Optimal. Leaf size=321 \[ \frac {5 i \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}+\frac {2 \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {2 \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d}+\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {\log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d} \]
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Rubi [A] time = 0.42, antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3552, 3529, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ \frac {5 i \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}+\frac {2 \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {5 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}+\frac {2 \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d}+\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {\log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{3 a d} \]
Antiderivative was successfully verified.
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Rule 31
Rule 203
Rule 204
Rule 209
Rule 275
Rule 292
Rule 329
Rule 618
Rule 628
Rule 634
Rule 3476
Rule 3529
Rule 3538
Rule 3552
Rubi steps
\begin {align*} \int \frac {1}{\tan ^{\frac {5}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx &=\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {\int \frac {-\frac {8 a}{3}+\frac {5}{3} i a \tan (c+d x)}{\tan ^{\frac {5}{3}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {\int \frac {\frac {5 i a}{3}+\frac {8}{3} a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{2 a^2}\\ &=-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {(5 i) \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{6 a}-\frac {4 \int \sqrt [3]{\tan (c+d x)} \, dx}{3 a}\\ &=-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{6 a d}-\frac {4 \operatorname {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{3 a d}\\ &=-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}-\frac {4 \operatorname {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}\\ &=-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1+\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {2 \operatorname {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{a d}\\ &=-\frac {5 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{24 a d}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {2 \operatorname {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {(5 i) \operatorname {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 {(5 i) \operatorname {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 {5 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 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 {5 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 {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {\operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{a d}\\ &=\frac {5 i \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 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 {5 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 )}{3 a d}-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{a d}\\ &=\frac {5 i \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}+\frac {2 \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{\sqrt {3} a d}-\frac {5 i \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a d}+\frac {5 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 {5 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 )}{3 a d}-\frac {2}{a d \tan ^{\frac {2}{3}}(c+d x)}+\frac {1}{2 d \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))}\\ \end {align*}
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Mathematica [C] time = 1.55, size = 201, normalized size = 0.63 \[ \frac {i \sqrt [3]{\tan (c+d x)} \csc (c+d x) \sec (c+d x) \left (2 \left (13 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (i \sin (2 (c+d x))+\cos (2 (c+d x))-1)+8 i \sin (2 (c+d x))+6 \cos (2 (c+d x))+6\right )-3\ 2^{2/3} \left (-1+e^{2 i (c+d x)}\right ) \sqrt [3]{1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right )}{16 a d (\tan (c+d x)-i)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.64, size = 644, normalized size = 2.01 \[ \frac {{\left (\sqrt {3} {\left (12 i \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - 12 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 12 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 12 \, 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 ) + {\left (\sqrt {3} {\left (-12 i \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 12 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 12 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 12 \, 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 ) + {\left (\sqrt {\frac {1}{3}} {\left (-156 i \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + 156 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 52 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 52 \, 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 ) + {\left (\sqrt {\frac {1}{3}} {\left (156 i \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - 156 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {1}{a^{2} d^{2}}} - 52 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 52 \, 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 ) + 104 \, {\left (e^{\left (4 i \, d x + 4 i \, c\right )} - e^{\left (2 i \, d x + 2 i \, c\right )}\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 ) + 24 \, {\left (e^{\left (4 i \, d x + 4 i \, c\right )} - e^{\left (2 i \, d x + 2 i \, c\right )}\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 ) + 4 \, \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}} {\left (-42 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 36 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )}}{96 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )} \tan \left (d x + c\right )^{\frac {5}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 273, normalized size = 0.85 \[ -\frac {i}{6 d a \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {13 \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12 d a}-\frac {\ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8 d a}+\frac {i \sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4 d a}-\frac {3}{2 a d \tan \left (d x +c \right )^{\frac {2}{3}}}+\frac {i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {1}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {13 \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24 d a}-\frac {13 i \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12 d a}+\frac {\ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4 d a} \]
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 = 5.36, size = 630, normalized size = 1.96 \[ -\frac {\frac {3}{2\,a\,d}+\frac {\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{a\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^{2/3}+{\mathrm {tan}\left (c+d\,x\right )}^{5/3}\,1{}\mathrm {i}}+\ln \left (\left (331776\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{2/3}+a^3\,d^3\,312480{}\mathrm {i}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}-83304\,a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}+\frac {13\,\ln \left (\frac {13\,\left (389376\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{a^3\,d^3}\right )}^{2/3}+a^3\,d^3\,312480{}\mathrm {i}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{12}-83304\,a^2\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{12}+\frac {13\,\ln \left (\frac {13\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,312480{}\mathrm {i}+97344\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (\frac {1}{a^3\,d^3}\right )}^{2/3}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{24}-83304\,a^2\,d^2\,{\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 {13\,\ln \left (\frac {13\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (a^3\,d^3\,312480{}\mathrm {i}+97344\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (\frac {1}{a^3\,d^3}\right )}^{2/3}\right )\,{\left (\frac {1}{a^3\,d^3}\right )}^{1/3}}{24}+83304\,a^2\,d^2\,{\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 )\,\left (a^3\,d^3\,312480{}\mathrm {i}+331776\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}-83304\,a^2\,d^2\,{\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 )\,\left (a^3\,d^3\,312480{}\mathrm {i}+331776\,a^5\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{2/3}\right )\,{\left (\frac {1}{64\,a^3\,d^3}\right )}^{1/3}+83304\,a^2\,d^2\,{\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} \]
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 \[ - \frac {i \int \frac {1}{\tan ^{\frac {8}{3}}{\left (c + d x \right )} - i \tan ^{\frac {5}{3}}{\left (c + d x \right )}}\, dx}{a} \]
Verification of antiderivative is not currently implemented for this CAS.
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