3.14 \(\int \tan ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=93 \[ -\frac{a^2 \tan ^4(c+d x)}{4 d}+\frac{2 i a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan ^2(c+d x)}{d}-\frac{2 i a^2 \tan (c+d x)}{d}+\frac{2 a^2 \log (\cos (c+d x))}{d}+2 i a^2 x \]

[Out]

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

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Rubi [A]  time = 0.114256, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3543, 3528, 3525, 3475} \[ -\frac{a^2 \tan ^4(c+d x)}{4 d}+\frac{2 i a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan ^2(c+d x)}{d}-\frac{2 i a^2 \tan (c+d x)}{d}+\frac{2 a^2 \log (\cos (c+d x))}{d}+2 i a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 0])

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3525

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

Rubi steps

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

Mathematica [A]  time = 0.204116, size = 73, normalized size = 0.78 \[ \frac{a^2 \left (-3 \tan ^4(c+d x)+8 i \tan ^3(c+d x)+12 \tan ^2(c+d x)+24 i \tan ^{-1}(\tan (c+d x))-24 i \tan (c+d x)+24 \log (\cos (c+d x))\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*((24*I)*ArcTan[Tan[c + d*x]] + 24*Log[Cos[c + d*x]] - (24*I)*Tan[c + d*x] + 12*Tan[c + d*x]^2 + (8*I)*Tan
[c + d*x]^3 - 3*Tan[c + d*x]^4))/(12*d)

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Maple [A]  time = 0.004, size = 100, normalized size = 1.1 \begin{align*}{\frac{-2\,i{a}^{2}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}+{\frac{{\frac{2\,i}{3}}{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{{a}^{2} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}-{\frac{{a}^{2}\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}+{\frac{2\,i{a}^{2}\arctan \left ( \tan \left ( dx+c \right ) \right ) }{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]

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

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Maxima [A]  time = 1.95062, size = 111, normalized size = 1.19 \begin{align*} -\frac{3 \, a^{2} \tan \left (d x + c\right )^{4} - 8 i \, a^{2} \tan \left (d x + c\right )^{3} - 12 \, a^{2} \tan \left (d x + c\right )^{2} - 24 i \,{\left (d x + c\right )} a^{2} + 12 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 24 i \, a^{2} \tan \left (d x + c\right )}{12 \, 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="maxima")

[Out]

-1/12*(3*a^2*tan(d*x + c)^4 - 8*I*a^2*tan(d*x + c)^3 - 12*a^2*tan(d*x + c)^2 - 24*I*(d*x + c)*a^2 + 12*a^2*log
(tan(d*x + c)^2 + 1) + 24*I*a^2*tan(d*x + c))/d

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Fricas [B]  time = 2.25002, size = 479, normalized size = 5.15 \begin{align*} \frac{2 \,{\left (21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{2} + 3 \,{\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \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]

2/3*(21*a^2*e^(6*I*d*x + 6*I*c) + 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2*I*d*x + 2*I*c) + 8*a^2 + 3*(a^2*e^(
8*I*d*x + 8*I*c) + 4*a^2*e^(6*I*d*x + 6*I*c) + 6*a^2*e^(4*I*d*x + 4*I*c) + 4*a^2*e^(2*I*d*x + 2*I*c) + a^2)*lo
g(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) + 4*d*e
^(2*I*d*x + 2*I*c) + d)

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Sympy [B]  time = 5.73199, size = 175, normalized size = 1.88 \begin{align*} \frac{2 a^{2} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{14 a^{2} e^{- 2 i c} e^{6 i d x}}{d} + \frac{24 a^{2} e^{- 4 i c} e^{4 i d x}}{d} + \frac{58 a^{2} e^{- 6 i c} e^{2 i d x}}{3 d} + \frac{16 a^{2} e^{- 8 i c}}{3 d}}{e^{8 i d x} + 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} + 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \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]

2*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (14*a**2*exp(-2*I*c)*exp(6*I*d*x)/d + 24*a**2*exp(-4*I*c)*exp(4*I*d
*x)/d + 58*a**2*exp(-6*I*c)*exp(2*I*d*x)/(3*d) + 16*a**2*exp(-8*I*c)/(3*d))/(exp(8*I*d*x) + 4*exp(-2*I*c)*exp(
6*I*d*x) + 6*exp(-4*I*c)*exp(4*I*d*x) + 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

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Giac [B]  time = 1.82529, size = 300, normalized size = 3.23 \begin{align*} \frac{2 \,{\left (3 \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 12 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a^{2}\right )}}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \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]

2/3*(3*a^2*e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 12*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*
c) + 1) + 18*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 12*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x
+ 2*I*c) + 1) + 21*a^2*e^(6*I*d*x + 6*I*c) + 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2*I*d*x + 2*I*c) + 3*a^2*l
og(e^(2*I*d*x + 2*I*c) + 1) + 8*a^2)/(d*e^(8*I*d*x + 8*I*c) + 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c
) + 4*d*e^(2*I*d*x + 2*I*c) + d)