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

Optimal. Leaf size=24 \[ -\tanh ^{-1}\left (\frac {2 x-2}{\sqrt {x^3+x^2-5 x+3}}\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.67, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2067, 2064, 63, 206} \begin {gather*} \frac {(1-x) \sqrt {x+3} \tanh ^{-1}\left (\frac {\sqrt {x+3}}{2}\right )}{\sqrt {x^3+x^2-5 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[3 - 5*x + x^2 + x^3],x]

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 37, normalized size = 1.54 \begin {gather*} -\frac {(x-1) \sqrt {x+3} \tanh ^{-1}\left (\frac {\sqrt {x+3}}{2}\right )}{\sqrt {(x-1)^2 (x+3)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.05, size = 24, normalized size = 1.00 \begin {gather*} -\tanh ^{-1}\left (\frac {-2+2 x}{\sqrt {3-5 x+x^2+x^3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 58, normalized size = 2.42 \begin {gather*} -\frac {1}{2} \, \log \left (\frac {2 \, x + \sqrt {x^{3} + x^{2} - 5 \, x + 3} - 2}{x - 1}\right ) + \frac {1}{2} \, \log \left (-\frac {2 \, x - \sqrt {x^{3} + x^{2} - 5 \, x + 3} - 2}{x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.34, size = 34, normalized size = 1.42 \begin {gather*} -\frac {\log \left (\sqrt {x + 3} + 2\right )}{2 \, \mathrm {sgn}\left (x - 1\right )} + \frac {\log \left ({\left | \sqrt {x + 3} - 2 \right |}\right )}{2 \, \mathrm {sgn}\left (x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 33, normalized size = 1.38

method	result	size
trager	\(-\frac {\ln \left (\frac {x^{2}+4 \sqrt {x^{3}+x^{2}-5 x +3}+6 x -7}{\left (-1+x \right )^{2}}\right )}{2}\)	\(33\)
default	\(-\frac {\left (-1+x \right ) \sqrt {3+x}\, \left (\ln \left (\sqrt {3+x}+2\right )-\ln \left (\sqrt {3+x}-2\right )\right )}{2 \sqrt {x^{3}+x^{2}-5 x +3}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3+x^2-5*x+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*ln((x^2+4*(x^3+x^2-5*x+3)^(1/2)+6*x-7)/(-1+x)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x^{3} + x^{2} - 5 \, x + 3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{\sqrt {x^3+x^2-5\,x+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 - 5*x + x^3 + 3)^(1/2),x)

[Out]

int(1/(x^2 - 5*x + x^3 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x^{3} + x^{2} - 5 x + 3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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