Optimal. Leaf size=151 \[ -\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}-\frac {1}{4} (d-f) \log \left (x^2-x+1\right )+\frac {1}{4} (d-f) \log \left (x^2+x+1\right )+\frac {\tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g-i)}{2 \sqrt {3}}+\frac {1}{4} (g-i) \log \left (x^4+x^2+1\right )+h x+\frac {i x^2}{2} \]
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Rubi [A] time = 0.18, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1673, 1676, 1169, 634, 618, 204, 628, 1663, 1657} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) (d+f-2 h)}{2 \sqrt {3}}-\frac {1}{4} (d-f) \log \left (x^2-x+1\right )+\frac {1}{4} (d-f) \log \left (x^2+x+1\right )+\frac {\tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g-i)}{2 \sqrt {3}}+\frac {1}{4} (g-i) \log \left (x^4+x^2+1\right )+h x+\frac {i x^2}{2} \end {gather*}
Antiderivative was successfully verified.
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Rule 204
Rule 618
Rule 628
Rule 634
Rule 1169
Rule 1657
Rule 1663
Rule 1673
Rule 1676
Rubi steps
\begin {align*} \int \frac {d+e x+f x^2+g x^3+h x^4+19 x^5}{1+x^2+x^4} \, dx &=\int \frac {x \left (e+g x^2+19 x^4\right )}{1+x^2+x^4} \, dx+\int \frac {d+f x^2+h x^4}{1+x^2+x^4} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {e+g x+19 x^2}{1+x+x^2} \, dx,x,x^2\right )+\int \left (h+\frac {d-h+(f-h) x^2}{1+x^2+x^4}\right ) \, dx\\ &=h x+\frac {1}{2} \operatorname {Subst}\left (\int \left (19-\frac {19-e+(19-g) x}{1+x+x^2}\right ) \, dx,x,x^2\right )+\int \frac {d-h+(f-h) x^2}{1+x^2+x^4} \, dx\\ &=h x+\frac {19 x^2}{2}+\frac {1}{2} \int \frac {d-h-(d-f) x}{1-x+x^2} \, dx+\frac {1}{2} \int \frac {d-h+(d-f) x}{1+x+x^2} \, dx-\frac {1}{2} \operatorname {Subst}\left (\int \frac {19-e+(19-g) x}{1+x+x^2} \, dx,x,x^2\right )\\ &=h x+\frac {19 x^2}{2}+\frac {1}{4} (d-f) \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {1}{4} (-d+f) \int \frac {-1+2 x}{1-x+x^2} \, dx-\frac {1}{4} (19-g) \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,x^2\right )-\frac {1}{4} (19-2 e+g) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right )+\frac {1}{4} (d+f-2 h) \int \frac {1}{1-x+x^2} \, dx+\frac {1}{4} (d+f-2 h) \int \frac {1}{1+x+x^2} \, dx\\ &=h x+\frac {19 x^2}{2}-\frac {1}{4} (d-f) \log \left (1-x+x^2\right )+\frac {1}{4} (d-f) \log \left (1+x+x^2\right )-\frac {1}{4} (19-g) \log \left (1+x^2+x^4\right )-\frac {1}{2} (-19+2 e-g) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right )+\frac {1}{2} (-d-f+2 h) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {1}{2} (-d-f+2 h) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=h x+\frac {19 x^2}{2}-\frac {(d+f-2 h) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {(d+f-2 h) \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {(19-2 e+g) \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{4} (d-f) \log \left (1-x+x^2\right )+\frac {1}{4} (d-f) \log \left (1+x+x^2\right )-\frac {1}{4} (19-g) \log \left (1+x^2+x^4\right )\\ \end {align*}
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Mathematica [C] time = 0.58, size = 187, normalized size = 1.24 \begin {gather*} \frac {1}{12} \left (\left (1+i \sqrt {3}\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right ) \left (2 \sqrt {3} d-\left (\sqrt {3}+3 i\right ) f-\left (\sqrt {3}-3 i\right ) h\right )+\left (\sqrt {3}+i\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right ) \left (-2 i \sqrt {3} d+\left (3+i \sqrt {3}\right ) f+i \left (\sqrt {3}+3 i\right ) h\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3}}{2 x^2+1}\right ) (2 e-g-i)+3 (g-i) \log \left (x^4+x^2+1\right )+6 x (2 h+i x)\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d+e x+f x^2+g x^3+h x^4+i x^5}{1+x^2+x^4} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 18.84, size = 106, normalized size = 0.70 \begin {gather*} \frac {1}{2} \, i x^{2} + \frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f + g - 2 \, h + i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f - g - 2 \, h - i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + h x + \frac {1}{4} \, {\left (d - f + g - i\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f - g + i\right )} \log \left (x^{2} - x + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 108, normalized size = 0.72 \begin {gather*} \frac {1}{2} \, i x^{2} + \frac {1}{6} \, \sqrt {3} {\left (d + f + g - 2 \, h + i - 2 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + f - g - 2 \, h - i + 2 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + h x + \frac {1}{4} \, {\left (d - f + g - i\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f - g + i\right )} \log \left (x^{2} - x + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 303, normalized size = 2.01 \begin {gather*} \frac {i \,x^{2}}{2}+\frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {d \ln \left (x^{2}-x +1\right )}{4}+\frac {d \ln \left (x^{2}+x +1\right )}{4}-\frac {\sqrt {3}\, e \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, e \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {f \ln \left (x^{2}-x +1\right )}{4}-\frac {f \ln \left (x^{2}+x +1\right )}{4}+\frac {\sqrt {3}\, g \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, g \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {g \ln \left (x^{2}-x +1\right )}{4}+\frac {g \ln \left (x^{2}+x +1\right )}{4}+h x -\frac {\sqrt {3}\, h \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}-\frac {\sqrt {3}\, h \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, i \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}-\frac {\sqrt {3}\, i \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {i \ln \left (x^{2}-x +1\right )}{4}-\frac {i \ln \left (x^{2}+x +1\right )}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.37, size = 106, normalized size = 0.70 \begin {gather*} \frac {1}{2} \, i x^{2} + \frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f + g - 2 \, h + i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f - g - 2 \, h - i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + h x + \frac {1}{4} \, {\left (d - f + g - i\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f - g + i\right )} \log \left (x^{2} - x + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.80, size = 1509, normalized size = 9.99
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] 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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