Integrand size = 23, antiderivative size = 95 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-a^2 x+\frac {a^2 \tan (e+f x)}{f}-\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan ^5(e+f x)}{5 f}+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \]
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Time = 0.13 (sec) , antiderivative size = 95, 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} \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {a^2 \tan ^5(e+f x)}{5 f}-\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan (e+f x)}{f}-a^2 x+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \]
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Rule 209
Rule 1816
Rule 4226
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^6 \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+a^2 x^4+b (2 a+b) x^6+b^2 x^8-\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 {a^2 \tan ^5(e+f x)}{5 f}+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 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 {a^2 \tan ^5(e+f x)}{5 f}+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(275\) vs. \(2(95)=190\).
Time = 3.71 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.89 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {4 \left (b+a \cos ^2(e+f x)\right )^2 \sec ^9(e+f x) \left (315 a^2 f x \cos ^9(e+f x)-35 b^2 \sec (e) \sin (f x)-5 (18 a-19 b) b \cos ^2(e+f x) \sec (e) \sin (f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \cos ^4(e+f x) \sec (e) \sin (f x)+\left (231 a^2-270 a b+5 b^2\right ) \cos ^6(e+f x) \sec (e) \sin (f x)-\left (483 a^2-90 a b-10 b^2\right ) \cos ^8(e+f x) \sec (e) \sin (f x)-35 b^2 \cos (e+f x) \tan (e)-5 (18 a-19 b) b \cos ^3(e+f x) \tan (e)-3 \left (21 a^2-90 a b+25 b^2\right ) \cos ^5(e+f x) \tan (e)+\left (231 a^2-270 a b+5 b^2\right ) \cos ^7(e+f x) \tan (e)\right )}{315 f (a+2 b+a \cos (2 (e+f x)))^2} \]
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Time = 12.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.93
method | result | size |
parts | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {b^{2} \left (\frac {\tan \left (f x +e \right )^{9}}{9}+\frac {\tan \left (f x +e \right )^{7}}{7}\right )}{f}+\frac {2 a b \tan \left (f x +e \right )^{7}}{7 f}\) | \(88\) |
derivativedivides | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{7}}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{7}}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \sin \left (f x +e \right )^{7}}{63 \cos \left (f x +e \right )^{7}}\right )}{f}\) | \(105\) |
default | \(\frac {a^{2} \left (\frac {\tan \left (f x +e \right )^{5}}{5}-\frac {\tan \left (f x +e \right )^{3}}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \sin \left (f x +e \right )^{7}}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin \left (f x +e \right )^{7}}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \sin \left (f x +e \right )^{7}}{63 \cos \left (f x +e \right )^{7}}\right )}{f}\) | \(105\) |
risch | \(-a^{2} x -\frac {2 i \left (90 a b -483 a^{2}+10 b^{2}-28350 a^{2} {\mathrm e}^{10 i \left (f x +e \right )}+5040 a b \,{\mathrm e}^{8 i \left (f x +e \right )}+3780 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+1980 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+180 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-32508 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}-1890 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-24402 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}+1890 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-11718 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-270 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-3402 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+90 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-5670 a^{2} {\mathrm e}^{14 i \left (f x +e \right )}-16170 a^{2} {\mathrm e}^{12 i \left (f x +e \right )}-1050 b^{2} {\mathrm e}^{12 i \left (f x +e \right )}+630 a b \,{\mathrm e}^{16 i \left (f x +e \right )}+1260 a b \,{\mathrm e}^{14 i \left (f x +e \right )}+3780 a b \,{\mathrm e}^{12 i \left (f x +e \right )}-945 a^{2} {\mathrm e}^{16 i \left (f x +e \right )}+630 b^{2} {\mathrm e}^{14 i \left (f x +e \right )}+6300 a b \,{\mathrm e}^{10 i \left (f x +e \right )}+3150 b^{2} {\mathrm e}^{10 i \left (f x +e \right )}\right )}{315 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{9}}\) | \(356\) |
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Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.44 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=-\frac {315 \, a^{2} f x \cos \left (f x + e\right )^{9} - {\left ({\left (483 \, a^{2} - 90 \, a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{8} - {\left (231 \, a^{2} - 270 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (21 \, a^{2} - 90 \, a b + 25 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \, {\left (18 \, a b - 19 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \tan ^{6}{\left (e + f x \right )}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 45 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
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Time = 2.52 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 90 \, a b \tan \left (f x + e\right )^{7} + 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
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Time = 19.69 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {{\left (a+b\right )}^2}{3}+\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {{\left (a+b\right )}^2}{5}+\frac {b^2}{5}-\frac {2\,b\,\left (a+b\right )}{5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (\frac {b^2}{7}-\frac {2\,b\,\left (a+b\right )}{7}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9}{9}-a^2\,f\,x}{f} \]
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