Integrand size = 12, antiderivative size = 50 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=\frac {4 x}{289}+\frac {15 \log (5 \cos (c+d x)+3 \sin (c+d x))}{578 d}-\frac {3}{34 d (5+3 \tan (c+d x))} \]
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=-\frac {(15+8 i) \log (i-\tan (c+d x))+(15-8 i) \log (i+\tan (c+d x))-30 \log (5+3 \tan (c+d x))+\frac {102}{5+3 \tan (c+d x)}}{1156 d} \]
-1/1156*((15 + 8*I)*Log[I - Tan[c + d*x]] + (15 - 8*I)*Log[I + Tan[c + d*x ]] - 30*Log[5 + 3*Tan[c + d*x]] + 102/(5 + 3*Tan[c + d*x]))/d
Time = 0.37 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3964, 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.
\(\displaystyle \int \frac {1}{(3 \tan (c+d x)+5)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(3 \tan (c+d x)+5)^2}dx\) |
\(\Big \downarrow \) 3964 |
\(\displaystyle \frac {1}{34} \int \frac {5-3 \tan (c+d x)}{3 \tan (c+d x)+5}dx-\frac {3}{34 d (3 \tan (c+d x)+5)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{34} \int \frac {5-3 \tan (c+d x)}{3 \tan (c+d x)+5}dx-\frac {3}{34 d (3 \tan (c+d x)+5)}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {1}{34} \left (\frac {15}{17} \int \frac {3-5 \tan (c+d x)}{3 \tan (c+d x)+5}dx+\frac {8 x}{17}\right )-\frac {3}{34 d (3 \tan (c+d x)+5)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{34} \left (\frac {15}{17} \int \frac {3-5 \tan (c+d x)}{3 \tan (c+d x)+5}dx+\frac {8 x}{17}\right )-\frac {3}{34 d (3 \tan (c+d x)+5)}\) |
\(\Big \downarrow \) 4013 |
\(\displaystyle \frac {1}{34} \left (\frac {15 \log (3 \sin (c+d x)+5 \cos (c+d x))}{17 d}+\frac {8 x}{17}\right )-\frac {3}{34 d (3 \tan (c+d x)+5)}\) |
((8*x)/17 + (15*Log[5*Cos[c + d*x] + 3*Sin[c + d*x]])/(17*d))/34 - 3/(34*d *(5 + 3*Tan[c + d*x]))
3.5.100.3.1 Defintions of rubi rules used
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[((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 = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {-\frac {3}{34 \left (5+3 \tan \left (d x +c \right )\right )}+\frac {15 \ln \left (5+3 \tan \left (d x +c \right )\right )}{578}-\frac {15 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{1156}+\frac {4 \arctan \left (\tan \left (d x +c \right )\right )}{289}}{d}\) | \(55\) |
default | \(\frac {-\frac {3}{34 \left (5+3 \tan \left (d x +c \right )\right )}+\frac {15 \ln \left (5+3 \tan \left (d x +c \right )\right )}{578}-\frac {15 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{1156}+\frac {4 \arctan \left (\tan \left (d x +c \right )\right )}{289}}{d}\) | \(55\) |
norman | \(\frac {\frac {20 x}{289}+\frac {12 x \tan \left (d x +c \right )}{289}-\frac {3}{34 d}}{5+3 \tan \left (d x +c \right )}+\frac {15 \ln \left (5+3 \tan \left (d x +c \right )\right )}{578 d}-\frac {15 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{1156 d}\) | \(65\) |
risch | \(\frac {4 x}{289}-\frac {15 i x}{578}-\frac {15 i c}{289 d}-\frac {135}{578 d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}+8+15 i\right )}+\frac {36 i}{289 d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}+8+15 i\right )}+\frac {15 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {8}{17}+\frac {15 i}{17}\right )}{578 d}\) | \(80\) |
parallelrisch | \(\frac {144 \tan \left (d x +c \right ) x d -306+270 \ln \left (\frac {5}{3}+\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )-135 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )+240 d x +450 \ln \left (\frac {5}{3}+\tan \left (d x +c \right )\right )-225 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{3468 d \left (5+3 \tan \left (d x +c \right )\right )}\) | \(94\) |
1/d*(-3/34/(5+3*tan(d*x+c))+15/578*ln(5+3*tan(d*x+c))-15/1156*ln(1+tan(d*x +c)^2)+4/289*arctan(tan(d*x+c)))
Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=\frac {80 \, d x + 15 \, {\left (3 \, \tan \left (d x + c\right ) + 5\right )} \log \left (\frac {9 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 25}{\tan \left (d x + c\right )^{2} + 1}\right ) + 3 \, {\left (16 \, d x + 15\right )} \tan \left (d x + c\right ) - 27}{1156 \, {\left (3 \, d \tan \left (d x + c\right ) + 5 \, d\right )}} \]
1/1156*(80*d*x + 15*(3*tan(d*x + c) + 5)*log((9*tan(d*x + c)^2 + 30*tan(d* x + c) + 25)/(tan(d*x + c)^2 + 1)) + 3*(16*d*x + 15)*tan(d*x + c) - 27)/(3 *d*tan(d*x + c) + 5*d)
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (42) = 84\).
Time = 0.27 (sec) , antiderivative size = 190, normalized size of antiderivative = 3.80 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=\begin {cases} \frac {48 d x \tan {\left (c + d x \right )}}{3468 d \tan {\left (c + d x \right )} + 5780 d} + \frac {80 d x}{3468 d \tan {\left (c + d x \right )} + 5780 d} + \frac {90 \log {\left (3 \tan {\left (c + d x \right )} + 5 \right )} \tan {\left (c + d x \right )}}{3468 d \tan {\left (c + d x \right )} + 5780 d} + \frac {150 \log {\left (3 \tan {\left (c + d x \right )} + 5 \right )}}{3468 d \tan {\left (c + d x \right )} + 5780 d} - \frac {45 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{3468 d \tan {\left (c + d x \right )} + 5780 d} - \frac {75 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{3468 d \tan {\left (c + d x \right )} + 5780 d} - \frac {102}{3468 d \tan {\left (c + d x \right )} + 5780 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (3 \tan {\left (c \right )} + 5\right )^{2}} & \text {otherwise} \end {cases} \]
Piecewise((48*d*x*tan(c + d*x)/(3468*d*tan(c + d*x) + 5780*d) + 80*d*x/(34 68*d*tan(c + d*x) + 5780*d) + 90*log(3*tan(c + d*x) + 5)*tan(c + d*x)/(346 8*d*tan(c + d*x) + 5780*d) + 150*log(3*tan(c + d*x) + 5)/(3468*d*tan(c + d *x) + 5780*d) - 45*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(3468*d*tan(c + d *x) + 5780*d) - 75*log(tan(c + d*x)**2 + 1)/(3468*d*tan(c + d*x) + 5780*d) - 102/(3468*d*tan(c + d*x) + 5780*d), Ne(d, 0)), (x/(3*tan(c) + 5)**2, Tr ue))
Time = 0.40 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=\frac {16 \, d x + 16 \, c - \frac {102}{3 \, \tan \left (d x + c\right ) + 5} - 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 30 \, \log \left (3 \, \tan \left (d x + c\right ) + 5\right )}{1156 \, d} \]
1/1156*(16*d*x + 16*c - 102/(3*tan(d*x + c) + 5) - 15*log(tan(d*x + c)^2 + 1) + 30*log(3*tan(d*x + c) + 5))/d
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=\frac {16 \, d x + 16 \, c - \frac {18 \, {\left (5 \, \tan \left (d x + c\right ) + 14\right )}}{3 \, \tan \left (d x + c\right ) + 5} - 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 30 \, \log \left ({\left | 3 \, \tan \left (d x + c\right ) + 5 \right |}\right )}{1156 \, d} \]
1/1156*(16*d*x + 16*c - 18*(5*tan(d*x + c) + 14)/(3*tan(d*x + c) + 5) - 15 *log(tan(d*x + c)^2 + 1) + 30*log(abs(3*tan(d*x + c) + 5)))/d
Time = 5.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(5+3 \tan (c+d x))^2} \, dx=\frac {15\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {5}{3}\right )}{578\,d}-\frac {1}{34\,d\,\left (\mathrm {tan}\left (c+d\,x\right )+\frac {5}{3}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {15}{1156}-\frac {2}{289}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-\frac {15}{1156}+\frac {2}{289}{}\mathrm {i}\right )}{d} \]