\(\int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [71]

Optimal result

Integrand size = 22, antiderivative size = 84 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i x}{8 a^3}-\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {1}{8 a d (a+i a \tan (c+d x))^2}+\frac {1}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

-1/8*I*x/a^3-1/6/d/(a+I*a*tan(d*x+c))^3+1/8/a/d/(a+I*a*tan(d*x+c))^2+1/8/d/(a^3+I*a^3*tan(d*x+c))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3607, 3560, 8} \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {1}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {i x}{8 a^3}+\frac {1}{8 a d (a+i a \tan (c+d x))^2}-\frac {1}{6 d (a+i a \tan (c+d x))^3} \]

[In]

Int[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((-1/8*I)*x)/a^3 - 1/(6*d*(a + I*a*Tan[c + d*x])^3) + 1/(8*a*d*(a + I*a*Tan[c + d*x])^2) + 1/(8*d*(a^3 + I*a^3
*Tan[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[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]

Rule 3607

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{6 d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{2 a} \\ & = -\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {1}{8 a d (a+i a \tan (c+d x))^2}-\frac {i \int \frac {1}{a+i a \tan (c+d x)} \, dx}{4 a^2} \\ & = -\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {1}{8 a d (a+i a \tan (c+d x))^2}+\frac {1}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {i \int 1 \, dx}{8 a^3} \\ & = -\frac {i x}{8 a^3}-\frac {1}{6 d (a+i a \tan (c+d x))^3}+\frac {1}{8 a d (a+i a \tan (c+d x))^2}+\frac {1}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\sec ^3(c+d x) (3 i \cos (c+d x)+2 (-i+6 d x) \cos (3 (c+d x))-9 \sin (c+d x)-2 \sin (3 (c+d x))+12 i d x \sin (3 (c+d x)))}{96 a^3 d (-i+\tan (c+d x))^3} \]

[In]

Integrate[Tan[c + d*x]/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*((3*I)*Cos[c + d*x] + 2*(-I + 6*d*x)*Cos[3*(c + d*x)] - 9*Sin[c + d*x] - 2*Sin[3*(c + d*x)] +
(12*I)*d*x*Sin[3*(c + d*x)]))/(96*a^3*d*(-I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {i x}{8 a^{3}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\)	\(60\)
derivativedivides	\(-\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}\)	\(77\)
default	\(-\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {i}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}\)	\(77\)
norman	\(\frac {-\frac {i x}{8 a}+\frac {1}{12 a d}+\frac {i \tan \left (d x +c \right )}{8 d a}-\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}-\frac {i \left (\tan ^{5}\left (d x +c \right )\right )}{8 a d}-\frac {3 i x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}-\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {i x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{4 a d}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}\)	\(143\)

[In]

int(tan(d*x+c)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-1/8*I*x/a^3+1/16/a^3/d*exp(-2*I*(d*x+c))-1/32/a^3/d*exp(-4*I*(d*x+c))-1/48/a^3/d*exp(-6*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.64 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (-12 i \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(-12*I*d*x*e^(6*I*d*x + 6*I*c) + 6*e^(4*I*d*x + 4*I*c) - 3*e^(2*I*d*x + 2*I*c) - 2)*e^(-6*I*d*x - 6*I*c)/
(a^3*d)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.82 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (1536 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 768 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} - 512 a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- i e^{6 i c} - i e^{4 i c} + i e^{2 i c} + i\right ) e^{- 6 i c}}{8 a^{3}} + \frac {i}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {i x}{8 a^{3}} \]

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((1536*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) - 768*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) - 512*a**6*d**2*
exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(a**9*d**3*exp(12*I*c), 0)), (x*((-I*exp(6*I*c) -
I*exp(4*I*c) + I*exp(2*I*c) + I)*exp(-6*I*c)/(8*a**3) + I/(8*a**3)), True)) - I*x/(8*a**3)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.60 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {\frac {6 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {6 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {11 \, \tan \left (d x + c\right )^{3} - 45 i \, \tan \left (d x + c\right )^{2} - 69 \, \tan \left (d x + c\right ) + 19 i}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]

[In]

integrate(tan(d*x+c)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(6*log(tan(d*x + c) + I)/a^3 - 6*log(tan(d*x + c) - I)/a^3 + (11*tan(d*x + c)^3 - 45*I*tan(d*x + c)^2 - 6
9*tan(d*x + c) + 19*I)/(a^3*(tan(d*x + c) - I)^3))/d

Mupad [B] (verification not implemented)

Time = 4.44 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.58 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {x\,1{}\mathrm {i}}{8\,a^3}+\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{8}+\frac {\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{8}+\frac {1}{12}}{a^3\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \]

[In]

int(tan(c + d*x)/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

((tan(c + d*x)*3i)/8 - tan(c + d*x)^2/8 + 1/12)/(a^3*d*(tan(c + d*x)*1i + 1)^3) - (x*1i)/(8*a^3)