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

Optimal. Leaf size=71 \[ \frac {a^3 \tan (e+f x)}{c f}-\frac {4 i a^3}{f (c-i c \tan (e+f x))}+\frac {4 i a^3 \log (\cos (e+f x))}{c f}-\frac {4 a^3 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3522, 3487, 43} \[ \frac {a^3 \tan (e+f x)}{c f}-\frac {4 i a^3}{f (c-i c \tan (e+f x))}+\frac {4 i a^3 \log (\cos (e+f x))}{c f}-\frac {4 a^3 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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)} \, dx &=\left (a^3 c^3\right ) \int \frac {\sec ^6(e+f x)}{(c-i c \tan (e+f x))^4} \, dx\\ &=\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \frac {(c-x)^2}{(c+x)^2} \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=\frac {\left (i a^3\right ) \operatorname {Subst}\left (\int \left (1+\frac {4 c^2}{(c+x)^2}-\frac {4 c}{c+x}\right ) \, dx,x,-i c \tan (e+f x)\right )}{c^2 f}\\ &=-\frac {4 a^3 x}{c}+\frac {4 i a^3 \log (\cos (e+f x))}{c f}+\frac {a^3 \tan (e+f x)}{c f}-\frac {4 i a^3}{f (c-i c \tan (e+f x))}\\ \end {align*}

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Mathematica [B]  time = 2.70, size = 214, normalized size = 3.01 \[ \frac {a^3 \sec (e) (\tan (e+f x)-i) \left (-2 f x \sin (e+2 f x)+2 i \sin (e+2 f x)-2 f x \sin (3 e+2 f x)+i \sin (3 e+2 f x)-2 i f x \cos (3 e+2 f x)+\cos (3 e+2 f x)-\cos (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )+\cos (e) \left (-2 \log \left (\cos ^2(e+f x)\right )-4 i f x+3\right )+\cos (e+2 f x) \left (-\log \left (\cos ^2(e+f x)\right )-2 i f x\right )+i \sin (e+2 f x) \log \left (\cos ^2(e+f x)\right )+i \sin (3 e+2 f x) \log \left (\cos ^2(e+f x)\right )-i \sin (e)\right )}{2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e]*(Cos[3*e + 2*f*x] - (2*I)*f*x*Cos[3*e + 2*f*x] + Cos[e]*(3 - (4*I)*f*x - 2*Log[Cos[e + f*x]^2]) +
Cos[e + 2*f*x]*((-2*I)*f*x - Log[Cos[e + f*x]^2]) - Cos[3*e + 2*f*x]*Log[Cos[e + f*x]^2] - I*Sin[e] + (2*I)*Si
n[e + 2*f*x] - 2*f*x*Sin[e + 2*f*x] + I*Log[Cos[e + f*x]^2]*Sin[e + 2*f*x] + I*Sin[3*e + 2*f*x] - 2*f*x*Sin[3*
e + 2*f*x] + I*Log[Cos[e + f*x]^2]*Sin[3*e + 2*f*x])*(-I + Tan[e + f*x]))/(2*c*f)

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fricas [A]  time = 0.48, size = 86, normalized size = 1.21 \[ \frac {-2 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 2 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, a^{3} + {\left (4 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, a^{3}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}{c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.03, size = 183, normalized size = 2.58 \[ \frac {2 \, {\left (\frac {2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} - \frac {4 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}{c} + \frac {2 i \, a^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{c} + \frac {-2 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 i \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c} + \frac {6 i \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 16 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 i \, a^{3}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + i\right )}^{2}}\right )}}{f} \]

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)),x, algorithm="giac")

[Out]

2*(2*I*a^3*log(tan(1/2*f*x + 1/2*e) + 1)/c - 4*I*a^3*log(tan(1/2*f*x + 1/2*e) + I)/c + 2*I*a^3*log(tan(1/2*f*x
+ 1/2*e) - 1)/c + (-2*I*a^3*tan(1/2*f*x + 1/2*e)^2 - a^3*tan(1/2*f*x + 1/2*e) + 2*I*a^3)/((tan(1/2*f*x + 1/2*
e)^2 - 1)*c) + (6*I*a^3*tan(1/2*f*x + 1/2*e)^2 - 16*a^3*tan(1/2*f*x + 1/2*e) - 6*I*a^3)/(c*(tan(1/2*f*x + 1/2*
e) + I)^2))/f

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maple [A]  time = 0.20, size = 62, normalized size = 0.87 \[ \frac {a^{3} \tan \left (f x +e \right )}{c f}+\frac {4 a^{3}}{f c \left (\tan \left (f x +e \right )+i\right )}-\frac {4 i a^{3} \ln \left (\tan \left (f x +e \right )+i\right )}{f c} \]

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)),x)

[Out]

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

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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)),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 = 4.62, size = 61, normalized size = 0.86 \[ \frac {a^3\,\mathrm {tan}\left (e+f\,x\right )}{c\,f}+\frac {4\,a^3}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {a^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f} \]

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),x)

[Out]

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

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sympy [A]  time = 0.39, size = 102, normalized size = 1.44 \[ - \frac {2 i a^{3}}{- c f e^{2 i e} e^{2 i f x} - c f} + \frac {4 i a^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{c f} + \begin {cases} - \frac {2 i a^{3} e^{2 i e} e^{2 i f x}}{c f} & \text {for}\: c f \neq 0 \\\frac {4 a^{3} x e^{2 i e}}{c} & \text {otherwise} \end {cases} \]

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)),x)

[Out]

-2*I*a**3/(-c*f*exp(2*I*e)*exp(2*I*f*x) - c*f) + 4*I*a**3*log(exp(2*I*f*x) + exp(-2*I*e))/(c*f) + Piecewise((-
2*I*a**3*exp(2*I*e)*exp(2*I*f*x)/(c*f), Ne(c*f, 0)), (4*a**3*x*exp(2*I*e)/c, True))

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