3.10.11 \(\int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx\) [911]

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

[Out]

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

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Rubi [A]
time = 0.08, 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 = {3603, 3568, 37} \begin {gather*} \frac {i c^3 \left (a^2-i a^2 \tan (e+f x)\right )^3}{6 f \left (a^3+i a^3 \tan (e+f x)\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/6)*c^3*(a^2 - I*a^2*Tan[e + f*x])^3)/(f*(a^3 + I*a^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 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*} \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(a+i a \tan (e+f x))^6} \, dx\\ &=-\frac {\left (i c^3\right ) \text {Subst}\left (\int \frac {(a-x)^2}{(a+x)^4} \, dx,x,i a \tan (e+f x)\right )}{a^2 f}\\ &=\frac {i c^3 (1-i \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 34, normalized size = 0.68 \begin {gather*} \frac {c^3 (i \cos (6 (e+f x))+\sin (6 (e+f x)))}{6 a^3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.19, size = 50, normalized size = 1.00

method	result	size
risch	\(\frac {i c^{3} {\mathrm e}^{-6 i \left (f x +e \right )}}{6 a^{3} f}\)	\(22\)
derivativedivides	\(\frac {c^{3} \left (\frac {2 i}{\left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {1}{\tan \left (f x +e \right )-i}\right )}{f \,a^{3}}\)	\(50\)
default	\(\frac {c^{3} \left (\frac {2 i}{\left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4}{3 \left (\tan \left (f x +e \right )-i\right )^{3}}+\frac {1}{\tan \left (f x +e \right )-i}\right )}{f \,a^{3}}\)	\(50\)
norman	\(\frac {\frac {c^{3} \left (\tan ^{5}\left (f x +e \right )\right )}{a f}+\frac {c^{3} \tan \left (f x +e \right )}{a f}-\frac {2 i c^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{a f}+\frac {i c^{3}}{3 a f}-\frac {10 c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{3 a f}+\frac {3 i c^{3} \left (\tan ^{4}\left (f x +e \right )\right )}{a f}}{a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^3/(a+I*a*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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Fricas [A]
time = 1.30, size = 21, normalized size = 0.42 \begin {gather*} \frac {i \, c^{3} e^{\left (-6 i \, f x - 6 i \, e\right )}}{6 \, a^{3} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.19, size = 51, normalized size = 1.02 \begin {gather*} \begin {cases} \frac {i c^{3} e^{- 6 i e} e^{- 6 i f x}}{6 a^{3} f} & \text {for}\: a^{3} f e^{6 i e} \neq 0 \\\frac {c^{3} x e^{- 6 i e}}{a^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.81, size = 72, normalized size = 1.44 \begin {gather*} -\frac {2 \, {\left (3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{3 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.82, size = 59, normalized size = 1.18 \begin {gather*} -\frac {c^3\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}-\frac {1}{3}{}\mathrm {i}\right )}{a^3\,f\,\left (-{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,3{}\mathrm {i}+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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