Optimal. Leaf size=61 \[ \frac {i}{4 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {x}{4 a^2}+\frac {i}{4 d (a+i a \tan (c+d x))^2} \]
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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.
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Rule 8
Rule 3479
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.
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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.
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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.
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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.
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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.
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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.
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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.
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