3.119 \(\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=61 \[ \frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {x}{4 a^2}+\frac {i}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3479, 8} \[ \frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {x}{4 a^2}+\frac {i}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(-2),x]

[Out]

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

Rule 8

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

Rule 3479

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))^2} \, dx &=\frac {i}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {1}{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=\frac {i}{4 d (a+i a \tan (c+d x))^2}+\frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\int 1 \, dx}{4 a^2}\\ &=\frac {x}{4 a^2}+\frac {i}{4 d (a+i a \tan (c+d x))^2}+\frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 68, normalized size = 1.11 \[ -\frac {\sec ^2(c+d x) ((1+4 i d x) \sin (2 (c+d x))+(4 d x+i) \cos (2 (c+d x))+4 i)}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(Sec[c + d*x]^2*(4*I + (I + 4*d*x)*Cos[2*(c + d*x)] + (1 + (4*I)*d*x)*Sin[2*(c + d*x)]))/(a^2*d*(-I + Ta
n[c + d*x])^2)

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fricas [A]  time = 0.53, size = 43, normalized size = 0.70 \[ \frac {{\left (4 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(4*d*x*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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giac [A]  time = 0.54, size = 72, normalized size = 1.18 \[ -\frac {\frac {2 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {2 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{2}} + \frac {-3 i \, \tan \left (d x + c\right )^{2} - 10 \, \tan \left (d x + c\right ) + 11 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(2*I*log(I*tan(d*x + c) + 1)/a^2 - 2*I*log(I*tan(d*x + c) - 1)/a^2 + (-3*I*tan(d*x + c)^2 - 10*tan(d*x +
c) + 11*I)/(a^2*(tan(d*x + c) - I)^2))/d

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maple [A]  time = 0.10, size = 79, normalized size = 1.30 \[ \frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8 a^{2} d}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*I/d/a^2*ln(tan(d*x+c)+I)-1/8*I/d/a^2*ln(tan(d*x+c)-I)-1/4*I/d/a^2/(tan(d*x+c)-I)^2+1/4/d/a^2/(tan(d*x+c)-I
)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(d*x+c))^2,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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mupad [B]  time = 3.40, size = 39, normalized size = 0.64 \[ \frac {x}{4\,a^2}-\frac {\frac {\mathrm {tan}\left (c+d\,x\right )}{4}-\frac {1}{2}{}\mathrm {i}}{a^2\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)*1i)^2,x)

[Out]

x/(4*a^2) - (tan(c + d*x)/4 - 1i/2)/(a^2*d*(tan(c + d*x)*1i + 1)^2)

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sympy [A]  time = 0.23, size = 119, normalized size = 1.95 \[ \begin {cases} \frac {\left (16 i a^{2} d e^{4 i c} e^{- 2 i d x} + 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {1}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((16*I*a**2*d*exp(4*I*c)*exp(-2*I*d*x) + 4*I*a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(64*a**4*d
**2), Ne(64*a**4*d**2*exp(6*I*c), 0)), (x*((exp(4*I*c) + 2*exp(2*I*c) + 1)*exp(-4*I*c)/(4*a**2) - 1/(4*a**2)),
True)) + x/(4*a**2)

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