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

Optimal. Leaf size=27 \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \]

[Out]

a*arctanh(sin(d*x+c))/d+I*a*sec(d*x+c)/d

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3486, 3770} \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (I*a*Sec[c + d*x])/d

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx &=\frac {i a \sec (c+d x)}{d}+a \int \sec (c+d x) \, dx\\ &=\frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (I*a*Sec[c + d*x])/d

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fricas [B]  time = 0.51, size = 82, normalized size = 3.04 \[ \frac {2 i \, a e^{\left (i \, d x + i \, c\right )} + {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.96, size = 52, normalized size = 1.93 \[ \frac {a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 i \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, size = 36, normalized size = 1.33 \[ \frac {i a}{d \cos \left (d x +c \right )}+\frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/d*a/cos(d*x+c)+1/d*a*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.35, size = 32, normalized size = 1.19 \[ \frac {a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {i \, a}{\cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

(a*log(sec(d*x + c) + tan(d*x + c)) + I*a/cos(d*x + c))/d

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mupad [B]  time = 3.35, size = 39, normalized size = 1.44 \[ \frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,2{}\mathrm {i}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)/cos(c + d*x),x)

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (a*2i)/(d*(tan(c/2 + (d*x)/2)^2 - 1))

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sympy [A]  time = 4.54, size = 41, normalized size = 1.52 \[ \begin {cases} \frac {a \log {\left (\tan {\left (c + d x \right )} + \sec {\left (c + d x \right )} \right )} + i a \sec {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\relax (c )} + a\right ) \sec {\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((a*log(tan(c + d*x) + sec(c + d*x)) + I*a*sec(c + d*x))/d, Ne(d, 0)), (x*(I*a*tan(c) + a)*sec(c), T
rue))

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