Integrand size = 23, antiderivative size = 77 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=a^2 x-\frac {a^2 \tan (e+f x)}{f}+\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \]
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Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4226, 1816, 209} \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {a^2 \tan ^3(e+f x)}{3 f}-\frac {a^2 \tan (e+f x)}{f}+a^2 x+\frac {b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \]
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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 )^2}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (-a^2+a^2 x^2+b (2 a+b) x^4+b^2 x^6+\frac {a^2}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {a^2 \tan (e+f x)}{f}+\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f}+\frac {a^2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = a^2 x-\frac {a^2 \tan (e+f x)}{f}+\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {b (2 a+b) \tan ^5(e+f x)}{5 f}+\frac {b^2 \tan ^7(e+f x)}{7 f} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(395\) vs. \(2(77)=154\).
Time = 2.30 (sec) , antiderivative size = 395, normalized size of antiderivative = 5.13 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {\sec (e) \sec ^7(e+f x) \left (3675 a^2 f x \cos (f x)+3675 a^2 f x \cos (2 e+f x)+2205 a^2 f x \cos (2 e+3 f x)+2205 a^2 f x \cos (4 e+3 f x)+735 a^2 f x \cos (4 e+5 f x)+735 a^2 f x \cos (6 e+5 f x)+105 a^2 f x \cos (6 e+7 f x)+105 a^2 f x \cos (8 e+7 f x)-5320 a^2 \sin (f x)+1680 a b \sin (f x)+840 b^2 \sin (f x)+4480 a^2 \sin (2 e+f x)-1260 a b \sin (2 e+f x)+420 b^2 \sin (2 e+f x)-3780 a^2 \sin (2 e+3 f x)+924 a b \sin (2 e+3 f x)-168 b^2 \sin (2 e+3 f x)+2100 a^2 \sin (4 e+3 f x)-840 a b \sin (4 e+3 f x)-420 b^2 \sin (4 e+3 f x)-1540 a^2 \sin (4 e+5 f x)+168 a b \sin (4 e+5 f x)+84 b^2 \sin (4 e+5 f x)+420 a^2 \sin (6 e+5 f x)-420 a b \sin (6 e+5 f x)-280 a^2 \sin (6 e+7 f x)+84 a b \sin (6 e+7 f x)+12 b^2 \sin (6 e+7 f x)\right )}{13440 f} \]
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Time = 6.53 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01
method | result | size |
parts | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{7}}{7}+\frac {\tan \left (f x +e \right )^{5}}{5}\right )}{f}+\frac {2 a b \tan \left (f x +e \right )^{5}}{5 f}\) | \(78\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {2 a b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{5}}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \sin \left (f x +e \right )^{5}}{35 \cos \left (f x +e \right )^{5}}\right )}{f}\) | \(94\) |
default | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{3}}{3}-\tan \left (f x +e \right )+f x +e \right )+\frac {2 a b \sin \left (f x +e \right )^{5}}{5 \cos \left (f x +e \right )^{5}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{5}}{7 \cos \left (f x +e \right )^{7}}+\frac {2 \sin \left (f x +e \right )^{5}}{35 \cos \left (f x +e \right )^{5}}\right )}{f}\) | \(94\) |
risch | \(a^{2} x +\frac {4 i \left (-105 a^{2} {\mathrm e}^{12 i \left (f x +e \right )}+105 a b \,{\mathrm e}^{12 i \left (f x +e \right )}-525 a^{2} {\mathrm e}^{10 i \left (f x +e \right )}+210 a b \,{\mathrm e}^{10 i \left (f x +e \right )}+105 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}-1120 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+315 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-105 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-1330 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+420 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+210 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-945 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+231 a b \,{\mathrm e}^{4 i \left (f x +e \right )}-42 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-385 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+42 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+21 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-70 a^{2}+21 a b +3 b^{2}\right )}{105 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{7}}\) | \(273\) |
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Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.47 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {105 \, a^{2} f x \cos \left (f x + e\right )^{7} - {\left (2 \, {\left (70 \, a^{2} - 21 \, a b - 3 \, b^{2}\right )} \cos \left (f x + e\right )^{6} - {\left (35 \, a^{2} - 84 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 6 \, {\left (7 \, a b - 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 15 \, b^{2}\right )} \sin \left (f x + e\right )}{105 \, f \cos \left (f x + e\right )^{7}} \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \tan ^{4}{\left (e + f x \right )}\, dx \]
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Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.92 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 21 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 35 \, a^{2} \tan \left (f x + e\right )^{3} + 105 \, {\left (f x + e\right )} a^{2} - 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \]
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Time = 1.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {15 \, b^{2} \tan \left (f x + e\right )^{7} + 42 \, a b \tan \left (f x + e\right )^{5} + 21 \, b^{2} \tan \left (f x + e\right )^{5} + 35 \, a^{2} \tan \left (f x + e\right )^{3} + 105 \, {\left (f x + e\right )} a^{2} - 105 \, a^{2} \tan \left (f x + e\right )}{105 \, f} \]
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Time = 19.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.26 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^4(e+f x) \, dx=\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {{\left (a+b\right )}^2}{3}+\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )-\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {b^2}{5}-\frac {2\,b\,\left (a+b\right )}{5}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^7}{7}+a^2\,f\,x}{f} \]
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