3.70 \(\int \frac{\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0982562, antiderivative size = 88, 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 = {3540, 3526, 3479, 8} \[ -\frac{i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{x}{8 a^3}+\frac{3 i}{8 a d (a+i a \tan (c+d x))^2}-\frac{i}{6 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^2/(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) + ((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 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 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{\tan ^2(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{i}{6 d (a+i a \tan (c+d x))^3}+\frac{\int \frac{a-2 i a \tan (c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{2 a^2}\\ &=-\frac{i}{6 d (a+i a \tan (c+d x))^3}+\frac{3 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{3 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{3 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 = 0.441476, size = 91, normalized size = 1.03 \[ \frac{\sec ^3(c+d x) (-3 i \sin (c+d x)+12 d x \sin (3 (c+d x))-2 i \sin (3 (c+d x))-9 \cos (c+d x)+2 (1-6 i d x) \cos (3 (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(-9*Cos[c + d*x] + 2*(1 - (6*I)*d*x)*Cos[3*(c + d*x)] - (3*I)*Sin[c + d*x] - (2*I)*Sin[3*(c +
d*x)] + 12*d*x*Sin[3*(c + d*x)]))/(96*a^3*d*(-I + Tan[c + d*x])^3)

________________________________________________________________________________________

Maple [A]  time = 0.026, 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{3\,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(tan(d*x+c)^2/(a+I*a*tan(d*x+c))^3,x)

[Out]

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

________________________________________________________________________________________

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)^2/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [A]  time = 2.1566, size = 166, normalized size = 1.89 \begin{align*} -\frac{{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 3 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(tan(d*x+c)^2/(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) - 6*I*e^(4*I*d*x + 4*I*c) - 3*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(-6*I*d*x - 6*I
*c)/(a^3*d)

________________________________________________________________________________________

Sympy [A]  time = 1.11927, size = 153, normalized size = 1.74 \begin{align*} \begin{cases} \frac{\left (1536 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 768 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} - e^{4 i c} - 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(tan(d*x+c)**2/(a+I*a*tan(d*x+c))**3,x)

[Out]

Piecewise(((1536*I*a**6*d**2*exp(10*I*c)*exp(-2*I*d*x) + 768*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) - exp(4*I*c) - exp(2*I*c) + 1)*exp(-6*I*c)/(8*a**3) + 1/(8*a**3)), True)) - x/(8*a**3)

________________________________________________________________________________________

Giac [A]  time = 1.51162, 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} - 21 i \, \tan \left (d x + c\right ) - 3}{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)^2/(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 - 21*I*tan(d*x + c) - 3)/(a^3*(tan(d*x + c) - I)^3))/d