Optimal. Leaf size=147 \[ -\frac {3 \left (117-47 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (117+47 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {3 (4 x+1)^{m+1}}{4 (m+1)} \]
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Rubi [A] time = 0.15, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1628, 68} \[ -\frac {3 \left (117-47 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (m+1)}-\frac {3 \left (117+47 \sqrt {13}\right ) (4 x+1)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {3 (4 x+1)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (m+1)}+\frac {3 (4 x+1)^{m+1}}{4 (m+1)} \]
Antiderivative was successfully verified.
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Rule 68
Rule 1628
Rubi steps
\begin {align*} \int \frac {(2+3 x)^2 (1+4 x)^m}{1-5 x+3 x^2} \, dx &=\int \left (3 (1+4 x)^m+\frac {\left (27+\frac {141}{\sqrt {13}}\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (27-\frac {141}{\sqrt {13}}\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx\\ &=\frac {3 (1+4 x)^{1+m}}{4 (1+m)}+\frac {1}{13} \left (3 \left (117-47 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx+\frac {1}{13} \left (3 \left (117+47 \sqrt {13}\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx\\ &=\frac {3 (1+4 x)^{1+m}}{4 (1+m)}-\frac {3 \left (117-47 \sqrt {13}\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{26 \left (13-2 \sqrt {13}\right ) (1+m)}-\frac {3 \left (117+47 \sqrt {13}\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{26 \left (13+2 \sqrt {13}\right ) (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 91, normalized size = 0.62 \[ \frac {(4 x+1)^{m+1} \left (\left (58 \sqrt {13}-46\right ) \, _2F_1\left (1,m+1;m+2;\frac {12 x+3}{13-2 \sqrt {13}}\right )-2 \left (23+29 \sqrt {13}\right ) \, _2F_1\left (1,m+1;m+2;\frac {12 x+3}{13+2 \sqrt {13}}\right )+117\right )}{156 (m+1)} \]
Antiderivative was successfully verified.
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fricas [F] time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (9 \, x^{2} + 12 \, x + 4\right )} {\left (4 \, x + 1\right )}^{m}}{3 \, x^{2} - 5 \, x + 1}, x\right ) \]
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 \[ \int \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{2}}{3 \, x^{2} - 5 \, x + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (3 x +2\right )^{2} \left (4 x +1\right )^{m}}{3 x^{2}-5 x +1}\, dx \]
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 \[ \int \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}^{2}}{3 \, x^{2} - 5 \, x + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (3\,x+2\right )}^2\,{\left (4\,x+1\right )}^m}{3\,x^2-5\,x+1} \,d x \]
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 \[ \int \frac {\left (3 x + 2\right )^{2} \left (4 x + 1\right )^{m}}{3 x^{2} - 5 x + 1}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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