Optimal. Leaf size=176 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {151 i}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 0.37, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3558, 3595, 3592, 3526, 3480, 206} \[ \frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {151 i}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 206
Rule 3480
Rule 3526
Rule 3558
Rule 3592
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan ^2(c+d x) \left (-3 a+\frac {11}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan (c+d x) \left (-17 i a^2-\frac {83}{4} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {\int \frac {\frac {83 a^2}{4}-17 i a^2 \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {151 i}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}+\frac {\int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {151 i}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}-\frac {i \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {\tan ^3(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {17 i \tan ^2(c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {151 i}{60 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {83 i \sqrt {a+i a \tan (c+d x)}}{30 a^3 d}\\ \end {align*}
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Mathematica [A] time = 1.57, size = 149, normalized size = 0.85 \[ \frac {e^{-6 i (c+d x)} \left (i \left (1+e^{2 i (c+d x)}\right ) \left (-26 e^{2 i (c+d x)}+194 e^{4 i (c+d x)}+463 e^{6 i (c+d x)}+3\right ) \sec ^2(c+d x)-\frac {60 i e^{7 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{240 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.44, size = 283, normalized size = 1.61 \[ \frac {{\left (-15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 15 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (463 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 194 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 26 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
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 \[ \int \frac {\tan \left (d x + c\right )^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.16, size = 111, normalized size = 0.63 \[ \frac {2 i \left (\sqrt {a +i a \tan \left (d x +c \right )}+\frac {17 a}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {7 a^{2}}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{3}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16}\right )}{d \,a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.60, size = 139, normalized size = 0.79 \[ \frac {i \, {\left (15 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 480 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2} + \frac {4 \, {\left (255 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} - 70 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} + 12 \, a^{5}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{5} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.25, size = 129, normalized size = 0.73 \[ \frac {\frac {1{}\mathrm {i}}{5\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,17{}\mathrm {i}}{4\,a^2\,d}-\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{6\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a^3\,d}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d} \]
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 \[ \int \frac {\tan ^{4}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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