3.2.4 \(\int \frac {1}{a+i a \tan (c+d x)} \, dx\) [104]

Optimal. Leaf size=33 \[ \frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3560, 8} \begin {gather*} \frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-1),x]

[Out]

x/(2*a) + (I/2)/(d*(a + I*a*Tan[c + d*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3560

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] +
Dist[1/(2*a), Int[(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0]

Rubi steps

\begin {align*} \int \frac {1}{a+i a \tan (c+d x)} \, dx &=\frac {i}{2 d (a+i a \tan (c+d x))}+\frac {\int 1 \, dx}{2 a}\\ &=\frac {x}{2 a}+\frac {i}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 45, normalized size = 1.36 \begin {gather*} \frac {1-2 i d x+(-i+2 d x) \tan (c+d x)}{4 a d (-i+\tan (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(-1),x]

[Out]

(1 - (2*I)*d*x + (-I + 2*d*x)*Tan[c + d*x])/(4*a*d*(-I + Tan[c + d*x]))

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Maple [A]
time = 0.09, size = 48, normalized size = 1.45

method	result	size
risch	\(\frac {x}{2 a}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d a}\)	\(26\)
derivativedivides	\(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{4}+\frac {1}{2 \tan \left (d x +c \right )-2 i}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4}}{d a}\)	\(48\)
default	\(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{4}+\frac {1}{2 \tan \left (d x +c \right )-2 i}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4}}{d a}\)	\(48\)
norman	\(\frac {\frac {x}{2 a}+\frac {i}{2 d a}+\frac {x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}+\frac {\tan \left (d x +c \right )}{2 d a}}{1+\tan ^{2}\left (d x +c \right )}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/4*I*ln(tan(d*x+c)-I)+1/2/(tan(d*x+c)-I)+1/4*I*ln(tan(d*x+c)+I))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 0.36, size = 32, normalized size = 0.97 \begin {gather*} \frac {{\left (2 \, d x e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.08, size = 60, normalized size = 1.82 \begin {gather*} \begin {cases} \frac {i e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (\frac {\left (e^{2 i c} + 1\right ) e^{- 2 i c}}{2 a} - \frac {1}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {x}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c)),x)

[Out]

Piecewise((I*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*exp(2*I*c), 0)), (x*((exp(2*I*c) + 1)*exp(-2*I*c)/(2*a)
- 1/(2*a)), True)) + x/(2*a)

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (25) = 50\).
time = 0.47, size = 60, normalized size = 1.82 \begin {gather*} -\frac {\frac {i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac {i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} + \frac {-i \, \tan \left (d x + c\right ) - 3}{a {\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/4*(I*log(tan(d*x + c) - I)/a - I*log(-I*tan(d*x + c) + 1)/a + (-I*tan(d*x + c) - 3)/(a*(tan(d*x + c) - I)))
/d

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Mupad [B]
time = 3.36, size = 29, normalized size = 0.88 \begin {gather*} \frac {x}{2\,a}+\frac {1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)*1i),x)

[Out]

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

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