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 = 0.21, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 68
Rule 822
Rule 830
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*}
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Mathematica [A] time = 0.24, 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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fricas [F] time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{9 \, x^{4} - 30 \, x^{3} + 31 \, x^{2} - 10 \, x + 1}, x\right ) \]
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 \[ \int \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (3 x +2\right ) \left (4 x +1\right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}\, dx \]
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 \[ \int \frac {{\left (4 \, x + 1\right )}^{m} {\left (3 \, x + 2\right )}}{{\left (3 \, x^{2} - 5 \, x + 1\right )}^{2}}\,{d x} \]
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 \[ \int \frac {\left (3\,x+2\right )\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \]
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 \[ \int \frac {\left (3 x + 2\right ) \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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