Optimal. Leaf size=343 \[ \frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {5 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d}+\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {5 i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{12 a d} \]
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Rubi [A] time = 0.46, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3550, 3528, 3538, 3476, 329, 209, 634, 618, 204, 628, 203, 275, 292, 31} \[ -\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}+\frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{6 a d}+\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}-\frac {7 \log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{8 \sqrt {3} a d}+\frac {5 i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{12 a d} \]
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 3528
Rule 3538
Rule 3550
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {10}{3}}(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan ^{\frac {4}{3}}(c+d x) \left (\frac {7 a}{3}-\frac {10}{3} i a \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \sqrt [3]{\tan (c+d x)} \left (\frac {10 i a}{3}+\frac {7}{3} a \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \frac {-\frac {7 a}{3}+\frac {10}{3} i a \tan (c+d x)}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{2 a^2}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \int \sqrt [3]{\tan (c+d x)} \, dx}{3 a}-\frac {7 \int \frac {1}{\tan ^{\frac {2}{3}}(c+d x)} \, dx}{6 a}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {\sqrt [3]{x}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{3 a d}-\frac {7 \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 {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {x^3}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{2 a d}\\ &=\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {7 \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 {7 \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 {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1+x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}-\frac {7 \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 {7 \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 {7 \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 {7 \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 {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \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 {7 \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 {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{12 a d}+\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{4 a d}+\frac {7 \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 {7 \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 {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \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 {7 \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-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(5 i) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \tan ^{\frac {2}{3}}(c+d x)\right )}{2 a d}\\ &=\frac {7 \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {7 \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{\tan (c+d x)}\right )}{12 a d}-\frac {5 i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{2 \sqrt {3} a d}-\frac {7 \tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{6 a d}-\frac {5 i \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a d}+\frac {7 \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 {7 \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-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )}{12 a d}+\frac {7 \sqrt [3]{\tan (c+d x)}}{2 a d}-\frac {5 i \tan ^{\frac {4}{3}}(c+d x)}{4 a d}-\frac {\tan ^{\frac {7}{3}}(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}
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Mathematica [C] time = 1.68, size = 181, normalized size = 0.53 \[ -\frac {i \sqrt [3]{\tan (c+d x)} \sec ^2(c+d x) \left (3\ 2^{2/3} \left (1+e^{2 i (c+d x)}\right )^{4/3} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )-34 \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (i \sin (2 (c+d x))+\cos (2 (c+d x))+1)+18 i \sin (2 (c+d x))+22 \cos (2 (c+d x))+34\right )}{16 a d (\tan (c+d x)-i)} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.64, size = 643, normalized size = 1.87 \[ \frac {{\left (3 \, \sqrt {3} {\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 )} \sqrt {\frac {1}{a^{2} d^{2}}} + 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, 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 (3 \, \sqrt {3} {\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 )} \sqrt {\frac {1}{a^{2} d^{2}}} - 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 3 i \, 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 (51 \, \sqrt {\frac {1}{3}} {\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 )} \sqrt {\frac {1}{a^{2} d^{2}}} - 17 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 17 i \, 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 (51 \, \sqrt {\frac {1}{3}} {\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 )} \sqrt {\frac {1}{a^{2} d^{2}}} + 17 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 i \, 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 (-34 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 34 i \, 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 ) + {\left (-6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 6 i \, 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 ) + 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 {1}{3}} {\left (10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 17 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{24 \, {\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 \[ \int \frac {\tan \left (d x + c\right )^{\frac {10}{3}}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.32, size = 291, normalized size = 0.85 \[ \frac {3 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{d a}-\frac {3 i \left (\tan ^{\frac {4}{3}}\left (d x +c \right )\right )}{4 d a}-\frac {17 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{12 d a}+\frac {1}{6 d a \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{8 d a}+\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{4 d a}-\frac {\tan ^{\frac {1}{3}}\left (d x +c \right )}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}-\frac {i}{6 d a \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}+\frac {17 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{24 d a}-\frac {17 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{12 d a}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{4 d a} \]
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 = 5.72, size = 655, normalized size = 1.91 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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