Integrand size = 15, antiderivative size = 66 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=-32 \arctan \left (\frac {1}{2} \sqrt {1+5 \tan ^2(x)}\right )+16 \sqrt {1+5 \tan ^2(x)}-\frac {4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2} \]
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Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3751, 455, 52, 65, 209} \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=-32 \arctan \left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right )+\frac {1}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-\frac {4}{3} \left (5 \tan ^2(x)+1\right )^{3/2}+16 \sqrt {5 \tan ^2(x)+1} \]
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Rule 52
Rule 65
Rule 209
Rule 455
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x \left (1+5 x^2\right )^{5/2}}{1+x^2} \, dx,x,\tan (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {(1+5 x)^{5/2}}{1+x} \, dx,x,\tan ^2(x)\right ) \\ & = \frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}-2 \text {Subst}\left (\int \frac {(1+5 x)^{3/2}}{1+x} \, dx,x,\tan ^2(x)\right ) \\ & = -\frac {4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}+8 \text {Subst}\left (\int \frac {\sqrt {1+5 x}}{1+x} \, dx,x,\tan ^2(x)\right ) \\ & = 16 \sqrt {1+5 \tan ^2(x)}-\frac {4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}-32 \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {1+5 x}} \, dx,x,\tan ^2(x)\right ) \\ & = 16 \sqrt {1+5 \tan ^2(x)}-\frac {4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2}-\frac {64}{5} \text {Subst}\left (\int \frac {1}{\frac {4}{5}+\frac {x^2}{5}} \, dx,x,\sqrt {1+5 \tan ^2(x)}\right ) \\ & = -32 \arctan \left (\frac {1}{2} \sqrt {1+5 \tan ^2(x)}\right )+16 \sqrt {1+5 \tan ^2(x)}-\frac {4}{3} \left (1+5 \tan ^2(x)\right )^{3/2}+\frac {1}{5} \left (1+5 \tan ^2(x)\right )^{5/2} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.74 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {5 \sqrt {5} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {5}{2},-\frac {3}{2},\frac {4 \cos ^2(x)}{5}\right ) \left (1+5 \tan ^2(x)\right )^{5/2}}{(3-2 \cos (2 x))^{5/2}} \]
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Time = 0.07 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {223 \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{15}+5 \left (\tan ^{4}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}-\frac {14 \left (\tan ^{2}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{3}-32 \arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{2}\right )\) | \(61\) |
default | \(\frac {223 \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{15}+5 \left (\tan ^{4}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}-\frac {14 \left (\tan ^{2}\left (x \right )\right ) \sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{3}-32 \arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\left (x \right )\right )}}{2}\right )\) | \(61\) |
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Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.76 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {1}{15} \, {\left (75 \, \tan \left (x\right )^{4} - 70 \, \tan \left (x\right )^{2} + 223\right )} \sqrt {5 \, \tan \left (x\right )^{2} + 1} - 16 \, \arctan \left (\frac {5 \, \tan \left (x\right )^{2} - 3}{4 \, \sqrt {5 \, \tan \left (x\right )^{2} + 1}}\right ) \]
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\[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\int \left (5 \tan ^{2}{\left (x \right )} + 1\right )^{\frac {5}{2}} \tan {\left (x \right )}\, dx \]
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\[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\int { {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {5}{2}} \tan \left (x\right ) \,d x } \]
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Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.79 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {1}{5} \, {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {5}{2}} - \frac {4}{3} \, {\left (5 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} + 16 \, \sqrt {5 \, \tan \left (x\right )^{2} + 1} - 32 \, \arctan \left (\frac {1}{2} \, \sqrt {5 \, \tan \left (x\right )^{2} + 1}\right ) \]
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Time = 1.62 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.36 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx=\frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}\,\left (25\,{\mathrm {tan}\left (x\right )}^4-\frac {70\,{\mathrm {tan}\left (x\right )}^2}{3}+\frac {223}{3}\right )}{5}-\ln \left (\mathrm {tan}\left (x\right )-\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{5}+\frac {1}{5}{}\mathrm {i}\right )\,16{}\mathrm {i}-\ln \left (\mathrm {tan}\left (x\right )+\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\left (x\right )}^2+\frac {1}{5}}}{5}-\frac {1}{5}{}\mathrm {i}\right )\,16{}\mathrm {i}+\ln \left (\mathrm {tan}\left (x\right )-\mathrm {i}\right )\,16{}\mathrm {i}+\ln \left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i} \]
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