Optimal. Leaf size=179 \[ -\frac{\left (2 \left (5+7 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{13 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (2 \left (5-7 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{13 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{(20-21 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \]
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Rubi [A] time = 0.212097, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {822, 830, 68} \[ -\frac{\left (2 \left (5+7 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13-2 \sqrt{13}}\right )}{13 \sqrt{13} \left (13-2 \sqrt{13}\right ) (m+1)}+\frac{\left (2 \left (5-7 \sqrt{13}\right ) m+81\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{3 (4 x+1)}{13+2 \sqrt{13}}\right )}{13 \sqrt{13} \left (13+2 \sqrt{13}\right ) (m+1)}+\frac{(20-21 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \]
Antiderivative was successfully verified.
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Rule 822
Rule 830
Rule 68
Rubi steps
\begin{align*} \int \frac{(2+3 x) (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx &=\frac{(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac{1}{507} \int \frac{(1+4 x)^m (13 (81+80 m)-1092 m x)}{1-5 x+3 x^2} \, dx\\ &=\frac{(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac{1}{507} \int \left (\frac{\left (-1092 m+6 \sqrt{13} (81+10 m)\right ) (1+4 x)^m}{-5-\sqrt{13}+6 x}+\frac{\left (-1092 m-6 \sqrt{13} (81+10 m)\right ) (1+4 x)^m}{-5+\sqrt{13}+6 x}\right ) \, dx\\ &=\frac{(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}+\frac{1}{169} \left (2 \left (182 m+\sqrt{13} (81+10 m)\right )\right ) \int \frac{(1+4 x)^m}{-5+\sqrt{13}+6 x} \, dx-\frac{1}{507} \left (-1092 m+6 \sqrt{13} (81+10 m)\right ) \int \frac{(1+4 x)^m}{-5-\sqrt{13}+6 x} \, dx\\ &=\frac{(20-21 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac{\left (182 m+\sqrt{13} (81+10 m)\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{3 (1+4 x)}{13-2 \sqrt{13}}\right )}{169 \left (13-2 \sqrt{13}\right ) (1+m)}+\frac{\left (81+2 \left (5-7 \sqrt{13}\right ) m\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{3 (1+4 x)}{13+2 \sqrt{13}}\right )}{13 \sqrt{13} \left (13+2 \sqrt{13}\right ) (1+m)}\\ \end{align*}
Mathematica [A] time = 0.177787, size = 149, normalized size = 0.83 \[ \frac{1}{507} (4 x+1)^{m+1} \left (\frac{3 \left (182 m+\sqrt{13} (10 m+81)\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13-2 \sqrt{13}}\right )}{\left (2 \sqrt{13}-13\right ) (m+1)}-\frac{3 \left (182 m-\sqrt{13} (10 m+81)\right ) \, _2F_1\left (1,m+1;m+2;\frac{12 x+3}{13+2 \sqrt{13}}\right )}{\left (13+2 \sqrt{13}\right ) (m+1)}+\frac{260-273 x}{3 x^2-5 x+1}\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 1.311, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( 4\,x+1 \right ) ^{m} \left ( 2+3\,x \right ) }{ \left ( 3\,{x}^{2}-5\,x+1 \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 \frac{{\left (4 \, x + 1\right )}^{m}{\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}}\,{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 (\frac{{\left (4 \, x + 1\right )}^{m}{\left (3 \, x + 2\right )}}{9 \, x^{4} - 30 \, x^{3} + 31 \, x^{2} - 10 \, x + 1}, 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 \frac{\left (3 x + 2\right ) \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\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 \frac{{\left (4 \, x + 1\right )}^{m}{\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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