Optimal. Leaf size=140 \[ \frac{\left (a^2-b^2\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac{2 a b (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac{b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)} \]
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Rubi [A] time = 0.147638, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3543, 3538, 3476, 364} \[ \frac{\left (a^2-b^2\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}+\frac{2 a b (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac{b^2 (d \tan (e+f x))^{n+1}}{d f (n+1)} \]
Antiderivative was successfully verified.
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Rule 3543
Rule 3538
Rule 3476
Rule 364
Rubi steps
\begin{align*} \int (d \tan (e+f x))^n (a+b \tan (e+f x))^2 \, dx &=\frac{b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\int (d \tan (e+f x))^n \left (a^2-b^2+2 a b \tan (e+f x)\right ) \, dx\\ &=\frac{b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\left (a^2-b^2\right ) \int (d \tan (e+f x))^n \, dx+\frac{(2 a b) \int (d \tan (e+f x))^{1+n} \, dx}{d}\\ &=\frac{b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac{(2 a b) \operatorname{Subst}\left (\int \frac{x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac{\left (\left (a^2-b^2\right ) d\right ) \operatorname{Subst}\left (\int \frac{x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=\frac{b^2 (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac{\left (a^2-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{2 a b \, _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)}\\ \end{align*}
Mathematica [A] time = 0.36639, size = 116, normalized size = 0.83 \[ \frac{\tan (e+f x) (d \tan (e+f x))^n \left ((n+2) \left (a^2-b^2\right ) \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )+b \left (2 a (n+1) \tan (e+f x) \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )+b (n+2)\right )\right )}{f (n+1) (n+2)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.288, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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 \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan{\left (e + f x \right )}\right )^{n} \left (a + b \tan{\left (e + f x \right )}\right )^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right ) + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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