Optimal. Leaf size=88 \[ \frac {x}{8 a^3}+\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Rubi [A]
time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps
used = 4, 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 {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {x}{8 a^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{6 d (a+i a \tan (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3560
Rubi steps
\begin {align*} \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx &=\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {\int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{2 a}\\ &=\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {\int \frac {1}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {\int 1 \, dx}{8 a^3}\\ &=\frac {x}{8 a^3}+\frac {i}{6 d (a+i a \tan (c+d x))^3}+\frac {i}{8 a d (a+i a \tan (c+d x))^2}+\frac {i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.25, size = 93, normalized size = 1.06 \begin {gather*} \frac {i \sec ^3(c+d x) (27 i \cos (c+d x)+2 (i+6 d x) \cos (3 (c+d x))-9 \sin (c+d x)+2 \sin (3 (c+d x))+12 i d x \sin (3 (c+d x)))}{96 a^3 d (-i+\tan (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 75, normalized size = 0.85
method | result | size |
risch | \(\frac {x}{8 a^{3}}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}+\frac {3 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}\) | \(62\) |
derivativedivides | \(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\) | \(75\) |
default | \(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{6 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {1}{8 \tan \left (d x +c \right )-8 i}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{16}}{d \,a^{3}}\) | \(75\) |
norman | \(\frac {\frac {x}{8 a}+\frac {\tan ^{5}\left (d x +c \right )}{8 d a}+\frac {3 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}+\frac {3 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {5 i}{12 d a}+\frac {7 \tan \left (d x +c \right )}{8 d a}+\frac {\tan ^{3}\left (d x +c \right )}{3 d a}-\frac {i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}\) | \(138\) |
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 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 54, normalized size = 0.61 \begin {gather*} \frac {{\left (12 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.18, size = 155, normalized size = 1.76 \begin {gather*} \begin {cases} \frac {\left (4608 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} + 2304 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} - \frac {1}{8 a^{3}}\right ) & \text {otherwise} \end {cases} + \frac {x}{8 a^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.60, size = 80, normalized size = 0.91 \begin {gather*} -\frac {\frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac {6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac {-11 i \, \tan \left (d x + c\right )^{3} - 45 \, \tan \left (d x + c\right )^{2} + 69 i \, \tan \left (d x + c\right ) + 51}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.75, size = 50, normalized size = 0.57 \begin {gather*} \frac {x}{8\,a^3}-\frac {\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{8}+\frac {3\,\mathrm {tan}\left (c+d\,x\right )}{8}-\frac {5}{12}{}\mathrm {i}}{a^3\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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