3.5 \(\int \tan (c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=34 \[ \frac{i a \tan (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d}-i a x \]

[Out]

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

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Rubi [A]  time = 0.0175204, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3525, 3475} \[ \frac{i a \tan (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d}-i a x \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(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

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 (c+d x) (a+i a \tan (c+d x)) \, dx &=-i a x+\frac{i a \tan (c+d x)}{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}\\ \end{align*}

Mathematica [A]  time = 0.0173237, size = 43, normalized size = 1.26 \[ -\frac{i a \tan ^{-1}(\tan (c+d x))}{d}+\frac{i a \tan (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, size = 46, normalized size = 1.4 \begin{align*}{\frac{ia\tan \left ( dx+c \right ) }{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)*(a+I*a*tan(d*x+c)),x)

[Out]

I*a*tan(d*x+c)/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 = 2.10057, size = 50, normalized size = 1.47 \begin{align*} -\frac{2 i \,{\left (d x + c\right )} a - a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 i \, 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)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.51703, size = 128, normalized size = 3.76 \begin{align*} -\frac{{\left (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 ) + 2 \, a}{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)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.832894, size = 51, normalized size = 1.5 \begin{align*} - \frac{a \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} - \frac{2 a e^{- 2 i c}}{d \left (e^{2 i d x} + e^{- 2 i c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.17392, size = 78, normalized size = 2.29 \begin{align*} -\frac{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 ) + a \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 2 \, a}{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)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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