Integrand size = 26, antiderivative size = 213 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {x}{8 \sqrt [3]{2} a^{4/3}}-\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \]
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Time = 0.26 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3621, 3607, 3562, 57, 631, 210, 31} \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {x}{8 \sqrt [3]{2} a^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}} \]
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Rule 31
Rule 57
Rule 210
Rule 631
Rule 3562
Rule 3607
Rule 3621
Rubi steps \begin{align*} \text {integral}& = -\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {\int \frac {a-2 i a \tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2} \\ & = -\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {i \text {Subst}\left (\int \frac {1}{(a-x) \sqrt [3]{a+x}} \, dx,x,i a \tan (c+d x)\right )}{4 a d} \\ & = \frac {x}{8 \sqrt [3]{2} a^{4/3}}-\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-x} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {(3 i) \text {Subst}\left (\int \frac {1}{2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} x+x^2} \, dx,x,\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 a d} \\ & = \frac {x}{8 \sqrt [3]{2} a^{4/3}}-\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {(3 i) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d} \\ & = \frac {x}{8 \sqrt [3]{2} a^{4/3}}-\frac {i \sqrt {3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} a^{4/3} d}-\frac {i \log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3 i}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {9 i}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.38 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.41 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {3 \left (7+\operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {1}{2} (1+i \tan (c+d x))\right ) (-2-2 i \tan (c+d x))+8 i \tan (c+d x)\right )}{8 a d (-i+\tan (c+d x)) \sqrt [3]{a+i a \tan (c+d x)}} \]
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Time = 0.62 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {3 i \left (-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{24 a^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{48 a^{\frac {1}{3}}}-\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{24 a^{\frac {1}{3}}}+\frac {3}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {a}{8 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\right )}{d a}\) | \(168\) |
default | \(\frac {3 i \left (-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{24 a^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{48 a^{\frac {1}{3}}}-\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{24 a^{\frac {1}{3}}}+\frac {3}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {a}{8 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}\right )}{d a}\) | \(168\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (146) = 292\).
Time = 0.25 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.62 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {{\left (32 \, a^{2} d \left (\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (32 \, a^{3} d^{2} \left (\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (-11 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 16 \, {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-16 \, {\left (i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 16 \, {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-16 \, {\left (-i \, \sqrt {3} a^{3} d^{2} + a^{3} d^{2}\right )} \left (\frac {i}{128 \, a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]
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\[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]
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none
Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.81 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - \frac {6 \, {\left (6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\right )}}{16 \, a^{3} d} \]
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\[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int { \frac {\tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
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Time = 0.78 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.02 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=-\frac {\frac {3{}\mathrm {i}}{8\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,9{}\mathrm {i}}{4\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}+\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (9\,{\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}+9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,a^{1/3}\right )}{a^{4/3}\,d}-\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d}+\frac {{\left (\frac {1}{128}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{16\,a^2\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{32\,a^{5/3}\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{a^{4/3}\,d} \]
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