3.85 \(\int \frac{1}{1+\pi x-2 x^2} \, dx\)

Optimal. Leaf size=27 \[ -\frac{2 \tanh ^{-1}\left (\frac{\pi -4 x}{\sqrt{8+\pi ^2}}\right )}{\sqrt{8+\pi ^2}} \]

[Out]

(-2*ArcTanh[(Pi - 4*x)/Sqrt[8 + Pi^2]])/Sqrt[8 + Pi^2]

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Rubi [A]  time = 0.01862, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {618, 206} \[ -\frac{2 \tanh ^{-1}\left (\frac{\pi -4 x}{\sqrt{8+\pi ^2}}\right )}{\sqrt{8+\pi ^2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Pi*x - 2*x^2)^(-1),x]

[Out]

(-2*ArcTanh[(Pi - 4*x)/Sqrt[8 + Pi^2]])/Sqrt[8 + Pi^2]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{1+\pi x-2 x^2} \, dx &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{8+\pi ^2-x^2} \, dx,x,\pi -4 x\right )\right )\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\pi -4 x}{\sqrt{8+\pi ^2}}\right )}{\sqrt{8+\pi ^2}}\\ \end{align*}

Mathematica [A]  time = 0.0082617, size = 29, normalized size = 1.07 \[ \frac{2 \tanh ^{-1}\left (\frac{4 x-\pi }{\sqrt{8+\pi ^2}}\right )}{\sqrt{8+\pi ^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Pi*x - 2*x^2)^(-1),x]

[Out]

(2*ArcTanh[(-Pi + 4*x)/Sqrt[8 + Pi^2]])/Sqrt[8 + Pi^2]

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Maple [A]  time = 0.044, size = 26, normalized size = 1. \begin{align*} 2\,{\frac{1}{\sqrt{{\pi }^{2}+8}}{\it Artanh} \left ({\frac{4\,x-\pi }{\sqrt{{\pi }^{2}+8}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(Pi*x-2*x^2+1),x)

[Out]

2/(Pi^2+8)^(1/2)*arctanh((4*x-Pi)/(Pi^2+8)^(1/2))

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Maxima [A]  time = 1.14767, size = 53, normalized size = 1.96 \begin{align*} -\frac{\log \left (\frac{\pi - 4 \, x + \sqrt{\pi ^{2} + 8}}{\pi - 4 \, x - \sqrt{\pi ^{2} + 8}}\right )}{\sqrt{\pi ^{2} + 8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(pi*x-2*x^2+1),x, algorithm="maxima")

[Out]

-log((pi - 4*x + sqrt(pi^2 + 8))/(pi - 4*x - sqrt(pi^2 + 8)))/sqrt(pi^2 + 8)

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Fricas [B]  time = 2.18887, size = 131, normalized size = 4.85 \begin{align*} \frac{\log \left (-\frac{\pi ^{2} - 4 \, \pi x + 8 \, x^{2} -{\left (\pi - 4 \, x\right )} \sqrt{\pi ^{2} + 8} + 4}{\pi x - 2 \, x^{2} + 1}\right )}{\sqrt{\pi ^{2} + 8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(pi*x-2*x^2+1),x, algorithm="fricas")

[Out]

log(-(pi^2 - 4*pi*x + 8*x^2 - (pi - 4*x)*sqrt(pi^2 + 8) + 4)/(pi*x - 2*x^2 + 1))/sqrt(pi^2 + 8)

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Sympy [B]  time = 0.320016, size = 76, normalized size = 2.81 \begin{align*} - \frac{\log{\left (x - \frac{\pi }{4} - \frac{\pi ^{2}}{4 \sqrt{8 + \pi ^{2}}} - \frac{2}{\sqrt{8 + \pi ^{2}}} \right )}}{\sqrt{8 + \pi ^{2}}} + \frac{\log{\left (x - \frac{\pi }{4} + \frac{2}{\sqrt{8 + \pi ^{2}}} + \frac{\pi ^{2}}{4 \sqrt{8 + \pi ^{2}}} \right )}}{\sqrt{8 + \pi ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(pi*x-2*x**2+1),x)

[Out]

-log(x - pi/4 - pi**2/(4*sqrt(8 + pi**2)) - 2/sqrt(8 + pi**2))/sqrt(8 + pi**2) + log(x - pi/4 + 2/sqrt(8 + pi*
*2) + pi**2/(4*sqrt(8 + pi**2)))/sqrt(8 + pi**2)

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Giac [A]  time = 1.27299, size = 61, normalized size = 2.26 \begin{align*} -\frac{\log \left (\frac{{\left | -\pi + 4 \, x - \sqrt{\pi ^{2} + 8} \right |}}{{\left | -\pi + 4 \, x + \sqrt{\pi ^{2} + 8} \right |}}\right )}{\sqrt{\pi ^{2} + 8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(pi*x-2*x^2+1),x, algorithm="giac")

[Out]

-log(abs(-pi + 4*x - sqrt(pi^2 + 8))/abs(-pi + 4*x + sqrt(pi^2 + 8)))/sqrt(pi^2 + 8)