\(\int \frac {1}{(a+i a \tan (c+d x))^4} \, dx\) [155]

Optimal result

Integrand size = 15, antiderivative size = 116 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {x}{16 a^4}+\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \]

[Out]

1/16*x/a^4+1/8*I/d/(a+I*a*tan(d*x+c))^4+1/12*I/a/d/(a+I*a*tan(d*x+c))^3+1/16*I/d/(a^2+I*a^2*tan(d*x+c))^2+1/16
*I/d/(a^4+I*a^4*tan(d*x+c))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3560, 8} \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {x}{16 a^4}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{8 d (a+i a \tan (c+d x))^4} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(-4),x]

[Out]

x/(16*a^4) + (I/8)/(d*(a + I*a*Tan[c + d*x])^4) + (I/12)/(a*d*(a + I*a*Tan[c + d*x])^3) + (I/16)/(d*(a^2 + I*a
^2*Tan[c + d*x])^2) + (I/16)/(d*(a^4 + I*a^4*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]

Rubi steps \begin{align*} \text {integral}& = \frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^3} \, dx}{2 a} \\ & = \frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2} \\ & = \frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {\int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac {\int 1 \, dx}{16 a^4} \\ & = \frac {x}{16 a^4}+\frac {i}{8 d (a+i a \tan (c+d x))^4}+\frac {i}{12 a d (a+i a \tan (c+d x))^3}+\frac {i}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {i}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i a \left (\frac {i \arctan (\tan (c+d x))}{16 a^5}-\frac {1}{8 a (a+i a \tan (c+d x))^4}-\frac {1}{12 a^2 (a+i a \tan (c+d x))^3}-\frac {1}{16 a^3 (a+i a \tan (c+d x))^2}-\frac {1}{16 a^4 (a+i a \tan (c+d x))}\right )}{d} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^(-4),x]

[Out]

((-I)*a*(((I/16)*ArcTan[Tan[c + d*x]])/a^5 - 1/(8*a*(a + I*a*Tan[c + d*x])^4) - 1/(12*a^2*(a + I*a*Tan[c + d*x
])^3) - 1/(16*a^3*(a + I*a*Tan[c + d*x])^2) - 1/(16*a^4*(a + I*a*Tan[c + d*x]))))/d

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {x}{16 a^{4}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{4} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{24 a^{4} d}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\)	\(80\)
derivativedivides	\(\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{16 a^{4} d}-\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{16 a^{4} d \left (\tan \left (d x +c \right )-i\right )}\)	\(95\)
default	\(\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {\arctan \left (\tan \left (d x +c \right )\right )}{16 a^{4} d}-\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{16 a^{4} d \left (\tan \left (d x +c \right )-i\right )}\)	\(95\)
norman	\(\frac {\frac {x}{16 a}+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{48 a d}+\frac {11 \left (\tan ^{5}\left (d x +c \right )\right )}{48 a d}+\frac {\tan ^{7}\left (d x +c \right )}{16 a d}+\frac {x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}+\frac {3 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}+\frac {x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {i}{3 a d}+\frac {15 \tan \left (d x +c \right )}{16 a d}-\frac {2 i \left (\tan ^{2}\left (d x +c \right )\right )}{3 a d}}{a^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{4}}\)	\(168\)

[In]

int(1/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/16*x/a^4+1/8*I/a^4/d*exp(-2*I*(d*x+c))+3/32*I/a^4/d*exp(-4*I*(d*x+c))+1/24*I/a^4/d*exp(-6*I*(d*x+c))+1/128*I
/a^4/d*exp(-8*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (24 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 48 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 36 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/384*(24*d*x*e^(8*I*d*x + 8*I*c) + 48*I*e^(6*I*d*x + 6*I*c) + 36*I*e^(4*I*d*x + 4*I*c) + 16*I*e^(2*I*d*x + 2*
I*c) + 3*I)*e^(-8*I*d*x - 8*I*c)/(a^4*d)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (98304 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} + 73728 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 32768 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} + 6144 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{786432 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 8 i c}}{16 a^{4}} - \frac {1}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {x}{16 a^{4}} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((98304*I*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) + 73728*I*a**12*d**3*exp(16*I*c)*exp(-4*I*d*x) + 3276
8*I*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) + 6144*I*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(786432*a
**16*d**4), Ne(a**16*d**4*exp(20*I*c), 0)), (x*((exp(8*I*c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c) + 1)*
exp(-8*I*c)/(16*a**4) - 1/(16*a**4)), True)) + x/(16*a**4)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=-\frac {-\frac {12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac {12 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} + \frac {-25 i \, \tan \left (d x + c\right )^{4} - 124 \, \tan \left (d x + c\right )^{3} + 246 i \, \tan \left (d x + c\right )^{2} + 252 \, \tan \left (d x + c\right ) - 153 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]

[In]

integrate(1/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

-1/384*(-12*I*log(tan(d*x + c) + I)/a^4 + 12*I*log(tan(d*x + c) - I)/a^4 + (-25*I*tan(d*x + c)^4 - 124*tan(d*x
+ c)^3 + 246*I*tan(d*x + c)^2 + 252*tan(d*x + c) - 153*I)/(a^4*(tan(d*x + c) - I)^4))/d

Mupad [B] (verification not implemented)

Time = 4.52 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a+i a \tan (c+d x))^4} \, dx=\frac {x}{16\,a^4}-\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{16}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4}+\frac {19\,\mathrm {tan}\left (c+d\,x\right )}{48}-\frac {1}{3}{}\mathrm {i}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \]

[In]

int(1/(a + a*tan(c + d*x)*1i)^4,x)

[Out]

x/(16*a^4) - ((19*tan(c + d*x))/48 + (tan(c + d*x)^2*1i)/4 - tan(c + d*x)^3/16 - 1i/3)/(a^4*d*(tan(c + d*x)*1i
+ 1)^4)