3.1082 \(\int \frac{(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=140 \[ \frac{(c+i d) (c-3 i d) (d+i c)}{8 a^3 f (1+i \tan (e+f x))}+\frac{x (c-i d)^3}{8 a^3}+\frac{i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac{(c+i d)^2 (d+i c)}{8 a f (a+i a \tan (e+f x))^2} \]

[Out]

((c - I*d)^3*x)/(8*a^3) + ((c + I*d)*(c - (3*I)*d)*(I*c + d))/(8*a^3*f*(1 + I*Tan[e + f*x])) + ((c + I*d)^2*(I
*c + d))/(8*a*f*(a + I*a*Tan[e + f*x])^2) + ((I/6)*(c + d*Tan[e + f*x])^3)/(f*(a + I*a*Tan[e + f*x])^3)

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Rubi [A]  time = 0.219159, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3546, 3540, 3526, 8} \[ \frac{(c+i d) (c-3 i d) (d+i c)}{8 a^3 f (1+i \tan (e+f x))}+\frac{x (c-i d)^3}{8 a^3}+\frac{i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac{(c+i d)^2 (d+i c)}{8 a f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((c - I*d)^3*x)/(8*a^3) + ((c + I*d)*(c - (3*I)*d)*(I*c + d))/(8*a^3*f*(1 + I*Tan[e + f*x])) + ((c + I*d)^2*(I
*c + d))/(8*a*f*(a + I*a*Tan[e + f*x])^2) + ((I/6)*(c + d*Tan[e + f*x])^3)/(f*(a + I*a*Tan[e + f*x])^3)

Rule 3546

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rule 3540

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
(b*(a*c + b*d)^2*(a + b*Tan[e + f*x])^m)/(2*a^3*f*m), x] + Dist[1/(2*a^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Si
mp[a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{(c+d \tan (e+f x))^3}{(a+i a \tan (e+f x))^3} \, dx &=\frac{i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac{(c-i d) \int \frac{(c+d \tan (e+f x))^2}{(a+i a \tan (e+f x))^2} \, dx}{2 a}\\ &=\frac{(c+i d)^2 (i c+d)}{8 a f (a+i a \tan (e+f x))^2}+\frac{i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac{(c-i d) \int \frac{a \left (c^2-2 i c d+d^2\right )-2 i a d^2 \tan (e+f x)}{a+i a \tan (e+f x)} \, dx}{4 a^3}\\ &=\frac{(c+i d) (c-3 i d) (i c+d)}{8 a^3 f (1+i \tan (e+f x))}+\frac{(c+i d)^2 (i c+d)}{8 a f (a+i a \tan (e+f x))^2}+\frac{i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}+\frac{(c-i d)^3 \int 1 \, dx}{8 a^3}\\ &=\frac{(c-i d)^3 x}{8 a^3}+\frac{(c+i d) (c-3 i d) (i c+d)}{8 a^3 f (1+i \tan (e+f x))}+\frac{(c+i d)^2 (i c+d)}{8 a f (a+i a \tan (e+f x))^2}+\frac{i (c+d \tan (e+f x))^3}{6 f (a+i a \tan (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 1.79059, size = 260, normalized size = 1.86 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (12 f x (c-i d)^3 (\cos (3 e)+i \sin (3 e))+18 i (c+i d) (c-i d)^2 (\cos (e)+i \sin (e)) \cos (2 f x)+18 (c+i d) (c-i d)^2 (\cos (e)+i \sin (e)) \sin (2 f x)+9 (c+i d)^2 (c-i d) (\cos (e)-i \sin (e)) \sin (4 f x)+9 (c+i d)^2 (d+i c) (\cos (e)-i \sin (e)) \cos (4 f x)+2 (c+i d)^3 (\sin (3 e)+i \cos (3 e)) \cos (6 f x)+2 (c+i d)^3 (\cos (3 e)-i \sin (3 e)) \sin (6 f x)\right )}{96 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*(9*(c + I*d)^2*(I*c + d)*Cos[4*f*x]*(Cos[e] - I*Sin[e]) + (18*I)*(c
- I*d)^2*(c + I*d)*Cos[2*f*x]*(Cos[e] + I*Sin[e]) + 12*(c - I*d)^3*f*x*(Cos[3*e] + I*Sin[3*e]) + 2*(c + I*d)^3
*Cos[6*f*x]*(I*Cos[3*e] + Sin[3*e]) + 18*(c - I*d)^2*(c + I*d)*(Cos[e] + I*Sin[e])*Sin[2*f*x] + 9*(c - I*d)*(c
+ I*d)^2*(Cos[e] - I*Sin[e])*Sin[4*f*x] + 2*(c + I*d)^3*(Cos[3*e] - I*Sin[3*e])*Sin[6*f*x]))/(96*f*(a + I*a*T
an[e + f*x])^3)

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Maple [B]  time = 0.036, size = 454, normalized size = 3.2 \begin{align*}{\frac{{c}^{3}}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{3\,c{d}^{2}}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{8}}{c}^{3}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{9\,i}{8}}c{d}^{2}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ){c}^{3}}{f{a}^{3}}}+{\frac{c{d}^{2}}{2\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}+{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ){c}^{3}}{f{a}^{3}}}-{\frac{{c}^{3}}{6\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{\frac{7\,i}{8}}{d}^{3}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}+{\frac{\ln \left ( \tan \left ( fx+e \right ) -i \right ){d}^{3}}{16\,f{a}^{3}}}+{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) -i \right ) c{d}^{2}}{f{a}^{3}}}-{\frac{3\,\ln \left ( \tan \left ( fx+e \right ) -i \right ){c}^{2}d}{16\,f{a}^{3}}}-{\frac{3\,{c}^{2}d}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{5\,{d}^{3}}{8\,f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{2}}}+{\frac{{\frac{i}{6}}{d}^{3}}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}}-{\frac{{\frac{3\,i}{16}}\ln \left ( \tan \left ( fx+e \right ) +i \right ) c{d}^{2}}{f{a}^{3}}}+{\frac{3\,\ln \left ( \tan \left ( fx+e \right ) +i \right ){c}^{2}d}{16\,f{a}^{3}}}-{\frac{\ln \left ( \tan \left ( fx+e \right ) +i \right ){d}^{3}}{16\,f{a}^{3}}}-{\frac{{\frac{3\,i}{8}}{c}^{2}d}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) }}-{\frac{{\frac{i}{2}}{c}^{2}d}{f{a}^{3} \left ( \tan \left ( fx+e \right ) -i \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/f/a^3/(tan(f*x+e)-I)*c^3-3/8/f/a^3/(tan(f*x+e)-I)*c*d^2-1/8*I/f/a^3/(tan(f*x+e)-I)^2*c^3-9/8*I/f/a^3/(tan(
f*x+e)-I)^2*c*d^2-1/16*I/f/a^3*ln(tan(f*x+e)-I)*c^3+1/2/f/a^3/(tan(f*x+e)-I)^3*c*d^2+1/16*I/f/a^3*ln(tan(f*x+e
)+I)*c^3-1/6/f/a^3/(tan(f*x+e)-I)^3*c^3-7/8*I/f/a^3/(tan(f*x+e)-I)*d^3+1/16/f/a^3*ln(tan(f*x+e)-I)*d^3+3/16*I/
f/a^3*ln(tan(f*x+e)-I)*c*d^2-3/16/f/a^3*ln(tan(f*x+e)-I)*c^2*d-3/8/f/a^3/(tan(f*x+e)-I)^2*c^2*d+5/8/f/a^3/(tan
(f*x+e)-I)^2*d^3+1/6*I/f/a^3/(tan(f*x+e)-I)^3*d^3-3/16*I/f/a^3*ln(tan(f*x+e)+I)*c*d^2+3/16/f/a^3*ln(tan(f*x+e)
+I)*c^2*d-1/16/f/a^3*ln(tan(f*x+e)+I)*d^3-3/8*I/f/a^3/(tan(f*x+e)-I)*c^2*d-1/2*I/f/a^3/(tan(f*x+e)-I)^3*c^2*d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.61577, size = 369, normalized size = 2.64 \begin{align*} \frac{{\left ({\left (12 \, c^{3} - 36 i \, c^{2} d - 36 \, c d^{2} + 12 i \, d^{3}\right )} f x e^{\left (6 i \, f x + 6 i \, e\right )} + 2 i \, c^{3} - 6 \, c^{2} d - 6 i \, c d^{2} + 2 \, d^{3} +{\left (18 i \, c^{3} + 18 \, c^{2} d + 18 i \, c d^{2} + 18 \, d^{3}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (9 i \, c^{3} - 9 \, c^{2} d + 9 i \, c d^{2} - 9 \, d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*((12*c^3 - 36*I*c^2*d - 36*c*d^2 + 12*I*d^3)*f*x*e^(6*I*f*x + 6*I*e) + 2*I*c^3 - 6*c^2*d - 6*I*c*d^2 + 2*
d^3 + (18*I*c^3 + 18*c^2*d + 18*I*c*d^2 + 18*d^3)*e^(4*I*f*x + 4*I*e) + (9*I*c^3 - 9*c^2*d + 9*I*c*d^2 - 9*d^3
)*e^(2*I*f*x + 2*I*e))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

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Sympy [A]  time = 1.61707, size = 554, normalized size = 3.96 \begin{align*} \begin{cases} \frac{\left (\left (512 i a^{6} c^{3} f^{2} e^{6 i e} - 1536 a^{6} c^{2} d f^{2} e^{6 i e} - 1536 i a^{6} c d^{2} f^{2} e^{6 i e} + 512 a^{6} d^{3} f^{2} e^{6 i e}\right ) e^{- 6 i f x} + \left (2304 i a^{6} c^{3} f^{2} e^{8 i e} - 2304 a^{6} c^{2} d f^{2} e^{8 i e} + 2304 i a^{6} c d^{2} f^{2} e^{8 i e} - 2304 a^{6} d^{3} f^{2} e^{8 i e}\right ) e^{- 4 i f x} + \left (4608 i a^{6} c^{3} f^{2} e^{10 i e} + 4608 a^{6} c^{2} d f^{2} e^{10 i e} + 4608 i a^{6} c d^{2} f^{2} e^{10 i e} + 4608 a^{6} d^{3} f^{2} e^{10 i e}\right ) e^{- 2 i f x}\right ) e^{- 12 i e}}{24576 a^{9} f^{3}} & \text{for}\: 24576 a^{9} f^{3} e^{12 i e} \neq 0 \\x \left (- \frac{c^{3} - 3 i c^{2} d - 3 c d^{2} + i d^{3}}{8 a^{3}} + \frac{\left (c^{3} e^{6 i e} + 3 c^{3} e^{4 i e} + 3 c^{3} e^{2 i e} + c^{3} - 3 i c^{2} d e^{6 i e} - 3 i c^{2} d e^{4 i e} + 3 i c^{2} d e^{2 i e} + 3 i c^{2} d - 3 c d^{2} e^{6 i e} + 3 c d^{2} e^{4 i e} + 3 c d^{2} e^{2 i e} - 3 c d^{2} + i d^{3} e^{6 i e} - 3 i d^{3} e^{4 i e} + 3 i d^{3} e^{2 i e} - i d^{3}\right ) e^{- 6 i e}}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x \left (c^{3} - 3 i c^{2} d - 3 c d^{2} + i d^{3}\right )}{8 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((((512*I*a**6*c**3*f**2*exp(6*I*e) - 1536*a**6*c**2*d*f**2*exp(6*I*e) - 1536*I*a**6*c*d**2*f**2*exp(
6*I*e) + 512*a**6*d**3*f**2*exp(6*I*e))*exp(-6*I*f*x) + (2304*I*a**6*c**3*f**2*exp(8*I*e) - 2304*a**6*c**2*d*f
**2*exp(8*I*e) + 2304*I*a**6*c*d**2*f**2*exp(8*I*e) - 2304*a**6*d**3*f**2*exp(8*I*e))*exp(-4*I*f*x) + (4608*I*
a**6*c**3*f**2*exp(10*I*e) + 4608*a**6*c**2*d*f**2*exp(10*I*e) + 4608*I*a**6*c*d**2*f**2*exp(10*I*e) + 4608*a*
*6*d**3*f**2*exp(10*I*e))*exp(-2*I*f*x))*exp(-12*I*e)/(24576*a**9*f**3), Ne(24576*a**9*f**3*exp(12*I*e), 0)),
(x*(-(c**3 - 3*I*c**2*d - 3*c*d**2 + I*d**3)/(8*a**3) + (c**3*exp(6*I*e) + 3*c**3*exp(4*I*e) + 3*c**3*exp(2*I*
e) + c**3 - 3*I*c**2*d*exp(6*I*e) - 3*I*c**2*d*exp(4*I*e) + 3*I*c**2*d*exp(2*I*e) + 3*I*c**2*d - 3*c*d**2*exp(
6*I*e) + 3*c*d**2*exp(4*I*e) + 3*c*d**2*exp(2*I*e) - 3*c*d**2 + I*d**3*exp(6*I*e) - 3*I*d**3*exp(4*I*e) + 3*I*
d**3*exp(2*I*e) - I*d**3)*exp(-6*I*e)/(8*a**3)), True)) + x*(c**3 - 3*I*c**2*d - 3*c*d**2 + I*d**3)/(8*a**3)

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Giac [B]  time = 1.99743, size = 386, normalized size = 2.76 \begin{align*} -\frac{\frac{6 \,{\left (i \, c^{3} + 3 \, c^{2} d - 3 i \, c d^{2} - d^{3}\right )} \log \left (\tan \left (f x + e\right ) - i\right )}{a^{3}} + \frac{6 \,{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \log \left (i \, \tan \left (f x + e\right ) - 1\right )}{a^{3}} + \frac{-11 i \, c^{3} \tan \left (f x + e\right )^{3} - 33 \, c^{2} d \tan \left (f x + e\right )^{3} + 33 i \, c d^{2} \tan \left (f x + e\right )^{3} + 11 \, d^{3} \tan \left (f x + e\right )^{3} - 45 \, c^{3} \tan \left (f x + e\right )^{2} + 135 i \, c^{2} d \tan \left (f x + e\right )^{2} + 135 \, c d^{2} \tan \left (f x + e\right )^{2} + 51 i \, d^{3} \tan \left (f x + e\right )^{2} + 69 i \, c^{3} \tan \left (f x + e\right ) + 207 \, c^{2} d \tan \left (f x + e\right ) - 63 i \, c d^{2} \tan \left (f x + e\right ) + 75 \, d^{3} \tan \left (f x + e\right ) + 51 \, c^{3} - 57 i \, c^{2} d - 9 \, c d^{2} - 29 i \, d^{3}}{a^{3}{\left (\tan \left (f x + e\right ) - i\right )}^{3}}}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*(I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*log(tan(f*x + e) - I)/a^3 + 6*(-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3
)*log(I*tan(f*x + e) - 1)/a^3 + (-11*I*c^3*tan(f*x + e)^3 - 33*c^2*d*tan(f*x + e)^3 + 33*I*c*d^2*tan(f*x + e)^
3 + 11*d^3*tan(f*x + e)^3 - 45*c^3*tan(f*x + e)^2 + 135*I*c^2*d*tan(f*x + e)^2 + 135*c*d^2*tan(f*x + e)^2 + 51
*I*d^3*tan(f*x + e)^2 + 69*I*c^3*tan(f*x + e) + 207*c^2*d*tan(f*x + e) - 63*I*c*d^2*tan(f*x + e) + 75*d^3*tan(
f*x + e) + 51*c^3 - 57*I*c^2*d - 9*c*d^2 - 29*I*d^3)/(a^3*(tan(f*x + e) - I)^3))/f