Integrand size = 24, antiderivative size = 88 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {x}{8 a^3}-\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i}{8 a d (a+i a \tan (c+d x))^2}-\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
-1/8*x/a^3-1/6*I/d/(a+I*a*tan(d*x+c))^3+3/8*I/a/d/(a+I*a*tan(d*x+c))^2-1/8 *I/d/(a^3+I*a^3*tan(d*x+c))
Time = 0.55 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (-9 \cos (c+d x)+2 (1-6 i d x) \cos (3 (c+d x))-3 i \sin (c+d x)-2 i \sin (3 (c+d x))+12 d x \sin (3 (c+d x)))}{96 a^3 d (-i+\tan (c+d x))^3} \]
(Sec[c + d*x]^3*(-9*Cos[c + d*x] + 2*(1 - (6*I)*d*x)*Cos[3*(c + d*x)] - (3 *I)*Sin[c + d*x] - (2*I)*Sin[3*(c + d*x)] + 12*d*x*Sin[3*(c + d*x)]))/(96* a^3*d*(-I + Tan[c + d*x])^3)
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 4023, 3042, 4009, 3042, 3960, 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.
\(\displaystyle \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^2}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 4023 |
\(\displaystyle \frac {\int \frac {a-2 i a \tan (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}-\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a-2 i a \tan (c+d x)}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}-\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle \frac {\frac {3 i a}{4 d (a+i a \tan (c+d x))^2}-\frac {1}{2} \int \frac {1}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 i a}{4 d (a+i a \tan (c+d x))^2}-\frac {1}{2} \int \frac {1}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {1}{2} \left (-\frac {\int 1dx}{2 a}-\frac {i}{2 d (a+i a \tan (c+d x))}\right )+\frac {3 i a}{4 d (a+i a \tan (c+d x))^2}}{2 a^2}-\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {3 i a}{4 d (a+i a \tan (c+d x))^2}+\frac {1}{2} \left (-\frac {x}{2 a}-\frac {i}{2 d (a+i a \tan (c+d x))}\right )}{2 a^2}-\frac {i}{6 d (a+i a \tan (c+d x))^3}\) |
(-1/6*I)/(d*(a + I*a*Tan[c + d*x])^3) + ((((3*I)/4)*a)/(d*(a + I*a*Tan[c + d*x])^2) + (-1/2*x/a - (I/2)/(d*(a + I*a*Tan[c + d*x])))/2)/(2*a^2)
3.1.70.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a) Int[(a + b*Tan[c + d*x])^ (n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
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/(2*a *f*m)), x] + Simp[(b*c + a*d)/(2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && LtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^ m/(2*a^3*f*m)), x] + Simp[1/(2*a^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp [a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]
Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70
method | result | size |
risch | \(-\frac {x}{8 a^{3}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\) | \(62\) |
derivativedivides | \(-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {3 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {1}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(75\) |
default | \(-\frac {\arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {3 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {1}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(75\) |
norman | \(\frac {-\frac {x}{8 a}+\frac {i}{12 a d}-\frac {\tan ^{5}\left (d x +c \right )}{8 a d}-\frac {3 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}-\frac {3 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {\tan \left (d x +c \right )}{8 a d}+\frac {2 \left (\tan ^{3}\left (d x +c \right )\right )}{3 a d}-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{2 d a}+\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}\) | \(155\) |
-1/8*x/a^3+1/16*I/a^3/d*exp(-2*I*(d*x+c))+1/32*I/a^3/d*exp(-4*I*(d*x+c))-1 /48*I/a^3/d*exp(-6*I*(d*x+c))
Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
-1/96*(12*d*x*e^(6*I*d*x + 6*I*c) - 6*I*e^(4*I*d*x + 4*I*c) - 3*I*e^(2*I*d *x + 2*I*c) + 2*I)*e^(-6*I*d*x - 6*I*c)/(a^3*d)
Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.72 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (1536 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 768 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- e^{6 i c} + e^{4 i c} + e^{2 i c} - 1\right ) e^{- 6 i c}}{8 a^{3}} + \frac {1}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {x}{8 a^{3}} \]
Piecewise(((1536*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 768*I*a**6*d**2*e xp(8*I*c)*exp(-4*I*d*x) - 512*I*a**6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-1 2*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((-exp(6*I*c) + exp(4*I*c) + exp(2*I*c) - 1)*exp(-6*I*c)/(8*a**3) + 1/(8*a**3)), True)) - x/(8*a**3)
Exception generated. \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.72 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.89 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {11 i \, \tan \left (d x + c\right )^{3} + 45 \, \tan \left (d x + c\right )^{2} - 21 i \, \tan \left (d x + c\right ) - 3}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]
-1/96*(6*I*log(tan(d*x + c) + I)/a^3 - 6*I*log(tan(d*x + c) - I)/a^3 + (11 *I*tan(d*x + c)^3 + 45*tan(d*x + c)^2 - 21*I*tan(d*x + c) - 3)/(a^3*(tan(d *x + c) - I)^3))/d
Time = 4.48 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.56 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {x}{8\,a^3}+\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{8}-\frac {\mathrm {tan}\left (c+d\,x\right )}{8}+\frac {1}{12}{}\mathrm {i}}{a^3\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \]