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

Optimal. Leaf size=64 \[ \frac{a^2 \tan (c+d x)}{d}+\frac{2 i a^2 \log (\cos (c+d x))}{d}-2 a^2 x-\frac{i (a+i a \tan (c+d x))^3}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.0546303, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3543, 3477, 3475} \[ \frac{a^2 \tan (c+d x)}{d}+\frac{2 i a^2 \log (\cos (c+d x))}{d}-2 a^2 x-\frac{i (a+i a \tan (c+d x))^3}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2*a^2*x + ((2*I)*a^2*Log[Cos[c + d*x]])/d + (a^2*Tan[c + d*x])/d - ((I/3)*(a + I*a*Tan[c + d*x])^3)/(a*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 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

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

Mathematica [A]  time = 0.236781, size = 76, normalized size = 1.19 \[ -\frac{a^2 \tan ^3(c+d x)}{3 d}-\frac{2 a^2 \tan ^{-1}(\tan (c+d x))}{d}+\frac{2 a^2 \tan (c+d x)}{d}+\frac{i a^2 \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \text{hanged} \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))^2,x)

[Out]

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^2,x)

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Maxima [A]  time = 2.00135, size = 92, normalized size = 1.44 \begin{align*} -\frac{a^{2} \tan \left (d x + c\right )^{3} - 3 i \, a^{2} \tan \left (d x + c\right )^{2} + 6 \,{\left (d x + c\right )} a^{2} + 3 i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 \, a^{2} \tan \left (d x + c\right )}{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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.16782, size = 393, normalized size = 6.14 \begin{align*} \frac{30 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 36 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 14 i \, a^{2} +{\left (6 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}{3 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^2*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(30*I*a^2*e^(4*I*d*x + 4*I*c) + 36*I*a^2*e^(2*I*d*x + 2*I*c) + 14*I*a^2 + (6*I*a^2*e^(6*I*d*x + 6*I*c) + 1
8*I*a^2*e^(4*I*d*x + 4*I*c) + 18*I*a^2*e^(2*I*d*x + 2*I*c) + 6*I*a^2)*log(e^(2*I*d*x + 2*I*c) + 1))/(d*e^(6*I*
d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [B]  time = 4.64087, size = 141, normalized size = 2.2 \begin{align*} \frac{2 i a^{2} \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{10 i a^{2} e^{- 2 i c} e^{4 i d x}}{d} + \frac{12 i a^{2} e^{- 4 i c} e^{2 i d x}}{d} + \frac{14 i a^{2} e^{- 6 i c}}{3 d}}{e^{6 i d x} + 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} + e^{- 6 i c}} \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))**2,x)

[Out]

2*I*a**2*log(exp(2*I*d*x) + exp(-2*I*c))/d + (10*I*a**2*exp(-2*I*c)*exp(4*I*d*x)/d + 12*I*a**2*exp(-4*I*c)*exp
(2*I*d*x)/d + 14*I*a**2*exp(-6*I*c)/(3*d))/(exp(6*I*d*x) + 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2*I*
d*x) + exp(-6*I*c))

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Giac [B]  time = 1.5541, size = 230, normalized size = 3.59 \begin{align*} \frac{6 i \, 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 i \, 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 ) + 18 i \, 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 ) + 30 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 36 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{2} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 i \, a^{2}}{3 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^2*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/3*(6*I*a^2*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*I*a^2*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x +
2*I*c) + 1) + 18*I*a^2*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 30*I*a^2*e^(4*I*d*x + 4*I*c) + 36*I*
a^2*e^(2*I*d*x + 2*I*c) + 6*I*a^2*log(e^(2*I*d*x + 2*I*c) + 1) + 14*I*a^2)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I
*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)