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

Optimal. Leaf size=79 \[ -\frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{\log (\cos (c+d x))}{a^2 d}-\frac{3 i x}{4 a^2}-\frac{\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(((-3*I)/4)*x)/a^2 + Log[Cos[c + d*x]]/(a^2*d) - 3/(4*a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^2/(4*d*(a + I
*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.140027, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3558, 3589, 3475, 12, 3526, 8} \[ -\frac{3}{4 a^2 d (1+i \tan (c+d x))}+\frac{\log (\cos (c+d x))}{a^2 d}-\frac{3 i x}{4 a^2}-\frac{\tan ^2(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/4)*x)/a^2 + Log[Cos[c + d*x]]/(a^2*d) - 3/(4*a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^2/(4*d*(a + I
*a*Tan[c + d*x])^2)

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0]

Rule 3475

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.344092, size = 135, normalized size = 1.71 \[ \frac{\sec ^2(c+d x) \left (4 d x \sin (2 (c+d x))+i \sin (2 (c+d x))+\cos (2 (c+d x)) \left (-8 \log \left (\cos ^2(c+d x)\right )-4 i d x-1\right )-8 i \sin (2 (c+d x)) \log \left (\cos ^2(c+d x)\right )+16 i \tan ^{-1}(\tan (d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+8\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^2*(8 + Cos[2*(c + d*x)]*(-1 - (4*I)*d*x - 8*Log[Cos[c + d*x]^2]) + (16*I)*ArcTan[Tan[d*x]]*(Cos[
2*(c + d*x)] + I*Sin[2*(c + d*x)]) + I*Sin[2*(c + d*x)] + 4*d*x*Sin[2*(c + d*x)] - (8*I)*Log[Cos[c + d*x]^2]*S
in[2*(c + d*x)]))/(16*a^2*d*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 0.023, size = 77, normalized size = 1. \begin{align*}{\frac{{\frac{5\,i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{1}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{7\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{8\,{a}^{2}d}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}} \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))^2,x)

[Out]

5/4*I/d/a^2/(tan(d*x+c)-I)-1/4/d/a^2/(tan(d*x+c)-I)^2-7/8/d/a^2*ln(tan(d*x+c)-I)-1/8/d/a^2*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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.19746, size = 201, normalized size = 2.54 \begin{align*} \frac{{\left (-28 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 16 \, 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 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/16*(-28*I*d*x*e^(4*I*d*x + 4*I*c) + 16*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 8*e^(2*I*d*x + 2*I
*c) + 1)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [A]  time = 1.9938, size = 150, normalized size = 1.9 \begin{align*} \begin{cases} \frac{\left (- 16 a^{2} d e^{4 i c} e^{- 2 i d x} + 2 a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{32 a^{4} d^{2}} & \text{for}\: 32 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (- \frac{\left (7 i e^{4 i c} - 4 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} + \frac{7 i}{4 a^{2}}\right ) & \text{otherwise} \end{cases} - \frac{7 i x}{4 a^{2}} + \frac{\log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} 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))**2,x)

[Out]

Piecewise(((-16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 2*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(32*a**4*d**2
), Ne(32*a**4*d**2*exp(6*I*c), 0)), (x*(-(7*I*exp(4*I*c) - 4*I*exp(2*I*c) + I)*exp(-4*I*c)/(4*a**2) + 7*I/(4*a
**2)), True)) - 7*I*x/(4*a**2) + log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)

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Giac [A]  time = 1.86398, size = 93, normalized size = 1.18 \begin{align*} -\frac{\frac{2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{14 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{21 \, \tan \left (d x + c\right )^{2} - 22 i \, \tan \left (d x + c\right ) - 5}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, 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))^2,x, algorithm="giac")

[Out]

-1/16*(2*log(tan(d*x + c) + I)/a^2 + 14*log(tan(d*x + c) - I)/a^2 - (21*tan(d*x + c)^2 - 22*I*tan(d*x + c) - 5
)/(a^2*(tan(d*x + c) - I)^2))/d