Optimal. Leaf size=106 \[ \frac {2^{n+3} \tan ^3(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+3} (a \sec (c+d x)+a)^n F_1\left (\frac {3}{2};n+2,1;\frac {5}{2};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]
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Rubi [A] time = 0.06, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {3889} \[ \frac {2^{n+3} \tan ^3(c+d x) \left (\frac {1}{\sec (c+d x)+1}\right )^{n+3} (a \sec (c+d x)+a)^n F_1\left (\frac {3}{2};n+2,1;\frac {5}{2};-\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac {a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]
Antiderivative was successfully verified.
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Rule 3889
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx &=\frac {2^{3+n} F_1\left (\frac {3}{2};2+n,1;\frac {5}{2};-\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac {a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac {1}{1+\sec (c+d x)}\right )^{3+n} (a+a \sec (c+d x))^n \tan ^3(c+d x)}{3 d}\\ \end {align*}
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Mathematica [B] time = 11.33, size = 910, normalized size = 8.58 \[ \frac {(a (\sec (c+d x)+1))^n \left (-\frac {\, _2F_1\left (1-n,n+2;2-n;\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n \left (\tan \left (\frac {1}{2} (c+d x)\right )-1\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right )+1\right )^n (\sec (c+d x)+1)^{-n}}{n-1}-\frac {4 \, _2F_1\left (-n-1,n;-n;\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^n \left (\tan \left (\frac {1}{2} (c+d x)\right )+1\right )^n (\sec (c+d x)+1)^{-n}}{(n+1) \left (\tan \left (\frac {1}{2} (c+d x)\right )-1\right )}-\frac {120 F_1\left (\frac {1}{2};n,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \sin (c+d x) \left (3 F_1\left (\frac {1}{2};n,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 \left (F_1\left (\frac {3}{2};n,2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-n F_1\left (\frac {3}{2};n+1,1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{45 F_1\left (\frac {1}{2};n,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ){}^2 (-2 \cos (c+d x) n+2 n+\cos (2 (c+d x))+1) \cos ^2\left (\frac {1}{2} (c+d x)\right )+40 \left (F_1\left (\frac {3}{2};n,2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-n F_1\left (\frac {3}{2};n+1,1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ){}^2 \cos (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+6 F_1\left (\frac {1}{2};n,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (-48 \left (2 F_1\left (\frac {5}{2};n,3;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 n F_1\left (\frac {5}{2};n+1,2;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+n (n+1) F_1\left (\frac {5}{2};n+2,1;\frac {7}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cot (c+d x) \csc (c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-5 F_1\left (\frac {3}{2};n,2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (2 n-2 (n+2) \cos (c+d x)+\cos (2 (c+d x))+1)+5 n F_1\left (\frac {3}{2};n+1,1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (2 n-2 (n+2) \cos (c+d x)+\cos (2 (c+d x))+1)\right )}\right )}{4 d} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2}, x\right ) \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.76, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\tan ^{2}\left (d x +c \right )\right )\, dx \]
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 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{2}\,{d x} \]
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 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]
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 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{2}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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