Optimal. Leaf size=205 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}} \]
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Rubi [A] time = 0.15, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3526, 3479, 3481, 55, 617, 204, 31} \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{4 \sqrt [3]{2} a^{4/3} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{8 \sqrt [3]{2} a^{4/3} d}+\frac {\log (\cos (c+d x))}{8 \sqrt [3]{2} a^{4/3} d}+\frac {i x}{8 \sqrt [3]{2} a^{4/3}}+\frac {3}{4 a d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3}{8 d (a+i a \tan (c+d x))^{4/3}} \]
Antiderivative was successfully verified.
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Rule 31
Rule 55
Rule 204
Rule 617
Rule 3479
Rule 3481
Rule 3526
Rubi steps
\begin {align*} \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{4/3}} \, dx &=-\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 {\operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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} \tan ^{-1}\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*}
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Mathematica [C] time = 0.66, size = 130, normalized size = 0.63 \[ \frac {3 i \sec ^2(c+d x) \left (\, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-2 i \sin (2 (c+d x))-\cos (2 (c+d x))-1\right )}{16 a d (\tan (c+d x)-i) \sqrt [3]{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.45, size = 360, normalized size = 1.76 \[ \frac {{\left (8 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a^{2} d \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a^{3} d^{2} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 4 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a^{2} d + a^{2} d\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a^{3} d^{2} - a^{3} d^{2}\right )} \left (\frac {1}{a^{4} d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) + 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]
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 \[ \int \frac {\tan \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 176, normalized size = 0.86 \[ \frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{16 d \,a^{\frac {4}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{8 d \,a^{\frac {4}{3}}}-\frac {3}{8 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}+\frac {3}{4 a d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.95, size = 171, normalized size = 0.83 \[ \frac {2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {6 \, {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}}}}{16 \, a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.97, size = 193, normalized size = 0.94 \[ \frac {\frac {3\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{4\,a}-\frac {3}{8}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-18\,4^{2/3}\,a^{4/3}\,d\right )}{8\,a^{4/3}\,d}+\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (-\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d}-\frac {4^{1/3}\,\ln \left (36\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-1152\,4^{2/3}\,a^{4/3}\,d\,{\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}^2\right )\,\left (\frac {1}{16}+\frac {\sqrt {3}\,1{}\mathrm {i}}{16}\right )}{a^{4/3}\,d} \]
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 \[ \int \frac {\tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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