Optimal. Leaf size=95 \[ \frac {a^2 \tan ^5(e+f x)}{5 f}-\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan (e+f x)}{f}-a^2 x+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \]
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Rubi [A] time = 0.11, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4141, 1802, 203} \[ \frac {a^2 \tan ^5(e+f x)}{5 f}-\frac {a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan (e+f x)}{f}-a^2 x+\frac {b (2 a+b) \tan ^7(e+f x)}{7 f}+\frac {b^2 \tan ^9(e+f x)}{9 f} \]
Antiderivative was successfully verified.
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Rule 203
Rule 1802
Rule 4141
Rubi steps
\begin {align*} \int \left (a+b \sec ^2(e+f x)\right )^2 \tan ^6(e+f x) \, dx &=\frac {\operatorname {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 {\operatorname {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 \operatorname {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*}
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Mathematica [B] time = 2.07, size = 275, normalized size = 2.89 \[ -\frac {4 \sec ^9(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (\left (231 a^2-270 a b+5 b^2\right ) \tan (e) \cos ^7(e+f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \tan (e) \cos ^5(e+f x)-\left (483 a^2-90 a b-10 b^2\right ) \sec (e) \sin (f x) \cos ^8(e+f x)+\left (231 a^2-270 a b+5 b^2\right ) \sec (e) \sin (f x) \cos ^6(e+f x)-3 \left (21 a^2-90 a b+25 b^2\right ) \sec (e) \sin (f x) \cos ^4(e+f x)+315 a^2 f x \cos ^9(e+f x)-5 b (18 a-19 b) \tan (e) \cos ^3(e+f x)-5 b (18 a-19 b) \sec (e) \sin (f x) \cos ^2(e+f x)-35 b^2 \tan (e) \cos (e+f x)-35 b^2 \sec (e) \sin (f x)\right )}{315 f (a \cos (2 (e+f x))+a+2 b)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.49, size = 137, normalized size = 1.44 \[ -\frac {315 \, a^{2} f x \cos \left (f x + e\right )^{9} - {\left ({\left (483 \, a^{2} - 90 \, a b - 10 \, b^{2}\right )} \cos \left (f x + e\right )^{8} - {\left (231 \, a^{2} - 270 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 3 \, {\left (21 \, a^{2} - 90 \, a b + 25 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 5 \, {\left (18 \, a b - 19 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 35 \, b^{2}\right )} \sin \left (f x + e\right )}{315 \, f \cos \left (f x + e\right )^{9}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 4.74, size = 98, normalized size = 1.03 \[ \frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 90 \, a b \tan \left (f x + e\right )^{7} + 45 \, b^{2} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.77, size = 105, normalized size = 1.11 \[ \frac {a^{2} \left (\frac {\left (\tan ^{5}\left (f x +e \right )\right )}{5}-\frac {\left (\tan ^{3}\left (f x +e \right )\right )}{3}+\tan \left (f x +e \right )-f x -e \right )+\frac {2 a b \left (\sin ^{7}\left (f x +e \right )\right )}{7 \cos \left (f x +e \right )^{7}}+b^{2} \left (\frac {\sin ^{7}\left (f x +e \right )}{9 \cos \left (f x +e \right )^{9}}+\frac {2 \left (\sin ^{7}\left (f x +e \right )\right )}{63 \cos \left (f x +e \right )^{7}}\right )}{f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 84, normalized size = 0.88 \[ \frac {35 \, b^{2} \tan \left (f x + e\right )^{9} + 45 \, {\left (2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{7} + 63 \, a^{2} \tan \left (f x + e\right )^{5} - 105 \, a^{2} \tan \left (f x + e\right )^{3} - 315 \, {\left (f x + e\right )} a^{2} + 315 \, a^{2} \tan \left (f x + e\right )}{315 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.54, size = 126, normalized size = 1.33 \[ \frac {\mathrm {tan}\left (e+f\,x\right )\,\left ({\left (a+b\right )}^2+b^2-2\,b\,\left (a+b\right )\right )-{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (\frac {{\left (a+b\right )}^2}{3}+\frac {b^2}{3}-\frac {2\,b\,\left (a+b\right )}{3}\right )+{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (\frac {{\left (a+b\right )}^2}{5}+\frac {b^2}{5}-\frac {2\,b\,\left (a+b\right )}{5}\right )-{\mathrm {tan}\left (e+f\,x\right )}^7\,\left (\frac {b^2}{7}-\frac {2\,b\,\left (a+b\right )}{7}\right )+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^9}{9}-a^2\,f\,x}{f} \]
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 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \tan ^{6}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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