Optimal. Leaf size=335 \[ \frac {\text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {\text {ArcTan}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {i \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {\text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{18 a^2 d}+\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A]
time = 0.35, antiderivative size = 335, 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 = {3639, 3677,
3619, 3557, 335, 215, 648, 632, 210, 642, 209, 281, 298, 31} \begin {gather*} \frac {i \text {ArcTan}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {\text {ArcTan}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {\text {ArcTan}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}-\frac {\text {ArcTan}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}+\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{18 a^2 d}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 31
Rule 209
Rule 210
Rule 215
Rule 281
Rule 298
Rule 335
Rule 632
Rule 642
Rule 648
Rule 3557
Rule 3619
Rule 3639
Rule 3677
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {4}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {-\frac {a}{3}+\frac {7}{3} i a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))} \, dx}{4 a^2}\\ &=\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\frac {2 a^2}{9}+\frac {16}{9} i a^2 \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{8 a^4}\\ &=\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{9 a^2}-\frac {\int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{36 a^2}\\ &=\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \text {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{x^{2/3} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}\\ &=\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \text {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}\\ &=\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {i \text {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {\text {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 )}{36 a^2 d}-\frac {\text {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 )}{36 a^2 d}\\ &=-\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}+\frac {i \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {i \text {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}-\frac {\text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{144 a^2 d}+\frac {\text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt {3} a^2 d}-\frac {\text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{48 \sqrt {3} a^2 d}\\ &=-\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}+\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}-\frac {i \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{18 a^2 d}-\frac {i \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a^2 d}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}\\ &=\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {\tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{18 a^2 d}+\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}+\frac {i \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d}\\ &=\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}-\frac {\tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {\log \left (1-\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (1+\sqrt {3} \sqrt [3]{\tan (c+d x)}+\tan ^{\frac {2}{3}}(c+d x)\right )}{48 \sqrt {3} a^2 d}-\frac {i \log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{18 a^2 d}+\frac {\sqrt [3]{\tan (c+d x)}}{3 a^2 d (1+i \tan (c+d x))}-\frac {\sqrt [3]{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 1.70, size = 190, normalized size = 0.57 \begin {gather*} \frac {\sec ^2(c+d x) \left (9\ 2^{2/3} e^{2 i (c+d x)} \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 )-2 \left (1+\cos (2 (c+d x))+7 \, _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 ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+4 i \sin (2 (c+d x))\right )\right ) \sqrt [3]{\tan (c+d x)}}{48 a^2 d (-i+\tan (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.18, size = 225, normalized size = 0.67
method | result | size |
derivativedivides | \(\frac {-\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {7 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}+\frac {1}{12 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+12 i}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {6 \tan \left (d x +c \right )+4 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+4 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-4 i}{72 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}-\frac {7 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {7 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}}{d \,a^{2}}\) | \(225\) |
default | \(\frac {-\frac {i}{36 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {7 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72}+\frac {1}{12 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+12 i}-\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8}-\frac {6 \tan \left (d x +c \right )+4 i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )+4 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )-4 i}{72 \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}-\frac {7 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144}+\frac {7 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72}+\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8}}{d \,a^{2}}\) | \(225\) |
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 = 0.62, size = 521, normalized size = 1.56 \begin {gather*} -\frac {{\left (9 \, {\left (\sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\frac {1}{2} \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} 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 ) - 9 \, {\left (\sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (-\frac {1}{2} \, \sqrt {3} a^{2} d \sqrt {\frac {1}{a^{4} 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 ) - 7 \, {\left (3 \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (\frac {3}{2} \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} 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 ) + 7 \, {\left (3 \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (-\frac {3}{2} \, \sqrt {\frac {1}{3}} a^{2} d \sqrt {\frac {1}{a^{4} 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 ) - 14 i \, e^{\left (4 i \, d x + 4 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 ) - 18 i \, e^{\left (4 i \, d x + 4 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 ) - 3 \, \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 (5 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{144 \, a^{2} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.76, size = 228, normalized size = 0.68 \begin {gather*} -\frac {7 \, \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 )}{144 \, a^{2} d} + \frac {\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 )}{16 \, a^{2} d} - \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} + i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{16 \, a^{2} d} - \frac {7 i \, \log \left (\tan \left (d x + c\right )^{\frac {2}{3}} - i \, \tan \left (d x + c\right )^{\frac {1}{3}} - 1\right )}{144 \, a^{2} d} + \frac {7 i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} + i\right )}{72 \, a^{2} d} + \frac {i \, \log \left (\tan \left (d x + c\right )^{\frac {1}{3}} - i\right )}{8 \, a^{2} d} + \frac {-4 i \, \tan \left (d x + c\right )^{\frac {4}{3}} - \tan \left (d x + c\right )^{\frac {1}{3}}}{12 \, a^{2} d {\left (\tan \left (d x + c\right ) - i\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.18, size = 653, normalized size = 1.95 \begin {gather*} -\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^{4/3}}{3\,a^2\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,1{}\mathrm {i}}{12\,a^2\,d}}{{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}}+\ln \left (\left (\frac {a^6\,d^3\,49408{}\mathrm {i}}{3}+3538944\,a^{10}\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}+\frac {a^4\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,14560{}\mathrm {i}}{3}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}+\ln \left (\left (\frac {a^6\,d^3\,49408{}\mathrm {i}}{3}+3538944\,a^{10}\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}+\frac {a^4\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,14560{}\mathrm {i}}{3}\right )\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}+\frac {\ln \left (\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {a^6\,d^3\,49408{}\mathrm {i}}{3}+884736\,a^{10}\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}}{2}+\frac {a^4\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,14560{}\mathrm {i}}{3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {a^6\,d^3\,49408{}\mathrm {i}}{3}+884736\,a^{10}\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}}{2}-\frac {a^4\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,14560{}\mathrm {i}}{3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {1{}\mathrm {i}}{512\,a^6\,d^3}\right )}^{1/3}}{2}+\frac {\ln \left (\frac {\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {a^6\,d^3\,49408{}\mathrm {i}}{3}+884736\,a^{10}\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}}{2}+\frac {a^4\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,14560{}\mathrm {i}}{3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}}{2}-\frac {\ln \left (\frac {\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (\frac {a^6\,d^3\,49408{}\mathrm {i}}{3}+884736\,a^{10}\,d^5\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{2/3}\right )\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}}{2}-\frac {a^4\,d^2\,{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,14560{}\mathrm {i}}{3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (-\frac {343{}\mathrm {i}}{373248\,a^6\,d^3}\right )}^{1/3}}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
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