Optimal. Leaf size=194 \[ -\frac {1}{8} \log \left (x^2-x+1\right ) (2 d-f+h)+\frac {1}{8} \log \left (x^2+x+1\right ) (2 d-f+h)+\frac {x \left (-\left (x^2 (d-2 f+h)\right )+d+f-2 h\right )}{6 \left (x^4+x^2+1\right )}-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right ) (4 d+f+h)}{12 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f+h)}{12 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g+2 i)}{3 \sqrt {3}}+\frac {x^2 (2 e-g-i)+e-2 g+i}{6 \left (x^4+x^2+1\right )} \]
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Rubi [A] time = 0.20, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1673, 1678, 1169, 634, 618, 204, 628, 1663, 1660, 12} \begin {gather*} \frac {x \left (x^2 (-(d-2 f+h))+d+f-2 h\right )}{6 \left (x^4+x^2+1\right )}-\frac {1}{8} \log \left (x^2-x+1\right ) (2 d-f+h)+\frac {1}{8} \log \left (x^2+x+1\right ) (2 d-f+h)-\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right ) (4 d+f+h)}{12 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f+h)}{12 \sqrt {3}}+\frac {x^2 (2 e-g-i)+e-2 g+i}{6 \left (x^4+x^2+1\right )}+\frac {\tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g+2 i)}{3 \sqrt {3}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 204
Rule 618
Rule 628
Rule 634
Rule 1169
Rule 1660
Rule 1663
Rule 1673
Rule 1678
Rubi steps
\begin {align*} \int \frac {d+e x+f x^2+g x^3+h x^4+35 x^5}{\left (1+x^2+x^4\right )^2} \, dx &=\int \frac {x \left (e+g x^2+35 x^4\right )}{\left (1+x^2+x^4\right )^2} \, dx+\int \frac {d+f x^2+h x^4}{\left (1+x^2+x^4\right )^2} \, dx\\ &=\frac {x \left (d+f-2 h-(d-2 f+h) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{6} \int \frac {5 d-f+2 h+(-d+2 f-h) x^2}{1+x^2+x^4} \, dx+\frac {1}{2} \operatorname {Subst}\left (\int \frac {e+g x+35 x^2}{\left (1+x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac {35+e-2 g-(35-2 e+g) x^2}{6 \left (1+x^2+x^4\right )}+\frac {x \left (d+f-2 h-(d-2 f+h) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{12} \int \frac {5 d-f+2 h-(6 d-3 f+3 h) x}{1-x+x^2} \, dx+\frac {1}{12} \int \frac {5 d-f+2 h+(6 d-3 f+3 h) x}{1+x+x^2} \, dx+\frac {1}{6} \operatorname {Subst}\left (\int \frac {70+2 e-g}{1+x+x^2} \, dx,x,x^2\right )\\ &=\frac {35+e-2 g-(35-2 e+g) x^2}{6 \left (1+x^2+x^4\right )}+\frac {x \left (d+f-2 h-(d-2 f+h) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {1}{6} (70+2 e-g) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right )+\frac {1}{8} (-2 d+f-h) \int \frac {-1+2 x}{1-x+x^2} \, dx+\frac {1}{8} (2 d-f+h) \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {1}{24} (4 d+f+h) \int \frac {1}{1-x+x^2} \, dx+\frac {1}{24} (4 d+f+h) \int \frac {1}{1+x+x^2} \, dx\\ &=\frac {35+e-2 g-(35-2 e+g) x^2}{6 \left (1+x^2+x^4\right )}+\frac {x \left (d+f-2 h-(d-2 f+h) x^2\right )}{6 \left (1+x^2+x^4\right )}-\frac {1}{8} (2 d-f+h) \log \left (1-x+x^2\right )+\frac {1}{8} (2 d-f+h) \log \left (1+x+x^2\right )+\frac {1}{3} (-70-2 e+g) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right )+\frac {1}{12} (-4 d-f-h) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac {1}{12} (-4 d-f-h) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac {35+e-2 g-(35-2 e+g) x^2}{6 \left (1+x^2+x^4\right )}+\frac {x \left (d+f-2 h-(d-2 f+h) x^2\right )}{6 \left (1+x^2+x^4\right )}-\frac {(4 d+f+h) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(4 d+f+h) \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(70+2 e-g) \tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{8} (2 d-f+h) \log \left (1-x+x^2\right )+\frac {1}{8} (2 d-f+h) \log \left (1+x+x^2\right )\\ \end {align*}
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Mathematica [C] time = 0.66, size = 243, normalized size = 1.25 \begin {gather*} \frac {1}{36} \left (\frac {6 \left (-d x^3+d x+2 e x^2+e+2 f x^3+f x-g \left (x^2+2\right )-h x^3-2 h x-i x^2+i\right )}{x^4+x^2+1}-\frac {\tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right ) \left (\left (\sqrt {3}-11 i\right ) d-2 \left (\sqrt {3}-2 i\right ) f+\left (\sqrt {3}-5 i\right ) h\right )}{\sqrt {\frac {1}{6} \left (1+i \sqrt {3}\right )}}-\frac {\tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right ) \left (\left (\sqrt {3}+11 i\right ) d-2 \left (\sqrt {3}+2 i\right ) f+\left (\sqrt {3}+5 i\right ) h\right )}{\sqrt {\frac {1}{6} \left (1-i \sqrt {3}\right )}}-4 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3}}{2 x^2+1}\right ) (2 e-g+2 i)\right ) \end {gather*}
Warning: Unable to verify antiderivative.
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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}{\left (1+x^2+x^4\right )^2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 23.83, size = 279, normalized size = 1.44 \begin {gather*} -\frac {12 \, {\left (d - 2 \, f + h\right )} x^{3} - 12 \, {\left (2 \, e - g - i\right )} x^{2} - 2 \, \sqrt {3} {\left ({\left (4 \, d - 8 \, e + f + 4 \, g + h - 8 \, i\right )} x^{4} + {\left (4 \, d - 8 \, e + f + 4 \, g + h - 8 \, i\right )} x^{2} + 4 \, d - 8 \, e + f + 4 \, g + h - 8 \, i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt {3} {\left ({\left (4 \, d + 8 \, e + f - 4 \, g + h + 8 \, i\right )} x^{4} + {\left (4 \, d + 8 \, e + f - 4 \, g + h + 8 \, i\right )} x^{2} + 4 \, d + 8 \, e + f - 4 \, g + h + 8 \, i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 12 \, {\left (d + f - 2 \, h\right )} x - 9 \, {\left ({\left (2 \, d - f + h\right )} x^{4} + {\left (2 \, d - f + h\right )} x^{2} + 2 \, d - f + h\right )} \log \left (x^{2} + x + 1\right ) + 9 \, {\left ({\left (2 \, d - f + h\right )} x^{4} + {\left (2 \, d - f + h\right )} x^{2} + 2 \, d - f + h\right )} \log \left (x^{2} - x + 1\right ) - 12 \, e + 24 \, g - 12 \, i}{72 \, {\left (x^{4} + x^{2} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.31, size = 169, normalized size = 0.87 \begin {gather*} \frac {1}{36} \, \sqrt {3} {\left (4 \, d + f + 4 \, g + h - 8 \, i - 8 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{36} \, \sqrt {3} {\left (4 \, d + f - 4 \, g + h + 8 \, i + 8 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, {\left (2 \, d - f + h\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, {\left (2 \, d - f + h\right )} \log \left (x^{2} - x + 1\right ) - \frac {d x^{3} - 2 \, f x^{3} + h x^{3} + g x^{2} + i x^{2} - 2 \, x^{2} e - d x - f x + 2 \, h x + 2 \, g - i - e}{6 \, {\left (x^{4} + x^{2} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.02, size = 374, normalized size = 1.93 \begin {gather*} \frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}-\frac {d \ln \left (x^{2}-x +1\right )}{4}+\frac {d \ln \left (x^{2}+x +1\right )}{4}-\frac {2 \sqrt {3}\, e \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {2 \sqrt {3}\, e \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{36}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{36}+\frac {f \ln \left (x^{2}-x +1\right )}{8}-\frac {f \ln \left (x^{2}+x +1\right )}{8}+\frac {\sqrt {3}\, g \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}-\frac {\sqrt {3}\, g \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+\frac {\sqrt {3}\, h \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{36}+\frac {\sqrt {3}\, h \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{36}-\frac {h \ln \left (x^{2}-x +1\right )}{8}+\frac {h \ln \left (x^{2}+x +1\right )}{8}-\frac {2 \sqrt {3}\, i \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {2 \sqrt {3}\, i \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+\frac {-\frac {2 d}{3}+\frac {e}{3}+\frac {f}{3}-\frac {2 g}{3}+\frac {h}{3}+\frac {i}{3}+\left (-\frac {d}{3}-\frac {e}{3}+\frac {2 f}{3}-\frac {g}{3}-\frac {h}{3}+\frac {2 i}{3}\right ) x}{4 x^{2}+4 x +4}-\frac {-\frac {2 d}{3}-\frac {e}{3}+\frac {f}{3}+\frac {2 g}{3}+\frac {h}{3}-\frac {i}{3}+\left (\frac {d}{3}-\frac {e}{3}-\frac {2 f}{3}-\frac {g}{3}+\frac {h}{3}+\frac {2 i}{3}\right ) x}{4 \left (x^{2}-x +1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.63, size = 155, normalized size = 0.80 \begin {gather*} \frac {1}{36} \, \sqrt {3} {\left (4 \, d - 8 \, e + f + 4 \, g + h - 8 \, i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{36} \, \sqrt {3} {\left (4 \, d + 8 \, e + f - 4 \, g + h + 8 \, i\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, {\left (2 \, d - f + h\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, {\left (2 \, d - f + h\right )} \log \left (x^{2} - x + 1\right ) - \frac {{\left (d - 2 \, f + h\right )} x^{3} - {\left (2 \, e - g - i\right )} x^{2} - {\left (d + f - 2 \, h\right )} x - e + 2 \, g - i}{6 \, {\left (x^{4} + x^{2} + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 8.18, size = 1894, normalized size = 9.76
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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