Integrand size = 21, antiderivative size = 48 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=a x-\frac {a \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \]
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Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4226, 1816, 209} \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {a \tan ^3(e+f x)}{3 f}-\frac {a \tan (e+f x)}{f}+a x+\frac {b \tan ^5(e+f x)}{5 f} \]
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Rule 209
Rule 1816
Rule 4226
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4 \left (a+b \left (1+x^2\right )\right )}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (-a+a x^2+b x^4+\frac {a}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {a \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f}+\frac {a \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = a x-\frac {a \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {a \arctan (\tan (e+f x))}{f}-\frac {a \tan (e+f x)}{f}+\frac {a \tan ^3(e+f x)}{3 f}+\frac {b \tan ^5(e+f x)}{5 f} \]
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Time = 1.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98
method | result | size |
parts | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b \tan \left (f x +e \right )^{5}}{5 f}\) | \(47\) |
derivativedivides | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}}{f}\) | \(50\) |
default | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}}{f}\) | \(50\) |
risch | \(a x +\frac {2 i \left (-30 a \,{\mathrm e}^{8 i \left (f x +e \right )}+15 b \,{\mathrm e}^{8 i \left (f x +e \right )}-90 a \,{\mathrm e}^{6 i \left (f x +e \right )}-110 a \,{\mathrm e}^{4 i \left (f x +e \right )}+30 b \,{\mathrm e}^{4 i \left (f x +e \right )}-70 a \,{\mathrm e}^{2 i \left (f x +e \right )}-20 a +3 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\) | \(103\) |
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Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.50 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {15 \, a f x \cos \left (f x + e\right )^{5} - {\left ({\left (20 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{4} - {\left (5 \, a - 6 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, b\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \]
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Time = 0.83 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.12 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=a \left (\begin {cases} x + \frac {\tan ^{3}{\left (e + f x \right )}}{3 f} - \frac {\tan {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \tan ^{4}{\left (e \right )} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} x \tan ^{4}{\left (e \right )} \sec ^{2}{\left (e \right )} & \text {for}\: f = 0 \\\frac {\tan ^{5}{\left (e + f x \right )}}{5 f} & \text {otherwise} \end {cases}\right ) \]
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Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \, {\left (f x + e\right )} a - 15 \, a \tan \left (f x + e\right )}{15 \, f} \]
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Time = 0.94 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 15 \, {\left (f x + e\right )} a - 15 \, a \tan \left (f x + e\right )}{15 \, f} \]
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Time = 19.46 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.83 \[ \int \left (a+b \sec ^2(e+f x)\right ) \tan ^4(e+f x) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5}+\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^3}{3}-a\,\mathrm {tan}\left (e+f\,x\right )+a\,f\,x}{f} \]
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