Integrand size = 14, antiderivative size = 79 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\frac {5 x}{16 a^3}+\frac {5 \cos (c+d x) \sin (c+d x)}{16 a^3 d}+\frac {5 \cos ^3(c+d x) \sin (c+d x)}{24 a^3 d}+\frac {\cos ^5(c+d x) \sin (c+d x)}{6 a^3 d} \]
5/16*x/a^3+5/16*cos(d*x+c)*sin(d*x+c)/a^3/d+5/24*cos(d*x+c)^3*sin(d*x+c)/a ^3/d+1/6*cos(d*x+c)^5*sin(d*x+c)/a^3/d
Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.58 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\frac {60 c+60 d x+45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))}{192 a^3 d} \]
Time = 0.37 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4140, 27, 3042, 3115, 3042, 3115, 3042, 3115, 24}
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.
\(\displaystyle \int \frac {1}{\left (a \tan ^2(c+d x)+a\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \tan (c+d x)^2+a\right )^3}dx\) |
\(\Big \downarrow \) 4140 |
\(\displaystyle \int \frac {\cos ^6(c+d x)}{a^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \cos ^6(c+d x)dx}{a^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^6dx}{a^3}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {5}{6} \int \cos ^4(c+d x)dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5}{6} \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a^3}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {5}{6} \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {5}{6} \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a^3}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {\frac {5}{6} \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}}{a^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {5}{6} \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^3}\) |
((Cos[c + d*x]^5*Sin[c + d*x])/(6*d) + (5*((Cos[c + d*x]^3*Sin[c + d*x])/( 4*d) + (3*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4))/6)/a^3
3.2.84.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*sec[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a, b]
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {5 x}{16 a^{3}}+\frac {\sin \left (6 d x +6 c \right )}{192 a^{3} d}+\frac {3 \sin \left (4 d x +4 c \right )}{64 a^{3} d}+\frac {15 \sin \left (2 d x +2 c \right )}{64 a^{3} d}\) | \(59\) |
derivativedivides | \(\frac {\frac {\tan \left (d x +c \right )}{6 \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}+\frac {5 \tan \left (d x +c \right )}{24 \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {5 \tan \left (d x +c \right )}{16 \left (1+\tan \left (d x +c \right )^{2}\right )}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{16}}{d \,a^{3}}\) | \(78\) |
default | \(\frac {\frac {\tan \left (d x +c \right )}{6 \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}+\frac {5 \tan \left (d x +c \right )}{24 \left (1+\tan \left (d x +c \right )^{2}\right )^{2}}+\frac {5 \tan \left (d x +c \right )}{16 \left (1+\tan \left (d x +c \right )^{2}\right )}+\frac {5 \arctan \left (\tan \left (d x +c \right )\right )}{16}}{d \,a^{3}}\) | \(78\) |
parallelrisch | \(\frac {15 \tan \left (d x +c \right )^{6} x d +45 \tan \left (d x +c \right )^{4} x d +15 \tan \left (d x +c \right )^{5}+45 \tan \left (d x +c \right )^{2} x d +40 \tan \left (d x +c \right )^{3}+15 d x +33 \tan \left (d x +c \right )}{48 d \,a^{3} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}\) | \(90\) |
norman | \(\frac {\frac {5 x}{16 a}+\frac {11 \tan \left (d x +c \right )}{16 a d}+\frac {5 \tan \left (d x +c \right )^{3}}{6 a d}+\frac {5 \tan \left (d x +c \right )^{5}}{16 a d}+\frac {15 x \tan \left (d x +c \right )^{2}}{16 a}+\frac {15 x \tan \left (d x +c \right )^{4}}{16 a}+\frac {5 x \tan \left (d x +c \right )^{6}}{16 a}}{a^{2} \left (1+\tan \left (d x +c \right )^{2}\right )^{3}}\) | \(112\) |
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.52 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\frac {15 \, d x \tan \left (d x + c\right )^{6} + 45 \, d x \tan \left (d x + c\right )^{4} + 15 \, \tan \left (d x + c\right )^{5} + 45 \, d x \tan \left (d x + c\right )^{2} + 40 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 33 \, \tan \left (d x + c\right )}{48 \, {\left (a^{3} d \tan \left (d x + c\right )^{6} + 3 \, a^{3} d \tan \left (d x + c\right )^{4} + 3 \, a^{3} d \tan \left (d x + c\right )^{2} + a^{3} d\right )}} \]
1/48*(15*d*x*tan(d*x + c)^6 + 45*d*x*tan(d*x + c)^4 + 15*tan(d*x + c)^5 + 45*d*x*tan(d*x + c)^2 + 40*tan(d*x + c)^3 + 15*d*x + 33*tan(d*x + c))/(a^3 *d*tan(d*x + c)^6 + 3*a^3*d*tan(d*x + c)^4 + 3*a^3*d*tan(d*x + c)^2 + a^3* d)
Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (75) = 150\).
Time = 0.50 (sec) , antiderivative size = 454, normalized size of antiderivative = 5.75 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\begin {cases} \frac {15 d x \tan ^{6}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {45 d x \tan ^{4}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {45 d x \tan ^{2}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {15 d x}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {15 \tan ^{5}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {40 \tan ^{3}{\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} + \frac {33 \tan {\left (c + d x \right )}}{48 a^{3} d \tan ^{6}{\left (c + d x \right )} + 144 a^{3} d \tan ^{4}{\left (c + d x \right )} + 144 a^{3} d \tan ^{2}{\left (c + d x \right )} + 48 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \tan ^{2}{\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((15*d*x*tan(c + d*x)**6/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d* tan(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 45*d*x*tan(c + d*x)**4/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a** 3*d*tan(c + d*x)**2 + 48*a**3*d) + 45*d*x*tan(c + d*x)**2/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48* a**3*d) + 15*d*x/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 15*tan(c + d*x)**5/(48*a**3*d*t an(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 40*tan(c + d*x)**3/(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*t an(c + d*x)**4 + 144*a**3*d*tan(c + d*x)**2 + 48*a**3*d) + 33*tan(c + d*x) /(48*a**3*d*tan(c + d*x)**6 + 144*a**3*d*tan(c + d*x)**4 + 144*a**3*d*tan( c + d*x)**2 + 48*a**3*d), Ne(d, 0)), (x/(a*tan(c)**2 + a)**3, True))
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\frac {\frac {15 \, \tan \left (d x + c\right )^{5} + 40 \, \tan \left (d x + c\right )^{3} + 33 \, \tan \left (d x + c\right )}{a^{3} \tan \left (d x + c\right )^{6} + 3 \, a^{3} \tan \left (d x + c\right )^{4} + 3 \, a^{3} \tan \left (d x + c\right )^{2} + a^{3}} + \frac {15 \, {\left (d x + c\right )}}{a^{3}}}{48 \, d} \]
1/48*((15*tan(d*x + c)^5 + 40*tan(d*x + c)^3 + 33*tan(d*x + c))/(a^3*tan(d *x + c)^6 + 3*a^3*tan(d*x + c)^4 + 3*a^3*tan(d*x + c)^2 + a^3) + 15*(d*x + c)/a^3)/d
Time = 0.48 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.77 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\frac {\frac {15 \, {\left (d x + c\right )}}{a^{3}} + \frac {15 \, \tan \left (d x + c\right )^{5} + 40 \, \tan \left (d x + c\right )^{3} + 33 \, \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, d} \]
1/48*(15*(d*x + c)/a^3 + (15*tan(d*x + c)^5 + 40*tan(d*x + c)^3 + 33*tan(d *x + c))/((tan(d*x + c)^2 + 1)^3*a^3))/d
Time = 11.56 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\left (a+a \tan ^2(c+d x)\right )^3} \, dx=\frac {5\,x}{16\,a^3}+\frac {{\cos \left (c+d\,x\right )}^6\,\left (\frac {5\,{\mathrm {tan}\left (c+d\,x\right )}^5}{16}+\frac {5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{6}+\frac {11\,\mathrm {tan}\left (c+d\,x\right )}{16}\right )}{a^3\,d} \]