3.3.30 \(\int \frac {1}{\sqrt {9+3 x-5 x^2+x^3}} \, dx\) [230]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, 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 = {2092, 2089, 65, 212} \begin {gather*} \frac {(3-x) \sqrt {x+1} \tanh ^{-1}\left (\frac {\sqrt {x+1}}{2}\right )}{\sqrt {x^3-5 x^2+3 x+9}} \end {gather*}

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 65

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 212

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

Rule 2089

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 2092

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 {9+3 x-5 x^2+x^3}} \, dx &=\text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 37, normalized size = 0.88 \begin {gather*} -\frac {(-3+x) \sqrt {1+x} \tanh ^{-1}\left (\frac {\sqrt {1+x}}{2}\right )}{\sqrt {(-3+x)^2 (1+x)}} \end {gather*}

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 = 0.05, size = 45, normalized size = 1.07

method	result	size
trager	\(-\frac {\ln \left (\frac {x^{2}+4 \sqrt {x^{3}-5 x^{2}+3 x +9}+2 x -15}{\left (-3+x \right )^{2}}\right )}{2}\)	\(35\)
default	\(\frac {\left (-3+x \right ) \sqrt {1+x}\, \left (\ln \left (\sqrt {1+x}-2\right )-\ln \left (\sqrt {1+x}+2\right )\right )}{2 \sqrt {x^{3}-5 x^{2}+3 x +9}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.41, size = 62, normalized size = 1.48 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {x^{3} - 5 x^{2} + 3 x + 9}}\, dx \end {gather*}

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 = 0.53, size = 34, normalized size = 0.81 \begin {gather*} -\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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {x^3-5\,x^2+3\,x+9}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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