3.230 \(\int \frac{1}{\sqrt{9+3 x-5 x^2+x^3}} \, dx\)

Optimal. Leaf size=42 \[ \frac{(3-x) \sqrt{x+1} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{2}\right )}{\sqrt{x^3-5 x^2+3 x+9}} \]

[Out]

((3 - x)*Sqrt[1 + x]*ArcTanh[Sqrt[1 + x]/2])/Sqrt[9 + 3*x - 5*x^2 + x^3]

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Rubi [A]  time = 0.0426458, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2067, 2064, 63, 206} \[ \frac{(3-x) \sqrt{x+1} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{2}\right )}{\sqrt{x^3-5 x^2+3 x+9}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[9 + 3*x - 5*x^2 + x^3],x]

[Out]

((3 - x)*Sqrt[1 + x]*ArcTanh[Sqrt[1 + x]/2])/Sqrt[9 + 3*x - 5*x^2 + x^3]

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rule 2064

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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}{\sqrt{9+3 x-5 x^2+x^3}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{128}{27}-\frac{16 x}{3}+x^3}} \, dx,x,-\frac{5}{3}+x\right )\\ &=\frac{\left (128 (3-x) \sqrt{1+x}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{128}{9}-\frac{32 x}{3}\right ) \sqrt{\frac{128}{9}+\frac{16 x}{3}}} \, dx,x,-\frac{5}{3}+x\right )}{3 \sqrt{3} \sqrt{9+3 x-5 x^2+x^3}}\\ &=\frac{\left (16 (3-x) \sqrt{1+x}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{128}{3}-2 x^2} \, dx,x,\frac{4 \sqrt{1+x}}{\sqrt{3}}\right )}{\sqrt{3} \sqrt{9+3 x-5 x^2+x^3}}\\ &=\frac{(3-x) \sqrt{1+x} \tanh ^{-1}\left (\frac{\sqrt{1+x}}{2}\right )}{\sqrt{9+3 x-5 x^2+x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0082524, size = 37, normalized size = 0.88 \[ -\frac{(x-3) \sqrt{x+1} \tanh ^{-1}\left (\frac{\sqrt{x+1}}{2}\right )}{\sqrt{(x-3)^2 (x+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[9 + 3*x - 5*x^2 + x^3],x]

[Out]

-(((-3 + x)*Sqrt[1 + x]*ArcTanh[Sqrt[1 + x]/2])/Sqrt[(-3 + x)^2*(1 + x)])

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Maple [A]  time = 0.008, size = 45, normalized size = 1.1 \begin{align*} -{\frac{-3+x}{2}\sqrt{1+x} \left ( \ln \left ( \sqrt{1+x}+2 \right ) -\ln \left ( \sqrt{1+x}-2 \right ) \right ){\frac{1}{\sqrt{{x}^{3}-5\,{x}^{2}+3\,x+9}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3-5*x^2+3*x+9)^(1/2),x)

[Out]

-1/2*(-3+x)*(1+x)^(1/2)*(ln((1+x)^(1/2)+2)-ln((1+x)^(1/2)-2))/(x^3-5*x^2+3*x+9)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{3} - 5 \, x^{2} + 3 \, x + 9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(x^3 - 5*x^2 + 3*x + 9), x)

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Fricas [A]  time = 1.72879, size = 161, normalized size = 3.83 \begin{align*} -\frac{1}{2} \, \log \left (\frac{2 \, x + \sqrt{x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) + \frac{1}{2} \, \log \left (-\frac{2 \, x - \sqrt{x^{3} - 5 \, x^{2} + 3 \, x + 9} - 6}{x - 3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(1/2),x, algorithm="fricas")

[Out]

-1/2*log((2*x + sqrt(x^3 - 5*x^2 + 3*x + 9) - 6)/(x - 3)) + 1/2*log(-(2*x - sqrt(x^3 - 5*x^2 + 3*x + 9) - 6)/(
x - 3))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{3} - 5 x^{2} + 3 x + 9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3-5*x**2+3*x+9)**(1/2),x)

[Out]

Integral(1/sqrt(x**3 - 5*x**2 + 3*x + 9), x)

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Giac [A]  time = 1.07318, size = 46, normalized size = 1.1 \begin{align*} -\frac{\log \left (\sqrt{x + 1} + 2\right )}{2 \, \mathrm{sgn}\left (x - 3\right )} + \frac{\log \left ({\left | \sqrt{x + 1} - 2 \right |}\right )}{2 \, \mathrm{sgn}\left (x - 3\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3-5*x^2+3*x+9)^(1/2),x, algorithm="giac")

[Out]

-1/2*log(sqrt(x + 1) + 2)/sgn(x - 3) + 1/2*log(abs(sqrt(x + 1) - 2))/sgn(x - 3)