Optimal. Leaf size=220 \[ \frac{2^{-\frac{i n}{2}} \left (-6 a^2-6 a n-n^2+2\right ) (-i a-i b x+1)^{1+\frac{i n}{2}} \, _2F_1\left (\frac{i n}{2}+1,\frac{i n}{2};\frac{i n}{2}+2;\frac{1}{2} (-i a-i b x+1)\right )}{3 b^3 (-n+2 i)}-\frac{(4 a+n) (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{6 b^3}+\frac{x (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{3 b^2} \]
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Rubi [A] time = 0.139089, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5095, 90, 80, 69} \[ \frac{2^{-\frac{i n}{2}} \left (-6 a^2-6 a n-n^2+2\right ) (-i a-i b x+1)^{1+\frac{i n}{2}} \, _2F_1\left (\frac{i n}{2}+1,\frac{i n}{2};\frac{i n}{2}+2;\frac{1}{2} (-i a-i b x+1)\right )}{3 b^3 (-n+2 i)}-\frac{(4 a+n) (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{6 b^3}+\frac{x (-i a-i b x+1)^{1+\frac{i n}{2}} (i a+i b x+1)^{1-\frac{i n}{2}}}{3 b^2} \]
Antiderivative was successfully verified.
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Rule 5095
Rule 90
Rule 80
Rule 69
Rubi steps
\begin{align*} \int e^{n \tan ^{-1}(a+b x)} x^2 \, dx &=\int x^2 (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \, dx\\ &=\frac{x (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{3 b^2}+\frac{\int (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \left (-1-a^2-b (4 a+n) x\right ) \, dx}{3 b^2}\\ &=-\frac{(4 a+n) (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{6 b^3}+\frac{x (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{3 b^2}-\frac{\left (2-6 a^2-6 a n-n^2\right ) \int (1-i a-i b x)^{\frac{i n}{2}} (1+i a+i b x)^{-\frac{i n}{2}} \, dx}{6 b^2}\\ &=-\frac{(4 a+n) (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{6 b^3}+\frac{x (1-i a-i b x)^{1+\frac{i n}{2}} (1+i a+i b x)^{1-\frac{i n}{2}}}{3 b^2}+\frac{2^{-\frac{i n}{2}} \left (2-6 a^2-6 a n-n^2\right ) (1-i a-i b x)^{1+\frac{i n}{2}} \, _2F_1\left (1+\frac{i n}{2},\frac{i n}{2};2+\frac{i n}{2};\frac{1}{2} (1-i a-i b x)\right )}{3 b^3 (2 i-n)}\\ \end{align*}
Mathematica [A] time = 0.135236, size = 160, normalized size = 0.73 \[ \frac{(-i (a+b x+i))^{1+\frac{i n}{2}} \left (\frac{2^{1-\frac{i n}{2}} \left (6 a^2+6 a n+n^2-2\right ) \, _2F_1\left (\frac{i n}{2}+1,\frac{i n}{2};\frac{i n}{2}+2;-\frac{1}{2} i (a+b x+i)\right )}{n-2 i}-(4 a+n) (i a+i b x+1)^{1-\frac{i n}{2}}+2 b x (i a+i b x+1)^{1-\frac{i n}{2}}\right )}{6 b^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{n\arctan \left ( bx+a \right ) }}{x}^{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 x^{2} e^{\left (n \arctan \left (b x + a\right )\right )}\,{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 (x^{2} e^{\left (n \arctan \left (b x + a\right )\right )}, 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 x^{2} e^{n \operatorname{atan}{\left (a + b x \right )}}\, 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 x^{2} e^{\left (n \arctan \left (b x + a\right )\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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