Optimal. Leaf size=233 \[ \frac {4 x}{5 e^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (423 d^2-2734 d e+293 e^2\right ) \tan ^{-1}\left (\frac {1+5 x}{\sqrt {14}}\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (3+2 x+5 x^2\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2} \]
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Rubi [A]
time = 0.15, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {1642, 648, 632,
210, 642} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {5 x+1}{\sqrt {14}}\right ) \left (423 d^2-2734 d e+293 e^2\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (5 x^2+2 x+3\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {4 x}{5 e^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1642
Rubi steps
\begin {align*} \int \frac {2+x+3 x^2-5 x^3+4 x^4}{(d+e x)^2 \left (3+2 x+5 x^2\right )} \, dx &=\int \left (\frac {4}{5 e^2}+\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^2 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)^2}+\frac {-40 d^5-d^4 e-28 d^3 e^2-44 d^2 e^3+2 d e^4-e^5}{e^2 \left (5 d^2-2 d e+3 e^2\right )^2 (d+e x)}+\frac {7 d^2+544 d e-113 e^2+2 \left (229 d^2-7 d e-136 e^2\right ) x}{5 \left (5 d^2-2 d e+3 e^2\right )^2 \left (3+2 x+5 x^2\right )}\right ) \, dx\\ &=\frac {4 x}{5 e^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\int \frac {7 d^2+544 d e-113 e^2+2 \left (229 d^2-7 d e-136 e^2\right ) x}{3+2 x+5 x^2} \, dx}{5 \left (5 d^2-2 d e+3 e^2\right )^2}\\ &=\frac {4 x}{5 e^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \int \frac {2+10 x}{3+2 x+5 x^2} \, dx}{25 \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {\left (423 d^2-2734 d e+293 e^2\right ) \int \frac {1}{3+2 x+5 x^2} \, dx}{25 \left (5 d^2-2 d e+3 e^2\right )^2}\\ &=\frac {4 x}{5 e^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (3+2 x+5 x^2\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (2 \left (423 d^2-2734 d e+293 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{-56-x^2} \, dx,x,2+10 x\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2}\\ &=\frac {4 x}{5 e^2}-\frac {4 d^4+5 d^3 e+3 d^2 e^2-d e^3+2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}-\frac {\left (423 d^2-2734 d e+293 e^2\right ) \tan ^{-1}\left (\frac {1+5 x}{\sqrt {14}}\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}-\frac {\left (40 d^5+d^4 e+28 d^3 e^2+44 d^2 e^3-2 d e^4+e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (3+2 x+5 x^2\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2}\\ \end {align*}
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Mathematica [A]
time = 0.11, size = 233, normalized size = 1.00 \begin {gather*} \frac {4 x}{5 e^2}+\frac {-4 d^4-5 d^3 e-3 d^2 e^2+d e^3-2 e^4}{e^3 \left (5 d^2-2 d e+3 e^2\right ) (d+e x)}+\frac {\left (-423 d^2+2734 d e-293 e^2\right ) \tan ^{-1}\left (\frac {1+5 x}{\sqrt {14}}\right )}{25 \sqrt {14} \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (-40 d^5-d^4 e-28 d^3 e^2-44 d^2 e^3+2 d e^4-e^5\right ) \log (d+e x)}{e^3 \left (5 d^2-2 d e+3 e^2\right )^2}+\frac {\left (229 d^2-7 d e-136 e^2\right ) \log \left (3+2 x+5 x^2\right )}{25 \left (5 d^2-2 d e+3 e^2\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.26, size = 213, normalized size = 0.91
method | result | size |
default | \(-\frac {4 d^{4}+5 d^{3} e +3 d^{2} e^{2}-d \,e^{3}+2 e^{4}}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right ) \left (e x +d \right )}+\frac {\left (-40 d^{5}-d^{4} e -28 d^{3} e^{2}-44 d^{2} e^{3}+2 d \,e^{4}-e^{5}\right ) \ln \left (e x +d \right )}{e^{3} \left (5 d^{2}-2 d e +3 e^{2}\right )^{2}}+\frac {4 x}{5 e^{2}}+\frac {\frac {\left (458 d^{2}-14 d e -272 e^{2}\right ) \ln \left (5 x^{2}+2 x +3\right )}{10}+\frac {\left (-\frac {423}{5} d^{2}+\frac {2734}{5} d e -\frac {293}{5} e^{2}\right ) \sqrt {14}\, \arctan \left (\frac {\left (10 x +2\right ) \sqrt {14}}{28}\right )}{14}}{5 \left (5 d^{2}-2 d e +3 e^{2}\right )^{2}}\) | \(213\) |
risch | \(\text {Expression too large to display}\) | \(14526\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.54, size = 274, normalized size = 1.18 \begin {gather*} \frac {4}{5} \, x e^{\left (-2\right )} - \frac {\sqrt {14} {\left (423 \, d^{2} - 2734 \, d e + 293 \, e^{2}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right )}{350 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} + \frac {{\left (229 \, d^{2} - 7 \, d e - 136 \, e^{2}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right )}{25 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {{\left (40 \, d^{5} + d^{4} e + 28 \, d^{3} e^{2} + 44 \, d^{2} e^{3} - 2 \, d e^{4} + e^{5}\right )} \log \left (x e + d\right )}{25 \, d^{4} e^{3} - 20 \, d^{3} e^{4} + 34 \, d^{2} e^{5} - 12 \, d e^{6} + 9 \, e^{7}} - \frac {4 \, d^{4} + 5 \, d^{3} e + 3 \, d^{2} e^{2} - d e^{3} + 2 \, e^{4}}{5 \, d^{3} e^{3} - 2 \, d^{2} e^{4} + {\left (5 \, d^{2} e^{4} - 2 \, d e^{5} + 3 \, e^{6}\right )} x + 3 \, d e^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.44, size = 398, normalized size = 1.71 \begin {gather*} -\frac {7000 \, d^{6} + \sqrt {14} {\left (423 \, d^{3} e^{3} + 293 \, x e^{6} - {\left (2734 \, d x - 293 \, d\right )} e^{5} + {\left (423 \, d^{2} x - 2734 \, d^{2}\right )} e^{4}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, x + 1\right )}\right ) - 420 \, {\left (6 \, x^{2} - 5\right )} e^{6} + 70 \, {\left (48 \, d x^{2} - 36 \, d x - 35 \, d\right )} e^{5} - 70 \, {\left (136 \, d^{2} x^{2} - 48 \, d^{2} x - 105 \, d^{2}\right )} e^{4} + 280 \, {\left (20 \, d^{3} x^{2} - 34 \, d^{3} x + 5 \, d^{3}\right )} e^{3} - 350 \, {\left (20 \, d^{4} x^{2} - 16 \, d^{4} x - 17 \, d^{4}\right )} e^{2} - 350 \, {\left (20 \, d^{5} x - 17 \, d^{5}\right )} e - 14 \, {\left (229 \, d^{3} e^{3} - 136 \, x e^{6} - {\left (7 \, d x + 136 \, d\right )} e^{5} + {\left (229 \, d^{2} x - 7 \, d^{2}\right )} e^{4}\right )} \log \left (5 \, x^{2} + 2 \, x + 3\right ) + 350 \, {\left (40 \, d^{6} + x e^{6} - {\left (2 \, d x - d\right )} e^{5} + 2 \, {\left (22 \, d^{2} x - d^{2}\right )} e^{4} + 4 \, {\left (7 \, d^{3} x + 11 \, d^{3}\right )} e^{3} + {\left (d^{4} x + 28 \, d^{4}\right )} e^{2} + {\left (40 \, d^{5} x + d^{5}\right )} e\right )} \log \left (x e + d\right )}{350 \, {\left (25 \, d^{5} e^{3} + 9 \, x e^{8} - 3 \, {\left (4 \, d x - 3 \, d\right )} e^{7} + 2 \, {\left (17 \, d^{2} x - 6 \, d^{2}\right )} e^{6} - 2 \, {\left (10 \, d^{3} x - 17 \, d^{3}\right )} e^{5} + 5 \, {\left (5 \, d^{4} x - 4 \, d^{4}\right )} e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 4.43, size = 355, normalized size = 1.52 \begin {gather*} \frac {1}{25} \, {\left (40 \, d + 33 \, e\right )} e^{\left (-3\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - \frac {\sqrt {14} {\left (423 \, d^{2} e^{2} - 2734 \, d e^{3} + 293 \, e^{4}\right )} \arctan \left (\frac {1}{14} \, \sqrt {14} {\left (5 \, d - \frac {5 \, d^{2}}{x e + d} + \frac {2 \, d e}{x e + d} - \frac {3 \, e^{2}}{x e + d} - e\right )} e^{\left (-1\right )}\right ) e^{\left (-2\right )}}{350 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} + \frac {4}{5} \, {\left (x e + d\right )} e^{\left (-3\right )} + \frac {{\left (229 \, d^{2} - 7 \, d e - 136 \, e^{2}\right )} \log \left (-\frac {10 \, d}{x e + d} + \frac {5 \, d^{2}}{{\left (x e + d\right )}^{2}} + \frac {2 \, e}{x e + d} - \frac {2 \, d e}{{\left (x e + d\right )}^{2}} + \frac {3 \, e^{2}}{{\left (x e + d\right )}^{2}} + 5\right )}{25 \, {\left (25 \, d^{4} - 20 \, d^{3} e + 34 \, d^{2} e^{2} - 12 \, d e^{3} + 9 \, e^{4}\right )}} - \frac {\frac {4 \, d^{4} e^{3}}{x e + d} + \frac {5 \, d^{3} e^{4}}{x e + d} + \frac {3 \, d^{2} e^{5}}{x e + d} - \frac {d e^{6}}{x e + d} + \frac {2 \, e^{7}}{x e + d}}{5 \, d^{2} e^{6} - 2 \, d e^{7} + 3 \, e^{8}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 4.67, size = 312, normalized size = 1.34 \begin {gather*} \frac {4\,x}{5\,e^2}-\frac {\ln \left (x+\frac {1}{5}-\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {423\,\sqrt {14}}{700}-\frac {229}{25}{}\mathrm {i}\right )\,d^2+\left (-\frac {1367\,\sqrt {14}}{350}+\frac {7}{25}{}\mathrm {i}\right )\,d\,e+\left (\frac {293\,\sqrt {14}}{700}+\frac {136}{25}{}\mathrm {i}\right )\,e^2\right )}{d^4\,25{}\mathrm {i}-d^3\,e\,20{}\mathrm {i}+d^2\,e^2\,34{}\mathrm {i}-d\,e^3\,12{}\mathrm {i}+e^4\,9{}\mathrm {i}}+\frac {\ln \left (x+\frac {1}{5}+\frac {\sqrt {14}\,1{}\mathrm {i}}{5}\right )\,\left (\left (\frac {423\,\sqrt {14}}{700}+\frac {229}{25}{}\mathrm {i}\right )\,d^2+\left (-\frac {1367\,\sqrt {14}}{350}-\frac {7}{25}{}\mathrm {i}\right )\,d\,e+\left (\frac {293\,\sqrt {14}}{700}-\frac {136}{25}{}\mathrm {i}\right )\,e^2\right )}{d^4\,25{}\mathrm {i}-d^3\,e\,20{}\mathrm {i}+d^2\,e^2\,34{}\mathrm {i}-d\,e^3\,12{}\mathrm {i}+e^4\,9{}\mathrm {i}}-\frac {5\,\left (4\,d^4+5\,d^3\,e+3\,d^2\,e^2-d\,e^3+2\,e^4\right )}{e\,\left (5\,x\,e^3+5\,d\,e^2\right )\,\left (5\,d^2-2\,d\,e+3\,e^2\right )}-\frac {\ln \left (d+e\,x\right )\,\left (40\,d^5+d^4\,e+28\,d^3\,e^2+44\,d^2\,e^3-2\,d\,e^4+e^5\right )}{e^3\,{\left (5\,d^2-2\,d\,e+3\,e^2\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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