3.1065 \(\int (a+i a \tan (e+f x))^3 (c+d \tan (e+f x)) \, dx\)

Optimal. Leaf size=110 \[ -\frac{2 a^3 (c-i d) \tan (e+f x)}{f}-\frac{4 a^3 (d+i c) \log (\cos (e+f x))}{f}+4 a^3 x (c-i d)+\frac{a (d+i c) (a+i a \tan (e+f x))^2}{2 f}+\frac{d (a+i a \tan (e+f x))^3}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0952098, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3527, 3478, 3477, 3475} \[ -\frac{2 a^3 (c-i d) \tan (e+f x)}{f}-\frac{4 a^3 (d+i c) \log (\cos (e+f x))}{f}+4 a^3 x (c-i d)+\frac{a (d+i c) (a+i a \tan (e+f x))^2}{2 f}+\frac{d (a+i a \tan (e+f x))^3}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3527

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

Rule 3478

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

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 4.29233, size = 331, normalized size = 3.01 \[ \frac{a^3 \sec (e) \sec ^3(e+f x) \left (3 \cos (f x) \left ((-3 d-3 i c) \log \left (\cos ^2(e+f x)\right )+6 c f x-i c-6 i d f x-3 d\right )+3 \cos (2 e+f x) \left ((-3 d-3 i c) \log \left (\cos ^2(e+f x)\right )+6 c f x-i c-6 i d f x-3 d\right )+9 c \sin (2 e+f x)-9 c \sin (2 e+3 f x)+6 c f x \cos (2 e+3 f x)+6 c f x \cos (4 e+3 f x)-3 i c \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 i c \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )-18 c \sin (f x)-15 i d \sin (2 e+f x)+13 i d \sin (2 e+3 f x)-6 i d f x \cos (2 e+3 f x)-6 i d f x \cos (4 e+3 f x)-3 d \cos (2 e+3 f x) \log \left (\cos ^2(e+f x)\right )-3 d \cos (4 e+3 f x) \log \left (\cos ^2(e+f x)\right )+24 i d \sin (f x)\right )}{12 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[e]*Sec[e + f*x]^3*(6*c*f*x*Cos[2*e + 3*f*x] - (6*I)*d*f*x*Cos[2*e + 3*f*x] + 6*c*f*x*Cos[4*e + 3*f*x]
- (6*I)*d*f*x*Cos[4*e + 3*f*x] - (3*I)*c*Cos[2*e + 3*f*x]*Log[Cos[e + f*x]^2] - 3*d*Cos[2*e + 3*f*x]*Log[Cos[
e + f*x]^2] - (3*I)*c*Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] - 3*d*Cos[4*e + 3*f*x]*Log[Cos[e + f*x]^2] + 3*Cos[
f*x]*((-I)*c - 3*d + 6*c*f*x - (6*I)*d*f*x + ((-3*I)*c - 3*d)*Log[Cos[e + f*x]^2]) + 3*Cos[2*e + f*x]*((-I)*c
- 3*d + 6*c*f*x - (6*I)*d*f*x + ((-3*I)*c - 3*d)*Log[Cos[e + f*x]^2]) - 18*c*Sin[f*x] + (24*I)*d*Sin[f*x] + 9*
c*Sin[2*e + f*x] - (15*I)*d*Sin[2*e + f*x] - 9*c*Sin[2*e + 3*f*x] + (13*I)*d*Sin[2*e + 3*f*x]))/(12*f)

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Maple [A]  time = 0.011, size = 160, normalized size = 1.5 \begin{align*}{\frac{-{\frac{i}{3}}{a}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{3}d}{f}}-{\frac{{\frac{i}{2}}{a}^{3}c \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{f}}+{\frac{4\,i{a}^{3}d\tan \left ( fx+e \right ) }{f}}-{\frac{3\,{a}^{3} \left ( \tan \left ( fx+e \right ) \right ) ^{2}d}{2\,f}}-3\,{\frac{{a}^{3}c\tan \left ( fx+e \right ) }{f}}+{\frac{2\,i{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) c}{f}}+2\,{\frac{{a}^{3}\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) d}{f}}-{\frac{4\,i{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) d}{f}}+4\,{\frac{{a}^{3}\arctan \left ( \tan \left ( fx+e \right ) \right ) c}{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.54406, size = 147, normalized size = 1.34 \begin{align*} -\frac{2 i \, a^{3} d \tan \left (f x + e\right )^{3} -{\left (-3 i \, a^{3} c - 9 \, a^{3} d\right )} \tan \left (f x + e\right )^{2} - 24 \,{\left (a^{3} c - i \, a^{3} d\right )}{\left (f x + e\right )} - 6 \,{\left (2 i \, a^{3} c + 2 \, a^{3} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 6 \,{\left (3 \, a^{3} c - 4 i \, a^{3} d\right )} \tan \left (f x + e\right )}{6 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*I*a^3*d*tan(f*x + e)^3 - (-3*I*a^3*c - 9*a^3*d)*tan(f*x + e)^2 - 24*(a^3*c - I*a^3*d)*(f*x + e) - 6*(2
*I*a^3*c + 2*a^3*d)*log(tan(f*x + e)^2 + 1) + 6*(3*a^3*c - 4*I*a^3*d)*tan(f*x + e))/f

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Fricas [B]  time = 1.58663, size = 508, normalized size = 4.62 \begin{align*} -\frac{2 \,{\left (9 i \, a^{3} c + 13 \, a^{3} d + 12 \,{\left (i \, a^{3} c + 2 \, a^{3} d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \,{\left (7 i \, a^{3} c + 11 \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 6 \,{\left (i \, a^{3} c + a^{3} d +{\left (i \, a^{3} c + a^{3} d\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \,{\left (i \, a^{3} c + a^{3} d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \,{\left (i \, a^{3} c + a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )\right )}}{3 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(9*I*a^3*c + 13*a^3*d + 12*(I*a^3*c + 2*a^3*d)*e^(4*I*f*x + 4*I*e) + 3*(7*I*a^3*c + 11*a^3*d)*e^(2*I*f*x
+ 2*I*e) + 6*(I*a^3*c + a^3*d + (I*a^3*c + a^3*d)*e^(6*I*f*x + 6*I*e) + 3*(I*a^3*c + a^3*d)*e^(4*I*f*x + 4*I*e
) + 3*(I*a^3*c + a^3*d)*e^(2*I*f*x + 2*I*e))*log(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I
*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [A]  time = 5.78776, size = 172, normalized size = 1.56 \begin{align*} - \frac{4 a^{3} \left (i c + d\right ) \log{\left (e^{2 i f x} + e^{- 2 i e} \right )}}{f} + \frac{- \frac{\left (8 i a^{3} c + 16 a^{3} d\right ) e^{- 2 i e} e^{4 i f x}}{f} - \frac{\left (14 i a^{3} c + 22 a^{3} d\right ) e^{- 4 i e} e^{2 i f x}}{f} - \frac{\left (18 i a^{3} c + 26 a^{3} d\right ) e^{- 6 i e}}{3 f}}{e^{6 i f x} + 3 e^{- 2 i e} e^{4 i f x} + 3 e^{- 4 i e} e^{2 i f x} + e^{- 6 i e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*a**3*(I*c + d)*log(exp(2*I*f*x) + exp(-2*I*e))/f + (-(8*I*a**3*c + 16*a**3*d)*exp(-2*I*e)*exp(4*I*f*x)/f -
(14*I*a**3*c + 22*a**3*d)*exp(-4*I*e)*exp(2*I*f*x)/f - (18*I*a**3*c + 26*a**3*d)*exp(-6*I*e)/(3*f))/(exp(6*I*f
*x) + 3*exp(-2*I*e)*exp(4*I*f*x) + 3*exp(-4*I*e)*exp(2*I*f*x) + exp(-6*I*e))

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Giac [B]  time = 1.53138, size = 450, normalized size = 4.09 \begin{align*} \frac{-12 i \, a^{3} c e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 12 \, a^{3} d e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 i \, a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 \, a^{3} d e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 i \, a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 36 \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 24 i \, a^{3} c e^{\left (4 i \, f x + 4 i \, e\right )} - 48 \, a^{3} d e^{\left (4 i \, f x + 4 i \, e\right )} - 42 i \, a^{3} c e^{\left (2 i \, f x + 2 i \, e\right )} - 66 \, a^{3} d e^{\left (2 i \, f x + 2 i \, e\right )} - 12 i \, a^{3} c \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 12 \, a^{3} d \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 18 i \, a^{3} c - 26 \, a^{3} d}{3 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(-12*I*a^3*c*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 12*a^3*d*e^(6*I*f*x + 6*I*e)*log(e^(2*I*f*
x + 2*I*e) + 1) - 36*I*a^3*c*e^(4*I*f*x + 4*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 36*a^3*d*e^(4*I*f*x + 4*I*e)*l
og(e^(2*I*f*x + 2*I*e) + 1) - 36*I*a^3*c*e^(2*I*f*x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 36*a^3*d*e^(2*I*f*
x + 2*I*e)*log(e^(2*I*f*x + 2*I*e) + 1) - 24*I*a^3*c*e^(4*I*f*x + 4*I*e) - 48*a^3*d*e^(4*I*f*x + 4*I*e) - 42*I
*a^3*c*e^(2*I*f*x + 2*I*e) - 66*a^3*d*e^(2*I*f*x + 2*I*e) - 12*I*a^3*c*log(e^(2*I*f*x + 2*I*e) + 1) - 12*a^3*d
*log(e^(2*I*f*x + 2*I*e) + 1) - 18*I*a^3*c - 26*a^3*d)/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*
e^(2*I*f*x + 2*I*e) + f)