\(\int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx\) [952]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [F(-2)]
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 29, antiderivative size = 25 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=-\frac {i a}{4 f (c-i c \tan (e+f x))^4} \]

[Out]

-1/4*I*a/f/(c-I*c*tan(f*x+e))^4

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3603, 3568, 32} \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=-\frac {i a}{4 f (c-i c \tan (e+f x))^4} \]

[In]

Int[(a + I*a*Tan[e + f*x])/(c - I*c*Tan[e + f*x])^4,x]

[Out]

((-1/4*I)*a)/(f*(c - I*c*Tan[e + f*x])^4)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 3603

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

Rubi steps \begin{align*} \text {integral}& = (a c) \int \frac {\sec ^2(e+f x)}{(c-i c \tan (e+f x))^5} \, dx \\ & = \frac {(i a) \text {Subst}\left (\int \frac {1}{(c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{f} \\ & = -\frac {i a}{4 f (c-i c \tan (e+f x))^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=-\frac {i a}{4 c^4 f (i+\tan (e+f x))^4} \]

[In]

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

[Out]

((-1/4*I)*a)/(c^4*f*(I + Tan[e + f*x])^4)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method result size
derivativedivides \(-\frac {i a}{4 f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}\) \(22\)
default \(-\frac {i a}{4 f \,c^{4} \left (\tan \left (f x +e \right )+i\right )^{4}}\) \(22\)
risch \(-\frac {i a \,{\mathrm e}^{8 i \left (f x +e \right )}}{64 c^{4} f}-\frac {i a \,{\mathrm e}^{6 i \left (f x +e \right )}}{16 c^{4} f}-\frac {3 i a \,{\mathrm e}^{4 i \left (f x +e \right )}}{32 c^{4} f}-\frac {i a \,{\mathrm e}^{2 i \left (f x +e \right )}}{16 c^{4} f}\) \(78\)
norman \(\frac {\frac {a \tan \left (f x +e \right )}{c f}-\frac {i a}{4 c f}-\frac {a \left (\tan ^{3}\left (f x +e \right )\right )}{c f}-\frac {i a \left (\tan ^{4}\left (f x +e \right )\right )}{4 c f}+\frac {3 i a \left (\tan ^{2}\left (f x +e \right )\right )}{2 c f}}{\left (1+\tan ^{2}\left (f x +e \right )\right )^{4} c^{3}}\) \(95\)

[In]

int((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

-1/4*I/f*a/c^4/(tan(f*x+e)+I)^4

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (19) = 38\).

Time = 0.23 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.28 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=\frac {-i \, a e^{\left (8 i \, f x + 8 i \, e\right )} - 4 i \, a e^{\left (6 i \, f x + 6 i \, e\right )} - 6 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} - 4 i \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{64 \, c^{4} f} \]

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="fricas")

[Out]

1/64*(-I*a*e^(8*I*f*x + 8*I*e) - 4*I*a*e^(6*I*f*x + 6*I*e) - 6*I*a*e^(4*I*f*x + 4*I*e) - 4*I*a*e^(2*I*f*x + 2*
I*e))/(c^4*f)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (20) = 40\).

Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 6.68 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=\begin {cases} \frac {- 8192 i a c^{12} f^{3} e^{8 i e} e^{8 i f x} - 32768 i a c^{12} f^{3} e^{6 i e} e^{6 i f x} - 49152 i a c^{12} f^{3} e^{4 i e} e^{4 i f x} - 32768 i a c^{12} f^{3} e^{2 i e} e^{2 i f x}}{524288 c^{16} f^{4}} & \text {for}\: c^{16} f^{4} \neq 0 \\\frac {x \left (a e^{8 i e} + 3 a e^{6 i e} + 3 a e^{4 i e} + a e^{2 i e}\right )}{8 c^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**4,x)

[Out]

Piecewise(((-8192*I*a*c**12*f**3*exp(8*I*e)*exp(8*I*f*x) - 32768*I*a*c**12*f**3*exp(6*I*e)*exp(6*I*f*x) - 4915
2*I*a*c**12*f**3*exp(4*I*e)*exp(4*I*f*x) - 32768*I*a*c**12*f**3*exp(2*I*e)*exp(2*I*f*x))/(524288*c**16*f**4),
Ne(c**16*f**4, 0)), (x*(a*exp(8*I*e) + 3*a*exp(6*I*e) + 3*a*exp(4*I*e) + a*exp(2*I*e))/(8*c**4), True))

Maxima [F(-2)]

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

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (19) = 38\).

Time = 0.62 (sec) , antiderivative size = 117, normalized size of antiderivative = 4.68 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=-\frac {2 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 3 i \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 7 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 8 i \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 7 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 i \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{8}} \]

[In]

integrate((a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^4,x, algorithm="giac")

[Out]

-2*(a*tan(1/2*f*x + 1/2*e)^7 + 3*I*a*tan(1/2*f*x + 1/2*e)^6 - 7*a*tan(1/2*f*x + 1/2*e)^5 - 8*I*a*tan(1/2*f*x +
 1/2*e)^4 + 7*a*tan(1/2*f*x + 1/2*e)^3 + 3*I*a*tan(1/2*f*x + 1/2*e)^2 - a*tan(1/2*f*x + 1/2*e))/(c^4*f*(tan(1/
2*f*x + 1/2*e) + I)^8)

Mupad [B] (verification not implemented)

Time = 5.73 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {a+i a \tan (e+f x)}{(c-i c \tan (e+f x))^4} \, dx=-\frac {a\,1{}\mathrm {i}}{4\,c^4\,f\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^4} \]

[In]

int((a + a*tan(e + f*x)*1i)/(c - c*tan(e + f*x)*1i)^4,x)

[Out]

-(a*1i)/(4*c^4*f*(tan(e + f*x) + 1i)^4)