Optimal. Leaf size=255 \[ \frac{\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{m+1}}{125 e^3 (m+1)}-\frac{(40 d+33 e) (d+e x)^{m+2}}{25 e^3 (m+2)}+\frac{4 (d+e x)^{m+3}}{5 e^3 (m+3)}-\frac{\left (-423 \sqrt{14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{5 (d+e x)}{5 d+i \sqrt{14} e-e}\right )}{3500 (m+1) \left (5 i d-\left (\sqrt{14}+i\right ) e\right )}-\frac{\left (423 \sqrt{14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{5 (d+e x)}{5 d-\left (1+i \sqrt{14}\right ) e}\right )}{3500 (m+1) \left (5 i d-\left (-\sqrt{14}+i\right ) e\right )} \]
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Rubi [A] time = 0.480741, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {1628, 68} \[ \frac{\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{m+1}}{125 e^3 (m+1)}-\frac{(40 d+33 e) (d+e x)^{m+2}}{25 e^3 (m+2)}+\frac{4 (d+e x)^{m+3}}{5 e^3 (m+3)}-\frac{\left (-423 \sqrt{14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{5 (d+e x)}{5 d+i \sqrt{14} e-e}\right )}{3500 (m+1) \left (5 i d-\left (\sqrt{14}+i\right ) e\right )}-\frac{\left (423 \sqrt{14}+6412 i\right ) (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{5 (d+e x)}{5 d-\left (1+i \sqrt{14}\right ) e}\right )}{3500 (m+1) \left (5 i d-\left (-\sqrt{14}+i\right ) e\right )} \]
Antiderivative was successfully verified.
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Rule 1628
Rule 68
Rubi steps
\begin{align*} \int \frac{(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx &=\int \left (\frac{\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^m}{125 e^2}+\frac{\left (\frac{458}{125}+\frac{423 i}{125 \sqrt{14}}\right ) (d+e x)^m}{2-2 i \sqrt{14}+10 x}+\frac{\left (\frac{458}{125}-\frac{423 i}{125 \sqrt{14}}\right ) (d+e x)^m}{2+2 i \sqrt{14}+10 x}+\frac{(-40 d-33 e) (d+e x)^{1+m}}{25 e^2}+\frac{4 (d+e x)^{2+m}}{5 e^2}\right ) \, dx\\ &=\frac{\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{1+m}}{125 e^3 (1+m)}-\frac{(40 d+33 e) (d+e x)^{2+m}}{25 e^3 (2+m)}+\frac{4 (d+e x)^{3+m}}{5 e^3 (3+m)}+\frac{\left (6412-423 i \sqrt{14}\right ) \int \frac{(d+e x)^m}{2+2 i \sqrt{14}+10 x} \, dx}{1750}+\frac{\left (6412+423 i \sqrt{14}\right ) \int \frac{(d+e x)^m}{2-2 i \sqrt{14}+10 x} \, dx}{1750}\\ &=\frac{\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{1+m}}{125 e^3 (1+m)}-\frac{(40 d+33 e) (d+e x)^{2+m}}{25 e^3 (2+m)}+\frac{4 (d+e x)^{3+m}}{5 e^3 (3+m)}-\frac{\left (6412 i-423 \sqrt{14}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{5 (d+e x)}{5 d-e+i \sqrt{14} e}\right )}{3500 \left (5 i d-\left (i+\sqrt{14}\right ) e\right ) (1+m)}-\frac{\left (6412 i+423 \sqrt{14}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{5 (d+e x)}{5 d-\left (1+i \sqrt{14}\right ) e}\right )}{3500 \left (5 i d-\left (i-\sqrt{14}\right ) e\right ) (1+m)}\\ \end{align*}
Mathematica [A] time = 0.672754, size = 221, normalized size = 0.87 \[ \frac{(d+e x)^{m+1} \left (\frac{28 \left (100 d^2+165 d e+81 e^2\right )}{e^3 (m+1)}+\frac{2800 (d+e x)^2}{e^3 (m+3)}-\frac{140 (40 d+33 e) (d+e x)}{e^3 (m+2)}-\frac{\left (423 \sqrt{14}+6412 i\right ) \, _2F_1\left (1,m+1;m+2;\frac{5 (d+e x)}{5 d+\left (-1-i \sqrt{14}\right ) e}\right )}{(m+1) \left (5 i d+\left (\sqrt{14}-i\right ) e\right )}-\frac{\left (423 \sqrt{14}-6412 i\right ) \, _2F_1\left (1,m+1;m+2;\frac{5 (d+e x)}{5 d+i \left (i+\sqrt{14}\right ) e}\right )}{(m+1) \left (\left (\sqrt{14}+i\right ) e-5 i d\right )}\right )}{3500} \]
Antiderivative was successfully verified.
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Maple [F] time = 3.62, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m} \left ( 4\,{x}^{4}-5\,{x}^{3}+3\,{x}^{2}+x+2 \right ) }{5\,{x}^{2}+2\,x+3}}\, 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^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )}{\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3}\,{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^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )}{\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )}{\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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