Integrand size = 12, antiderivative size = 50 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=-\frac {4 x}{289}+\frac {15 \log (3 \cos (c+d x)+5 \sin (c+d x))}{578 d}-\frac {5}{34 d (3+5 \tan (c+d x))} \]
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=-\frac {(15-8 i) \log (i-\tan (c+d x))+(15+8 i) \log (i+\tan (c+d x))-30 \log (3+5 \tan (c+d x))+\frac {170}{3+5 \tan (c+d x)}}{1156 d} \]
-1/1156*((15 - 8*I)*Log[I - Tan[c + d*x]] + (15 + 8*I)*Log[I + Tan[c + d*x ]] - 30*Log[3 + 5*Tan[c + d*x]] + 170/(3 + 5*Tan[c + d*x]))/d
Time = 0.38 (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.
\(\displaystyle \int \frac {1}{(5 \tan (c+d x)+3)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(5 \tan (c+d x)+3)^2}dx\) |
\(\Big \downarrow \) 3964 |
\(\displaystyle \frac {1}{34} \int \frac {3-5 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {5}{34 d (5 \tan (c+d x)+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{34} \int \frac {3-5 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {5}{34 d (5 \tan (c+d x)+3)}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {1}{34} \left (\frac {15}{17} \int \frac {5-3 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {8 x}{17}\right )-\frac {5}{34 d (5 \tan (c+d x)+3)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{34} \left (\frac {15}{17} \int \frac {5-3 \tan (c+d x)}{5 \tan (c+d x)+3}dx-\frac {8 x}{17}\right )-\frac {5}{34 d (5 \tan (c+d x)+3)}\) |
\(\Big \downarrow \) 4013 |
\(\displaystyle \frac {1}{34} \left (\frac {15 \log (5 \sin (c+d x)+3 \cos (c+d x))}{17 d}-\frac {8 x}{17}\right )-\frac {5}{34 d (5 \tan (c+d x)+3)}\) |
((-8*x)/17 + (15*Log[3*Cos[c + d*x] + 5*Sin[c + d*x]])/(17*d))/34 - 5/(34* d*(3 + 5*Tan[c + d*x]))
3.5.96.3.1 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 = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {-\frac {5}{34 \left (3+5 \tan \left (d x +c \right )\right )}+\frac {15 \ln \left (3+5 \tan \left (d x +c \right )\right )}{578}-\frac {15 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{1156}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right )}{289}}{d}\) | \(55\) |
default | \(\frac {-\frac {5}{34 \left (3+5 \tan \left (d x +c \right )\right )}+\frac {15 \ln \left (3+5 \tan \left (d x +c \right )\right )}{578}-\frac {15 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{1156}-\frac {4 \arctan \left (\tan \left (d x +c \right )\right )}{289}}{d}\) | \(55\) |
norman | \(\frac {-\frac {12 x}{289}-\frac {20 x \tan \left (d x +c \right )}{289}+\frac {25 \tan \left (d x +c \right )}{102 d}}{3+5 \tan \left (d x +c \right )}+\frac {15 \ln \left (3+5 \tan \left (d x +c \right )\right )}{578 d}-\frac {15 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{1156 d}\) | \(71\) |
risch | \(-\frac {4 x}{289}-\frac {15 i x}{578}-\frac {15 i c}{289 d}-\frac {375}{578 d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}-8+15 i\right )}-\frac {100 i}{289 d \left (17 \,{\mathrm e}^{2 i \left (d x +c \right )}-8+15 i\right )}+\frac {15 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {8}{17}+\frac {15 i}{17}\right )}{578 d}\) | \(80\) |
parallelrisch | \(\frac {-240 \tan \left (d x +c \right ) x d +450 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right ) \tan \left (d x +c \right )-225 \ln \left (1+\tan ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )-144 d x +270 \ln \left (\frac {3}{5}+\tan \left (d x +c \right )\right )-135 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+850 \tan \left (d x +c \right )}{3468 d \left (3+5 \tan \left (d x +c \right )\right )}\) | \(101\) |
1/d*(-5/34/(3+5*tan(d*x+c))+15/578*ln(3+5*tan(d*x+c))-15/1156*ln(1+tan(d*x +c)^2)-4/289*arctan(tan(d*x+c)))
Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.66 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=-\frac {48 \, d x - 15 \, {\left (5 \, \tan \left (d x + c\right ) + 3\right )} \log \left (\frac {25 \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right ) + 9}{\tan \left (d x + c\right )^{2} + 1}\right ) + 5 \, {\left (16 \, d x - 15\right )} \tan \left (d x + c\right ) + 125}{1156 \, {\left (5 \, d \tan \left (d x + c\right ) + 3 \, d\right )}} \]
-1/1156*(48*d*x - 15*(5*tan(d*x + c) + 3)*log((25*tan(d*x + c)^2 + 30*tan( d*x + c) + 9)/(tan(d*x + c)^2 + 1)) + 5*(16*d*x - 15)*tan(d*x + c) + 125)/ (5*d*tan(d*x + c) + 3*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 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=\begin {cases} - \frac {80 d x \tan {\left (c + d x \right )}}{5780 d \tan {\left (c + d x \right )} + 3468 d} - \frac {48 d x}{5780 d \tan {\left (c + d x \right )} + 3468 d} + \frac {150 \log {\left (5 \tan {\left (c + d x \right )} + 3 \right )} \tan {\left (c + d x \right )}}{5780 d \tan {\left (c + d x \right )} + 3468 d} + \frac {90 \log {\left (5 \tan {\left (c + d x \right )} + 3 \right )}}{5780 d \tan {\left (c + d x \right )} + 3468 d} - \frac {75 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )} \tan {\left (c + d x \right )}}{5780 d \tan {\left (c + d x \right )} + 3468 d} - \frac {45 \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{5780 d \tan {\left (c + d x \right )} + 3468 d} - \frac {170}{5780 d \tan {\left (c + d x \right )} + 3468 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (5 \tan {\left (c \right )} + 3\right )^{2}} & \text {otherwise} \end {cases} \]
Piecewise((-80*d*x*tan(c + d*x)/(5780*d*tan(c + d*x) + 3468*d) - 48*d*x/(5 780*d*tan(c + d*x) + 3468*d) + 150*log(5*tan(c + d*x) + 3)*tan(c + d*x)/(5 780*d*tan(c + d*x) + 3468*d) + 90*log(5*tan(c + d*x) + 3)/(5780*d*tan(c + d*x) + 3468*d) - 75*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(5780*d*tan(c + d*x) + 3468*d) - 45*log(tan(c + d*x)**2 + 1)/(5780*d*tan(c + d*x) + 3468*d ) - 170/(5780*d*tan(c + d*x) + 3468*d), Ne(d, 0)), (x/(5*tan(c) + 3)**2, T rue))
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=-\frac {16 \, d x + 16 \, c + \frac {170}{5 \, \tan \left (d x + c\right ) + 3} + 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 30 \, \log \left (5 \, \tan \left (d x + c\right ) + 3\right )}{1156 \, d} \]
-1/1156*(16*d*x + 16*c + 170/(5*tan(d*x + c) + 3) + 15*log(tan(d*x + c)^2 + 1) - 30*log(5*tan(d*x + c) + 3))/d
Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=-\frac {16 \, d x + 16 \, c + \frac {10 \, {\left (15 \, \tan \left (d x + c\right ) + 26\right )}}{5 \, \tan \left (d x + c\right ) + 3} + 15 \, \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 30 \, \log \left ({\left | 5 \, \tan \left (d x + c\right ) + 3 \right |}\right )}{1156 \, d} \]
-1/1156*(16*d*x + 16*c + 10*(15*tan(d*x + c) + 26)/(5*tan(d*x + c) + 3) + 15*log(tan(d*x + c)^2 + 1) - 30*log(abs(5*tan(d*x + c) + 3)))/d
Time = 5.36 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(3+5 \tan (c+d x))^2} \, dx=\frac {15\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+\frac {3}{5}\right )}{578\,d}-\frac {1}{34\,d\,\left (\mathrm {tan}\left (c+d\,x\right )+\frac {3}{5}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-\frac {15}{1156}+\frac {2}{289}{}\mathrm {i}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (-\frac {15}{1156}-\frac {2}{289}{}\mathrm {i}\right )}{d} \]