Optimal. Leaf size=179 \[ \frac {(61-87 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {2 \left (153-\left (23-29 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13-2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13-2 \sqrt {13}\right ) (1+m)}+\frac {2 \left (153-\left (23+29 \sqrt {13}\right ) m\right ) (1+4 x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {3 (1+4 x)}{13+2 \sqrt {13}}\right )}{13 \sqrt {13} \left (13+2 \sqrt {13}\right ) (1+m)} \]
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Rubi [A]
time = 0.16, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1662, 844, 70}
\begin {gather*} -\frac {2 \left (153-\left (23-29 \sqrt {13}\right ) m\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 {2 \left (153-\left (23+29 \sqrt {13}\right ) m\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 {(61-87 x) (4 x+1)^{m+1}}{39 \left (3 x^2-5 x+1\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 844
Rule 1662
Rubi steps
\begin {align*} \int \frac {(2+3 x)^2 (1+4 x)^m}{\left (1-5 x+3 x^2\right )^2} \, dx &=\frac {(61-87 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \frac {(1+4 x)^m (26 (153+122 m)-4524 m x)}{1-5 x+3 x^2} \, dx\\ &=\frac {(61-87 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {1}{507} \int \left (\frac {\left (-4524 m-12 \sqrt {13} (-153+23 m)\right ) (1+4 x)^m}{-5-\sqrt {13}+6 x}+\frac {\left (-4524 m+12 \sqrt {13} (-153+23 m)\right ) (1+4 x)^m}{-5+\sqrt {13}+6 x}\right ) \, dx\\ &=\frac {(61-87 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}+\frac {\left (4 \left (153-\left (23-29 \sqrt {13}\right ) m\right )\right ) \int \frac {(1+4 x)^m}{-5+\sqrt {13}+6 x} \, dx}{13 \sqrt {13}}-\frac {\left (4 \left (153-\left (23+29 \sqrt {13}\right ) m\right )\right ) \int \frac {(1+4 x)^m}{-5-\sqrt {13}+6 x} \, dx}{13 \sqrt {13}}\\ &=\frac {(61-87 x) (1+4 x)^{1+m}}{39 \left (1-5 x+3 x^2\right )}-\frac {2 \left (153-\left (23-29 \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)}+\frac {2 \left (153-\left (23+29 \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 = 0.33, size = 156, normalized size = 0.87 \begin {gather*} \frac {1}{507} (1+4 x)^{1+m} \left (\frac {793-1131 x}{1-5 x+3 x^2}-\frac {6 \left (-153 \sqrt {13}+\left (-377+23 \sqrt {13}\right ) m\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13-2 \sqrt {13}}\right )}{\left (-13+2 \sqrt {13}\right ) (1+m)}-\frac {6 \left (-153 \sqrt {13}+\left (377+23 \sqrt {13}\right ) m\right ) \, _2F_1\left (1,1+m;2+m;\frac {3+12 x}{13+2 \sqrt {13}}\right )}{\left (13+2 \sqrt {13}\right ) (1+m)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (2+3 x \right )^{2} \left (1+4 x \right )^{m}}{\left (3 x^{2}-5 x +1\right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (3 x + 2\right )^{2} \left (4 x + 1\right )^{m}}{\left (3 x^{2} - 5 x + 1\right )^{2}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\left (3\,x+2\right )}^2\,{\left (4\,x+1\right )}^m}{{\left (3\,x^2-5\,x+1\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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