\(\int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx\) [31]

Optimal result

Integrand size = 21, antiderivative size = 209 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=-\frac {11 i d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

-11/96*I*d*x/a^3/f-1/16*d*x^2/a^3+1/8*x*(d*x+c)/a^3+1/36*d/f^2/(a+I*a*tan(f*x+e))^3+1/6*I*(d*x+c)/f/(a+I*a*tan
(f*x+e))^3+5/96*d/a/f^2/(a+I*a*tan(f*x+e))^2+1/8*I*(d*x+c)/a/f/(a+I*a*tan(f*x+e))^2+11/96*d/f^2/(a^3+I*a^3*tan
(f*x+e))+1/8*I*(d*x+c)/f/(a^3+I*a^3*tan(f*x+e))

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3560, 8, 3811} \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {x (c+d x)}{8 a^3}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {11 i d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3} \]

[In]

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

[Out]

(((-11*I)/96)*d*x)/(a^3*f) - (d*x^2)/(16*a^3) + (x*(c + d*x))/(8*a^3) + d/(36*f^2*(a + I*a*Tan[e + f*x])^3) +
((I/6)*(c + d*x))/(f*(a + I*a*Tan[e + f*x])^3) + (5*d)/(96*a*f^2*(a + I*a*Tan[e + f*x])^2) + ((I/8)*(c + d*x))
/(a*f*(a + I*a*Tan[e + f*x])^2) + (11*d)/(96*f^2*(a^3 + I*a^3*Tan[e + f*x])) + ((I/8)*(c + d*x))/(f*(a^3 + I*a
^3*Tan[e + f*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 3811

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{u = IntHide[(a
+ b*Tan[e + f*x])^n, x]}, Dist[(c + d*x)^m, u, x] - Dist[d*m, Int[Dist[(c + d*x)^(m - 1), u, x], x], x]] /; Fr
eeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 + b^2, 0] && ILtQ[n, -1] && GtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {x (c+d x)}{8 a^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-d \int \left (\frac {x}{8 a^3}+\frac {i}{6 f (a+i a \tan (e+f x))^3}+\frac {i}{8 a f (a+i a \tan (e+f x))^2}+\frac {i}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}\right ) \, dx \\ & = -\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int \frac {1}{a^3+i a^3 \tan (e+f x)} \, dx}{8 f}-\frac {(i d) \int \frac {1}{(a+i a \tan (e+f x))^3} \, dx}{6 f}-\frac {(i d) \int \frac {1}{(a+i a \tan (e+f x))^2} \, dx}{8 a f} \\ & = -\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {d}{32 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {d}{16 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int 1 \, dx}{16 a^3 f}-\frac {(i d) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{16 a^2 f}-\frac {(i d) \int \frac {1}{(a+i a \tan (e+f x))^2} \, dx}{12 a f} \\ & = -\frac {i d x}{16 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {3 d}{32 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int 1 \, dx}{32 a^3 f}-\frac {(i d) \int \frac {1}{a+i a \tan (e+f x)} \, dx}{24 a^2 f} \\ & = -\frac {3 i d x}{32 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac {(i d) \int 1 \, dx}{48 a^3 f} \\ & = -\frac {11 i d x}{96 a^3 f}-\frac {d x^2}{16 a^3}+\frac {x (c+d x)}{8 a^3}+\frac {d}{36 f^2 (a+i a \tan (e+f x))^3}+\frac {i (c+d x)}{6 f (a+i a \tan (e+f x))^3}+\frac {5 d}{96 a f^2 (a+i a \tan (e+f x))^2}+\frac {i (c+d x)}{8 a f (a+i a \tan (e+f x))^2}+\frac {11 d}{96 f^2 \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {i (c+d x)}{8 f \left (a^3+i a^3 \tan (e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.52 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \sec ^3(e+f x) \left (27 (12 i c f+d (5+12 i f x)) \cos (e+f x)+4 \left (6 c f (i+6 f x)+d \left (1+6 i f x+18 f^2 x^2\right )\right ) \cos (3 (e+f x))+81 i d \sin (e+f x)-108 c f \sin (e+f x)-108 d f x \sin (e+f x)-4 i d \sin (3 (e+f x))+24 c f \sin (3 (e+f x))+24 d f x \sin (3 (e+f x))+144 i c f^2 x \sin (3 (e+f x))+72 i d f^2 x^2 \sin (3 (e+f x))\right )}{1152 a^3 f^2 (-i+\tan (e+f x))^3} \]

[In]

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

[Out]

((I/1152)*Sec[e + f*x]^3*(27*((12*I)*c*f + d*(5 + (12*I)*f*x))*Cos[e + f*x] + 4*(6*c*f*(I + 6*f*x) + d*(1 + (6
*I)*f*x + 18*f^2*x^2))*Cos[3*(e + f*x)] + (81*I)*d*Sin[e + f*x] - 108*c*f*Sin[e + f*x] - 108*d*f*x*Sin[e + f*x
] - (4*I)*d*Sin[3*(e + f*x)] + 24*c*f*Sin[3*(e + f*x)] + 24*d*f*x*Sin[3*(e + f*x)] + (144*I)*c*f^2*x*Sin[3*(e
+ f*x)] + (72*I)*d*f^2*x^2*Sin[3*(e + f*x)]))/(a^3*f^2*(-I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.55

method	result	size
risch	\(\frac {d \,x^{2}}{16 a^{3}}+\frac {c x}{8 a^{3}}+\frac {3 i \left (2 d f x +2 c f -i d \right ) {\mathrm e}^{-2 i \left (f x +e \right )}}{32 a^{3} f^{2}}+\frac {3 i \left (4 d f x +4 c f -i d \right ) {\mathrm e}^{-4 i \left (f x +e \right )}}{128 a^{3} f^{2}}+\frac {i \left (6 d f x +6 c f -i d \right ) {\mathrm e}^{-6 i \left (f x +e \right )}}{288 a^{3} f^{2}}\)	\(114\)

[In]

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

[Out]

1/16*d*x^2/a^3+1/8/a^3*c*x+3/32*I*(2*d*f*x-I*d+2*c*f)/a^3/f^2*exp(-2*I*(f*x+e))+3/128*I*(4*d*f*x-I*d+4*c*f)/a^
3/f^2*exp(-4*I*(f*x+e))+1/288*I*(6*d*f*x-I*d+6*c*f)/a^3/f^2*exp(-6*I*(f*x+e))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.50 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (24 i \, d f x + 24 i \, c f + 72 \, {\left (d f^{2} x^{2} + 2 \, c f^{2} x\right )} e^{\left (6 i \, f x + 6 i \, e\right )} - 108 \, {\left (-2 i \, d f x - 2 i \, c f - d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - 27 \, {\left (-4 i \, d f x - 4 i \, c f - d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, d\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1152 \, a^{3} f^{2}} \]

[In]

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

[Out]

1/1152*(24*I*d*f*x + 24*I*c*f + 72*(d*f^2*x^2 + 2*c*f^2*x)*e^(6*I*f*x + 6*I*e) - 108*(-2*I*d*f*x - 2*I*c*f - d
)*e^(4*I*f*x + 4*I*e) - 27*(-4*I*d*f*x - 4*I*c*f - d)*e^(2*I*f*x + 2*I*e) + 4*d)*e^(-6*I*f*x - 6*I*e)/(a^3*f^2
)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.49 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\begin {cases} \frac {\left (\left (24576 i a^{6} c f^{5} e^{6 i e} + 24576 i a^{6} d f^{5} x e^{6 i e} + 4096 a^{6} d f^{4} e^{6 i e}\right ) e^{- 6 i f x} + \left (110592 i a^{6} c f^{5} e^{8 i e} + 110592 i a^{6} d f^{5} x e^{8 i e} + 27648 a^{6} d f^{4} e^{8 i e}\right ) e^{- 4 i f x} + \left (221184 i a^{6} c f^{5} e^{10 i e} + 221184 i a^{6} d f^{5} x e^{10 i e} + 110592 a^{6} d f^{4} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{1179648 a^{9} f^{6}} & \text {for}\: a^{9} f^{6} e^{12 i e} \neq 0 \\\frac {x^{2} \cdot \left (3 d e^{4 i e} + 3 d e^{2 i e} + d\right ) e^{- 6 i e}}{16 a^{3}} + \frac {x \left (3 c e^{4 i e} + 3 c e^{2 i e} + c\right ) e^{- 6 i e}}{8 a^{3}} & \text {otherwise} \end {cases} + \frac {c x}{8 a^{3}} + \frac {d x^{2}}{16 a^{3}} \]

[In]

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

[Out]

Piecewise((((24576*I*a**6*c*f**5*exp(6*I*e) + 24576*I*a**6*d*f**5*x*exp(6*I*e) + 4096*a**6*d*f**4*exp(6*I*e))*
exp(-6*I*f*x) + (110592*I*a**6*c*f**5*exp(8*I*e) + 110592*I*a**6*d*f**5*x*exp(8*I*e) + 27648*a**6*d*f**4*exp(8
*I*e))*exp(-4*I*f*x) + (221184*I*a**6*c*f**5*exp(10*I*e) + 221184*I*a**6*d*f**5*x*exp(10*I*e) + 110592*a**6*d*
f**4*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/(1179648*a**9*f**6), Ne(a**9*f**6*exp(12*I*e), 0)), (x**2*(3*d*e
xp(4*I*e) + 3*d*exp(2*I*e) + d)*exp(-6*I*e)/(16*a**3) + x*(3*c*exp(4*I*e) + 3*c*exp(2*I*e) + c)*exp(-6*I*e)/(8
*a**3), True)) + c*x/(8*a**3) + d*x**2/(16*a**3)

Maxima [F(-2)]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.68 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (72 \, d f^{2} x^{2} e^{\left (6 i \, f x + 6 i \, e\right )} + 144 \, c f^{2} x e^{\left (6 i \, f x + 6 i \, e\right )} + 216 i \, d f x e^{\left (4 i \, f x + 4 i \, e\right )} + 108 i \, d f x e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, d f x + 216 i \, c f e^{\left (4 i \, f x + 4 i \, e\right )} + 108 i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + 24 i \, c f + 108 \, d e^{\left (4 i \, f x + 4 i \, e\right )} + 27 \, d e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, d\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1152 \, a^{3} f^{2}} \]

[In]

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

[Out]

1/1152*(72*d*f^2*x^2*e^(6*I*f*x + 6*I*e) + 144*c*f^2*x*e^(6*I*f*x + 6*I*e) + 216*I*d*f*x*e^(4*I*f*x + 4*I*e) +
108*I*d*f*x*e^(2*I*f*x + 2*I*e) + 24*I*d*f*x + 216*I*c*f*e^(4*I*f*x + 4*I*e) + 108*I*c*f*e^(2*I*f*x + 2*I*e)
+ 24*I*c*f + 108*d*e^(4*I*f*x + 4*I*e) + 27*d*e^(2*I*f*x + 2*I*e) + 4*d)*e^(-6*I*f*x - 6*I*e)/(a^3*f^2)

Mupad [B] (verification not implemented)

Time = 3.94 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.70 \[ \int \frac {c+d x}{(a+i a \tan (e+f x))^3} \, dx=\frac {d\,x^2}{16\,a^3}-{\mathrm {e}}^{-e\,4{}\mathrm {i}-f\,x\,4{}\mathrm {i}}\,\left (\frac {\left (-12\,c\,f+d\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{128\,a^3\,f^2}-\frac {d\,x\,3{}\mathrm {i}}{32\,a^3\,f}\right )-{\mathrm {e}}^{-e\,6{}\mathrm {i}-f\,x\,6{}\mathrm {i}}\,\left (\frac {\left (-6\,c\,f+d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{288\,a^3\,f^2}-\frac {d\,x\,1{}\mathrm {i}}{48\,a^3\,f}\right )-{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,\left (\frac {\left (-6\,c\,f+d\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^3\,f^2}-\frac {d\,x\,3{}\mathrm {i}}{16\,a^3\,f}\right )+\frac {c\,x}{8\,a^3} \]

[In]

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

[Out]

(d*x^2)/(16*a^3) - exp(- e*4i - f*x*4i)*(((d*3i - 12*c*f)*1i)/(128*a^3*f^2) - (d*x*3i)/(32*a^3*f)) - exp(- e*6
i - f*x*6i)*(((d*1i - 6*c*f)*1i)/(288*a^3*f^2) - (d*x*1i)/(48*a^3*f)) - exp(- e*2i - f*x*2i)*(((d*3i - 6*c*f)*
1i)/(32*a^3*f^2) - (d*x*3i)/(16*a^3*f)) + (c*x)/(8*a^3)