3.941 \(\int \frac {(a+i a \tan (e+f x))^3}{(c-i c \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=50 \[ -\frac {i a^3 \left (c^2+i c^2 \tan (e+f x)\right )^3}{6 f \left (c^3-i c^3 \tan (e+f x)\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.11, 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} \[ -\frac {i a^3 \left (c^2+i c^2 \tan (e+f x)\right )^3}{6 f \left (c^3-i c^3 \tan (e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/6)*a^3*(c^2 + I*c^2*Tan[e + f*x])^3)/(f*(c^3 - I*c^3*Tan[e + f*x])^3)

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))^3}{(c-i c \tan (e+f x))^3} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^6} \, dx\\ &=\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {(c-x)^2}{(c+x)^4} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=-\frac {i a^3 (c+i c \tan (e+f x))^3}{6 f \left (c^2-i c^2 \tan (e+f x)\right )^3}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 34, normalized size = 0.68 \[ \frac {a^3 (\sin (6 (e+f x))-i \cos (6 (e+f x)))}{6 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*((-I)*Cos[6*(e + f*x)] + Sin[6*(e + f*x)]))/(6*c^3*f)

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fricas [A]  time = 0.49, size = 20, normalized size = 0.40 \[ -\frac {i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{6 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*I*a^3*e^(6*I*f*x + 6*I*e)/(c^3*f)

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giac [A]  time = 1.65, size = 72, normalized size = 1.44 \[ -\frac {2 \, {\left (3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*a^3*tan(1/2*f*x + 1/2*e)^5 - 10*a^3*tan(1/2*f*x + 1/2*e)^3 + 3*a^3*tan(1/2*f*x + 1/2*e))/(c^3*f*(tan(1
/2*f*x + 1/2*e) + I)^6)

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maple [A]  time = 0.23, size = 50, normalized size = 1.00 \[ \frac {a^{3} \left (-\frac {2 i}{\left (\tan \left (f x +e \right )+i\right )^{2}}+\frac {1}{\tan \left (f x +e \right )+i}-\frac {4}{3 \left (\tan \left (f x +e \right )+i\right )^{3}}\right )}{f \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^3,x)

[Out]

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

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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))^3/(c-I*c*tan(f*x+e))^3,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 = 4.71, size = 55, normalized size = 1.10 \[ -\frac {a^3\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2-\frac {1}{3}\right )}{c^3\,f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.32, size = 48, normalized size = 0.96 \[ \begin {cases} - \frac {i a^{3} e^{6 i e} e^{6 i f x}}{6 c^{3} f} & \text {for}\: 6 c^{3} f \neq 0 \\\frac {a^{3} x e^{6 i e}}{c^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*a**3*exp(6*I*e)*exp(6*I*f*x)/(6*c**3*f), Ne(6*c**3*f, 0)), (a**3*x*exp(6*I*e)/c**3, True))

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