Optimal. Leaf size=245 \[ \frac {a^3 (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {3 a^2 b \sec (e+f x) \cos ^2(e+f x)^{\frac {n+2}{2}} (d \tan (e+f x))^{n+1} \, _2F_1\left (\frac {n+1}{2},\frac {n+2}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1)}+\frac {3 a b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac {b^3 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac {n+4}{2}} (d \tan (e+f x))^{n+1} \, _2F_1\left (\frac {n+1}{2},\frac {n+4}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1)} \]
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Rubi [A] time = 0.27, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3886, 3476, 364, 2617, 2607, 32} \[ \frac {3 a^2 b \sec (e+f x) \cos ^2(e+f x)^{\frac {n+2}{2}} (d \tan (e+f x))^{n+1} \, _2F_1\left (\frac {n+1}{2},\frac {n+2}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1)}+\frac {a^3 (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac {3 a b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)}+\frac {b^3 \sec ^3(e+f x) \cos ^2(e+f x)^{\frac {n+4}{2}} (d \tan (e+f x))^{n+1} \, _2F_1\left (\frac {n+1}{2},\frac {n+4}{2};\frac {n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1)} \]
Antiderivative was successfully verified.
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Rule 32
Rule 364
Rule 2607
Rule 2617
Rule 3476
Rule 3886
Rubi steps
\begin {align*} \int (a+b \sec (e+f x))^3 (d \tan (e+f x))^n \, dx &=\int \left (a^3 (d \tan (e+f x))^n+3 a^2 b \sec (e+f x) (d \tan (e+f x))^n+3 a b^2 \sec ^2(e+f x) (d \tan (e+f x))^n+b^3 \sec ^3(e+f x) (d \tan (e+f x))^n\right ) \, dx\\ &=a^3 \int (d \tan (e+f x))^n \, dx+\left (3 a^2 b\right ) \int \sec (e+f x) (d \tan (e+f x))^n \, dx+\left (3 a b^2\right ) \int \sec ^2(e+f x) (d \tan (e+f x))^n \, dx+b^3 \int \sec ^3(e+f x) (d \tan (e+f x))^n \, dx\\ &=\frac {3 a^2 b \cos ^2(e+f x)^{\frac {2+n}{2}} \, _2F_1\left (\frac {1+n}{2},\frac {2+n}{2};\frac {3+n}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b^3 \cos ^2(e+f x)^{\frac {4+n}{2}} \, _2F_1\left (\frac {1+n}{2},\frac {4+n}{2};\frac {3+n}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {\left (3 a b^2\right ) \operatorname {Subst}\left (\int (d x)^n \, dx,x,\tan (e+f x)\right )}{f}+\frac {\left (a^3 d\right ) \operatorname {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac {3 a b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {a^3 \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {3 a^2 b \cos ^2(e+f x)^{\frac {2+n}{2}} \, _2F_1\left (\frac {1+n}{2},\frac {2+n}{2};\frac {3+n}{2};\sin ^2(e+f x)\right ) \sec (e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b^3 \cos ^2(e+f x)^{\frac {4+n}{2}} \, _2F_1\left (\frac {1+n}{2},\frac {4+n}{2};\frac {3+n}{2};\sin ^2(e+f x)\right ) \sec ^3(e+f x) (d \tan (e+f x))^{1+n}}{d f (1+n)}\\ \end {align*}
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Mathematica [A] time = 3.52, size = 238, normalized size = 0.97 \[ \frac {d \left (-\tan ^2(e+f x)\right )^{-n/2} (d \tan (e+f x))^{n-1} \left (-3 a^3 \left (-\tan ^2(e+f x)\right )^{\frac {n+2}{2}} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )+9 a^2 b (n+1) \sqrt {-\tan ^2(e+f x)} \sec (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3}{2};\sec ^2(e+f x)\right )+9 a b^2 \left (\sqrt {-\tan ^2(e+f x)}-\left (-\tan ^2(e+f x)\right )^{\frac {n+2}{2}}\right )+b^3 (n+1) \sqrt {-\tan ^2(e+f x)} \sec ^3(e+f x) \, _2F_1\left (\frac {3}{2},\frac {1-n}{2};\frac {5}{2};\sec ^2(e+f x)\right )\right )}{3 f (n+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{3} \sec \left (f x + e\right )^{3} + 3 \, a b^{2} \sec \left (f x + e\right )^{2} + 3 \, a^{2} b \sec \left (f x + e\right ) + a^{3}\right )} \left (d \tan \left (f x + e\right )\right )^{n}, x\right ) \]
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 \[ \int {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.80, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (f x +e \right )\right )^{3} \left (d \tan \left (f x +e \right )\right )^{n}\, dx \]
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 \[ \int {\left (b \sec \left (f x + e\right ) + a\right )}^{3} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^3 \,d x \]
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 \[ \int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \sec {\left (e + f x \right )}\right )^{3}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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