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

Optimal. Leaf size=50 \[ -\frac {i a^4 \left (c^2+i c^2 \tan (e+f x)\right )^4}{8 f \left (c^3-i c^3 \tan (e+f x)\right )^4} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3522, 3487, 37} \[ -\frac {i a^4 \left (c^2+i c^2 \tan (e+f x)\right )^4}{8 f \left (c^3-i c^3 \tan (e+f x)\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*a^4*(c^2 + I*c^2*Tan[e + f*x])^4)/(f*(c^3 - I*c^3*Tan[e + f*x])^4)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 3487

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 3522

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*} \int \frac {(a+i a \tan (e+f x))^4}{(c-i c \tan (e+f x))^4} \, dx &=\left (a^4 c^4\right ) \int \frac {\sec ^8(e+f x)}{(c-i c \tan (e+f x))^8} \, dx\\ &=\frac {\left (i a^4\right ) \operatorname {Subst}\left (\int \frac {(c-x)^3}{(c+x)^5} \, dx,x,-i c \tan (e+f x)\right )}{c^3 f}\\ &=-\frac {i a^4 (c+i c \tan (e+f x))^4}{8 f \left (c^2-i c^2 \tan (e+f x)\right )^4}\\ \end {align*}

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Mathematica [A]  time = 0.41, size = 34, normalized size = 0.68 \[ \frac {a^4 (\sin (8 (e+f x))-i \cos (8 (e+f x)))}{8 c^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*((-I)*Cos[8*(e + f*x)] + Sin[8*(e + f*x)]))/(8*c^4*f)

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fricas [A]  time = 0.54, size = 20, normalized size = 0.40 \[ -\frac {i \, a^{4} e^{\left (8 i \, f x + 8 i \, e\right )}}{8 \, c^{4} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 2.31, size = 88, normalized size = 1.76 \[ -\frac {2 \, {\left (a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 7 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 7 \, a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - a^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.21, size = 66, normalized size = 1.32 \[ \frac {a^{4} \left (-\frac {1}{\tan \left (f x +e \right )+i}+\frac {4}{\left (\tan \left (f x +e \right )+i\right )^{3}}-\frac {2 i}{\left (\tan \left (f x +e \right )+i\right )^{4}}+\frac {3 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}\right )}{f \,c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 4.72, size = 69, normalized size = 1.38 \[ -\frac {a^4\,\mathrm {tan}\left (e+f\,x\right )\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2-1\right )}{c^4\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+{\mathrm {tan}\left (e+f\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (e+f\,x\right )}^2-\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^4*tan(e + f*x)*(tan(e + f*x)^2 - 1))/(c^4*f*(tan(e + f*x)^3*4i - 6*tan(e + f*x)^2 - tan(e + f*x)*4i + tan(
e + f*x)^4 + 1))

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sympy [A]  time = 0.43, size = 48, normalized size = 0.96 \[ \begin {cases} - \frac {i a^{4} e^{8 i e} e^{8 i f x}}{8 c^{4} f} & \text {for}\: 8 c^{4} f \neq 0 \\\frac {a^{4} x e^{8 i e}}{c^{4}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a**4*exp(8*I*e)*exp(8*I*f*x)/(8*c**4*f), Ne(8*c**4*f, 0)), (a**4*x*exp(8*I*e)/c**4, True))

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