Optimal. Leaf size=147 \[ \frac {d \left (b^2-a^2 (m+1)\right ) \sin (e+f x) (d \sec (e+f x))^{m-1} \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt {\sin ^2(e+f x)}}+\frac {a b (m+2) (d \sec (e+f x))^m}{f m (m+1)}+\frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)} \]
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Rubi [A] time = 0.17, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3508, 3486, 3772, 2643} \[ \frac {d \left (b^2-a^2 (m+1)\right ) \sin (e+f x) (d \sec (e+f x))^{m-1} \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(e+f x)\right )}{f (1-m) (m+1) \sqrt {\sin ^2(e+f x)}}+\frac {a b (m+2) (d \sec (e+f x))^m}{f m (m+1)}+\frac {b (a+b \tan (e+f x)) (d \sec (e+f x))^m}{f (m+1)} \]
Antiderivative was successfully verified.
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Rule 2643
Rule 3486
Rule 3508
Rule 3772
Rubi steps
\begin {align*} \int (d \sec (e+f x))^m (a+b \tan (e+f x))^2 \, dx &=\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\frac {\int (d \sec (e+f x))^m \left (-b^2+a^2 (1+m)+a b (2+m) \tan (e+f x)\right ) \, dx}{1+m}\\ &=\frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\left (a^2-\frac {b^2}{1+m}\right ) \int (d \sec (e+f x))^m \, dx\\ &=\frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}+\left (\left (a^2-\frac {b^2}{1+m}\right ) \left (\frac {\cos (e+f x)}{d}\right )^m (d \sec (e+f x))^m\right ) \int \left (\frac {\cos (e+f x)}{d}\right )^{-m} \, dx\\ &=\frac {a b (2+m) (d \sec (e+f x))^m}{f m (1+m)}-\frac {\left (a^2-\frac {b^2}{1+m}\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(e+f x)\right ) (d \sec (e+f x))^m \sin (e+f x)}{f (1-m) \sqrt {\sin ^2(e+f x)}}+\frac {b (d \sec (e+f x))^m (a+b \tan (e+f x))}{f (1+m)}\\ \end {align*}
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Mathematica [C] time = 26.50, size = 11095, normalized size = 75.48 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \left (d \sec \left (f x + e\right )\right )^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.04, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx \]
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 \[ \int {\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \sec \left (f x + e\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]
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 \[ \int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \tan {\left (e + f x \right )}\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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