Optimal. Leaf size=303 \[ \frac {i \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {i \text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {\text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {i \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))} \]
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Rubi [A]
time = 0.32, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps
used = 23, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3633, 3619,
3557, 335, 281, 206, 31, 648, 632, 210, 642, 301, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {i \text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {i \text {ArcTan}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}-\frac {i \text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d}-\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {\log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 206
Rule 209
Rule 210
Rule 281
Rule 301
Rule 335
Rule 632
Rule 642
Rule 648
Rule 3557
Rule 3619
Rule 3633
Rubi steps
\begin {align*} \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a+i a \tan (c+d x))} \, dx &=\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\int \frac {-\frac {4 a}{3}+\frac {1}{3} i a \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{2 a^2}\\ &=\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \int \tan ^{\frac {2}{3}}(c+d x) \, dx}{6 a}+\frac {2 \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{3 a}\\ &=\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{6 a d}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{3 a d}\\ &=\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}+\frac {2 \text {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}\\ &=\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {i \text {Subst}\left (\int \frac {-\frac {1}{2}+\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {i \text {Subst}\left (\int \frac {-\frac {1}{2}-\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\text {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{a d}\\ &=-\frac {i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {i \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}-\frac {i \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {\text {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {i \text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}+\frac {i \text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{8 \sqrt {3} a d}\\ &=-\frac {i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {i \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {i \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {\text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {\text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}\\ &=\frac {i \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {i \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{a d}\\ &=\frac {i \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {i \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {\tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {\log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {i \log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}+\frac {i \log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{8 \sqrt {3} a d}-\frac {\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{6 a d}+\frac {\tan ^{\frac {2}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 1.15, size = 161, normalized size = 0.53 \begin {gather*} \frac {e^{-2 i (c+d x)} \left (4+4 e^{2 i (c+d x)}+3 \sqrt [3]{2} e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+10 e^{2 i (c+d x)} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right ) \tan ^{\frac {2}{3}}(c+d x)}{16 a d} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.17, size = 189, normalized size = 0.62
method | result | size |
derivativedivides | \(\frac {-\frac {i}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {5 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}-\frac {i \sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}+\frac {-4 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2}{-12 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+12 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-12}-\frac {5 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}+\frac {5 i \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}}{d a}\) | \(189\) |
default | \(\frac {-\frac {i}{6 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {5 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8}-\frac {i \sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4}+\frac {-4 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-2}{-12 i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+12 \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )-12}-\frac {5 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24}+\frac {5 i \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4}}{d a}\) | \(189\) |
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 [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 486 vs. \(2 (238) = 476\).
time = 0.70, size = 486, normalized size = 1.60 \begin {gather*} -\frac {{\left (3 \, {\left (i \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + 3 \, {\left (-i \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + \frac {1}{2} i\right ) + 5 \, {\left (-3 i \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) + 5 \, {\left (3 i \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a d \sqrt {\frac {1}{a^{2} d^{2}}} + \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - \frac {1}{2} i\right ) - 10 \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} + i\right ) - 6 \, e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} - i\right ) - 6 \, \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{24 \, a d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\tan ^{\frac {4}{3}}{\left (c + d x \right )} - i \sqrt [3]{\tan {\left (c + d x \right )}}}\, dx}{a} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.64, size = 215, normalized size = 0.71 \begin {gather*} -\frac {5 i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}\right )}{24 \, a d} + \frac {i \, \sqrt {3} \log \left (-\frac {\sqrt {3} - 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} - i}{\sqrt {3} + 2 \, \tan \left (d x + c\right )^{\frac {1}{3}} + i}\right )}{8 \, a d} - \frac {\log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{8 \, a d} - \frac {5 \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{24 \, a d} + \frac {5 \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{12 \, a d} + \frac {\log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{4 \, a d} - \frac {i \, \tan \left (d x + c\right )^{\frac {2}{3}}}{2 \, a d {\left (\tan \left (d x + c\right ) - i\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.31, size = 587, normalized size = 1.94 \begin {gather*} \frac {5\,\ln \left (\frac {25\,\left (24480\,a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}+a^3\,d^3\,14112{}\mathrm {i}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{2/3}}{144}-1800\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{12}+\ln \left (\left (58752\,a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}+a^3\,d^3\,14112{}\mathrm {i}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{2/3}-1800\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}+\frac {5\,\ln \left (\frac {25\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,14112{}\mathrm {i}+12240\,a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{2/3}}{576}-1800\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{24}-\frac {5\,\ln \left (\frac {25\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (a^3\,d^3\,14112{}\mathrm {i}-12240\,a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{2/3}}{576}-1800\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{24}+\ln \left ({\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (a^3\,d^3\,14112{}\mathrm {i}+58752\,a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{2/3}-1800\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}-\ln \left ({\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,\left (a^3\,d^3\,14112{}\mathrm {i}-58752\,a^4\,d^4\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{2/3}-1800\,a\,d\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{2/3}}{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.
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