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

Optimal. Leaf size=67 \[ \frac{i a \tan ^3(c+d x)}{3 d}+\frac{a \tan ^2(c+d x)}{2 d}-\frac{i a \tan (c+d x)}{d}+\frac{a \log (\cos (c+d x))}{d}+i a x \]

[Out]

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

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Rubi [A]  time = 0.0601947, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3528, 3525, 3475} \[ \frac{i a \tan ^3(c+d x)}{3 d}+\frac{a \tan ^2(c+d x)}{2 d}-\frac{i a \tan (c+d x)}{d}+\frac{a \log (\cos (c+d x))}{d}+i a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.153555, size = 74, normalized size = 1.1 \[ \frac{i a \tan ^3(c+d x)}{3 d}+\frac{i a \tan ^{-1}(\tan (c+d x))}{d}-\frac{i a \tan (c+d x)}{d}+\frac{a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 75, normalized size = 1.1 \begin{align*}{\frac{-ia\tan \left ( dx+c \right ) }{d}}+{\frac{{\frac{i}{3}}a \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d}}+{\frac{ia\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)),x)

[Out]

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

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Maxima [A]  time = 1.9417, size = 80, normalized size = 1.19 \begin{align*} -\frac{-2 i \, a \tan \left (d x + c\right )^{3} - 3 \, a \tan \left (d x + c\right )^{2} - 6 i \,{\left (d x + c\right )} a + 3 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 i \, a \tan \left (d x + c\right )}{6 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.20169, size = 348, normalized size = 5.19 \begin{align*} \frac{18 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \,{\left (a e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a}{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)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(18*a*e^(4*I*d*x + 4*I*c) + 18*a*e^(2*I*d*x + 2*I*c) + 3*(a*e^(6*I*d*x + 6*I*c) + 3*a*e^(4*I*d*x + 4*I*c)
+ 3*a*e^(2*I*d*x + 2*I*c) + a)*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a)/(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 = 4.6724, size = 126, normalized size = 1.88 \begin{align*} \frac{a \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{6 a e^{- 2 i c} e^{4 i d x}}{d} + \frac{6 a e^{- 4 i c} e^{2 i d x}}{d} + \frac{8 a 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)**3*(a+I*a*tan(d*x+c)),x)

[Out]

a*log(exp(2*I*d*x) + exp(-2*I*c))/d + (6*a*exp(-2*I*c)*exp(4*I*d*x)/d + 6*a*exp(-4*I*c)*exp(2*I*d*x)/d + 8*a*e
xp(-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 = 1.85808, size = 211, normalized size = 3.15 \begin{align*} \frac{3 \, a e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, a e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 9 \, a e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 18 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 8 \, a}{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)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/3*(3*a*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 9*a*e^(4*I*d*x + 4*I*c)*log(e^(2*I*d*x + 2*I*c) +
1) + 9*a*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 18*a*e^(4*I*d*x + 4*I*c) + 18*a*e^(2*I*d*x + 2*I*c
) + 3*a*log(e^(2*I*d*x + 2*I*c) + 1) + 8*a)/(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)