3.33 \(\int \frac{1+2 x}{\left (3+x+x^2\right ) \sqrt{5+x+x^2}} \, dx\)

Optimal. Leaf size=24 \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{x^2+x+5}}{\sqrt{2}}\right ) \]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[5 + x + x^2]/Sqrt[2]])

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Rubi [A]  time = 0.0567864, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083 \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{x^2+x+5}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]  Int[(1 + 2*x)/((3 + x + x^2)*Sqrt[5 + x + x^2]),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[5 + x + x^2]/Sqrt[2]])

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Rubi in Sympy [A]  time = 14.8139, size = 24, normalized size = 1. \[ - \sqrt{2} \operatorname{atanh}{\left (\frac{\sqrt{2} \sqrt{x^{2} + x + 5}}{2} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((1+2*x)/(x**2+x+3)/(x**2+x+5)**(1/2),x)

[Out]

-sqrt(2)*atanh(sqrt(2)*sqrt(x**2 + x + 5)/2)

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Mathematica [A]  time = 0.0192771, size = 24, normalized size = 1. \[ -\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{x^2+x+5}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[(1 + 2*x)/((3 + x + x^2)*Sqrt[5 + x + x^2]),x]

[Out]

-(Sqrt[2]*ArcTanh[Sqrt[5 + x + x^2]/Sqrt[2]])

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Maple [A]  time = 0.018, size = 20, normalized size = 0.8 \[ -{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{{x}^{2}+x+5}} \right ) \sqrt{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((1+2*x)/(x^2+x+3)/(x^2+x+5)^(1/2),x)

[Out]

-arctanh(1/2*(x^2+x+5)^(1/2)*2^(1/2))*2^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{2 \, x + 1}{\sqrt{x^{2} + x + 5}{\left (x^{2} + x + 3\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*x + 1)/(sqrt(x^2 + x + 5)*(x^2 + x + 3)),x, algorithm="maxima")

[Out]

integrate((2*x + 1)/(sqrt(x^2 + x + 5)*(x^2 + x + 3)), x)

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Fricas [A]  time = 0.272334, size = 180, normalized size = 7.5 \[ \frac{1}{2} \, \sqrt{2} \log \left (\frac{8 \, x^{4} + 16 \, x^{3} + 85 \, x^{2} + 8 \, \sqrt{2}{\left (2 \, x^{3} + 3 \, x^{2} + 11 \, x + 5\right )} - 2 \,{\left (4 \, x^{3} + 6 \, x^{2} + \sqrt{2}{\left (8 \, x^{2} + 8 \, x + 21\right )} + 30 \, x + 14\right )} \sqrt{x^{2} + x + 5} + 77 \, x + 147}{8 \, x^{4} + 16 \, x^{3} + 53 \, x^{2} - 4 \,{\left (2 \, x^{3} + 3 \, x^{2} + 7 \, x + 3\right )} \sqrt{x^{2} + x + 5} + 45 \, x + 63}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*x + 1)/(sqrt(x^2 + x + 5)*(x^2 + x + 3)),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*log((8*x^4 + 16*x^3 + 85*x^2 + 8*sqrt(2)*(2*x^3 + 3*x^2 + 11*x + 5)
- 2*(4*x^3 + 6*x^2 + sqrt(2)*(8*x^2 + 8*x + 21) + 30*x + 14)*sqrt(x^2 + x + 5) +
77*x + 147)/(8*x^4 + 16*x^3 + 53*x^2 - 4*(2*x^3 + 3*x^2 + 7*x + 3)*sqrt(x^2 + x
+ 5) + 45*x + 63))

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Sympy [A]  time = 2.30833, size = 68, normalized size = 2.83 \[ 2 \left (\begin{cases} - \frac{\sqrt{2} \operatorname{acoth}{\left (\frac{\sqrt{2}}{\sqrt{x^{2} + x + 5}} \right )}}{2} & \text{for}\: \frac{1}{x^{2} + x + 5} > \frac{1}{2} \\- \frac{\sqrt{2} \operatorname{atanh}{\left (\frac{\sqrt{2}}{\sqrt{x^{2} + x + 5}} \right )}}{2} & \text{for}\: \frac{1}{x^{2} + x + 5} < \frac{1}{2} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((1+2*x)/(x**2+x+3)/(x**2+x+5)**(1/2),x)

[Out]

2*Piecewise((-sqrt(2)*acoth(sqrt(2)/sqrt(x**2 + x + 5))/2, 1/(x**2 + x + 5) > 1/
2), (-sqrt(2)*atanh(sqrt(2)/sqrt(x**2 + x + 5))/2, 1/(x**2 + x + 5) < 1/2))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{2 \, x + 1}{\sqrt{x^{2} + x + 5}{\left (x^{2} + x + 3\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*x + 1)/(sqrt(x^2 + x + 5)*(x^2 + x + 3)),x, algorithm="giac")

[Out]

integrate((2*x + 1)/(sqrt(x^2 + x + 5)*(x^2 + x + 3)), x)