Optimal. Leaf size=198 \[ \frac {a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac {a \left (a^2-3 b^2\right ) \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac {b \left (3 a^2-b^2\right ) \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)} \]
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Rubi [A]
time = 0.21, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3647, 3711,
3619, 3557, 371} \begin {gather*} \frac {b \left (3 a^2-b^2\right ) (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac {a \left (a^2-3 b^2\right ) (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 {a b^2 (2 n+5) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+2)}+\frac {b^2 (a+b \tan (e+f x)) (d \tan (e+f x))^{n+1}}{d f (n+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 371
Rule 3557
Rule 3619
Rule 3647
Rule 3711
Rubi steps
\begin {align*} \int (d \tan (e+f x))^n (a+b \tan (e+f x))^3 \, dx &=\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\frac {\int (d \tan (e+f x))^n \left (-a d \left (b^2 (1+n)-a^2 (2+n)\right )+b \left (3 a^2-b^2\right ) d (2+n) \tan (e+f x)+a b^2 d (5+2 n) \tan ^2(e+f x)\right ) \, dx}{d (2+n)}\\ &=\frac {a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\frac {\int (d \tan (e+f x))^n \left (a \left (a^2-3 b^2\right ) d (2+n)+b \left (3 a^2-b^2\right ) d (2+n) \tan (e+f x)\right ) \, dx}{d (2+n)}\\ &=\frac {a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\left (a \left (a^2-3 b^2\right )\right ) \int (d \tan (e+f x))^n \, dx+\frac {\left (b \left (3 a^2-b^2\right )\right ) \int (d \tan (e+f x))^{1+n} \, dx}{d}\\ &=\frac {a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}+\frac {\left (b \left (3 a^2-b^2\right )\right ) \text {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac {\left (a \left (a^2-3 b^2\right ) d\right ) \text {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac {a b^2 (5+2 n) (d \tan (e+f x))^{1+n}}{d f (1+n) (2+n)}+\frac {a \left (a^2-3 b^2\right ) \, _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 {b \left (3 a^2-b^2\right ) \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}+\frac {b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))}{d f (2+n)}\\ \end {align*}
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Mathematica [A]
time = 0.80, size = 141, normalized size = 0.71 \begin {gather*} \frac {\tan (e+f x) (d \tan (e+f x))^n \left (a \left (a^2-3 b^2\right ) (2+n) \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right )+b \left (\left (3 a^2-b^2\right ) (1+n) \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) \tan (e+f x)+b (3 a (2+n)+b (1+n) \tan (e+f x))\right )\right )}{f (1+n) (2+n)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.43, size = 0, normalized size = 0.00 \[\int \left (d \tan \left (f x +e \right )\right )^{n} \left (a +b \tan \left (f x +e \right )\right )^{3}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \left (d \tan {\left (e + f x \right )}\right )^{n} \left (a + b \tan {\left (e + f x \right )}\right )^{3}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int {\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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