5.45 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(3+5 \tan (c+d x))^4} \, dx=-\frac {60\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {3}{5}\right )}{83521\,d}-\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{19652}+\frac {57\,\mathrm {tan}\left (c+d\,x\right )}{98260}+\frac {266}{368475}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^3+\frac {9\,{\mathrm {tan}\left (c+d\,x\right )}^2}{5}+\frac {27\,\mathrm {tan}\left (c+d\,x\right )}{25}+\frac {27}{125}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (\frac {30}{83521}+\frac {161}{668168}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (\frac {30}{83521}-\frac {161}{668168}{}\mathrm {i}\right )}{d} \]