Optimal. Leaf size=66 \[ \frac {1}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-\frac {4}{3} \left (5 \tan ^2(x)+1\right )^{3/2}+16 \sqrt {5 \tan ^2(x)+1}-32 \tan ^{-1}\left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right ) \]
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Rubi [A] time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3670, 444, 50, 63, 203} \[ \frac {1}{5} \left (5 \tan ^2(x)+1\right )^{5/2}-\frac {4}{3} \left (5 \tan ^2(x)+1\right )^{3/2}+16 \sqrt {5 \tan ^2(x)+1}-32 \tan ^{-1}\left (\frac {1}{2} \sqrt {5 \tan ^2(x)+1}\right ) \]
Antiderivative was successfully verified.
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Rule 50
Rule 63
Rule 203
Rule 444
Rule 3670
Rubi steps
\begin {align*} \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx &=\operatorname {Subst}\left (\int \frac {x \left (1+5 x^2\right )^{5/2}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {1}{2} \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{\frac {4}{5}+\frac {x^2}{5}} \, dx,x,\sqrt {1+5 \tan ^2(x)}\right )\\ &=-32 \tan ^{-1}\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*}
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Mathematica [C] time = 0.07, size = 49, normalized size = 0.74 \[ \frac {5 \sqrt {5} \left (5 \tan ^2(x)+1\right )^{5/2} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};\frac {4 \cos ^2(x)}{5}\right )}{(3-2 \cos (2 x))^{5/2}} \]
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \tan (x) \left (1+5 \tan ^2(x)\right )^{5/2} \, dx \]
Verification is Not applicable to the result.
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fricas [A] time = 1.56, size = 50, normalized size = 0.76 \[ \frac {1}{15} \, {\left (75 \, \tan \relax (x)^{4} - 70 \, \tan \relax (x)^{2} + 223\right )} \sqrt {5 \, \tan \relax (x)^{2} + 1} - 16 \, \arctan \left (\frac {5 \, \tan \relax (x)^{2} - 3}{4 \, \sqrt {5 \, \tan \relax (x)^{2} + 1}}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.62, size = 52, normalized size = 0.79 \[ \frac {1}{5} \, {\left (5 \, \tan \relax (x)^{2} + 1\right )}^{\frac {5}{2}} - \frac {4}{3} \, {\left (5 \, \tan \relax (x)^{2} + 1\right )}^{\frac {3}{2}} + 16 \, \sqrt {5 \, \tan \relax (x)^{2} + 1} - 32 \, \arctan \left (\frac {1}{2} \, \sqrt {5 \, \tan \relax (x)^{2} + 1}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 61, normalized size = 0.92
method | result | size |
derivativedivides | \(5 \left (\tan ^{4}\relax (x )\right ) \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}-\frac {14 \left (\tan ^{2}\relax (x )\right ) \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{3}+\frac {223 \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{15}-32 \arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{2}\right )\) | \(61\) |
default | \(5 \left (\tan ^{4}\relax (x )\right ) \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}-\frac {14 \left (\tan ^{2}\relax (x )\right ) \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{3}+\frac {223 \sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{15}-32 \arctan \left (\frac {\sqrt {1+5 \left (\tan ^{2}\relax (x )\right )}}{2}\right )\) | \(61\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (5 \, \tan \relax (x)^{2} + 1\right )}^{\frac {5}{2}} \tan \relax (x)\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.57, size = 90, normalized size = 1.36 \[ \frac {\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}\,\left (25\,{\mathrm {tan}\relax (x)}^4-\frac {70\,{\mathrm {tan}\relax (x)}^2}{3}+\frac {223}{3}\right )}{5}-\ln \left (\mathrm {tan}\relax (x)-\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}}{5}+\frac {1}{5}{}\mathrm {i}\right )\,16{}\mathrm {i}-\ln \left (\mathrm {tan}\relax (x)+\frac {2\,\sqrt {5}\,\sqrt {{\mathrm {tan}\relax (x)}^2+\frac {1}{5}}}{5}-\frac {1}{5}{}\mathrm {i}\right )\,16{}\mathrm {i}+\ln \left (\mathrm {tan}\relax (x)-\mathrm {i}\right )\,16{}\mathrm {i}+\ln \left (\mathrm {tan}\relax (x)+1{}\mathrm {i}\right )\,16{}\mathrm {i} \]
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 \[ \int \left (5 \tan ^{2}{\relax (x )} + 1\right )^{\frac {5}{2}} \tan {\relax (x )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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