Optimal. Leaf size=337 \[ -\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {\tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}-\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}+\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{18 a^2 d}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.59, antiderivative size = 337, 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 = {3559, 3595, 3538, 3476, 329, 275, 200, 31, 634, 618, 204, 628, 295, 203} \[ \frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{\tan (c+d x)}\right )}{72 a^2 d}+\frac {\tan ^{-1}\left (2 \sqrt [3]{\tan (c+d x)}+\sqrt {3}\right )}{72 a^2 d}+\frac {i \tan ^{-1}\left (\frac {1-2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )}{3 \sqrt {3} a^2 d}+\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}-\frac {i \log \left (\tan ^{\frac {2}{3}}(c+d x)+1\right )}{9 a^2 d}+\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)-\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}-\frac {\log \left (\tan ^{\frac {2}{3}}(c+d x)+\sqrt {3} \sqrt [3]{\tan (c+d x)}+1\right )}{48 \sqrt {3} a^2 d}+\frac {i \log \left (\tan ^{\frac {4}{3}}(c+d x)-\tan ^{\frac {2}{3}}(c+d x)+1\right )}{18 a^2 d}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 31
Rule 200
Rule 203
Rule 204
Rule 275
Rule 295
Rule 329
Rule 618
Rule 628
Rule 634
Rule 3476
Rule 3538
Rule 3559
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^{\frac {2}{3}}(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^{\frac {2}{3}}(c+d x) \left (\frac {7 a}{3}-\frac {1}{3} i a \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\frac {16 i a^2}{9}-\frac {2}{9} a^2 \tan (c+d x)}{\sqrt [3]{\tan (c+d x)}} \, dx}{8 a^4}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \int \frac {1}{\sqrt [3]{\tan (c+d x)}} \, dx}{9 a^2}+\frac {\int \tan ^{\frac {2}{3}}(c+d x) \, dx}{36 a^2}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{9 a^2 d}+\frac {\operatorname {Subst}\left (\int \frac {x^{2/3}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{36 a^2 d}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {x}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{3 a^2 d}+\frac {\operatorname {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{12 a^2 d}\\ &=\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {i \operatorname {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{3 a^2 d}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {\operatorname {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 )}{36 a^2 d}+\frac {\operatorname {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 )}{36 a^2 d}\\ &=\frac {\tan ^{-1}\left (\sqrt [3]{\tan (c+d x)}\right )}{36 a^2 d}+\frac {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {i \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}-\frac {i \operatorname {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{9 a^2 d}+\frac {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {i \operatorname {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 \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\tan ^{\frac {2}{3}}(c+d x)\right )}{6 a^2 d}-\frac {\operatorname {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 {\operatorname {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 {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {i \operatorname {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 {i \tan ^{\frac {2}{3}}(c+d x)}{3 a^2 d (1+i \tan (c+d x))}+\frac {\tan ^{\frac {5}{3}}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [C] time = 1.45, size = 194, normalized size = 0.58 \[ \frac {i \tan ^{\frac {2}{3}}(c+d x) \sec ^2(c+d x) \left (9 \sqrt [3]{2} e^{2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+14 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-4 (i \sin (2 (c+d x))+4 \cos (2 (c+d x))+4)\right )}{96 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 521, normalized size = 1.55 \[ -\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 ) - \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 (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 24 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 9 i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{144 \, a^{2} d} \]
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 \[ \int \frac {\tan \left (d x + c\right )^{\frac {2}{3}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.32, size = 357, normalized size = 1.06 \[ -\frac {7 i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}{72 d \,a^{2}}+\frac {i}{36 d \,a^{2} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )^{2}}+\frac {1}{36 d \,a^{2} \left (\tan ^{\frac {1}{3}}\left (d x +c \right )+i\right )}+\frac {i \ln \left (i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{16 d \,a^{2}}-\frac {\sqrt {3}\, \arctanh \left (\frac {\left (i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{8 d \,a^{2}}-\frac {i \left (\tan ^{\frac {2}{3}}\left (d x +c \right )\right )}{9 d \,a^{2} \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}-\frac {7 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )}{36 d \,a^{2} \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}+\frac {\tan \left (d x +c \right )}{18 d \,a^{2} \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}+\frac {i}{18 d \,a^{2} \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )^{2}}+\frac {7 i \ln \left (-i \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )+\tan ^{\frac {2}{3}}\left (d x +c \right )-1\right )}{144 d \,a^{2}}+\frac {7 \sqrt {3}\, \arctanh \left (\frac {\left (-i+2 \left (\tan ^{\frac {1}{3}}\left (d x +c \right )\right )\right ) \sqrt {3}}{3}\right )}{72 d \,a^{2}}-\frac {i \ln \left (\tan ^{\frac {1}{3}}\left (d x +c \right )-i\right )}{8 d \,a^{2}} \]
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 = 5.06, size = 640, normalized size = 1.90 \[ \text {result too large to display} \]
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 \[ - \frac {\int \frac {\tan ^{\frac {2}{3}}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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