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

Optimal. Leaf size=49 \[ \frac{i a \tan ^2(c+d x)}{2 d}+\frac{a \tan (c+d x)}{d}+\frac{i a \log (\cos (c+d x))}{d}-a x \]

[Out]

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

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Rubi [A]  time = 0.0389246, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3528, 3525, 3475} \[ \frac{i a \tan ^2(c+d x)}{2 d}+\frac{a \tan (c+d x)}{d}+\frac{i a \log (\cos (c+d x))}{d}-a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*x) + (I*a*Log[Cos[c + d*x]])/d + (a*Tan[c + d*x])/d + ((I/2)*a*Tan[c + d*x]^2)/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 ^2(c+d x) (a+i a \tan (c+d x)) \, dx &=\frac{i a \tan ^2(c+d x)}{2 d}+\int \tan (c+d x) (-i a+a \tan (c+d x)) \, dx\\ &=-a x+\frac{a \tan (c+d x)}{d}+\frac{i a \tan ^2(c+d x)}{2 d}-(i a) \int \tan (c+d x) \, dx\\ &=-a x+\frac{i a \log (\cos (c+d x))}{d}+\frac{a \tan (c+d x)}{d}+\frac{i a \tan ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0966326, size = 53, normalized size = 1.08 \[ -\frac{a \tan ^{-1}(\tan (c+d x))}{d}+\frac{a \tan (c+d x)}{d}+\frac{i a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 59, normalized size = 1.2 \begin{align*}{\frac{{\frac{i}{2}}a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{a\tan \left ( dx+c \right ) }{d}}-{\frac{{\frac{i}{2}}a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d}}-{\frac{a\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)^2*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 2.34358, size = 65, normalized size = 1.33 \begin{align*} -\frac{-i \, a \tan \left (d x + c\right )^{2} + 2 \,{\left (d x + c\right )} a + i \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, a \tan \left (d x + c\right )}{2 \, 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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.3079, size = 246, normalized size = 5.02 \begin{align*} \frac{4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 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)),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 2.03375, size = 92, normalized size = 1.88 \begin{align*} \frac{i a \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac{\frac{4 i a e^{- 2 i c} e^{2 i d x}}{d} + \frac{2 i a e^{- 4 i c}}{d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 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)),x)

[Out]

I*a*log(exp(2*I*d*x) + exp(-2*I*c))/d + (4*I*a*exp(-2*I*c)*exp(2*I*d*x)/d + 2*I*a*exp(-4*I*c)/d)/(exp(4*I*d*x)
+ 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

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Giac [B]  time = 1.47232, size = 144, normalized size = 2.94 \begin{align*} \frac{i \, 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 ) + 2 i \, 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 ) + 4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 i \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 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)),x, algorithm="giac")

[Out]

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