Integrand size = 12, antiderivative size = 69 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=-\frac {5 x}{19652}+\frac {99 \log (5 \cos (c+d x)+3 \sin (c+d x))}{19652 d}-\frac {3}{68 d (5+3 \tan (c+d x))^2}-\frac {15}{578 d (5+3 \tan (c+d x))} \] Output:
-5/19652*x+99/19652*ln(5*cos(d*x+c)+3*sin(d*x+c))/d-3/68/d/(5+3*tan(d*x+c) )^2-15/578/d/(5+3*tan(d*x+c))
Result contains complex when optimal does not.
Time = 0.46 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.25 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=\frac {\left (\frac {1}{39304}+\frac {i}{39304}\right ) \left ((-47+52 i) \log (i-\tan (c+d x))-(52-47 i) \log (i+\tan (c+d x))+(3-3 i) \left (33 \log (5+3 \tan (c+d x))-\frac {17 (67+30 \tan (c+d x))}{(5+3 \tan (c+d x))^2}\right )\right )}{d} \] Input:
Integrate[(5 + 3*Tan[c + d*x])^(-3),x]
Output:
((1/39304 + I/39304)*((-47 + 52*I)*Log[I - Tan[c + d*x]] - (52 - 47*I)*Log [I + Tan[c + d*x]] + (3 - 3*I)*(33*Log[5 + 3*Tan[c + d*x]] - (17*(67 + 30* Tan[c + d*x]))/(5 + 3*Tan[c + d*x])^2)))/d
Time = 0.53 (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.
| \(\displaystyle \int \frac {1}{(3 \tan (c+d x)+5)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(3 \tan (c+d x)+5)^3}dx\) |
| \(\Big \downarrow \) 3964 |
| \(\displaystyle \frac {1}{34} \int \frac {5-3 \tan (c+d x)}{(3 \tan (c+d x)+5)^2}dx-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \int \frac {5-3 \tan (c+d x)}{(3 \tan (c+d x)+5)^2}dx-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{34} \int \frac {2 (8-15 \tan (c+d x))}{3 \tan (c+d x)+5}dx-\frac {15}{17 d (3 \tan (c+d x)+5)}\right )-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \int \frac {8-15 \tan (c+d x)}{3 \tan (c+d x)+5}dx-\frac {15}{17 d (3 \tan (c+d x)+5)}\right )-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \int \frac {8-15 \tan (c+d x)}{3 \tan (c+d x)+5}dx-\frac {15}{17 d (3 \tan (c+d x)+5)}\right )-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {99}{34} \int \frac {3-5 \tan (c+d x)}{3 \tan (c+d x)+5}dx-\frac {5 x}{34}\right )-\frac {15}{17 d (3 \tan (c+d x)+5)}\right )-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {99}{34} \int \frac {3-5 \tan (c+d x)}{3 \tan (c+d x)+5}dx-\frac {5 x}{34}\right )-\frac {15}{17 d (3 \tan (c+d x)+5)}\right )-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
| \(\Big \downarrow \) 4013 |
| \(\displaystyle \frac {1}{34} \left (\frac {1}{17} \left (\frac {99 \log (3 \sin (c+d x)+5 \cos (c+d x))}{34 d}-\frac {5 x}{34}\right )-\frac {15}{17 d (3 \tan (c+d x)+5)}\right )-\frac {3}{68 d (3 \tan (c+d x)+5)^2}\) |
Input:
Int[(5 + 3*Tan[c + d*x])^(-3),x]
Output:
-3/(68*d*(5 + 3*Tan[c + d*x])^2) + (((-5*x)/34 + (99*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(34*d))/17 - 15/(17*d*(5 + 3*Tan[c + d*x])))/34
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 = 0.47 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {-\frac {3}{68 \left (5+3 \tan \left (d x +c \right )\right )^{2}}-\frac {15}{578 \left (5+3 \tan \left (d x +c \right )\right )}+\frac {99 \ln \left (5+3 \tan \left (d x +c \right )\right )}{19652}-\frac {99 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{39304}-\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{19652}}{d}\) | \(69\) |
| default | \(\frac {-\frac {3}{68 \left (5+3 \tan \left (d x +c \right )\right )^{2}}-\frac {15}{578 \left (5+3 \tan \left (d x +c \right )\right )}+\frac {99 \ln \left (5+3 \tan \left (d x +c \right )\right )}{19652}-\frac {99 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{39304}-\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{19652}}{d}\) | \(69\) |
| risch | \(-\frac {5 x}{19652}-\frac {99 i x}{19652}-\frac {99 i c}{9826 d}+\frac {\left (-\frac {2241}{510952}+\frac {675 i}{510952}\right ) \left (442 \,{\mathrm e}^{2 i \left (d x +c \right )}+275+335 i\right )}{d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}+8+15 i\right )^{2}}+\frac {99 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {8}{17}+\frac {15 i}{17}\right )}{19652 d}\) | \(74\) |
| norman | \(\frac {-\frac {125 x}{19652}-\frac {75 x \tan \left (d x +c \right )}{9826}-\frac {45 x \tan \left (d x +c \right )^{2}}{19652}-\frac {201}{1156 d}-\frac {45 \tan \left (d x +c \right )}{578 d}}{\left (5+3 \tan \left (d x +c \right )\right )^{2}}+\frac {99 \ln \left (5+3 \tan \left (d x +c \right )\right )}{19652 d}-\frac {99 \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{39304 d}\) | \(87\) |
| parallelrisch | \(\frac {-810 \tan \left (d x +c \right )^{2} x d -61506+16038 \ln \left (\frac {5}{3}+\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )^{2}-8019 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )^{2}-2700 \tan \left (d x +c \right ) x d +53460 \ln \left (\frac {5}{3}+\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )-26730 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) \tan \left (d x +c \right )-2250 d x +44550 \ln \left (\frac {5}{3}+\tan \left (d x +c \right )\right )-22275 \ln \left (1+\tan \left (d x +c \right )^{2}\right )-27540 \tan \left (d x +c \right )}{353736 d \left (5+3 \tan \left (d x +c \right )\right )^{2}}\) | \(154\) |
Input:
int(1/(5+3*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-3/68/(5+3*tan(d*x+c))^2-15/578/(5+3*tan(d*x+c))+99/19652*ln(5+3*tan( d*x+c))-99/39304*ln(1+tan(d*x+c)^2)-5/19652*arctan(tan(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.74 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=-\frac {18 \, {\left (5 \, d x - 87\right )} \tan \left (d x + c\right )^{2} + 250 \, d x - 99 \, {\left (9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25\right )} \log \left (\frac {9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25}{\tan \left (d x + c\right )^{2} + 1}\right ) + 60 \, {\left (5 \, d x - 36\right )} \tan \left (d x + c\right ) + 2484}{39304 \, {\left (9 \, d \tan \left (d x + c\right )^{2} + 30 \, d \tan \left (d x + c\right ) + 25 \, d\right )}} \] Input:
integrate(1/(5+3*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-1/39304*(18*(5*d*x - 87)*tan(d*x + c)^2 + 250*d*x - 99*(9*tan(d*x + c)^2 + 30*tan(d*x + c) + 25)*log((9*tan(d*x + c)^2 + 30*tan(d*x + c) + 25)/(tan (d*x + c)^2 + 1)) + 60*(5*d*x - 36)*tan(d*x + c) + 2484)/(9*d*tan(d*x + c) ^2 + 30*d*tan(d*x + c) + 25*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}{(5+3 \tan (c+d x))^3} \, dx=\begin {cases} - \frac {90 d x \tan ^{2}{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {300 d x \tan {\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {250 d x}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} + \frac {1782 \log {\left (3 \tan {\left (c + d x \right )} + 5 \right )} \tan ^{2}{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} + \frac {5940 \log {\left (3 \tan {\left (c + d x \right )} + 5 \right )} \tan {\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} + \frac {4950 \log {\left (3 \tan {\left (c + d x \right )} + 5 \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {891 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan ^{2}{\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {2970 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {2475 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {3060 \tan {\left (c + d x \right )}}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} - \frac {6834}{353736 d \tan ^{2}{\left (c + d x \right )} + 1179120 d \tan {\left (c + d x \right )} + 982600 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (3 \tan {\left (c \right )} + 5\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(5+3*tan(d*x+c))**3,x)
Output:
Piecewise((-90*d*x*tan(c + d*x)**2/(353736*d*tan(c + d*x)**2 + 1179120*d*t an(c + d*x) + 982600*d) - 300*d*x*tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 250*d*x/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) + 1782*log(3*tan(c + d*x) + 5)*tan(c + d*x)**2/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) + 5 940*log(3*tan(c + d*x) + 5)*tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 11791 20*d*tan(c + d*x) + 982600*d) + 4950*log(3*tan(c + d*x) + 5)/(353736*d*tan (c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 891*log(tan(c + d*x)** 2 + 1)*tan(c + d*x)**2/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 2970*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 2475*log(tan(c + d*x)**2 + 1)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d) - 3060* tan(c + d*x)/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d ) - 6834/(353736*d*tan(c + d*x)**2 + 1179120*d*tan(c + d*x) + 982600*d), N e(d, 0)), (x/(3*tan(c) + 5)**3, True))
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=-\frac {10 \, d x + 10 \, c + \frac {102 \, {\left (30 \, \tan \left (d x + c\right ) + 67\right )}}{9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25} + 99 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 198 \, \log \left (3 \, \tan \left (d x + c\right ) + 5\right )}{39304 \, d} \] Input:
integrate(1/(5+3*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/39304*(10*d*x + 10*c + 102*(30*tan(d*x + c) + 67)/(9*tan(d*x + c)^2 + 3 0*tan(d*x + c) + 25) + 99*log(tan(d*x + c)^2 + 1) - 198*log(3*tan(d*x + c) + 5))/d
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=-\frac {5 \, {\left (d x + c\right )}}{19652 \, d} - \frac {99 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{39304 \, d} + \frac {99 \, \log \left ({\left | 3 \, \tan \left (d x + c\right ) + 5 \right |}\right )}{19652 \, d} - \frac {3 \, {\left (30 \, \tan \left (d x + c\right ) + 67\right )}}{1156 \, d {\left (3 \, \tan \left (d x + c\right ) + 5\right )}^{2}} \] Input:
integrate(1/(5+3*tan(d*x+c))^3,x, algorithm="giac")
Output:
-5/19652*(d*x + c)/d - 99/39304*log(tan(d*x + c)^2 + 1)/d + 99/19652*log(a bs(3*tan(d*x + c) + 5))/d - 3/1156*(30*tan(d*x + c) + 67)/(d*(3*tan(d*x + c) + 5)^2)
Time = 1.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=\frac {99\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {5}{3}\right )}{19652\,d}-\frac {\frac {5\,\mathrm {tan}\left (c+d\,x\right )}{578}+\frac {67}{3468}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2+\frac {10\,\mathrm {tan}\left (c+d\,x\right )}{3}+\frac {25}{9}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {99}{39304}+\frac {5}{39304}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-\frac {99}{39304}-\frac {5}{39304}{}\mathrm {i}\right )}{d} \] Input:
int(1/(3*tan(c + d*x) + 5)^3,x)
Output:
(99*log(tan(c + d*x) + 5/3))/(19652*d) - (log(tan(c + d*x) + 1i)*(99/39304 + 5i/39304))/d - (log(tan(c + d*x) - 1i)*(99/39304 - 5i/39304))/d - ((5*t an(c + d*x))/578 + 67/3468)/(d*((10*tan(c + d*x))/3 + tan(c + d*x)^2 + 25/ 9))
Time = 0.24 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.48 \[ \int \frac {1}{(5+3 \tan (c+d x))^3} \, dx=\frac {-891 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right )^{2}-2970 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) \tan \left (d x +c \right )-2475 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )+1782 \,\mathrm {log}\left (3 \tan \left (d x +c \right )+5\right ) \tan \left (d x +c \right )^{2}+5940 \,\mathrm {log}\left (3 \tan \left (d x +c \right )+5\right ) \tan \left (d x +c \right )+4950 \,\mathrm {log}\left (3 \tan \left (d x +c \right )+5\right )-90 \tan \left (d x +c \right )^{2} d x +918 \tan \left (d x +c \right )^{2}-300 \tan \left (d x +c \right ) d x -250 d x -4284}{39304 d \left (9 \tan \left (d x +c \right )^{2}+30 \tan \left (d x +c \right )+25\right )} \] Input:
int(1/(5+3*tan(d*x+c))^3,x)
Output:
( - 891*log(tan(c + d*x)**2 + 1)*tan(c + d*x)**2 - 2970*log(tan(c + d*x)** 2 + 1)*tan(c + d*x) - 2475*log(tan(c + d*x)**2 + 1) + 1782*log(3*tan(c + d *x) + 5)*tan(c + d*x)**2 + 5940*log(3*tan(c + d*x) + 5)*tan(c + d*x) + 495 0*log(3*tan(c + d*x) + 5) - 90*tan(c + d*x)**2*d*x + 918*tan(c + d*x)**2 - 300*tan(c + d*x)*d*x - 250*d*x - 4284)/(39304*d*(9*tan(c + d*x)**2 + 30*t an(c + d*x) + 25))