Integrand size = 14, antiderivative size = 119 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}} \]
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Time = 0.05 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3739, 3554, 8} \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}+\frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}} \]
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Rule 8
Rule 3554
Rule 3739
Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^2(c+d x) \int \cot ^6(c+d x) \, dx}{b \sqrt {b \tan ^4(c+d x)}} \\ & = -\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan ^2(c+d x) \int \cot ^4(c+d x) \, dx}{b \sqrt {b \tan ^4(c+d x)}} \\ & = \frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}+\frac {\tan ^2(c+d x) \int \cot ^2(c+d x) \, dx}{b \sqrt {b \tan ^4(c+d x)}} \\ & = \frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan ^2(c+d x) \int 1 \, dx}{b \sqrt {b \tan ^4(c+d x)}} \\ & = \frac {\cot (c+d x)}{3 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\cot ^3(c+d x)}{5 b d \sqrt {b \tan ^4(c+d x)}}-\frac {\tan (c+d x)}{b d \sqrt {b \tan ^4(c+d x)}}-\frac {x \tan ^2(c+d x)}{b \sqrt {b \tan ^4(c+d x)}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(c+d x)\right ) \tan (c+d x)}{5 d \left (b \tan ^4(c+d x)\right )^{3/2}} \]
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Time = 0.07 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.53
method | result | size |
derivativedivides | \(-\frac {\tan \left (d x +c \right ) \left (15 \arctan \left (\tan \left (d x +c \right )\right ) \left (\tan ^{5}\left (d x +c \right )\right )+15 \left (\tan ^{4}\left (d x +c \right )\right )-5 \left (\tan ^{2}\left (d x +c \right )\right )+3\right )}{15 d {\left (\left (\tan ^{4}\left (d x +c \right )\right ) b \right )}^{\frac {3}{2}}}\) | \(63\) |
default | \(-\frac {\tan \left (d x +c \right ) \left (15 \arctan \left (\tan \left (d x +c \right )\right ) \left (\tan ^{5}\left (d x +c \right )\right )+15 \left (\tan ^{4}\left (d x +c \right )\right )-5 \left (\tan ^{2}\left (d x +c \right )\right )+3\right )}{15 d {\left (\left (\tan ^{4}\left (d x +c \right )\right ) b \right )}^{\frac {3}{2}}}\) | \(63\) |
risch | \(\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} x}{b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} b}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}}+\frac {2 i \left (45 \,{\mathrm e}^{8 i \left (d x +c \right )}-90 \,{\mathrm e}^{6 i \left (d x +c \right )}+140 \,{\mathrm e}^{4 i \left (d x +c \right )}-70 \,{\mathrm e}^{2 i \left (d x +c \right )}+23\right )}{15 b \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \sqrt {\frac {\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} b}{\left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}}\, d}\) | \(174\) |
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Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {{\left (15 \, d x \tan \left (d x + c\right )^{5} + 15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3\right )} \sqrt {b \tan \left (d x + c\right )^{4}}}{15 \, b^{2} d \tan \left (d x + c\right )^{7}} \]
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\[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{\left (b \tan ^{4}{\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
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Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {15 \, {\left (d x + c\right )}}{b^{\frac {3}{2}}} + \frac {15 \, \tan \left (d x + c\right )^{4} - 5 \, \tan \left (d x + c\right )^{2} + 3}{b^{\frac {3}{2}} \tan \left (d x + c\right )^{5}}}{15 \, d} \]
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Time = 0.65 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=-\frac {\frac {480 \, {\left (d x + c\right )}}{\sqrt {b}} - \frac {3 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 330 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{b^{\frac {5}{2}}} + \frac {330 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3}{\sqrt {b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{480 \, b d} \]
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Timed out. \[ \int \frac {1}{\left (b \tan ^4(c+d x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,{\mathrm {tan}\left (c+d\,x\right )}^4\right )}^{3/2}} \,d x \]
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