Integrand size = 24, antiderivative size = 125 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
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Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3607, 3560, 3561, 212} \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {1}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}} \]
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Rule 212
Rule 3560
Rule 3561
Rule 3607
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {i \int \frac {1}{(a+i a \tan (c+d x))^{3/2}} \, dx}{2 a} \\ & = -\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i \int \frac {1}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 a^2} \\ & = -\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\text {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 {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {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 = 0.50 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.50 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-6+\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {1}{2} (1+i \tan (c+d x))\right ) (5+5 i \tan (c+d x))}{30 d (a+i a \tan (c+d x))^{5/2}} \]
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Time = 1.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}-\frac {1}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{d}\) | \(91\) |
default | \(\frac {-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}-\frac {1}{5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{6 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}}{d}\) | \(91\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (94) = 188\).
Time = 0.24 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.27 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (17 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 16 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]
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\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.93 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {15 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 12 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}}{240 \, a^{2} d} \]
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\[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{6\,a}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}-\frac {1}{5}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]
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