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

Optimal. Leaf size=92 \[ \frac{3}{8 a^3 d (1+i \tan (c+d x))}+\frac{i x}{8 a^3}+\frac{i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{1}{8 a d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.129724, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3546, 3540, 3526, 8} \[ \frac{3}{8 a^3 d (1+i \tan (c+d x))}+\frac{i x}{8 a^3}+\frac{i \tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{1}{8 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.235596, size = 91, normalized size = 0.99 \[ -\frac{\sec ^3(c+d x) (27 \sin (c+d x)+12 i d x \sin (3 (c+d x))-2 \sin (3 (c+d x))-9 i \cos (c+d x)+2 (6 d x-i) \cos (3 (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.025, size = 97, normalized size = 1.1 \begin{align*}{\frac{{\frac{i}{6}}}{d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{{\frac{7\,i}{8}}}{d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{5}{8\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{16\,d{a}^{3}}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{16\,d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*I/d/a^3/(tan(d*x+c)-I)^3-7/8*I/d/a^3/(tan(d*x+c)-I)+5/8/d/a^3/(tan(d*x+c)-I)^2+1/16/d/a^3*ln(tan(d*x+c)-I)
-1/16/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(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.19763, size = 161, normalized size = 1.75 \begin{align*} \frac{{\left (12 i \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2\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(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.86485, size = 158, normalized size = 1.72 \begin{align*} \begin{cases} \frac{\left (4608 a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 2304 a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 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 (i e^{6 i c} - 3 i e^{4 i c} + 3 i e^{2 i c} - i\right ) e^{- 6 i c}}{8 a^{3}} - \frac{i}{8 a^{3}}\right ) & \text{otherwise} \end{cases} + \frac{i x}{8 a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.84574, size = 109, normalized size = 1.18 \begin{align*} \frac{\frac{6 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac{6 \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} - \frac{11 \, \tan \left (d x + c\right )^{3} + 51 i \, \tan \left (d x + c\right )^{2} + 75 \, \tan \left (d x + c\right ) - 29 i}{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(tan(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(6*log(tan(d*x + c) - I)/a^3 - 6*log(I*tan(d*x + c) - 1)/a^3 - (11*tan(d*x + c)^3 + 51*I*tan(d*x + c)^2 +
75*tan(d*x + c) - 29*I)/(a^3*(tan(d*x + c) - I)^3))/d