Integrand size = 24, antiderivative size = 205 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}} \]
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Time = 0.19 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3607, 3560, 3562, 57, 631, 210, 31} \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}} \]
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Rule 31
Rule 57
Rule 210
Rule 631
Rule 3560
Rule 3562
Rule 3607
Rubi steps \begin{align*} \text {integral}& = -\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}-\frac {i \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx}{2 a} \\ & = -\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \int (a+i a \tan (c+d x))^{2/3} \, dx}{4 a^2} \\ & = -\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\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 {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \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 \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 {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 \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}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \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 {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {\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 {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {3 \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}{8 d (a+i a \tan (c+d x))^{4/3}}+\frac {3}{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 = 0.34 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.31 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {3 \left (-1+\operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {1}{2} (1+i \tan (c+d x))\right ) (2+2 i \tan (c+d x))\right )}{8 d (a+i a \tan (c+d x))^{4/3}} \]
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Time = 0.63 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {3}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\) | \(176\) |
default | \(\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {3}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\) | \(176\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (146) = 292\).
Time = 0.25 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.76 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {{\left (8 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a^{2} d \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a^{3} d^{2} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]
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\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.83 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {6 \, {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}}{16 \, a^{2} d} \]
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\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\int { \frac {\tan \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
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Time = 4.47 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.94 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx=\frac {\frac {3\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{4\,a}-\frac {3}{8}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-18\,4^{2/3}\,a^{4/3}\,d\right )}{8\,a^{4/3}\,d}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d}-\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d} \]
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