3.72 \(\int \frac{1}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=88 \[ \frac{i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{x}{8 a^3}+\frac{i}{8 a d (a+i a \tan (c+d x))^2}+\frac{i}{6 d (a+i a \tan (c+d x))^3} \]

[Out]

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

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Rubi [A]  time = 0.0499407, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3479, 8} \[ \frac{i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{x}{8 a^3}+\frac{i}{8 a d (a+i a \tan (c+d x))^2}+\frac{i}{6 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-3),x]

[Out]

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

Rule 3479

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(a + b*Tan[c + d*x])^n)/(2*b*d*n), x] +
Dist[1/(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] && LtQ[n
, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.192354, size = 93, normalized size = 1.06 \[ \frac{i \sec ^3(c+d x) (-9 \sin (c+d x)+12 i d x \sin (3 (c+d x))+2 \sin (3 (c+d x))+27 i \cos (c+d x)+2 (6 d x+i) \cos (3 (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/96)*Sec[c + d*x]^3*((27*I)*Cos[c + d*x] + 2*(I + 6*d*x)*Cos[3*(c + d*x)] - 9*Sin[c + d*x] + 2*Sin[3*(c + d
*x)] + (12*I)*d*x*Sin[3*(c + d*x)]))/(a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]  time = 0.023, size = 98, normalized size = 1.1 \begin{align*}{\frac{-{\frac{i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{3}}}-{\frac{{\frac{i}{8}}}{d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{1}{6\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{1}{8\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c))^3,x)

[Out]

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

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.08765, size = 166, normalized size = 1.89 \begin{align*} \frac{{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.725512, size = 156, normalized size = 1.77 \begin{align*} \begin{cases} \frac{\left (4608 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 2304 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text{for}\: 24576 a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac{\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac{1}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{x}{8 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((4608*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 2304*I*a**6*d**2*exp(8*I*c)*exp(-4*I*d*x) + 512*I*a**
6*d**2*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(24576*a**9*d**3), Ne(24576*a**9*d**3*exp(12*I*c), 0)), (x*((exp
(6*I*c) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) - 1/(8*a**3)), True)) + x/(8*a**3)

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Giac [A]  time = 1.27212, size = 108, normalized size = 1.23 \begin{align*} -\frac{\frac{6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac{6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac{-11 i \, \tan \left (d x + c\right )^{3} - 45 \, \tan \left (d x + c\right )^{2} + 69 i \, \tan \left (d x + c\right ) + 51}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/96*(6*I*log(tan(d*x + c) - I)/a^3 - 6*I*log(I*tan(d*x + c) - 1)/a^3 + (-11*I*tan(d*x + c)^3 - 45*tan(d*x +
c)^2 + 69*I*tan(d*x + c) + 51)/(a^3*(tan(d*x + c) - I)^3))/d