3.40 \(\int (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=63 \[ -\frac{2 a^3 \tan (c+d x)}{d}-\frac{4 i a^3 \log (\cos (c+d x))}{d}+4 a^3 x+\frac{i a (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0317299, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3478, 3477, 3475} \[ -\frac{2 a^3 \tan (c+d x)}{d}-\frac{4 i a^3 \log (\cos (c+d x))}{d}+4 a^3 x+\frac{i a (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^3,x]

[Out]

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

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

Mathematica [A]  time = 0.784299, size = 119, normalized size = 1.89 \[ \frac{a^3 \sec (c) \sec ^2(c+d x) \left (-3 \sin (c+2 d x)+2 d x \cos (3 c+2 d x)-i \cos (3 c+2 d x) \log \left (\cos ^2(c+d x)\right )+\cos (c+2 d x) \left (2 d x-i \log \left (\cos ^2(c+d x)\right )\right )+\cos (c) \left (-2 i \log \left (\cos ^2(c+d x)\right )+4 d x-i\right )+3 \sin (c)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Sec[c]*Sec[c + d*x]^2*(2*d*x*Cos[3*c + 2*d*x] + Cos[c + 2*d*x]*(2*d*x - I*Log[Cos[c + d*x]^2]) + Cos[c]*(
-I + 4*d*x - (2*I)*Log[Cos[c + d*x]^2]) - I*Cos[3*c + 2*d*x]*Log[Cos[c + d*x]^2] + 3*Sin[c] - 3*Sin[c + 2*d*x]
))/(2*d)

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Maple [A]  time = 0.004, size = 68, normalized size = 1.1 \begin{align*} -3\,{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}-{\frac{{\frac{i}{2}}{a}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{2\,i{a}^{3}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+4\,{\frac{{a}^{3}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.66684, size = 103, normalized size = 1.63 \begin{align*} a^{3} x + \frac{3 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{3}}{d} + \frac{i \, a^{3}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{2 \, d} + \frac{3 i \, a^{3} \log \left (\sec \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + 3*(d*x + c - tan(d*x + c))*a^3/d + 1/2*I*a^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1))/d + 3*
I*a^3*log(sec(d*x + c))/d

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Fricas [A]  time = 1.23346, size = 267, normalized size = 4.24 \begin{align*} \frac{-8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, a^{3} +{\left (-4 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-8*I*a^3*e^(2*I*d*x + 2*I*c) - 6*I*a^3 + (-4*I*a^3*e^(4*I*d*x + 4*I*c) - 8*I*a^3*e^(2*I*d*x + 2*I*c) - 4*I*a^
3)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 1.35666, size = 100, normalized size = 1.59 \begin{align*} - \frac{4 i a^{3} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{- \frac{8 i a^{3} e^{- 2 i c} e^{2 i d x}}{d} - \frac{6 i a^{3} e^{- 4 i c}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*I*a**3*log(exp(2*I*d*x) + exp(-2*I*c))/d + (-8*I*a**3*exp(-2*I*c)*exp(2*I*d*x)/d - 6*I*a**3*exp(-4*I*c)/d)/
(exp(4*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

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Giac [B]  time = 1.11637, size = 158, normalized size = 2.51 \begin{align*} \frac{-4 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 6 i \, a^{3}}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-4*I*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 8*I*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*
c) + 1) - 8*I*a^3*e^(2*I*d*x + 2*I*c) - 4*I*a^3*log(e^(2*I*d*x + 2*I*c) + 1) - 6*I*a^3)/(d*e^(4*I*d*x + 4*I*c)
+ 2*d*e^(2*I*d*x + 2*I*c) + d)