Optimal. Leaf size=158 \[ \frac {1}{2} \sqrt {x^2+2 x+4} (x+1)+\frac {1}{4} (2 x+1) \sqrt {x^2+x+1}-2 \sqrt {x^2+x+1}-2 \sqrt {x^2+2 x+4}-2 \sqrt {7} \tanh ^{-1}\left (\frac {5 x+1}{2 \sqrt {7} \sqrt {x^2+x+1}}\right )+2 \sqrt {7} \tanh ^{-1}\left (\frac {1-2 x}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )+\frac {11}{2} \sinh ^{-1}\left (\frac {x+1}{\sqrt {3}}\right )+\frac {43}{8} \sinh ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]
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Rubi [A] time = 0.53, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 36, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {6742, 734, 843, 619, 215, 724, 206, 612} \[ \frac {1}{2} \sqrt {x^2+2 x+4} (x+1)+\frac {1}{4} (2 x+1) \sqrt {x^2+x+1}-2 \sqrt {x^2+x+1}-2 \sqrt {x^2+2 x+4}-2 \sqrt {7} \tanh ^{-1}\left (\frac {5 x+1}{2 \sqrt {7} \sqrt {x^2+x+1}}\right )+2 \sqrt {7} \tanh ^{-1}\left (\frac {1-2 x}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )+\frac {11}{2} \sinh ^{-1}\left (\frac {x+1}{\sqrt {3}}\right )+\frac {43}{8} \sinh ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right ) \]
Antiderivative was successfully verified.
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Rule 206
Rule 215
Rule 612
Rule 619
Rule 724
Rule 734
Rule 843
Rule 6742
Rubi steps
\begin {align*} \int \frac {1+x}{-\sqrt {1+x+x^2}+\sqrt {4+2 x+x^2}} \, dx &=\int \left (-\frac {1}{\sqrt {1+x+x^2}-\sqrt {4+2 x+x^2}}-\frac {x}{\sqrt {1+x+x^2}-\sqrt {4+2 x+x^2}}\right ) \, dx\\ &=-\int \frac {1}{\sqrt {1+x+x^2}-\sqrt {4+2 x+x^2}} \, dx-\int \frac {x}{\sqrt {1+x+x^2}-\sqrt {4+2 x+x^2}} \, dx\\ &=-\int \left (-\frac {\sqrt {1+x+x^2}}{3+x}-\frac {\sqrt {4+2 x+x^2}}{3+x}\right ) \, dx-\int \left (-\sqrt {1+x+x^2}+\frac {3 \sqrt {1+x+x^2}}{3+x}-\sqrt {4+2 x+x^2}+\frac {3 \sqrt {4+2 x+x^2}}{3+x}\right ) \, dx\\ &=-\left (3 \int \frac {\sqrt {1+x+x^2}}{3+x} \, dx\right )-3 \int \frac {\sqrt {4+2 x+x^2}}{3+x} \, dx+\int \sqrt {1+x+x^2} \, dx+\int \frac {\sqrt {1+x+x^2}}{3+x} \, dx+\int \sqrt {4+2 x+x^2} \, dx+\int \frac {\sqrt {4+2 x+x^2}}{3+x} \, dx\\ &=-2 \sqrt {1+x+x^2}+\frac {1}{4} (1+2 x) \sqrt {1+x+x^2}-2 \sqrt {4+2 x+x^2}+\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {3}{8} \int \frac {1}{\sqrt {1+x+x^2}} \, dx-\frac {1}{2} \int \frac {1+5 x}{(3+x) \sqrt {1+x+x^2}} \, dx-\frac {1}{2} \int \frac {-2+4 x}{(3+x) \sqrt {4+2 x+x^2}} \, dx+\frac {3}{2} \int \frac {1+5 x}{(3+x) \sqrt {1+x+x^2}} \, dx+\frac {3}{2} \int \frac {1}{\sqrt {4+2 x+x^2}} \, dx+\frac {3}{2} \int \frac {-2+4 x}{(3+x) \sqrt {4+2 x+x^2}} \, dx\\ &=-2 \sqrt {1+x+x^2}+\frac {1}{4} (1+2 x) \sqrt {1+x+x^2}-2 \sqrt {4+2 x+x^2}+\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}-2 \int \frac {1}{\sqrt {4+2 x+x^2}} \, dx-\frac {5}{2} \int \frac {1}{\sqrt {1+x+x^2}} \, dx+6 \int \frac {1}{\sqrt {4+2 x+x^2}} \, dx+7 \int \frac {1}{(3+x) \sqrt {1+x+x^2}} \, dx+7 \int \frac {1}{(3+x) \sqrt {4+2 x+x^2}} \, dx+\frac {15}{2} \int \frac {1}{\sqrt {1+x+x^2}} \, dx-21 \int \frac {1}{(3+x) \sqrt {1+x+x^2}} \, dx-21 \int \frac {1}{(3+x) \sqrt {4+2 x+x^2}} \, dx+\frac {1}{8} \sqrt {3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,1+2 x\right )+\frac {1}{4} \sqrt {3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{12}}} \, dx,x,2+2 x\right )\\ &=-2 \sqrt {1+x+x^2}+\frac {1}{4} (1+2 x) \sqrt {1+x+x^2}-2 \sqrt {4+2 x+x^2}+\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {3}{2} \sinh ^{-1}\left (\frac {1+x}{\sqrt {3}}\right )+\frac {3}{8} \sinh ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )-14 \operatorname {Subst}\left (\int \frac {1}{28-x^2} \, dx,x,\frac {-1-5 x}{\sqrt {1+x+x^2}}\right )-14 \operatorname {Subst}\left (\int \frac {1}{28-x^2} \, dx,x,\frac {2-4 x}{\sqrt {4+2 x+x^2}}\right )+42 \operatorname {Subst}\left (\int \frac {1}{28-x^2} \, dx,x,\frac {-1-5 x}{\sqrt {1+x+x^2}}\right )+42 \operatorname {Subst}\left (\int \frac {1}{28-x^2} \, dx,x,\frac {2-4 x}{\sqrt {4+2 x+x^2}}\right )-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{12}}} \, dx,x,2+2 x\right )}{\sqrt {3}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,1+2 x\right )}{2 \sqrt {3}}+\sqrt {3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{12}}} \, dx,x,2+2 x\right )+\frac {1}{2} \left (5 \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,1+2 x\right )\\ &=-2 \sqrt {1+x+x^2}+\frac {1}{4} (1+2 x) \sqrt {1+x+x^2}-2 \sqrt {4+2 x+x^2}+\frac {1}{2} (1+x) \sqrt {4+2 x+x^2}+\frac {11}{2} \sinh ^{-1}\left (\frac {1+x}{\sqrt {3}}\right )+\frac {43}{8} \sinh ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )-2 \sqrt {7} \tanh ^{-1}\left (\frac {1+5 x}{2 \sqrt {7} \sqrt {1+x+x^2}}\right )+2 \sqrt {7} \tanh ^{-1}\left (\frac {1-2 x}{\sqrt {7} \sqrt {4+2 x+x^2}}\right )\\ \end {align*}
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Mathematica [A] time = 0.34, size = 151, normalized size = 0.96 \[ \frac {1}{8} \left (2 \left (2 \sqrt {x^2+x+1} x+2 \sqrt {x^2+2 x+4} x-7 \sqrt {x^2+x+1}-6 \sqrt {x^2+2 x+4}-8 \sqrt {7} \tanh ^{-1}\left (\frac {5 x+1}{2 \sqrt {7} \sqrt {x^2+x+1}}\right )+8 \sqrt {7} \tanh ^{-1}\left (\frac {1-2 x}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )\right )+44 \sinh ^{-1}\left (\frac {x+1}{\sqrt {3}}\right )+43 \sinh ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )\right ) \]
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 2.99, size = 165, normalized size = 1.04 \[ \frac {1}{2} \sqrt {x^2+2 x+4} (x-3)+\frac {1}{4} (2 x-7) \sqrt {x^2+x+1}-\frac {43}{8} \log \left (2 \sqrt {x^2+x+1}-2 x-1\right )-\frac {11}{2} \log \left (\sqrt {x^2+2 x+4}-x-1\right )-4 \sqrt {7} \tanh ^{-1}\left (-\frac {\sqrt {x^2+x+1}}{\sqrt {7}}+\frac {x}{\sqrt {7}}+\frac {3}{\sqrt {7}}\right )-4 \sqrt {7} \tanh ^{-1}\left (-\frac {\sqrt {x^2+2 x+4}}{\sqrt {7}}+\frac {x}{\sqrt {7}}+\frac {3}{\sqrt {7}}\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.73, size = 155, normalized size = 0.98 \[ \frac {1}{4} \, \sqrt {x^{2} + x + 1} {\left (2 \, x - 7\right )} + \frac {1}{2} \, \sqrt {x^{2} + 2 \, x + 4} {\left (x - 3\right )} + 2 \, \sqrt {7} \log \left (\frac {2 \, \sqrt {7} {\left (5 \, x + 1\right )} + 2 \, \sqrt {x^{2} + x + 1} {\left (5 \, \sqrt {7} - 14\right )} - 25 \, x - 5}{x + 3}\right ) + 2 \, \sqrt {7} \log \left (\frac {\sqrt {7} {\left (2 \, x - 1\right )} + \sqrt {x^{2} + 2 \, x + 4} {\left (2 \, \sqrt {7} - 7\right )} - 4 \, x + 2}{x + 3}\right ) - \frac {11}{2} \, \log \left (-x + \sqrt {x^{2} + 2 \, x + 4} - 1\right ) - \frac {43}{8} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]
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 \[ \int \frac {x + 1}{\sqrt {x^{2} + 2 \, x + 4} - \sqrt {x^{2} + x + 1}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.03, size = 140, normalized size = 0.89
method | result | size |
default | \(-2 \sqrt {\left (3+x \right )^{2}-5 x -8}+\frac {43 \arcsinh \left (\frac {2 \left (\frac {1}{2}+x \right ) \sqrt {3}}{3}\right )}{8}+2 \sqrt {7}\, \arctanh \left (\frac {\left (-1-5 x \right ) \sqrt {7}}{14 \sqrt {\left (3+x \right )^{2}-5 x -8}}\right )-2 \sqrt {\left (3+x \right )^{2}-4 x -5}+\frac {11 \arcsinh \left (\frac {\left (1+x \right ) \sqrt {3}}{3}\right )}{2}+2 \sqrt {7}\, \arctanh \left (\frac {\left (2-4 x \right ) \sqrt {7}}{14 \sqrt {\left (3+x \right )^{2}-4 x -5}}\right )+\frac {\left (1+2 x \right ) \sqrt {x^{2}+x +1}}{4}+\frac {\left (2 x +2\right ) \sqrt {x^{2}+2 x +4}}{4}\) | \(140\) |
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 \[ \int \frac {x + 1}{\sqrt {x^{2} + 2 \, x + 4} - \sqrt {x^{2} + x + 1}}\,{d x} \]
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 \[ \int -\frac {x+1}{\sqrt {x^2+x+1}-\sqrt {x^2+2\,x+4}} \,d x \]
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 \[ \int \frac {x + 1}{- \sqrt {x^{2} + x + 1} + \sqrt {x^{2} + 2 x + 4}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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