3.2.10 \(\int \frac {\sec (c+d x)}{a+i a \tan (c+d x)} \, dx\) [110]

Optimal. Leaf size=28 \[ \frac {i \sec (c+d x)}{d (a+i a \tan (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {3569} \begin {gather*} \frac {i \sec (c+d x)}{d (a+i a \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3569

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 25, normalized size = 0.89 \begin {gather*} \frac {\sec (c+d x)}{a d (-i+\tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.16, size = 23, normalized size = 0.82

method	result	size
risch	\(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{d a}\)	\(19\)
derivativedivides	\(\frac {2}{d a \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(23\)
default	\(\frac {2}{d a \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 29, normalized size = 1.04 \begin {gather*} \frac {2}{{\left (-i \, a + \frac {a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )} d} \end {gather*}

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]

2/((-I*a + a*sin(d*x + c)/(cos(d*x + c) + 1))*d)

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Fricas [A]
time = 0.43, size = 17, normalized size = 0.61 \begin {gather*} \frac {i \, e^{\left (-i \, d x - i \, c\right )}}{a d} \end {gather*}

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]

I*e^(-I*d*x - I*c)/(a*d)

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Sympy [A]
time = 0.35, size = 34, normalized size = 1.21 \begin {gather*} \begin {cases} \frac {\sec {\left (c + d x \right )}}{a d \tan {\left (c + d x \right )} - i a d} & \text {for}\: d \neq 0 \\\frac {x \sec {\left (c \right )}}{i a \tan {\left (c \right )} + a} & \text {otherwise} \end {cases} \end {gather*}

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((sec(c + d*x)/(a*d*tan(c + d*x) - I*a*d), Ne(d, 0)), (x*sec(c)/(I*a*tan(c) + a), True))

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Giac [A]
time = 0.54, size = 21, normalized size = 0.75 \begin {gather*} \frac {2}{a d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}} \end {gather*}

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]

2/(a*d*(tan(1/2*d*x + 1/2*c) - I))

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Mupad [B]
time = 3.35, size = 25, normalized size = 0.89 \begin {gather*} \frac {2{}\mathrm {i}}{a\,d\,\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2i/(a*d*(tan(c/2 + (d*x)/2)*1i + 1))

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